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		<title>Le temps de la maturité s’amorce enfin pour l’e-learning (2005)</title>
		<link>https://telechargercours.com/le-temps-de-la-maturite-samorce-enfin-pour-le-learning-2005/</link>
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		<dc:creator><![CDATA[Cours Zilla]]></dc:creator>
		<pubDate>Tue, 10 Oct 2023 19:35:10 +0000</pubDate>
				<category><![CDATA[Economie & Entreprise]]></category>
		<guid isPermaLink="false">https://telechargercours.com/?p=22154</guid>

					<description><![CDATA[Il faut reproduire en ligne les situations pédagogiques qui marchent Seulement 2% des entreprises françaises auraient opté pour l&#8217;e-learning. Quels sont les facteurs favorisant l&#8217;adoption ou le rejet de ce mode de formation ? Quelles sont ses plus récentes évolutions ? (30/05/2005) Alors que seulement 2% des entreprises françaises ont aujourd&#8217;hui opté pour l&#8217;e-learning &#8211; [&#8230;]]]></description>
										<content:encoded><![CDATA[
<h2 class="wp-block-heading">Il faut reproduire en ligne les situations pédagogiques qui marchent</h2>



<p class="wp-block-paragraph">Seulement 2% des entreprises françaises auraient opté pour l&rsquo;e-learning. Quels sont les facteurs favorisant l&rsquo;adoption ou le rejet de ce mode de formation ? Quelles sont ses plus récentes évolutions ? (30/05/2005)</p>



<p class="wp-block-paragraph">Alors que seulement 2% des entreprises françaises ont aujourd&rsquo;hui opté pour l&rsquo;e-learning &#8211; selon une étude menée début 2005 par Accountemps Intérim auprès de 1 500 directeurs financiers et DRH internationaux -, la question se pose de savoir quels critères favorisent aujourd&rsquo;hui l&rsquo;adoption &#8211;&nbsp;ou le rejet &#8211; de cet outil de formation.<br><br>« Tous les cours ne peuvent être mis en ligne. L&rsquo;e-learning est particulièrement bien adapté aux connaissances pures, telles que les langues, les outils bureautiques, le droit, l&rsquo;anatomie », explique Eric Labouchet, chef de projet <em>e-learning</em> et Web à l&rsquo;INSEP (Institut National du Sport et de l&rsquo;Education Physique)</p>



<p class="wp-block-paragraph">La plate-forme d&rsquo;enseignement à distance mise en oeuvre à l&rsquo;INSEP vise les sportifs de haut niveau qui, pendant leur vie de sportif, se forment, depuis les études secondaires jusqu&rsquo;au CAPES, en passant par les masters et les concours pour devenir professeur de sport.<br><br>« Ce qui nous facilite grandement la vie est par ailleurs l&rsquo;utilisation du logiciel Didactinet, qui permet de médiatiser des documents Powerpoint existants en y intégrant du son, de la vidéo et du Flash, tout en paramétrant le mode de navigation de l&rsquo;apprenant », ajoute Eric Labouchet, qui a basé sa plate-forme d&rsquo;enseignement à distance sur les solutions de l&rsquo;éditeur Syfadis. Un virage que l&rsquo;éditeur Macromedia a su également prendre en lançant son produit Breeze, qui permet à l&rsquo;entreprise de fabriquer toute seule ses contenus à partir de n&rsquo;importe quel document Powerpoint.<br><br>Pour Pierrre Boudignon, P-DG de la société Just a Link, les outils sont importants, mais le contact avec l&rsquo;apprenant reste la priorité des priorités. Sa méthode repose donc sur un mariage des deux. « Nous utilisons exclusivement le partage d&rsquo;applications &#8211; en mode Web-conférence &#8211; et l&rsquo;appel téléphonique. Le coté dynamique et interactif &#8211; en mode synchrone &#8211; permet de sentir le niveau de son stagiaire alors que les QCM restent grossiers », déclare le dirigeant.<br><br>Confronté à une baisse des effectifs formés et à une forte pression concurrencielle, le centre de formation agricole de la Chambre d&rsquo;agriculture du Morbihan a quant à lui récemment restructuré son offre, la découpant en trois parties distinctes : apprentissage en autonomie, en présentiel et en situation professionnelle.</p>


<h2 class="wp-block-post-title">Le temps de la maturité s&rsquo;amorce enfin pour l&rsquo;e-learning (2005)</h2>


<p class="wp-block-paragraph">« Nous avons affaire à des profils de populations très différents, éloignés géographiquement. Nous leur proposons de l&rsquo;autoformation semi-présentielle soit sur leur siège d&rsquo;exploitation, soit dans des points formation, avec toujours plus de souplesse. Il suffit pour cela d&rsquo;une simple connexion Internet », note Claire Lapouble, en charge de la commercialisation du dispositif de formation de la Chambre d&rsquo;agriculture du Morbihan.<br><br>Le témoignage de Claire Lapouble est instructif à plus d&rsquo;un titre. Il montre notamment la nécessité de reproduire en ligne les situations pédagogiques les mieux adaptées aux besoins des candidats à la formation, celles qui marchent.<br><br>« Il faut utiliser chaque type d&rsquo;outil pour ce qu&rsquo;il apporte dans la relation d&rsquo;apprentissage, dans une démarche de pédagogie mixte, incluant des jeux d&rsquo;entreprise &#8211; pour les aspects marketing, finance ou comportementaux -, de la simulation &#8211; très adaptée au déploiement d&rsquo;applications -, du multimédia, de l&rsquo;autoformation, des classes virtuelles lors de conférences Web &#8211; pour les phases de débriefing ou pour recueillir le savoir des gourous -, etc. », commente Jean-Jacques Bouet, expert e-learning chez Oracle France.</p>



<p class="wp-block-paragraph">L&rsquo;e-learning sort de la phase d&rsquo;adolescence, il devient plus adulte. On le voit tout d&rsquo;abord au fait qu&rsquo;un tiers de nos clients nous sollicitent pour des services hébergés, ce qui prouve qu&rsquo;ils veulent des solutions légères pour des projets concrets. De plus, ce sont désormais plutôt les directions opérationnelles qui mettent en ouvre et financent les projets de formation, et non les DSI. Enfin, nos partenaires intégrateurs intègrent de plus en plus Oracle i-Learning comme moyen d&rsquo;accompagnement du changement dans le déploiement de projets applicatifs, ce qui prouve qu&rsquo;on peut former l&rsquo;utilisateur final avec ces solutions », conclut Jean-Jacques Bouet.</p>



<p class="wp-block-paragraph"></p>
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			</item>
		<item>
		<title>Les caractéristique des Supports de Transmission</title>
		<link>https://telechargercours.com/les-caracteristique-des-supports-de-transmission-2/</link>
		
		<dc:creator><![CDATA[Cours Zilla]]></dc:creator>
		<pubDate>Sun, 13 Nov 2022 02:58:26 +0000</pubDate>
				<category><![CDATA[Science & Technologie]]></category>
		<guid isPermaLink="false">https://telechargercours.com/?p=4561</guid>

					<description><![CDATA[LES CARACTERISTIQUES DES SUPPORTS DE TRANSMISSION 2005 LES CARACTERISTIQUES DES SUPPORTS DE TRANSMISSION 1) Caractéristiques techniques des supports. L’infrastructure d’un réseau, la qualité de service offerte, les solutions logicielles mettre en oeuvre dépendent largement des supports de transmission utilisés. Les supports de transmission exploitent les propriétés de conductibilité des métaux (paires torsadées, coaxial&#8230;) ou celles [&#8230;]]]></description>
										<content:encoded><![CDATA[
<p class="wp-block-paragraph"></p>


<p></p>
<p>LES CARACTERISTIQUES DES SUPPORTS DE TRANSMISSION 2005</p>
<p><strong>LES CARACTERISTIQUES DES SUPPORTS DE TRANSMISSION</strong></p>
<p><strong>1) Caractéristiques techniques des supports.</strong></p>
<p>L’infrastructure d’un réseau, la qualité de service offerte, les solutions logicielles</p>
<ul>
<li>mettre en oeuvre dépendent largement des supports de transmission utilisés. Les supports de transmission exploitent les propriétés de conductibilité des métaux (paires torsadées, coaxial&#8230;) ou celles des ondes électromagnétiques (faisceau hertzien, fibres optiques, satellite&#8230;).</li>
</ul>
<p>Un support de transmission est essentiellement caractérisé par son <strong>impédance</strong></p>
<p><strong>caractéristique </strong>et sa<strong> bande passante </strong>. Ces paramètres conditionnent les possibilités de transmission en termes de débits et de distance franchissable.</p>
<p><strong>1.1) L’impédance caractéristique</strong></p>
<p><img decoding="async" class="wp-image-4564" src="https://telechargercours.com/wp-content/uploads/2020/05/word-image-94.png"></p>
<p>R<em>dl</em> L<em>dl</em></p>
<p>C<em>dl </em> <img decoding="async" class="wp-image-4565" src="https://telechargercours.com/wp-content/uploads/2020/05/word-image-95.png"> G<em>dl</em></p>
<p>Schéma équivalent d’un élément (<em>dl</em>) d’une ligne de transmission</p>
<p><em>Page 1</em></p>
<p></p>
<p>LES CARACTERISTIQUES DES SUPPORTS DE TRANSMISSION 2005</p>
<p>Une ligne de transmission est constituée de deux conducteurs de cuivre séparés par un isolant. La résistance linéique des conducteurs, la capacité linéique, l’inductance linéique et la conductance linéique sont liés par la relation:</p>
<table>
<tbody>
<tr>
<td>
<p><em>Z<sub>c</sub> </em>=</p>
</td>
<td>
<p><em>R </em>+<em> jLω</em></p>
</td>
<td>
<p>≈</p>
</td>
<td>
<p><em>L</em></p>
</td>
</tr>
<tr>
<td>&nbsp;</td>
<td>
<p><em>G </em>+<em> jCω</em></p>
</td>
<td>&nbsp;</td>
<td>
<p><em>C</em></p>
</td>
</tr>
</tbody>
</table>
<p><img decoding="async" class="wp-image-4566" src="https://telechargercours.com/wp-content/uploads/2020/05/word-image-96.png"></p>
<p>avec <em>ω</em> = 2 × <em>π</em> × <em>f</em></p>
<p>où ω est la pulsation du courant exprimé en rd/s et f la fréquence du signal en Hertz.</p>
<p>Zc, ou impédance caractéristique, est l’impédance d’une ligne de longueur infinie. On montre qu’une ligne de longueur finie refermée sur un récepteur d’impédance Zr, tel que Zc=Zr, se comporte comme une ligne infinie, on dit alors que la ligne est adaptée (adaptation d’impédance).</p>
<p>Toute rupture d’impédance (Zc≠Zr) provoque une réflexion d’une partie de l’énergie incidente. Cette énergie (onde réfléchie ou écho) se combine à l’énergie incidente pour fournir des ondes stationnaires. pour éviter ces réflexions parasites, il est nécessaire tout au long de la ligne et à chaque raccordement d’un nouvel élément de liaison de réaliser la continuité de l’impédance, c’est l’adaptation d’impédance.</p>
<p><strong>1.2) La bande passante</strong></p>
<p>A l’extrémité de la ligne, le récepteur doit identifier et décoder le signal. Cette fonction ne peut valablement être effectuée que si le signal n’a pas été exagérément modifié pendant la transmission. La bande passante est la grandeur de base qui renseigne sur les possibilités de transmission d’une ligne.</p>
<p><img decoding="async" class="wp-image-4567" src="https://telechargercours.com/wp-content/uploads/2020/05/word-image-97.png"></p>
<p>ligne de</p>
<p><img decoding="async" class="wp-image-4568" src="https://telechargercours.com/wp-content/uploads/2020/05/word-image-98.png"></p>
<p>transmission</p>
<p><img decoding="async" class="wp-image-4569" src="https://telechargercours.com/wp-content/uploads/2020/05/word-image-109.png"></p>
<p>une ligne de transmission déforme le signal</p>
<p><em>Page 2</em></p>
<p></p>
<p>LES CARACTERISTIQUES DES SUPPORTS DE TRANSMISSION 2005</p>
<ol>
<li><strong>Notion d’analyse spectrale.</strong></li>
</ol>
<p>D’après le mathématicien français Joseph FOURIER, une fonction périodique de fréquence f<sub>0</sub> peut être considérée comme la somme d’une constante (<strong><em>composante</em></strong> <strong><em>continue</em></strong>) et de fonctions sinusoïdales :</p>
<ul>
<li><strong><em>Le fondamentale </em></strong>de fréquence égale à celle du signal périodique ;</li>
</ul>
<p><em><sup>y</sup></em>(<em>t</em> )</p>
<p>où</p>
<ul>
<li>
<ul>
<li><strong><em>Les harmoniques </em></strong>de fréquence multiple à celle du signal périodique.</li>
</ul>
</li>
<li><em>A</em><sub>0</sub> +<em> A</em><sub>1</sub> cos(2<em>πf</em><sub>0</sub> +<em> ϕ</em><sub>1</sub> ) +<em> A</em><sub>2</sub> cos(2<em>π </em>2<em> f</em><sub>0</sub> +<em> ϕ </em><sub>2</sub> ) + &#8230; +<em> A<sub>n</sub> </em>cos(2<em>πnf </em><sub>0</sub> +<em> ϕ <sub>n</sub> </em>) A<sub>0</sub> représente une constante appelée composante continue</li>
</ul>
<p>A<sub>1</sub> est l’amplitude du signal de même fréquence que le signal d’origine appelé fondamental.</p>
<p>A<sub>2</sub>…An sont les amplitudes des termes harmoniques de fréquence 2f<sub>0</sub> …nf<sub>0</sub>.</p>
<p>Le tableau de la page suivante représente les décompositions en série de Fourier de différentes fonctions usuelles.</p>
<p><em>Page 3</em></p>
<p></p>
<p>LES CARACTERISTIQUES DES SUPPORTS DE TRANSMISSION 2005</p>
<table>
<tbody>
<tr>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td colspan="12">
<p><strong>Produit, par l’amplitude du fondamental,</strong></p>
</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
</tr>
<tr>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td colspan="8">
<p><strong>de l’harmonique de rang :</strong></p>
</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
</tr>
<tr>
<td colspan="5">
<p><strong>Représentation graphique Composante</strong></p>
</td>
<td colspan="3">
<p><strong>Amplitude du</strong></p>
</td>
<td>&nbsp;</td>
<td colspan="2" rowspan="2">
<p><strong>2</strong></p>
</td>
<td>&nbsp;</td>
<td colspan="2" rowspan="2">
<p><strong>3</strong></p>
</td>
<td>&nbsp;</td>
<td colspan="2" rowspan="2">
<p><strong>4</strong></p>
</td>
<td>&nbsp;</td>
<td colspan="2" rowspan="2">
<p><strong>5</strong></p>
</td>
<td>&nbsp;</td>
<td colspan="2" rowspan="2">
<p><strong>6</strong></p>
</td>
<td>&nbsp;</td>
<td colspan="2" rowspan="2">
<p><strong>7</strong></p>
</td>
<td>&nbsp;</td>
<td colspan="2" rowspan="2">
<p><strong>8</strong></p>
</td>
<td>&nbsp;</td>
<td colspan="2" rowspan="2">
<p><strong>9</strong></p>
</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
</tr>
<tr>
<td>&nbsp;</td>
<td>
<p><strong>de la fonction</strong></p>
</td>
<td>&nbsp;</td>
<td colspan="2">
<p><strong>continue</strong></p>
</td>
<td colspan="3">
<p><strong>fondamental</strong></p>
</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
</tr>
<tr>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td colspan="2">
<p><strong>(valeur</strong></p>
</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
</tr>
<tr>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td colspan="2">
<p><strong>moyenne)</strong></p>
</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
</tr>
<tr>
<td>
<p>VM</p>
</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td colspan="2">
<p><em>V<sub>M</sub></em></p>
</td>
<td>&nbsp;</td>
<td>
<p><em>V<sub>M</sub></em></p>
</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td colspan="2">
<p>1</p>
</td>
<td>&nbsp;</td>
<td colspan="2">
<p>1</p>
</td>
<td>&nbsp;</td>
<td colspan="2">
<p>1</p>
</td>
<td>&nbsp;</td>
<td colspan="2">
<p>1</p>
</td>
<td>&nbsp;</td>
<td colspan="2">
<p>1</p>
</td>
<td>&nbsp;</td>
<td colspan="2">
<p>1</p>
</td>
<td>&nbsp;</td>
<td colspan="2">
<p>1</p>
</td>
<td>&nbsp;</td>
<td colspan="2">
<p>1</p>
</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
</tr>
<tr>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td colspan="2">
<p>2</p>
</td>
<td>&nbsp;</td>
<td>
<p><em>π</em></p>
</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td colspan="2">
<p>2</p>
</td>
<td>&nbsp;</td>
<td colspan="2">
<p>3</p>
</td>
<td>&nbsp;</td>
<td colspan="2">
<p>4</p>
</td>
<td>&nbsp;</td>
<td colspan="2">
<p>5</p>
</td>
<td>&nbsp;</td>
<td colspan="2">
<p>6</p>
</td>
<td>&nbsp;</td>
<td colspan="2">
<p>7</p>
</td>
<td>&nbsp;</td>
<td colspan="2">
<p>8</p>
</td>
<td>&nbsp;</td>
<td colspan="2">
<p>9</p>
</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
</tr>
<tr>
<td>
<p>0</p>
</td>
<td>
<p>T</p>
</td>
<td>&nbsp;</td>
<td>
<p>t</p>
</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
</tr>
<tr>
<td>
<p>VM</p>
</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td colspan="2">
<p>0</p>
</td>
<td>&nbsp;</td>
<td>
<p><sup>8</sup><sub>2</sub><em> V<sub>M</sub></em></p>
</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td colspan="2">
<p>0</p>
</td>
<td>&nbsp;</td>
<td colspan="2">
<p>1</p>
</td>
<td>&nbsp;</td>
<td colspan="2">
<p>0</p>
</td>
<td>&nbsp;</td>
<td colspan="2">
<p>1</p>
</td>
<td>&nbsp;</td>
<td colspan="2">
<p>0</p>
</td>
<td>&nbsp;</td>
<td colspan="2">
<p>1</p>
</td>
<td>&nbsp;</td>
<td colspan="2">
<p>0</p>
</td>
<td>&nbsp;</td>
<td colspan="2">
<p>1</p>
</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
</tr>
<tr>
<td rowspan="2">
<p>0</p>
</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td rowspan="2">
<p>t</p>
</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>
<p><em>π</em></p>
</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td colspan="2">
<p>9</p>
</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td colspan="2">
<p>25</p>
</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td colspan="2">
<p>49</p>
</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td colspan="2">
<p>81</p>
</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
</tr>
<tr>
<td>&nbsp;</td>
<td rowspan="2">
<p>T</p>
</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
</tr>
<tr>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
</tr>
<tr>
<td rowspan="2">
<p>VM</p>
</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>
<p>4<em>V<sub>M</sub></em></p>
</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td colspan="2">
<p>1</p>
</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td colspan="2">
<p>1</p>
</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td colspan="2">
<p>1</p>
</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td colspan="2">
<p>1</p>
</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
</tr>
<tr>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td colspan="2" rowspan="2">
<p>0</p>
</td>
<td>&nbsp;</td>
<td rowspan="2">
<p><em>π</em></p>
</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td colspan="2" rowspan="2">
<p>0</p>
</td>
<td>&nbsp;</td>
<td colspan="2" rowspan="2">
<p>3</p>
</td>
<td>&nbsp;</td>
<td colspan="2" rowspan="2">
<p>0</p>
</td>
<td>&nbsp;</td>
<td colspan="2" rowspan="2">
<p>5</p>
</td>
<td>&nbsp;</td>
<td colspan="2" rowspan="2">
<p>0</p>
</td>
<td>&nbsp;</td>
<td colspan="2" rowspan="2">
<p>7</p>
</td>
<td>&nbsp;</td>
<td colspan="2" rowspan="2">
<p>0</p>
</td>
<td>&nbsp;</td>
<td colspan="2" rowspan="2">
<p>9</p>
</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
</tr>
<tr>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
</tr>
<tr>
<td>
<p>0</p>
</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>
<p>t</p>
</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
</tr>
<tr>
<td>
<p>-VM</p>
</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
</tr>
<tr>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>
<p><em>V</em></p>
</td>
<td rowspan="2">
<p><em>M</em></p>
</td>
<td>
<p>2</p>
</td>
<td>&nbsp;</td>
<td>
<p><em>π</em></p>
</td>
<td rowspan="2">
<p>1</p>
</td>
<td rowspan="2">&nbsp;</td>
<td>
<p><em>π</em></p>
</td>
<td rowspan="2">
<p>1</p>
</td>
<td rowspan="2">&nbsp;</td>
<td>
<p><em>π</em></p>
</td>
<td rowspan="2">
<p>1</p>
</td>
<td rowspan="2">&nbsp;</td>
<td>
<p><em>π</em></p>
</td>
<td rowspan="2">
<p>1</p>
</td>
<td rowspan="2">&nbsp;</td>
<td>
<p><em>π</em></p>
</td>
<td rowspan="2">
<p>1</p>
</td>
<td rowspan="2">&nbsp;</td>
<td>
<p><em>π</em></p>
</td>
<td rowspan="2">
<p>1</p>
</td>
<td rowspan="2">&nbsp;</td>
<td>
<p><em>π</em></p>
</td>
<td rowspan="2">
<p>1</p>
</td>
<td rowspan="2">&nbsp;</td>
<td>
<p><em>π</em></p>
</td>
<td rowspan="2">
<p>1</p>
</td>
<td rowspan="2">&nbsp;</td>
<td>
<p><em>π</em></p>
</td>
<td rowspan="2">&nbsp;</td>
<td>&nbsp;</td>
</tr>
<tr>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td rowspan="2">
<p><em>V<sub>M</sub> </em>sin</p>
</td>
<td>&nbsp;</td>
<td>
<p>2</p>
</td>
<td>
<p>3</p>
</td>
<td>
<p>4</p>
</td>
<td>
<p>5</p>
</td>
<td>
<p>6</p>
</td>
<td>
<p>7</p>
</td>
<td>
<p>8</p>
</td>
<td>
<p>9</p>
</td>
<td>&nbsp;</td>
</tr>
<tr>
<td rowspan="2">
<p>VM</p>
</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td colspan="2" rowspan="3">
<p><em>b</em></p>
</td>
<td rowspan="3">
<p><em>π</em></p>
</td>
<td colspan="2" rowspan="3">
<p><em>b </em>2</p>
</td>
<td>
<p>sin</p>
</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>
<p>sin</p>
</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>
<p>sin</p>
</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>
<p>sin</p>
</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>
<p>sin</p>
</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>
<p>sin</p>
</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>
<p>sin</p>
</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>
<p>sin</p>
</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
</tr>
<tr>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td colspan="3" rowspan="2">
<p><em>b </em>3</p>
</td>
<td colspan="3" rowspan="2">
<p><em>b </em>4</p>
</td>
<td colspan="3" rowspan="2">
<p><em>b </em>5</p>
</td>
<td colspan="3" rowspan="2">
<p><em>b </em>6</p>
</td>
<td colspan="3" rowspan="2">
<p><em>b </em>7</p>
</td>
<td colspan="3" rowspan="2">
<p><em>b </em>8</p>
</td>
<td colspan="3" rowspan="2">
<p><em>b </em>9</p>
</td>
<td colspan="3" rowspan="2">
<p><em>b</em></p>
</td>
<td>&nbsp;</td>
</tr>
<tr>
<td rowspan="2">
<p>0</p>
</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
</tr>
<tr>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td rowspan="2">
<p>t</p>
</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
</tr>
<tr>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
</tr>
<tr>
<td rowspan="2">
<p>VM</p>
</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td colspan="2">
<p><em>V<sub>M</sub></em></p>
</td>
<td>&nbsp;</td>
<td>
<p><em>V<sub>M</sub></em></p>
</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td colspan="2">
<p>4</p>
</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td colspan="2">
<p>4</p>
</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td colspan="2">
<p>4</p>
</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td colspan="2">
<p>4</p>
</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
</tr>
<tr>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
</tr>
<tr>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td colspan="2" rowspan="2">
<p><em>π</em></p>
</td>
<td>&nbsp;</td>
<td rowspan="2">
<p>2</p>
</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td colspan="2" rowspan="2">
<p>3<em>π</em></p>
</td>
<td>&nbsp;</td>
<td colspan="2">
<p>0</p>
</td>
<td>&nbsp;</td>
<td colspan="2" rowspan="2">
<p>15<em>π</em></p>
</td>
<td>&nbsp;</td>
<td colspan="2">
<p>0</p>
</td>
<td>&nbsp;</td>
<td colspan="2" rowspan="2">
<p>35<em>π</em></p>
</td>
<td>&nbsp;</td>
<td colspan="2">
<p>0</p>
</td>
<td>&nbsp;</td>
<td colspan="2" rowspan="2">
<p>63<em>π</em></p>
</td>
<td>&nbsp;</td>
<td colspan="2">
<p>0</p>
</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
</tr>
<tr>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
</tr>
<tr>
<td>
<p>0</p>
</td>
<td>
<p>T: période source</p>
</td>
<td>&nbsp;</td>
<td>
<p>t</p>
</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
</tr>
<tr>
<td rowspan="2">
<p>VM</p>
</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td colspan="2">
<p>2<em>V<sub>M</sub></em></p>
</td>
<td>&nbsp;</td>
<td>
<p>4<em>V<sub>M</sub></em></p>
</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td colspan="2">
<p>3</p>
</td>
<td>&nbsp;</td>
<td colspan="2">
<p>3</p>
</td>
<td>&nbsp;</td>
<td colspan="2">
<p>3</p>
</td>
<td>&nbsp;</td>
<td colspan="2">
<p>3</p>
</td>
<td>&nbsp;</td>
<td colspan="2">
<p>3</p>
</td>
<td>&nbsp;</td>
<td colspan="2">
<p>3</p>
</td>
<td>&nbsp;</td>
<td colspan="2">
<p>3</p>
</td>
<td>&nbsp;</td>
<td colspan="2">
<p>3</p>
</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
</tr>
<tr>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
</tr>
<tr>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td colspan="2">
<p><em>π</em></p>
</td>
<td>&nbsp;</td>
<td>
<p>3<em>π</em></p>
</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td colspan="2">
<p>15</p>
</td>
<td>&nbsp;</td>
<td colspan="2">
<p>35</p>
</td>
<td>&nbsp;</td>
<td colspan="2">
<p>63</p>
</td>
<td>&nbsp;</td>
<td colspan="2">
<p>99</p>
</td>
<td>&nbsp;</td>
<td colspan="2">
<p>143</p>
</td>
<td>&nbsp;</td>
<td colspan="2">
<p>195</p>
</td>
<td>&nbsp;</td>
<td colspan="2">
<p>255</p>
</td>
<td>&nbsp;</td>
<td colspan="2">
<p>323</p>
</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
</tr>
<tr>
<td>
<p>0</p>
</td>
<td>
<p>T: demi-période</p>
</td>
<td>&nbsp;</td>
<td>
<p>t</p>
</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
</tr>
</tbody>
</table>
<p><img decoding="async" class="wp-image-4570" src="https://telechargercours.com/wp-content/uploads/2020/05/word-image-130.png"></p>
<p><em>Page 4</em></p>
<p></p>
<p>LES CARACTERISTIQUES DES SUPPORTS DE TRANSMISSION 2005</p>
<p>Les figures suivantes illustrent la reconstitution d’un signal d’origine à partir de ses composantes. La représentation s’arrête à l’harmonique de rang5. Il faut savoir que plus le nombre d’harmoniques utilisé est important, plus le signal reconstitué est proche du signal d’origine.</p>
<table>
<tbody>
<tr>
<td>&nbsp;</td>
<td>
<p><strong>15</strong></p>
</td>
<td>&nbsp;</td>
</tr>
<tr>
<td rowspan="2">
<p><strong>amplitude</strong></p>
</td>
<td>
<p><strong>10</strong></p>
</td>
<td>&nbsp;</td>
</tr>
<tr>
<td>
<p><strong>-5</strong></p>
</td>
<td>&nbsp;</td>
</tr>
<tr>
<td>&nbsp;</td>
<td>
<p><strong>5</strong></p>
</td>
<td>&nbsp;</td>
</tr>
<tr>
<td>&nbsp;</td>
<td>
<p><strong>0</strong></p>
</td>
<td>&nbsp;</td>
</tr>
</tbody>
</table>
<p><img decoding="async" class="wp-image-4571" src="https://telechargercours.com/wp-content/uploads/2020/05/word-image-131.png"> <img decoding="async" class="wp-image-4572" src="https://telechargercours.com/wp-content/uploads/2020/05/word-image-132.png"></p>
<p><strong>-10</strong></p>
<p><strong>-15</strong></p>
<p><strong>temps</strong></p>
<p><strong>Signal périodique</strong></p>
<p><strong>15</strong></p>
<p><img decoding="async" class="wp-image-4573" src="https://telechargercours.com/wp-content/uploads/2020/05/word-image-133.png"></p>
<p><strong>10</strong></p>
<p><strong>5</strong></p>
<p><strong>0</strong></p>
<p><strong>-5</strong></p>
<p><strong>-10</strong></p>
<p><strong>-15</strong></p>
<p><strong>fréquence 3f0</strong></p>
<p><strong>15</strong></p>
<p><img decoding="async" class="wp-image-4574" src="https://telechargercours.com/wp-content/uploads/2020/05/word-image-145.png"></p>
<p><strong>10</strong></p>
<p><strong>5</strong></p>
<p><strong>0</strong></p>
<p><strong>-5</strong></p>
<p><strong>-10</strong></p>
<p><strong>-15</strong></p>
<p><strong>fréquence f0</strong></p>
<p><strong>Fondamentale</strong></p>
<p><strong>15</strong></p>
<p><img decoding="async" class="wp-image-4575" src="https://telechargercours.com/wp-content/uploads/2020/05/word-image-146.png"> <img decoding="async" class="wp-image-4576" src="https://telechargercours.com/wp-content/uploads/2020/05/word-image-158.png"></p>
<p><strong>10</strong></p>
<p><strong>5</strong></p>
<p><strong>0</strong></p>
<p><strong>-5</strong></p>
<p><strong>-10</strong></p>
<p><strong>-15</strong></p>
<p><strong>Fondamentale + harmonique de rang 3</strong></p>
<table>
<tbody>
<tr>
<td>
<p><strong>15</strong></p>
</td>
<td>
<p><strong>15</strong></p>
</td>
<td>&nbsp;</td>
</tr>
<tr>
<td>
<p><strong>10</strong></p>
</td>
<td>
<p><strong>10</strong></p>
</td>
<td>&nbsp;</td>
</tr>
<tr>
<td>
<p><strong>5</strong></p>
</td>
<td rowspan="2">
<p><strong>5</strong></p>
</td>
<td>&nbsp;</td>
</tr>
<tr>
<td rowspan="2">
<p><strong>0</strong></p>
</td>
<td>&nbsp;</td>
</tr>
<tr>
<td rowspan="2">
<p><strong>0</strong></p>
</td>
<td>&nbsp;</td>
</tr>
<tr>
<td rowspan="2">
<p><strong>-5</strong></p>
</td>
<td>&nbsp;</td>
</tr>
<tr>
<td rowspan="2">
<p><strong>-5</strong></p>
</td>
<td>&nbsp;</td>
</tr>
<tr>
<td rowspan="2">
<p><strong>-10</strong></p>
</td>
<td>&nbsp;</td>
</tr>
<tr>
<td>&nbsp;</td>
<td>&nbsp;</td>
</tr>
<tr>
<td>
<p><strong>-15</strong></p>
</td>
<td>
<p><strong>-10</strong></p>
</td>
<td>&nbsp;</td>
</tr>
<tr>
<td>
<p><strong>fréquence 5f0</strong></p>
</td>
<td>
<p><strong>-15</strong></p>
</td>
<td>&nbsp;</td>
</tr>
</tbody>
</table>
<p><img decoding="async" class="wp-image-4577" src="https://telechargercours.com/wp-content/uploads/2020/05/word-image-159.png"> <img decoding="async" class="wp-image-4578" src="https://telechargercours.com/wp-content/uploads/2020/05/word-image-160.png"></p>
<p><strong>Fondamentale + harmonique de rang 5</strong></p>
<p><em>Page 5</em></p>
<p></p>
<p>LES CARACTERISTIQUES DES SUPPORTS DE TRANSMISSION 2005</p>
<p>Un signal périodique quelconque est constitué d’une infinité de signaux sinusoïdaux. Chaque composante peut être représentée par l’énergie qu’elle contient. Cette représentation est appelée raie de fréquence (transformation de l’espace temps en espace fréquence). L’ensemble des raies de fréquence constitue le <strong><em>spectre de fréquence </em></strong>(spectre de raies) du signal. L’espace de fréquence occupé par le spectre est désigné par le terme de <strong><em>largeur de bande</em></strong>. En théorie, la largeur de bande d’un signal non sinusoïdal est infinie, cependant, dans la pratique, la largeur de bande exprime la largeur du spectre nécessaire à une reconstitution correcte (suffisante pour être interprétée) du signal d’origine.</p>
<table>
<tbody>
<tr>
<td>
<p><strong>Espace temps</strong></p>
</td>
<td>
<p><strong>Espace fréquence</strong></p>
</td>
<td>&nbsp;</td>
</tr>
<tr>
<td>
<p>amplitude</p>
</td>
<td>
<p>amplitude</p>
</td>
<td>&nbsp;</td>
</tr>
<tr>
<td rowspan="2">
<p>période T</p>
</td>
<td>
<p>raie</p>
</td>
<td>&nbsp;</td>
</tr>
<tr>
<td>&nbsp;</td>
<td>&nbsp;</td>
</tr>
</tbody>
</table>
<p><img decoding="async" class="wp-image-4579" src="https://telechargercours.com/wp-content/uploads/2020/05/word-image-172.png"></p>
<table>
<tbody>
<tr>
<td>
<p>0</p>
</td>
<td>
<p>t</p>
</td>
<td>
<p>1/T</p>
</td>
<td>
<p>F(Hz)</p>
</td>
</tr>
</tbody>
</table>
<table>
<tbody>
<tr>
<td>&nbsp;</td>
<td>
<p>signal sinusoïdal</p>
</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
</tr>
<tr>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td colspan="2">
<p>amplitude</p>
</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
</tr>
<tr>
<td>&nbsp;</td>
<td>
<p>période T</p>
</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
</tr>
<tr>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>
<p>F(Hz)</p>
</td>
<td>&nbsp;</td>
</tr>
<tr>
<td rowspan="2">
<p>0</p>
</td>
<td rowspan="2">
<p>θ</p>
</td>
<td rowspan="2">
<p>t</p>
</td>
<td>
<p>1/T</p>
</td>
<td>
<p>3/T 5/T</p>
</td>
<td>
<p>1/θ</p>
</td>
<td>
<p>2/θ</p>
</td>
<td>&nbsp;</td>
</tr>
<tr>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
</tr>
<tr>
<td>&nbsp;</td>
<td>
<p>signal carré</p>
</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>
<p>largeur de bande</p>
</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
</tr>
</tbody>
</table>
<p><img decoding="async" class="wp-image-4580" src="https://telechargercours.com/wp-content/uploads/2020/05/word-image-173.png"></p>
<p><em>Page 6</em></p>
<p></p>
<p>LES CARACTERISTIQUES DES SUPPORTS DE TRANSMISSION 2005</p>
<p><strong>3) Application à la transmission de données</strong></p>
<p>Les systèmes de transmission (lignes, amplificateurs…) ne transmettent pas toutes les harmoniques du signal de façon identique. Dans ces conditions, le signal en sortie du système n’est plus l’image de celui en entrée, on dit qu’il y a distorsion.</p>
<p><img decoding="async" class="wp-image-4581" src="https://telechargercours.com/wp-content/uploads/2020/05/word-image-184.png"></p>
<p>ligne de</p>
<p><img decoding="async" class="wp-image-4582" src="https://telechargercours.com/wp-content/uploads/2020/05/word-image-185.png"></p>
<p>transmission</p>
<p><img decoding="async" class="wp-image-4583" src="https://telechargercours.com/wp-content/uploads/2020/05/word-image-186.png"></p>
<p>La distorsion est dite en amplitude quand les éléments constitutifs du signal, fondamental et harmoniques, ne sont pas affaiblis de façon identique. La distorsion est dite de phase quand les différents éléments du signal ne sont pas tous transmis dans le même délai. La distorsion d’amplitude est plus importante que la distorsion de phase.</p>
<p>Dans un système de transmission, les signaux sont transmis avec une distorsion faible jusqu’à une certaine fréquence appelée fréquence de coupure. Au-delà de cette fréquence, toutes les harmoniques sont fortement atténuées (filtre passe bas). On appelle <strong><em>bande passante</em></strong> l’espace de fréquence tel que tout signal appartenant à cet intervalle, ne subisse, au plus, qu’un affaiblissement déterminé par rapport à un niveau de référence.</p>
<p><img decoding="async" class="wp-image-4584" src="https://telechargercours.com/wp-content/uploads/2020/05/word-image-189.png"></p>
<p>PM AX</p>
<p>3 dB</p>
<p>PM AX</p>
<p>2</p>
<table>
<tbody>
<tr>
<td>
<p>fi</p>
</td>
<td rowspan="3">
<p>bande</p>
</td>
<td>
<p>fs</p>
</td>
<td rowspan="2">
<p>fréquence</p>
</td>
<td>&nbsp;</td>
</tr>
<tr>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
</tr>
<tr>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
</tr>
<tr>
<td>&nbsp;</td>
<td>
<p>passante</p>
</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
</tr>
</tbody>
</table>
<p>L’affaiblissement, exprimé en décibel (dB), est donné par la relation suivante :</p>
<table>
<tbody>
<tr>
<td>&nbsp;</td>
<td>
<p><em>P</em></p>
</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
</tr>
<tr>
<td rowspan="2">&nbsp;</td>
<td>
<p><em>S</em></p>
</td>
<td rowspan="2">&nbsp;</td>
<td rowspan="3">
<p>où</p>
</td>
<td rowspan="3">
<p>P<sub>S</sub> est la puissance du signal en sortie ;</p>
</td>
<td>&nbsp;</td>
</tr>
<tr>
<td rowspan="2">
<p><em>P</em></p>
</td>
<td>&nbsp;</td>
</tr>
<tr>
<td>
<p><em>A </em>= 10 × log</p>
</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
</tr>
<tr>
<td>&nbsp;</td>
<td colspan="2">
<p><em>E</em></p>
</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
</tr>
</tbody>
</table>
<p>P<sub>E</sub> est la puissance du signal en entrée.</p>
<p>La bande passante est généralement définie à –3 dB, ce qui correspond à une atténuation en puissance de moitié.</p>
<p><em>Page 7</em></p>
<p></p>
<p>LES CARACTERISTIQUES DES SUPPORTS DE TRANSMISSION 2005</p>
<p><strong>4) Notion de rapidité de modulation</strong></p>
<p>Une ligne ou un canal de transmission se comporte comme un filtre passe-bas dont l’un des effets est l’étalement du signal. Dans les conditions limites, cet étalement a pour conséquence que la fin d’une impulsion transmise se confond avec le début de la suivante, les circuits électroniques ne peuvent, alors, distinguer les deux impulsions.</p>
<p>impulsion</p>
<p><img decoding="async" class="wp-image-4585" src="https://telechargercours.com/wp-content/uploads/2020/05/word-image-201.png"></p>
<p>émise</p>
<table>
<tbody>
<tr>
<td>
<p>impulsion</p>
</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
</tr>
<tr>
<td>
<p>déformée</p>
</td>
<td rowspan="2">
<p>étalement</p>
</td>
<td>&nbsp;</td>
</tr>
<tr>
<td>&nbsp;</td>
<td>&nbsp;</td>
</tr>
<tr>
<td>&nbsp;</td>
<td>
<p>du signal</p>
</td>
<td>&nbsp;</td>
</tr>
</tbody>
</table>
<p>temps</p>
<p>Plus la fréquence de coupure haute est faible, plus les impulsions successives devront être espacées dans le temps pour éviter ce phénomène. Le nombre maximal d’impulsions que peut transmettre un système, par unité de temps, est, au plus, égal au nombre de transitions que le système peut admettre.</p>
<p>Si on désigne par t<sub>e</sub> le temps élémentaire ou temps bit (durée du bit), une succession de 0 et 1 constitue un signal périodique de fréquence f et de période T conforme à la représentation suivante :</p>
<p><img decoding="async" class="wp-image-4586" src="https://telechargercours.com/wp-content/uploads/2020/05/word-image-202.png"></p>
<p>T</p>
<p>temps</p>
<p>te</p>
<p>te=T/2</p>
<table>
<tbody>
<tr>
<td rowspan="2">
<p><em>f </em>=</p>
</td>
<td>
<p>1</p>
</td>
<td rowspan="2">
<p>et <em>T</em> = 2 × <em>t<sub>e</sub></em></p>
</td>
<td>&nbsp;</td>
</tr>
<tr>
<td rowspan="2">
<p><em>T</em></p>
</td>
<td>&nbsp;</td>
</tr>
<tr>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
</tr>
</tbody>
</table>
<p><em>Page 8</em></p>
<p></p>
<p>LES CARACTERISTIQUES DES SUPPORTS DE TRANSMISSION 2005</p>
<p>Un bit compte pour une demi alternance, dans ces conditions, le temps élémentaire est lié à la fréquence par la relation :</p>
<table>
<tbody>
<tr>
<td rowspan="2">
<p><em>t<sub>e</sub> </em>=</p>
</td>
<td>
<p><em>T</em></p>
</td>
<td rowspan="2">
<p>=</p>
</td>
<td>
<p>1</p>
</td>
<td>&nbsp;</td>
</tr>
<tr>
<td>&nbsp;</td>
<td rowspan="2">
<p>2 × <em>f</em></p>
</td>
<td>&nbsp;</td>
</tr>
<tr>
<td colspan="2">
<p>2</p>
</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
</tr>
</tbody>
</table>
<p>Si on désigne par R le nombre de temps élémentaire par unité de temps, R et f sont liés par la relation :</p>
<p><em>R </em>= <sup>1</sup> = 2 ×<em> f</em></p>
<p><em>t<sub>e</sub></em></p>
<p>Si f<sub>max</sub> est la fréquence de coupure haute du canal, la relation devient :</p>
<p><em>R</em><sub>max</sub> = 2 ×<em> f</em><sub>max</sub> ou<em> R </em>≤ 2 ×<em> f</em><sub>max</sub></p>
<p>Si on assimile f<sub>max</sub> à la bande passante (BP) du canal, on obtient la relation :</p>
<p><em>R </em>≤ 2 ×<em> BP</em></p>
<p>R désigne le nombre maximal de transitions qu’un système peut supporter, et est appelé <strong><em>rapidité de modulation</em></strong>. La rapidité de modulation, grandeur analogue à une fréquence, s’exprime en <strong><em>Baud</em></strong> et représente le nombre de temps élémentaires du signal par unité de temps.</p>
<p><strong>5) Notion de débit binaire et de valence de signal</strong></p>
<p>Le <strong><em>débit binaire</em></strong> d’un canal quantifie la quantité d’information transportée sur le canal par unité de temps, il peut s’exprimer par la relation :</p>
<ol>
<li>= 2× <em>BP</em>× log<sub>2</sub> <em>n</em></li>
</ol>
<p>où n exprime la valence du signal, <strong><em>D est exprimé en bit/s</em></strong>.</p>
<p>Dans le cas du signal binaire (0,1) la valence du signal est de deux.. Le débit du canal est alors de :</p>
<ol>
<li>= 2 × <em>BP</em> × log<sub>2</sub> (2) = 2 × <em>BP</em> bit/s</li>
</ol>
<p>Dans ce cas là, la rapidité de modulation correspond au débit binaire.</p>
<p>Si durant le temps élémentaire, le signal peut prendre plusieurs valeurs, par exemple 4, la valence du signal est alors de quatre. Dans ces conditions, le débit du canal est :</p>
<ol>
<li>= 2 × <em>BP</em> × log<sub>2</sub> (4) = 4 × <em>BP</em> bit/s</li>
</ol>
<p>Dans ces conditions, le débit binaire est le double de la rapidité de modulation. C’est ainsi qu’il est possible d’augmenter, sur un canal de transmission de bande passante limitée, le nombre d’informations transmises.</p>
<p><em>Page 9</em></p>
<p></p>
<p>LES CARACTERISTIQUES DES SUPPORTS DE TRANSMISSION 2005</p>
<p>On appelle valence du signal et on la désigne par n, le nombre d’états que peut prendre le signal durant un temps élémentaire. Le débit s’exprime alors par la relation :</p>
<ol>
<li>= 2× <em>BP</em>× log<sub>2</sub> <em>n</em></li>
</ol>
<p><img decoding="async" class="wp-image-4587" src="https://telechargercours.com/wp-content/uploads/2020/05/word-image-203.png"></p>
<p>Signal bivalent ( 2 états)</p>
<p>1 bit par temps élémentaire</p>
<p>1</p>
<p>0</p>
<p>Signal tétravalent ( 4 états)</p>
<p>2 bits par temps élémentaire</p>
<p>11</p>
<p>10</p>
<p>01</p>
<p>00</p>
<p>L’opération qui consiste à faire correspondre à un ensemble de symboles binaires (00, 01…) un ensemble de valeurs représentatives (amplitude, fréquence ou phase), durant un intervalle de temps élémentaire, est effectué par un codeur.</p>
<p>En conclusion, rappelons que l’on peut augmenter le débit binaire, sur un canal de transmission donné, en agissant sur :</p>
<ul>
<li>La bande passante du canal ;</li>
<li>Et/ou la valence du signal transporté.</li>
</ul>
<p>La bande passante est limitée par le système de transmission (support…). On ne peut augmenter indéfiniment le nombre d’états du signal (valence), car les niveaux d’amplitude à discriminer deviennent si faibles qu’ils ne peuvent être distingués du bruit.</p>
<p><em>Page 10</em></p>
<p></p>
<p>LES CARACTERISTIQUES DES SUPPORTS DE TRANSMISSION 2005</p>
<p><strong>5) Notion de bruit</strong></p>
<p>Les signaux transmis sur un canal peuvent être perturbés par des phénomènes électriques ou électromagnétiques désignés sous le terme générique de bruit. On distingue essentiellement deux types de bruit : le bruit blanc et le bruit impulsionnel.</p>
<p><strong><em>Le bruit blanc </em></strong>provient de l’agitation thermique des électrons. Ses composantes (raies de fréquence) sont également réparties dans le spectre des fréquences, d’où son nom. D’amplitude généralement faible, il est peu gênant pour les transmissions.</p>
<p><strong><em>Le bruit impulsionnel </em></strong>est une perturbation brève qui à pour origine l’environnement physique du canal de transmission (parasite d’origine électromagnétique). D’intensité élevée et d’apparition erratique, il provoque des erreurs portant sur un ensemble de bits.</p>
<p>Le rapport entre la puissance du signal transmis et celle du signal de bruit qualifie le canal vis à vis du bruit. Ce rapport, appelé <strong><em>rapport signal sur bruit</em></strong> (S/N avec N pour Noise), s’exprime en dB :</p>
<table>
<tbody>
<tr>
<td>&nbsp;</td>
<td colspan="2">
<p><em>S</em></p>
</td>
<td>&nbsp;</td>
<td colspan="2">
<p><em>S</em></p>
</td>
<td>&nbsp;</td>
</tr>
<tr>
<td rowspan="2">&nbsp;</td>
<td>&nbsp;</td>
<td rowspan="2">&nbsp;</td>
<td rowspan="2">
<p>= 10 × log</p>
</td>
<td>&nbsp;</td>
<td rowspan="2">&nbsp;</td>
<td>&nbsp;</td>
</tr>
<tr>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
</tr>
<tr>
<td>&nbsp;</td>
<td colspan="2">
<p><em>N <sub>dB</sub></em></p>
</td>
<td>&nbsp;</td>
<td colspan="2">
<p><em><sup>N</sup> valeur</em></p>
</td>
<td>&nbsp;</td>
</tr>
</tbody>
</table>
<p>Shannon a montré, qu’en milieu perturbé, le nombre maximal d’états discernables (valence) est donné par la relation :</p>
<p><img decoding="async" class="wp-image-4588" src="https://telechargercours.com/wp-content/uploads/2020/05/word-image-204.png"></p>
<p><em>n </em>= <img decoding="async" class="wp-image-4589" src="https://telechargercours.com/wp-content/uploads/2020/05/word-image-205.png"> 1+ <em><sub>N</sub><sup>S</sup></em></p>
<p>La capacité maximale de transmission d’un canal est donc de :</p>
<table>
<tbody>
<tr>
<td rowspan="3">
<p><em>C </em>= 2×<em> BP</em>× log</p>
</td>
<td>&nbsp;</td>
<td rowspan="3">
<p><em>n </em>=<em> BP</em>× log</p>
</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td colspan="2">
<p><em>S</em></p>
</td>
<td>&nbsp;</td>
</tr>
<tr>
<td rowspan="3">
<p>2</p>
</td>
<td rowspan="3">
<p>2</p>
</td>
<td rowspan="2">
<p>1</p>
</td>
<td rowspan="2">
<p>+</p>
</td>
<td>&nbsp;</td>
<td rowspan="2">&nbsp;</td>
<td>&nbsp;</td>
</tr>
<tr>
<td>&nbsp;</td>
<td>&nbsp;</td>
</tr>
<tr>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td colspan="2">
<p><em>N</em></p>
</td>
<td>&nbsp;</td>
</tr>
</tbody>
</table>
<p><em>Page 11</em></p>]]></content:encoded>
					
		
		
			</item>
		<item>
		<title>LES SYSTEMES AUTOMATISES Limites du modèle : Systèmes linéaires continus et invariants partie2</title>
		<link>https://telechargercours.com/systemes-lineaires-auto-continus-invariants-2/</link>
		
		<dc:creator><![CDATA[Cours Zilla]]></dc:creator>
		<pubDate>Sat, 07 Aug 2021 14:38:43 +0000</pubDate>
				<category><![CDATA[Science & Technologie]]></category>
		<guid isPermaLink="false">https://telechargercours.com/?p=21438</guid>

					<description><![CDATA[Sciences Indusrielles Automatique : introduction Cours (2ème&#160; partie) &#160; 2&#160;&#160; Systèmes dynamiques, Représentations &#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;. 1 &#160; 2.1&#160;&#160; Systèmes dynamiques&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230; 1 2.1.1 Système instantané &#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;.. 1 2.1.2 Système dynamique&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;. 1 &#160; 2.2&#160;&#160; Systèmes continus, linéaires et invariants. &#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;. 3 2.2.1 Système continu &#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230; 3 2.2.2 Représentation des systèmes &#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;.. 3 2.2.3 Systèmes linéaires &#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230; 3 2.2.4 [&#8230;]]]></description>
										<content:encoded><![CDATA[
<h2 class="has-text-align-center wp-block-heading"><strong>Sciences Indusrielles</strong></h2>



<h2 class="has-text-align-center wp-block-heading"><strong>Automatique : introduction</strong></h2>



<h2 class="has-text-align-center wp-block-heading"><strong>Cours (2<sup>ème</sup>&nbsp; partie)</strong></h2>



<figure class="wp-block-table"><table><tbody><tr><td>&nbsp;</td><td>2&nbsp;&nbsp; Systèmes dynamiques, Représentations &#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;.</td><td><a href="#page1">1</a></td></tr><tr><td>&nbsp;</td><td>2.1&nbsp;&nbsp; Systèmes dynamiques&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;</td><td><a href="#page1">1</a></td></tr><tr><td><a href="#page1">2.1.1</a></td><td>Système instantané &#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;..</td><td><a href="#page1">1</a></td></tr><tr><td><a href="#page1">2.1.2</a></td><td>Système dynamique&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;.</td><td><a href="#page1">1</a></td></tr><tr><td>&nbsp;</td><td>2.2&nbsp;&nbsp; Systèmes continus, linéaires et invariants. &#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;.</td><td><a href="#page3">3</a></td></tr><tr><td><a href="#page3">2.2.1</a></td><td>Système continu &#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;</td><td><a href="#page3">3</a></td></tr><tr><td><a href="#page3">2.2.2</a></td><td>Représentation des systèmes &#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;..</td><td><a href="#page3">3</a></td></tr><tr><td><a href="#page3">2.2.3</a></td><td>Systèmes linéaires &#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;</td><td><a href="#page3">3</a></td></tr><tr><td><a href="#page7">2.2.4</a></td><td>Systèmes invariants&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;.</td><td><a href="#page7">7</a></td></tr><tr><td><a href="#page7">2.2.5</a></td><td>Les systèmes réels &#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;</td><td><a href="#page7">7</a></td></tr><tr><td>&nbsp;</td><td>2.3&nbsp;&nbsp; Représentation d’un système linéaire continu et invariant&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;.</td><td><a href="#page7">7</a></td></tr><tr><td><a href="#page8">2.3.1</a></td><td>Représentation par équations différentielles&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;..</td><td><a href="#page8">8</a></td></tr><tr><td><a href="#page9">2.3.2</a></td><td>Méthodes de résolution d&rsquo;équations intégro-différentielles par transformée de Laplace &#8230;&#8230;&#8230;</td><td><a href="#page9">9</a></td></tr><tr><td>&nbsp;</td><td>&nbsp;</td><td>**************</td><td>&nbsp;</td></tr></tbody></table></figure>



<h2 class="wp-block-heading"><strong>2 SYSTEMES DYNAMIQUES, REPRESENTATIONS</strong></h2>



<h3 class="wp-block-heading"><strong><em>2.1 Systèmes dynamiques.</em></strong></h3>



<h4 class="wp-block-heading">2.1.1&nbsp;&nbsp;&nbsp;&nbsp; SYSTEME INSTANTANE</h4>



<p class="wp-block-paragraph">On dit qu’un système est instantané si les grandeurs physiques de sortie dépendent uniquement et instantanément des grandeurs d’entrée, cette dépendance n’évoluant pas avec le temps.</p>



<p class="wp-block-paragraph">Exemples :</p>



<p class="wp-block-paragraph">La relation courant tension dans une résistance est un phénomène instantané, ne dépendant pas du temps. Le système est instantané, on peut qualifier les relations entrées sortie par des coefficients constants appelés gain ou gain pure.</p>



<p class="wp-block-paragraph">En réalité, il existe peu de systèmes instantanés car tout effet présente une certaine inertie ou mémoire. L’appellation « système instantané » relève donc souvent de l’approximation. Elle est justifiée lorsque le temps de réaction est négligeable devant la durée de transition de l’information dans les autres systèmes environnants (cette notion est donc excessivement relative). Par exemple, une résistance électrique (entrée tension, sortie en courant) sera modifiée par un coefficient constant pour une certaine gamme de fréquence. Lorsque la fréquence croît, il devient nécessaire de prendre en compte les effets d’inductance ou de capacité qui conduisent alors à représenter son comportement par une équation différentielle, donc à considérer un comportement dynamique.</p>



<p class="wp-block-paragraph"><a href="https://telechargercours.com/cours-sciences-indusrielles-systemes-automatises-limites-modele/" target="_blank" rel="noreferrer noopener">LES SYSTEMES AUTOMATISES Limites du modèle : Systèmes linéaires continus et invariants <strong>(1ère partie)</strong></a></p>



<h4 class="wp-block-heading">2.1.2&nbsp;&nbsp;&nbsp;&nbsp; SYSTEME DYNAMIQUE</h4>



<p class="wp-block-paragraph">La notion de système dynamique prend en compte ces phénomènes d’inertie et de mémoire (inertie mécanique, inertie thermique) et dans ce cas les grandeurs de sortie dépendent des valeurs présentes et passées des grandeurs d’entrées. C’est notamment le cas des systèmes régis par des équations intégro-différentielles.</p>



<p class="wp-block-paragraph">Cette mémoire passé est généralement de durée infinie, tout en s’atténuant le plus souvent selon les modes exponentiels. Ainsi, on considère en pratique qu’un système dynamique à une mémoire finie dont la durée fixe le temps de réponse.</p>



<h5 class="wp-block-heading"><strong>2.1.2.1</strong>  <strong>Exemple de système dynamique</strong></h5>



<p class="wp-block-paragraph">Il s’agit d’une masse M soumise à une action mécanique F(t) de direction l’axe y. Ce solide de masse M est lié à un support fixe par un ressort de raideur K et un amortisseur de coefficient de frottement visqueux f. Au repos, le solide a une position y<sub>0</sub> et que la norme de F(t)=f <sub>0</sub> est nulle.</p>



<p class="wp-block-paragraph">L’entrée et la sortie du système sont définie par le schéma fonctionnel ci-dessous.</p>


<div class="wp-block-image">
<figure class="aligncenter size-full" id="Systèmes-linéaires"><img decoding="async" src="https://telechargercours.com/wp-content/uploads/2021/08/systeme-dynamique.png" alt="LES SYSTEMES AUTOMATISES Systèmes linéaires continus et invariants partie2" class="wp-image-21441" title="LES SYSTEMES AUTOMATISES Systèmes linéaires "/></figure>
</div>


<p class="wp-block-paragraph">Lors de l’intégration, on montre qu’il suffit de connaître la position de la vitesse de la masse M à l’instant initial pour être en mesure de décrire l’évolution ultérieure de la position y(t),en supposant connue également connue F(t) pour t&gt;t <sub>0</sub>. Ces deux grandeurs de position et de vitesse à l’instant t<sub>0</sub> résument le passé du système. On peut illustrer l’effet de mémoire de ce système en appliquant une force impulsionnelle (un Dirac), le système oscille encore bien après la disparition de la force.</p>



<h3 class="wp-block-heading"><strong><em>2.2 Systèmes continus, linéaires et invariants.</em></strong></h3>



<h4 class="wp-block-heading">2.2.1&nbsp;&nbsp;&nbsp; SYSTEME CONTINU</h4>



<p class="wp-block-paragraph">Un système continu par opposition d’un système discret (hors programme en C.P.G.E.), lorsque les variations des grandeurs physiques le caractérisant sont des fonctions à temps continu et que l’on peut donc définir ces grandeurs à tout instant. On par aussi de système analogique.</p>



<p class="wp-block-paragraph">La plupart des systèmes sont continus du point de vue macroscopique. Mais, par exemple, un système informatique n&rsquo;est pas continu car il ne peut traiter que des échantillons des signaux continus qui lui sont soumis : on parle dans ce cas d&rsquo;un système échantillonné.</p>


<div class="wp-block-image">
<figure class="aligncenter size-full"><img decoding="async" src="https://telechargercours.com/wp-content/uploads/2021/08/systeme-continu.png" alt="Systèmes linéaires continus invariants" class="wp-image-21442" title="Systèmes linéaires continus invariants"/></figure>
</div>


<h4 class="wp-block-heading">2.2.2&nbsp;&nbsp;&nbsp;&nbsp; REPRESENTATION DES SYSTEMES</h4>



<p class="wp-block-paragraph">Dans l&rsquo;étude d&rsquo;un système, on cherche à établir les  » relations  » existant entre ses entrées et ses sorties. Des relations mathématiques peuvent représenter ces  » relations  » entrées-sorties du système. Cet</p>



<p class="wp-block-paragraph">ensemble de relations mathématiques représente un modèle mathématique du système.</p>



<p class="wp-block-paragraph">Dans le cas où le modèle mathématique du système est un ensemble ( ou système ) d&rsquo;équations différentielles linéaires et à coefficients constants on parle d&rsquo;un système continu, linéaire et invariant.</p>



<p class="wp-block-paragraph"><strong>Exemple :</strong>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; Soit le système monovariable :</p>


<div class="wp-block-image">
<figure class="aligncenter size-full"><img decoding="async" src="https://telechargercours.com/wp-content/uploads/2021/08/systeme-monovariable.png" alt="" class="wp-image-21443"/></figure>
</div>


<h5 class="wp-block-heading"><strong>2.2.2.1</strong> <strong>Exemples d&rsquo;équations différentielles :</strong></h5>



<p class="wp-block-paragraph"><strong>Continuité : </strong>les fonctions liées par l&rsquo;équation différentielle étant continues, le système qu&rsquo;elle représente est dit continu.</p>


<div class="wp-block-image">
<figure class="aligncenter size-full"><img decoding="async" src="https://telechargercours.com/wp-content/uploads/2021/08/equations-differentielles.png" alt="" class="wp-image-21444"/></figure>
</div>


<h4 class="wp-block-heading">2.2.3  SYSTEMES LINEAIRES</h4>



<p class="wp-block-paragraph">L&rsquo;hypothèse de linéarité, relative au comportement des systèmes, traduit simplement que  » <strong>l&rsquo;effet est</strong> <strong>proportionnel à la cause </strong>« .</p>



<p class="wp-block-paragraph">Cette notion, bien ancrée dans les esprits, repose donc en fait sur les principes de proportionnalité et de superposition.</p>



<p class="wp-block-paragraph">Dans le cas où le modèle mathématique représentant le système étudié forme un ensemble d&rsquo;équations différentielles linéaires on aurait affaire à un système continu et linéaire.</p>



<h5 class="wp-block-heading"><strong>2.2.3.1</strong>&nbsp; <strong>Propriétés</strong></h5>



<p class="wp-block-paragraph">Si à l&rsquo;entrée d&rsquo;un système linéaire on applique e<sub>1</sub>, une grandeur constante, et que l&rsquo;on obtienne, à la sortie du système, après un certain temps d&rsquo;attente, s<sub>1</sub>, une autre grandeur constante alors :</p>



<p class="wp-block-paragraph">En appliquant à l&rsquo;entrée du système k.e<sub>1</sub>, une grandeur constante proportionnelle à e<sub>1</sub>, on obtiendra, à la sortie du système, après un certain temps d&rsquo;attente, k.s<sub>1</sub>, grandeur proportionnelle à s<sub>1</sub>.</p>



<h6 class="wp-block-heading">2.2.3.1.1 Schéma traduisant le comportement, dans le temps, d&rsquo;un système linéaire.</h6>


<div class="wp-block-image">
<figure class="aligncenter size-full"><img decoding="async" src="https://telechargercours.com/wp-content/uploads/2021/08/systeme-lineaire.png" alt="" class="wp-image-21447"/></figure>
</div>


<h6 class="wp-block-heading"> 2.2.3.1.1 Schéma traduisant le comportement, dans le temps, d&rsquo;un système non linéaire. </h6>


<div class="wp-block-image">
<figure class="aligncenter size-full"><img decoding="async" src="https://telechargercours.com/wp-content/uploads/2021/08/systeme-non-lineaire.png" alt="" class="wp-image-21448"/></figure>
</div>

<div class="wp-block-image">
<figure class="aligncenter size-full"><img decoding="async" src="https://telechargercours.com/wp-content/uploads/2021/08/Screenshot-31.png" alt="" class="wp-image-21449"/></figure>
</div>


<h5 class="wp-block-heading"><strong>2.2.3.2</strong> &nbsp;&nbsp; <strong>Principe de superposition</strong></h5>



<p class="wp-block-paragraph">Si à l&rsquo;entrée du système linéaire on applique la fonction e<sub>1</sub>(t) et que l&rsquo;on obtienne, à la sortie s<sub>1</sub> (t). Si ensuite à l&rsquo;entrée du même système linéaire on applique e<sub>2</sub>(t) et que l&rsquo;on obtienne, à la sortie s<sub>2</sub>(t). Alors en appliquant à l&rsquo;entrée du système e<sub>1</sub> (t) et e<sub>2</sub> (t), on obtiendra, à la sortie du système s<sub>1</sub> (t) + s<sub>2</sub> (t).</p>



<p class="wp-block-paragraph">Représentation sous forme de schéma fonctionnel du Principe de superposition</p>


<div class="wp-block-image">
<figure class="aligncenter size-full"><img decoding="async" src="https://telechargercours.com/wp-content/uploads/2021/08/principe-de-superposition.png" alt="" class="wp-image-21452"/></figure>
</div>


<h5 class="wp-block-heading"><strong>2.2.3.3</strong> <strong>Conséquence pratique sur l’étude des systèmes multi-variables</strong></h5>



<p class="wp-block-paragraph">Cela permet de simplifier l&rsquo;étude d&rsquo;un système soumis à plusieurs entrées, par exemple une consigne et une perturbation.</p>


<div class="wp-block-image">
<figure class="aligncenter size-full"><img decoding="async" src="https://telechargercours.com/wp-content/uploads/2021/08/systemes-multi-variables.png" alt="" class="wp-image-21453"/></figure>
</div>


<p class="wp-block-paragraph">Ainsi pour une modélisation linéaire on pourra composer le schéma fonctionnel ci-dessus en deux études.</p>



<h6 class="wp-block-heading">2.2.3.3.1  Étude en poursuite</h6>



<p class="wp-block-paragraph">Dans cette étude, on considère que la perturbation P(t) et toujours nulle. On recherchera Sr(t) (effet de la consigne sur la sortie). Pour ce faire :</p>



<p class="wp-block-paragraph"><strong>Pour étudier le régime transitoire, </strong>on étudie<strong> la stabilité et le temps de réponse</strong>.</p>



<p class="wp-block-paragraph">Pour <strong>étudier le régime permanent</strong>, on recherchera <strong>la précision statique.</strong></p>


<div class="wp-block-image">
<figure class="aligncenter size-full"><img decoding="async" src="https://telechargercours.com/wp-content/uploads/2021/08/etude-en-porsuite.png" alt="" class="wp-image-21456"/></figure>
</div>


<h6 class="wp-block-heading">2.2.3.3.2 Étude en régulation</h6>



<p class="wp-block-paragraph">Dans cette étude, on considère que la consigne C(t) et toujours nulle. On recherchera Sc(t) (effet de la perturbation sur la sortie).</p>



<p class="wp-block-paragraph">Il est inutile d’étudier le régime transitoire en perturbation, si le système est stable en poursuite, alors, il l’est en régulation.</p>



<p class="wp-block-paragraph">De plus le critère de rapidité défini par le temps de réponse à 5% n’est évaluer qu’en poursuite.</p>



<p class="wp-block-paragraph">On étudiera donc que la précision, à savoir si dans le temps, la perturbation modifie la sortie. Un système bien régulé est un système tel que dans le temps, la sortie reste inchangée.</p>


<div class="wp-block-image">
<figure class="aligncenter size-full"><img decoding="async" src="https://telechargercours.com/wp-content/uploads/2021/08/Screenshot-34.png" alt="" class="wp-image-21457"/></figure>
</div>


<h6 class="wp-block-heading">2.2.3.3.3 Étude d’un système multi-variables</h6>



<p class="wp-block-paragraph">Puisque le système est linéaire, en appliquant le principe de superposition, on aura donc :</p>


<div class="wp-block-image">
<figure class="aligncenter size-full"><img decoding="async" src="https://telechargercours.com/wp-content/uploads/2021/08/Screenshot-33.png" alt="" class="wp-image-21458"/></figure>
</div>


<p class="wp-block-paragraph"><em><strong>2.2.3.3.3.1&nbsp;  Réponse indicielle d’un système multi-variables</strong></em></p>



<p class="wp-block-paragraph"><strong>Explication de la procédure de teste ayant conduit aux courbes ci-dessous.</strong></p>



<p class="wp-block-paragraph">On supposant que le système est stable. À l’instant t <sub>0</sub>, il est envoyé sur la consigne et échelon de poids E<sub>0</sub>, une durée de (tr – t0) plus tard, il est envoyé en perturbation un échelon de poids –P<sub>0</sub>.</p>



<p class="wp-block-paragraph"><strong>Système est dit mal régulé, alors on obtient la réponse suivante :</strong></p>


<div class="wp-block-image">
<figure class="aligncenter size-full"><img decoding="async" src="https://telechargercours.com/wp-content/uploads/2021/08/Screenshot-36.png" alt="" class="wp-image-21461"/></figure>
</div>


<p class="wp-block-paragraph">On remarque que le système ne revient pas en position après avoir appliqué la perturbation.</p>



<p class="wp-block-paragraph">Par exemple, la suspension d’une voiture Peugeot ne revient pas en position lorsque l’on met un sac de 50 kg de ciment dans le coffre.</p>



<p class="wp-block-paragraph"><strong>Système est dit bien régulé, alors on obtient la réponse suivante :</strong></p>


<div class="wp-block-image">
<figure class="aligncenter size-full"><img decoding="async" src="https://telechargercours.com/wp-content/uploads/2021/08/Screenshot-35.png" alt="" class="wp-image-21462"/></figure>
</div>


<p class="wp-block-paragraph">On remarque que le système revient en position après avoir appliqué la perturbation.</p>



<p class="wp-block-paragraph">Par exemple, la suspension d’une voiture Citroën (équipée du système SCMAC) revient en position lorsque l’on met un sac de 50 kg de ciment dans le coffre. Le système de maintient d’assiette (hauteur) du véhicule est donc bien régulé.</p>



<h4 class="wp-block-heading">2.2.4&nbsp;&nbsp;&nbsp; SYSTEMES INVARIANTS</h4>



<p class="wp-block-paragraph">Un système invariant a des caractéristiques qui ne se modifient pas dans le temps. Autrement dit c&rsquo;est un système ayant des composants qui ne vieillissent pas.</p>



<p class="wp-block-paragraph">Si à l&rsquo;entrée du système invariant on applique la fonction e (t)&nbsp; et que l&rsquo;on obtienne à la sortie s (t), alors :</p>



<p class="wp-block-paragraph">pour tout décalage temporel τ, si on applique e (t &#8211; τ )&nbsp; on obtiendra à la sortie s (t &#8211; τ ) .</p>



<h4 class="wp-block-heading">2.2.5&nbsp;&nbsp;&nbsp; LES SYSTEMES REELS</h4>



<p class="wp-block-paragraph">En toute théorie, les systèmes réels « dit physiques » ne sont :</p>



<ul class="wp-block-list"><li>ni continus ( point de vue microscopique )</li></ul>



<ul class="wp-block-list"><li>ni linéaires ( les courbes caractéristiques statiques ne sont jamais rigoureusement des droites)</li></ul>



<ul class="wp-block-list"><li>ni invariants ( vieillissement des composants )</li></ul>



<p class="wp-block-paragraph">Dans la pratique nous nous ramènerons au cas continu, linéaire et invariant par des hypothèses simplificatrices le plus souvent justifiées.</p>



<h3 class="wp-block-heading"><strong><em>2.3 </em>Représentation <em>d’un système linéaire continu et invariant</em></strong></h3>



<p class="wp-block-paragraph">Pour réaliser une commande automatique d&rsquo;un système, il est nécessaire d&rsquo;établir les relations existant entre les entrées et les sorties. L&rsquo;ensemble de ces relations s&rsquo;appelle « modèle mathématique » du système.</p>



<p class="wp-block-paragraph">En utilisant le principe de superposition, les systèmes linéaires multi-variables se ramène à plusieurs études de système monovariable. C’est pourquoi dans la suit de ce document, nous nous limiterons aux systèmes monovariables.</p>


<div class="wp-block-image">
<figure class="aligncenter size-full"><img decoding="async" src="https://telechargercours.com/wp-content/uploads/2021/08/Screenshot-38-1.png" alt="" class="wp-image-21466"/></figure>
</div>


<h4 class="wp-block-heading">2.3.1 REPRESENTATION PAR EQUATIONS DIFFERENTIELLES</h4>



<p class="wp-block-paragraph">On représente classiquement l&rsquo;évolution d&rsquo;un système dynamique linéaire et invariant par une équation différentielle linéaire à coefficients constants liant les grandeurs d&rsquo;entrée et de sortie :</p>


<div class="wp-block-image">
<figure class="aligncenter size-full"><img decoding="async" src="https://telechargercours.com/wp-content/uploads/2021/08/Screenshot-40.png" alt="" class="wp-image-21467"/></figure>
</div>


<p class="wp-block-paragraph">Les cas pratiques rencontrés imposent d&rsquo;avoir, <em>m</em> ≤ <em>n </em>est appelé l&rsquo;<strong>ordre</strong>du système</p>



<p class="wp-block-paragraph">La résolution d&rsquo;une telle équation différentielle à second membre s&rsquo;obtient en ajoutant une solution particulière à la solution générale sans second membre.</p>



<ul class="wp-block-list"><li>La solution générale s&rsquo;obtient par résolution de l&rsquo;équation sans second membre :</li></ul>


<div class="wp-block-image">
<figure class="aligncenter size-full"><img decoding="async" src="https://telechargercours.com/wp-content/uploads/2021/08/Screenshot-41.png" alt="" class="wp-image-21468"/></figure>
</div>


<ul class="wp-block-list"><li>Pour une solution particulière de l&rsquo;équation avec second membre, on choisit la forme de la solution</li></ul>



<p class="wp-block-paragraph">connaissant E(t), puis on détermine les constantes par identification.</p>



<p class="wp-block-paragraph">La réponse en régime établi (ou régime définitif ou permanent) d&rsquo;un système linéaire à une entrée de type donné est un signal de même nature que l&rsquo;entrée.</p>



<p class="wp-block-paragraph">Par exemple :</p>


<div class="wp-block-image">
<figure class="aligncenter size-full"><img decoding="async" src="https://telechargercours.com/wp-content/uploads/2021/08/Screenshot-42.png" alt="" class="wp-image-21471"/></figure>
</div>


<p class="wp-block-paragraph">Ainsi :</p>



<ul class="wp-block-list"><li>Le régime libre ou transitoire ne dépend que du système et de ses conditions initiales.</li></ul>



<ul class="wp-block-list"><li>Le régime forcé ou permanent est de même nature que celui appliqué à l&rsquo;entrée du système.</li></ul>



<ul class="wp-block-list"><li>Dans les systèmes stables le régime libre ou transitoire s&rsquo;annule au bout d&rsquo;un certain temps.</li></ul>



<h4 class="wp-block-heading">2.3.2 METHODES DE RESOLUTION D&rsquo;EQUATIONS INTEGRO-DIFFERENTIELLES PAR TRANSFORMEE DE LAPLACE</h4>



<p class="wp-block-paragraph">Nous allons maintenant introduire une technique permettant de se ramener à de simples opérations algébriques lors de la résolution. Cette technique suppose au départ une transformation dite de Laplace, puis à l&rsquo;arrivée une transformée inverse pour obtenir la solution temporelle :</p>


<div class="wp-block-image">
<figure class="aligncenter size-full"><img decoding="async" src="https://telechargercours.com/wp-content/uploads/2021/08/Screenshot-43.png" alt="" class="wp-image-21472"/></figure>
</div>


<p class="has-text-align-center wp-block-paragraph"><strong><a href="https://www.mediafire.com/file/kk23vbwnjymv2n1/Telechargercours.com+SI-CO02-AAS-AC_unlocked.pdf/file" target="_blank" rel="noreferrer noopener">Telecharger</a></strong></p>
]]></content:encoded>
					
		
		
			</item>
		<item>
		<title>LES SYSTEMES AUTOMATISES  Limites du modèle : Systèmes linéaires continus et invariants</title>
		<link>https://telechargercours.com/cours-sciences-indusrielles-systemes-automatises-limites-modele/</link>
		
		<dc:creator><![CDATA[Cours Zilla]]></dc:creator>
		<pubDate>Wed, 07 Jul 2021 16:25:16 +0000</pubDate>
				<category><![CDATA[Science & Technologie]]></category>
		<guid isPermaLink="false">https://telechargercours.com/?p=18367</guid>

					<description><![CDATA[Sciences Indusrielles Automatique : introduction Cours (1ère partie) Schéma fonctionnel de l’asservissement Plan&#160; &#160; &#160; &#160; ********************** &#160; &#160; 1&#160;&#160; INTRODUCTION ASPECTS GÉNÉRAUX &#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230; 2 1.1 généralités sur l’automatique&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;.. 2 1.2 Notion de systèmes&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230; 3 1.3 Rôle de l&#8217;automatique &#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230; 5 1.4 Bref historique&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;.. 7 1.5 Schémas fonctionnels &#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;.. 9 1.6 Structure d&#8217;un système de [&#8230;]]]></description>
										<content:encoded><![CDATA[
<p class="has-text-align-center wp-block-paragraph"><strong>Sciences Indusrielles</strong></p>



<p class="has-text-align-center wp-block-paragraph"><strong>Automatique : introduction</strong></p>



<p class="has-text-align-center wp-block-paragraph"><strong>Cours (1ère partie)</strong></p>


<div class="wp-block-image">
<figure class="aligncenter size-large"><img decoding="async" src="blob:https://telechargercours.com/c6d6a8ac-b044-4583-86df-f58e4d047428" alt=""/><figcaption><strong>Simulateur de vol 6 axes : Thomson CSF</strong></figcaption></figure>
</div>


<p class="has-text-align-center wp-block-paragraph"><strong><em><u>Schéma fonctionnel de l’asservissement</u></em></strong></p>



<figure class="wp-block-image size-large" id="les-systemes-automatises"><img decoding="async" src="https://telechargercours.com/wp-content/uploads/2021/07/les-systemes-automatises-Schema-fonctionnel-de-l-asservissement.png" alt="les systemes automatises" class="wp-image-18369" title="les systemes automatises"/></figure>



<figure class="wp-block-table"><table><tbody><tr><td><strong>Plan&nbsp;</strong></td><td>&nbsp;</td></tr><tr><td>&nbsp;</td><td>&nbsp;</td><td>**********************</td><td>&nbsp;</td></tr><tr><td>&nbsp;</td><td>1&nbsp;&nbsp; INTRODUCTION ASPECTS GÉNÉRAUX &#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;</td><td>2</td></tr><tr><td>1.1</td><td>généralités sur l’automatique&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;..</td><td>2</td></tr><tr><td>1.2</td><td>Notion de systèmes&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;</td><td>3</td></tr><tr><td>1.3</td><td>Rôle de l&rsquo;automatique &#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;</td><td>5</td></tr><tr><td>1.4</td><td>Bref historique&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;..</td><td>7</td></tr><tr><td>1.5</td><td>Schémas fonctionnels &#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;..</td><td>9</td></tr><tr><td>1.6</td><td>Structure d&rsquo;un système de commande ou système asservi..</td><td>11</td></tr><tr><td>&nbsp;</td><td>&nbsp;</td><td>**********************</td><td>&nbsp;</td></tr></tbody></table></figure>



<h2 class="wp-block-heading"><strong>1 INTRODUCTION ASPECTS GÉNÉRAUX</strong></h2>



<h3 class="wp-block-heading"><strong>1.1&nbsp; généralités sur l’automatique</strong></h3>



<p class="wp-block-paragraph">Le substantif <strong>« Automatique »</strong> a été utilisé pour la première fois en 1914 dans la revue académique royale des sciences de Madrid publié par Torres y Quevedo.</p>



<ul class="wp-block-list"><li>Le petit Larousse propose la définition suivante :</li></ul>



<p class="wp-block-paragraph"><strong>Automatique : </strong>Science et technique de l&rsquo;automatisation, qui étudient les méthodes scientifiques et lemoyen technologique utilisés pour la conception et la construction des systèmes automatiques.</p>



<p class="wp-block-paragraph"><strong>Automatisation : </strong>Exécution automatique de tâches industrielles, administratives ou scientifiques, sansintervention humaine.</p>



<p class="wp-block-paragraph">L&rsquo;automatique peut se définir comme un ensemble de théories mathématiques et une technique de raisonnement concernant la prise de décision et la commande des systèmes. La dénomination anglaise est plus explicite &lsquo;Automatique Control » puisqu&rsquo;elle précise la notion de commande. Le terme anglais « Control » est un faux ami à ne pas traduire comme contrôle, il signifie Commande.</p>



<p class="wp-block-paragraph"><strong>Définition du mot système : </strong>Nous le verrons dans d&rsquo;autre chapitre de cours mais nous pouvons ici retenir une définition littéraire : Un Système consiste en une combinaison de parties qui se coordonnent pour concourir à un résultat.</p>



<p class="wp-block-paragraph"><strong><a href="https://telechargercours.com/systemes-lineaires-auto-continus-invariants-2/" target="_blank" rel="noreferrer noopener">LES SYSTEMES AUTOMATISES Limites du modèle : Systèmes linéaires continus et invariants partie2</a></strong></p>



<p class="wp-block-paragraph">Le but d&rsquo;un système est d&rsquo;exécuter des actions qui sont regroupées en a activités. Chaque élément du système assume sa part d&rsquo;activité. Une action est le résultat de l&rsquo;organisation de matière, d&rsquo;énergie et d&rsquo;information. En quelque sorte, l&rsquo;information, avec le concours de l&rsquo;énergie, va modifier la matière en lui conférant un plus grand degré d&rsquo;organisation.</p>


<div class="wp-block-image">
<figure class="aligncenter size-large"><img decoding="async" src="https://telechargercours.com/wp-content/uploads/2021/07/sorties-entrees-systeme.png" alt="" class="wp-image-18370"/></figure>
</div>


<p class="wp-block-paragraph">Des objets agencés entre eux, c&rsquo;est à dire ayant une certaine dépendance, constituent un système réalisant une certaine fonction ou action.</p>



<p class="wp-block-paragraph">Les sorties-entrées sont les signaux qui apportent au système, les informations du milieu extérieur. Les sorties fournissent la réponse du système par dépendance des entrées; on peut parler de causes (entrées) et d&rsquo;effets (sorties).</p>



<p class="wp-block-paragraph">On utilise de manière synonyme le terme de processus : process en anglais. Théoriquement, un processus se défini comme un ensemble de lois d&rsquo;évolution de différentes grandeurs physiques, c&rsquo;est la façon dont évolue un système sous l&rsquo;effet des entrées. Dans le langage courant, on confond souvent le processus, notion abstraite, et l&rsquo;installation matérielle dont le fonctionnement est régi par ces lois.</p>



<p class="wp-block-paragraph">La commande d&rsquo;un système consiste à exercer, via les entrées, une influence sur le système de manière à obtenir en sortie un comportement déterminé. Cette influence s&rsquo;exerce souvent par l&rsquo;intermédiaire d&rsquo;une action sur le flux d&rsquo;énergie ou de matériaux injectés dans le système. Lorsque cette influence est exercée par l&rsquo;Homme, la commande est dite manuelle. Lorsque l&rsquo;Homme est remplacé par des dispositifs techniques autonomes, on parle de commande automatique.</p>



<p class="wp-block-paragraph">L&rsquo;automatique s&rsquo;est longtemps appliquée qu&rsquo;aux systèmes mécaniques, électroniques et électromécaniques, mais maintenant elle est utilisée en gestion, biologie, économie, etc.</p>



<p class="wp-block-paragraph">Les systèmes de commande automatique copient le plus souvent le comportement de l&rsquo;Homme, Un exemple introductif simple s&rsquo;obtient par observation d&rsquo;un conducteur au volant de son véhicule. Le système que l&rsquo;Homme commande est sa voiture dans un environnement de la route; le cerveau et les membres constituent les organes de commande. Les décisions concernant la direction l&rsquo;accélération et le freinage sont mises en oeuvre, à partir des mesures effectuées par l’œil, de manière à satisfaire un critère de performance qui peut -être -un compromis entre la durée du trajet, le confort, la consommation ou les réglementations.</p>



<p class="wp-block-paragraph">Cet exemple nous montre les trois opérations fondamentales accomplies par l&rsquo;Homme.</p>



<ul class="wp-block-list"><li><strong>l&rsquo;observation</strong></li></ul>



<ul class="wp-block-list"><li><strong>la réflexion</strong></li><li><strong>l&rsquo;action</strong></li></ul>



<p class="wp-block-paragraph">Il est intéressant de noter que des types d&rsquo;actions différentes peuvent être appliqués, suivant les critères intégrés lors de la phase de réflexion. Un conducteur pressé appliquera une succession de freinages et accélérations, au détriment de la consommation alors qu&rsquo;un autre conducteur économe adoptera une conduite plus souple.</p>



<p class="wp-block-paragraph">La structure à trois phases met en évidence une opération de bouclage (« Feed-back », nourrir en retour).</p>



<p class="wp-block-paragraph">Ce retour constitue l&rsquo;une des notions importantes de l&rsquo;automatique.</p>



<h3 class="wp-block-heading"><strong><em>1.2&nbsp; Notion de systèmes</em></strong></h3>



<p class="wp-block-paragraph">Un système est un ensemble d&rsquo;éléments liés entre eux dans le but de réaliser une tâche déterminée. Ce dispositif soumis aux lois physiques est caractérisé par des grandeurs de deux types : Les ENTREES et Les SORTIES.</p>



<p class="wp-block-paragraph">En général, les entrées et les sorties sont multiples. On définit ses systèmes comme multivariables.</p>



<h4 class="wp-block-heading">1.2.1&nbsp;&nbsp;&nbsp;&nbsp; REPRESENTATION D&rsquo;UN SYSTEME</h4>



<p class="wp-block-paragraph">Les entrées peuvent être classées en deux types :</p>



<ul class="wp-block-list"><li>Des grandeurs de commandes du système appelées communément Entrées de commandes. Les entrées de commandes correspondent aux signaux que l&rsquo;on pourra modifier afin d&rsquo;agir sur le système.</li></ul>



<ul class="wp-block-list"><li>Des signaux parasites appelés communément Perturbations.</li></ul>



<p class="wp-block-paragraph">Les perturbations correspondent aux signaux que l&rsquo;on ne pourra pas manipuler et que le système devra subir.</p>



<p class="wp-block-paragraph"><strong>Par exemple : </strong>Pour un four, les grandeurs suivantes peuvent être relevées</p>



<ul class="wp-block-list"><li>Entrée de commande: Débit de combustible</li></ul>



<ul class="wp-block-list"><li>Entrée de perturbation déperdition de chaleur</li></ul>



<ul class="wp-block-list"><li>Sortie : température à l&rsquo;intérieur du four</li></ul>



<p class="wp-block-paragraph">Il est devenu maintenant naturel de modéliser un système quelconque. Une représentation graphique de système se représente par un schéma fonctionnel dont on verra la normalisation plus loin dans ce document.</p>


<div class="wp-block-image">
<figure class="aligncenter size-large"><img decoding="async" src="https://telechargercours.com/wp-content/uploads/2021/07/schema-fonctionnel.png" alt="" class="wp-image-18373"/></figure>
</div>


<p class="wp-block-paragraph">Dans ce schéma fonctionnel sont indiquées les entrées accessibles à la commande, les entrées parasites subies (perturbations), et les sorties intéressantes. Par exemple pour qualifier le comportement d&rsquo;un véhicule automobile, on dresse une liste sommaire de grandeurs utiles et on précise leur nature avec le schéma ci-contre.</p>



<h4 class="wp-block-heading"><strong>1.2.2&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; SYSTEME DYNAMIQUE</strong></h4>



<p class="wp-block-paragraph">Un système dynamique est un système dont la réponse dépend simultanément de l&rsquo;excitation présente et des excitations et réponses passées. Les systèmes qui répercutent instantanément l&rsquo;effet d&rsquo;une entrée en sortie, ou sous une autre forme, dont la sortie à un instant ne dépend que de la valeur de l&rsquo;entrée au même instant, sont dits <strong><em>instantanés</em></strong>.</p>



<p class="wp-block-paragraph">On se l&rsquo;imitera dans un premier temps à des systèmes localisés (four, voiture, commande direction de bateau, etc.) dont le comportement est décrit par des équations différentielles ordinaires. Par opposition aux systèmes dits à paramètres distribués ou répartis dont la description du comportement met en oeuvre des phénomènes de propagation (ligne électrique) modélisés par des équations aux dérivées partielles.</p>



<p class="wp-block-paragraph">Les systèmes réels sont en général des systèmes multivariables (plusieurs entrées-sorties). Cependant, la difficulté de réaliser une commande multidimensionnelle est telle que l&rsquo;automaticien est incité à sélectionner un couple entrée-sortie conférant ainsi au système une représentation dite <strong><em>monovariable</em></strong>. L&rsquo;effet des autres entrées ou perturbations est analysé séparément. Si le système subit plusieurs perturbations, on analysera leur effet en ne considérant qu&rsquo;une seule perturbation présente à la fois puis on superposera les effets (Principe de superposition)</p>



<h4 class="wp-block-heading"><strong>1.2.3&nbsp;&nbsp;&nbsp;&nbsp; DEMARCHE D&rsquo;APPROXIMATION D&rsquo;UN SYSTEME COMPLEXE</strong></h4>



<p class="wp-block-paragraph"><em>Exemple : échangeur de chaleur.</em></p>


<div class="wp-block-image">
<figure class="aligncenter size-large"><img decoding="async" src="https://telechargercours.com/wp-content/uploads/2021/07/Screenshot-4.png" alt="" class="wp-image-18376"/></figure>
</div>


<p class="wp-block-paragraph">Un échangeur est constitué de deux enceintes dans lesquelles circulent des fluides de températures différentes. La proximité des fluides entraîne des échanges thermiques, permettant ainsi de refroidir ou réchauffer un fluide.</p>



<p class="wp-block-paragraph">Si la fonction de service de cet échangeur est de refroidir le fluide chaud, Tc est considéré comme grandeur unique de sortie. Il semble évident que les débits Qc et Qf agissent sur cette température. Ce système est donc naturellement multidimensionnel.</p>


<div class="wp-block-image">
<figure class="aligncenter size-large"><img decoding="async" src="https://telechargercours.com/wp-content/uploads/2021/07/echangeur.png" alt="" class="wp-image-18377"/></figure>
</div>


<p class="wp-block-paragraph">L&rsquo;automaticien pourra retenir comme seule variable d&rsquo;entrée le débit <strong>Qc</strong> et considérer en première approximation que <strong>Qf</strong> est constant. une variation de <strong>Qf</strong> serait considérée comme une perturbation, d&rsquo;où le schéma fonctionnel simplifier suivant :</p>


<div class="wp-block-image">
<figure class="aligncenter size-large"><img decoding="async" src="https://telechargercours.com/wp-content/uploads/2021/07/echangeur-2.png" alt="" class="wp-image-18378"/></figure>
</div>


<p class="wp-block-paragraph">Cet exemple révèle en fait que les grandeurs physiques agissant sur les sorties sont de deux natures :</p>



<ul class="wp-block-list"><li>Les grandeurs sur lesquelles il est exercé une action volontaire (le débit du fluide chaud réglé par la vanne).</li></ul>



<p class="wp-block-paragraph">Ces grandeurs seront appelées <strong><em>grandeurs de Commande</em></strong>.</p>



<ul class="wp-block-list"><li>Les grandeurs laissées libres de leur évolution (le débit du fluide froid pour l&rsquo;échangeur de chaleur), et cela généralement pour des raisons de coût économique prohibitif ou même, dans certain cas, en raison d&rsquo;une quasi-impossibilité technique à maîtriser ces variables.</li></ul>



<p class="wp-block-paragraph">Dans ces conditions, ces grandeurs sont regroupées sous le terme de <strong><em>Perturbation</em></strong><em>s</em>.</p>



<p class="wp-block-paragraph">Au cours de la synthèse d&rsquo;un système asservi, on est conduit à donner aux perturbations des caractéristiques particulières du type : perturbation en échelon, en rampe ou parfois de nature aléatoire (bruit). Ces caractéristiques tendent à simuler les allures réelles des signaux auxquels le système est soumis lors de son fonctionnement normal. Parmi ses signaux on a :</p>



<ul class="wp-block-list"><li>L&rsquo;échelon (perturbation constante)</li></ul>



<ul class="wp-block-list"><li>La rampe (perturbation linéaire en fonction du temps)</li></ul>



<ul class="wp-block-list"><li>La sinusoïde (perturbation sur fréquence)</li></ul>



<ul class="wp-block-list"><li>Aléatoire, faisant l&rsquo;objet de traitement probabiliste (perturbation sous forme de bruit)</li></ul>



<h3 class="wp-block-heading"><strong><em>1.3&nbsp; Rôle de l&rsquo;automatique</em></strong></h3>



<p class="wp-block-paragraph">Il est clair maintenant que réaliser un système automatique, c&rsquo;est concevoir un système capable d&rsquo;effectuer une ou plusieurs opérations sans l&rsquo;intervention de l&rsquo;homme.</p>



<p class="wp-block-paragraph">Les systèmes automatiques permettent :</p>



<ul class="wp-block-list"><li>de réaliser les opérations trop complexes, pénibles ou délicates et ne pouvant être confiées</li></ul>



<ul class="wp-block-list"><li>l&rsquo;homme. (Alunissage d&rsquo;un engin spatial)</li></ul>



<ul class="wp-block-list"><li>de substituer la machine à l&rsquo;homme dans des opérations trop répétitives ou dénuées d&rsquo;intérêt (BDV automatique, appareillage électroménager).</li></ul>



<ul class="wp-block-list"><li>d&rsquo;accroître la précision : notion de bouclage</li></ul>



<p class="wp-block-paragraph">Par exemple un obus tiré par un canon aura un point de chute dans un certain domaine lié aux imperfections et perturbations extérieures. A contrario, un missile téléguidé dont la position ,vitesse et accélération, par rapport à l&rsquo;objectif sont mesurée à chaque instant atteindra sa cible avec une bien meilleure précision.</p>



<ul class="wp-block-list"><li>de permettre une action de stabilité. Par exemple, un oscillateur entretenu sera surveillé par un dispositif qui lui évitera de diverger ou de cesser d&rsquo;osciller.</li></ul>



<p class="wp-block-paragraph"><strong>Il existe deux grands types de système automatiques :</strong></p>



<h4 class="wp-block-heading"><strong>1.3.1&nbsp;&nbsp;&nbsp;&nbsp; LES SYSTEMES LOGIQUES COMBINATOIRES ET SEQUENTIELS</strong></h4>



<p class="wp-block-paragraph">Les systèmes logiques combinatoires et séquentiels, câbles ou programmés, qui n&rsquo;ont pas nécessairement une structure bouclée. Cette automatisation porte sur un nombre fini d&rsquo;opérations prédéterminées dans leur déroulement. Par exemple un programmateur de machine à laver automatique est un tel système.</p>



<p class="wp-block-paragraph">De tels systèmes sont appelés systèmes à <strong><em>« événement discret »</em></strong> ou <strong><em>« automatismes séquentiels »</em></strong>. Les entrées et les sorties en automatisme séquentiel sont de type booléen (tout ou rien). Cette approche sera abordée en deuxième période.</p>



<p class="wp-block-paragraph">Exemple : La barrière automatique d&rsquo;un passage à niveau est un automatisme séquentiel.</p>


<div class="wp-block-image">
<figure class="aligncenter size-large" id="La-barrière-automatique"><img decoding="async" src="https://telechargercours.com/wp-content/uploads/2021/07/Screenshot-6.png" alt="La barrière automatique d'un passage à niveau est un automatisme séquentiel" class="wp-image-18381" title="La barrière automatique d'un passage à niveau est un automatisme séquentiel"/><figcaption>La barrière automatique d&rsquo;un passage à niveau est un automatisme séquentiel</figcaption></figure>
</div>


<p class="wp-block-paragraph">M=1 : moteur de relevage actionné</p>



<p class="wp-block-paragraph">D=1 : moteur descente actionné</p>



<p class="wp-block-paragraph">F=1 : Action du feu clignotant</p>



<figure class="wp-block-table"><table><tbody><tr><td>h=1 Barrière relevée</td><td>d=1 Arrivée d&rsquo;un train par la droite</td></tr><tr><td>b=1 Barrière baissée</td><td>g=1 Arrivée d&rsquo;un train par la gauche</td></tr><tr><td>&nbsp;</td><td>&nbsp;tr=1 Train passé</td></tr></tbody></table></figure>



<h4 class="wp-block-heading"><strong>1.3.2&nbsp;&nbsp;&nbsp;&nbsp; LES SYSTEMES ASSERVIS.</strong></h4>



<p class="wp-block-paragraph">Les systèmes asservis fonctionnent en régulation de maintient ou en poursuite d&rsquo;une loi de référence. Dans le cas des systèmes asservis, (toutes les situations possibles n&rsquo;étant pas prévisibles (perturbations)), le déroulement des opérations ne peut être prédéterminé à l&rsquo;avance. Les systèmes asservis sont nécessairement bouclés, C&rsquo;est à dire qu&rsquo;une mesure de la situation est en permanence prise en considération dans la détermination de la commande.</p>



<p class="wp-block-paragraph">Les entrées et les sorties sont des variables à variation continue en amplitude. Ces systèmes sont appelés <strong><em>système continu </em></strong>ou<strong><em> analogique</em>.</strong></p>



<p class="wp-block-paragraph">Exemple : La voiture est un système analogique.</p>



<p class="wp-block-paragraph">Ce système comporte une entrée τ , angle de rotation du volant, et une sortie d, distance du véhicule au bord de la route.</p>



<p class="wp-block-paragraph">Le système complet de commande comporte une distance d<sub>0</sub> qui fixe la distance désirée (consigne). d<sub>0</sub> n&rsquo;est pas un signal matérialisé mais une consigne présente dans le cerveau du conducteur. L&rsquo;angle <strong>τ</strong> est modifié pour que d soit en permanence égale à d<sub>0</sub>, ou tout au moins aussi proche que possible de cette consigne.</p>


<div class="wp-block-image">
<figure class="aligncenter size-large"><img decoding="async" src="https://telechargercours.com/wp-content/uploads/2021/07/La-voiture-est-un-systeme-analogique-1.png" alt="" class="wp-image-18384"/></figure>
</div>


<p class="wp-block-paragraph">Le schéma fonctionnel se représente sous la forme ci-dessous :</p>


<div class="wp-block-image">
<figure class="aligncenter size-large" id="Le-schéma-fonctionnel"><img decoding="async" src="https://telechargercours.com/wp-content/uploads/2021/07/Screenshot-8.png" alt="Le schéma fonctionnel" class="wp-image-18385" title="Le schéma fonctionnel"/></figure>
</div>


<h3 class="wp-block-heading"><strong><em>1.4 Bref historique.</em></strong></h3>



<p class="wp-block-paragraph">Trois époques divisent l&rsquo;histoire des systèmes automatiques :</p>



<h4 class="wp-block-heading">1.4.1&nbsp;&nbsp;&nbsp; LA PREMIERE EPOQUE</h4>



<p class="wp-block-paragraph">Qualifier de préhistoire, elle s&rsquo;étend de l&rsquo;antiquité au milieu du siècle dernier. Des inventeurs géniaux ont conçu des systèmes automatiques de manière intuitive.</p>


<div class="wp-block-image">
<figure class="aligncenter size-large"><img decoding="async" src="https://telechargercours.com/wp-content/uploads/2021/07/Horloge-a-eau-de-Ktesybios.png" alt="" class="wp-image-18388"/></figure>
</div>


<p class="wp-block-paragraph">Dès 50 avant J.C, des exemples de régulation de niveau existaient : horloge automatique à eau de Ktesybios (voir figure ci- contre), la lampe à huile de Philon de Byzance et la machine à doser le vin de Héron d&rsquo;Alexandrie.</p>



<p class="wp-block-paragraph"><strong>hc </strong>hauteur de référence pour obtenir le débit voulu en sortie. h hauteur courante dans le réservoir intermédiaire. v vitesse d&rsquo;ascension de l&rsquo;indicateur de temps. h&rsquo; hauteur d&rsquo;eau dans le réservoir de sortie.</p>



<p class="wp-block-paragraph">Le principe de cette horloge est d&rsquo;animer la tige qui sert d&rsquo;indicateur avec un mouvement à vitesse constante (amplitude proportionnelle au temps) . Pour obtenir le déplacement on fait varier le volume d&rsquo;eau d&rsquo;une manière proportionnelle au temps en utilisant le débit d&rsquo;alimentation rigoureusement constant. L&rsquo;intégrale d&rsquo;un débit constant donne bien un volume variant linéaire ment avec le temps.</p>



<p class="wp-block-paragraph">Le problème se ramène donc à générer un débit connu constant. Dans un réservoir à écoulement libre, le débit de sortie est en général fonction de la pression, donc de la hauteur d&rsquo;eau dans le réservoir. Pou obtenir un débit de sortie constant il suffit de garantir un niveau constant.</p>



<p class="wp-block-paragraph">Le niveau dans le réservoir intermédiaire est contrôlé à cause de une valve liée à un flotteur qui va garantir un niveau de référence.</p>



<p class="wp-block-paragraph">En fonctionnement normal, cette valve assure un niveau constant égale au niveau de référence souhaité, ce qui implique que le débit corresponde bien à la valeur souhaitée.</p>



<p class="wp-block-paragraph">Plus tard, Réaumur, Watt et son régulateur de vitesse (1788), Jacquard et son métier à cartes perforées, font progresser l&rsquo;automatisation.</p>



<p class="wp-block-paragraph">Le régulateur de Watt a pour but de maintenir constante la vitesse de rotation d&rsquo;une turbine à vapeur. La commande d&rsquo;admission de vapeur dans la turbine est contrôlée par une vanne dont on peut manœuvre le pointeau.</p>


<div class="wp-block-image">
<figure class="aligncenter size-large"><img decoding="async" src="https://telechargercours.com/wp-content/uploads/2021/07/Le-regulateur-de-Watt.png" alt="" class="wp-image-18390"/></figure>
</div>


<p class="wp-block-paragraph">Un ensemble mécanique déformable constitué de masselottes et de tringles permet une mesure de la vitesse de rotation par effet d&rsquo;inertie. Plus la turbine tourne vite, plus les masselottes sont écartées de l&rsquo;axe de rotation. Pour réaliser un asservissement en vitesse il suffit de transmettre mécaniquement une variation de cet écartement de commande de déplacement du pointeau de la vanne. Si la vitesse de rotation est trop faible, l&rsquo;écartement insuffisant des masselottes engendre une ouverture de la vanne d&rsquo;admission vapeur, entraînant une augmentation de la vitesse. Un comportement symétrique a lieu en cas de vitesse de rotation trop élevée.</p>



<h4 class="wp-block-heading">1.4.2&nbsp; LA SECONDE EPOQUE :</h4>



<p class="wp-block-paragraph">Elle débute du milieu du XIX<sup>ème</sup> siècle, et est caractérisée par la théorie du bouclage et des applications de l&rsquo;algèbre de Boole. Les premiers travaux sur le bouclage sont dus à Maxwell (1868); à Routh avec son critère algébrique (1872) et à Hurwitz (1890).</p>



<p class="wp-block-paragraph">L&rsquo;étude analytique du régulateur de Watt fut commencée par Maxwell en 1868 et complétée en 1876 par Wichnegradsky. L&rsquo;étude des systèmes bouclés doit beaucoup à l&rsquo;approche fréquentielle de Nyquist, Bode, Nichols, Hall, Evans, qui ont laissé leur nom à des représentations et qui ont publié la plupart de leurs résultats à la fin de la seconde guerre mondiale. Les premières implantations des systèmes de commandes à cette époque reposaient sur l&rsquo;utilisation de dispositif électronique à lampes.</p>



<h4 class="wp-block-heading">1.4.3&nbsp;&nbsp;&nbsp;&nbsp; LA TROISIEME EPOQUE :</h4>



<p class="wp-block-paragraph">Elle débute dans les années cinquante. L&rsquo;apparition de calculateurs numériques révolutionne le monde de l&rsquo;automatique. La puissance de calcul disponible fait naître les méthodes dites de <em>l&rsquo;automatique moderne</em> ou <em>avancée</em>. Parmi les faits marquant, on peut citer :</p>



<ul class="wp-block-list"><li>Introduction de la représentation d&rsquo;état, particulièrement bien adaptée à l&rsquo;utilisation des calculateurs numériques pour l&rsquo;étude et la commande des systèmes complexes et multivariables. (Kalman 1960).</li></ul>



<ul class="wp-block-list"><li>Développement des méthodes d&rsquo;étude des systèmes non-linéaires (Kochenburger, Cypkin) et des systèmes échantillonnés (Jury, Ragazzini).</li><li>Prise en compte des phénomènes aléatoires dans les théories récentes comme celles de Kalman</li></ul>



<p class="wp-block-paragraph">et De Bucy.</p>



<h3 class="wp-block-heading"><strong><em>1.5 Schémas fonctionnels</em></strong></h3>



<h4 class="wp-block-heading">1.5.1&nbsp;&nbsp;&nbsp;&nbsp; SYNTAXE DES SCHEMAS FONCTIONNELS</h4>



<p class="wp-block-paragraph">Le recours à un schéma est naturel dans la plupart des techniques, à tel point que cela fait l&rsquo;objet de symboles normalisés.</p>



<p class="wp-block-paragraph">Un schéma fonctionnel consiste en une représentation graphique abrégée des relations de cause à effet entre le signal d&rsquo;entrée et le signal de sortie d&rsquo;un <em>système de commande</em>.</p>



<p class="wp-block-paragraph">Définition : <strong>Un système de commande</strong> est un assemblage de constituants physiques branchés ou reliés les uns les autres de telle sorte qu&rsquo;il puisse se commander, se diriger ou se régler lui-même, ou bien commander, diriger ou régler un autre système.</p>



<p class="wp-block-paragraph">C&rsquo;est un moyen à la fois utile et aisé de caractériser les relations fonctionnelles existant entre les différents organes d&rsquo;un système de commande.</p>



<p class="wp-block-paragraph">On peut également utiliser le terme élément pour désigner un organe du système.</p>



<p class="wp-block-paragraph">Le schéma fonctionnel le plus simple est constitué d&rsquo;un seul élément avec un signal d&rsquo;entrée et un signal de sortie comme le montre la figure ci- dessous.</p>


<div class="wp-block-image">
<figure class="aligncenter size-large"><img decoding="async" src="https://telechargercours.com/wp-content/uploads/2021/07/Le-schema-fonctionnel-le-plus-simple.png" alt="Le schéma fonctionnel le plus simple " class="wp-image-18397"/></figure>
</div>


<p class="wp-block-paragraph">On inscrit à l&rsquo;intérieur du rectangle représentant l&rsquo;élément , la description ou le nom de l&rsquo;organe ainsi que le symbole de l&rsquo;opérateur mathématique à effectuer sur le signal d&rsquo;entrée pour obtenir le signal de sortie.</p>



<p class="wp-block-paragraph">Les flèches indiquent le sens dans lequel l&rsquo;information et le signal se transmettent.</p>



<p class="wp-block-paragraph">On représente les opérations d&rsquo;addition et soustraction d&rsquo;une manière particulière. L&rsquo;élément est figuré par un petit cercle, appelé <strong>Comparateur</strong> où aboutissent des flèches portant le signe + ou &#8211; selon les cas. Le signal de sortie est constitué par la somme algébrique des signaux d&rsquo;entrée. On peut faire aboutir au même comparateur un nombre quelconque de signaux d&rsquo;entrée.</p>


<div class="wp-block-image">
<figure class="aligncenter size-large"><img decoding="async" src="https://telechargercours.com/wp-content/uploads/2021/07/Screenshot-12.png" alt="" class="wp-image-18398"/></figure>
</div>


<p class="wp-block-paragraph">Certains auteurs marquent le cercle d&rsquo;une croix :</p>


<div class="wp-block-image">
<figure class="aligncenter size-large"><img decoding="async" src="https://telechargercours.com/wp-content/uploads/2021/07/marquent-le-cercle-dune-croix.png" alt="" class="wp-image-18399"/></figure>
</div>


<p class="wp-block-paragraph">On appelle Point de Dérivation le point où l&rsquo;on prélève un signal à destination de plusieurs organes d&rsquo;un système de commande.</p>


<div class="wp-block-image">
<figure class="aligncenter size-large"><img decoding="async" src="https://telechargercours.com/wp-content/uploads/2021/07/Point-de-Derivation.png" alt="" class="wp-image-18400"/></figure>
</div>


<h4 class="wp-block-heading">1.5.2 EXEMPLE DE SCHEMA FONCTIONNEL EN FONCTION DU SYSTEME ISOLE</h4>


<div class="wp-block-image">
<figure class="aligncenter size-large"><img decoding="async" src="https://telechargercours.com/wp-content/uploads/2021/07/SCHEMA-FONCTIONNEL-EN-FONCTION-DU-SYSTEME-ISOLE.png" alt="" class="wp-image-18403"/></figure>
</div>


<ul class="wp-block-list"><li><strong>Schéma fonctionnel du réservoir R2 seul non considéré dans l’environnement présenté ci-dessus </strong>Le réservoir R2 alimenté par un débit d&rsquo;entrée Q1(t) et dont on observe la hauteur H1(t) pour sortie peut être représenté par le schéma fonctionnel simple :</li></ul>


<div class="wp-block-image">
<figure class="aligncenter size-large is-resized"><img decoding="async" src="https://telechargercours.com/wp-content/uploads/2021/07/Schema-fonctionnel-du-reservoir.png" alt="" class="wp-image-18404" width="200" height="80"/><figcaption>Noter que le débit de soutirage Q2(t) est considéré comme perturbation<br></figcaption></figure>
</div>


<p class="wp-block-paragraph">Analysons maintenant le réservoir R2 dans sont environnement</p>



<ul class="wp-block-list"><li><strong>Schéma fonctionnel du réservoir R2 seul considéré dans l’environnement présenté ci-dessus</strong></li></ul>


<div class="wp-block-image">
<figure class="aligncenter size-large"><img decoding="async" src="https://telechargercours.com/wp-content/uploads/2021/07/Schema-fonctionnel-du-reservoir-R2-seul-considere-dans-l-environnement.png" alt="" class="wp-image-18405"/></figure>
</div>


<h3 class="wp-block-heading"><strong><em>1.6 Structure d&rsquo;un système de commande ou système asservi.</em></strong></h3>



<h4 class="wp-block-heading">1.6.1 COMMENTAIRE RAPIDE SUR LA MODELISATION DES SYSTEMES</h4>



<p class="wp-block-paragraph">Robert VALLEE (contemporain en bonne santé) nous donne une bonne définition de la modélisation des systèmes, que nous pouvons prendre à notre compte.</p>



<ul class="wp-block-list"><li><em>Le but de la modélisation est de fournir une image ou une représentation d&rsquo;un phénomène réel. S&rsquo;il est possible, à partir de la représentation de retrouver parfaitement le phénomène dans son évolution, il y a isomorphisme. Il est évident que ce cas extrême n&rsquo;est jamais réalisé. Dans le cas général, l y a dégradation dans</em></li></ul>



<p class="wp-block-paragraph"><em>le passage à la représentation et finalement simplement homomorphisme (dans le meilleur des cas). Mais il n&rsquo;y a pas là un défaut. Si selon KORZYBSKI, la carte n&rsquo;est pas le territoire, c&rsquo;est un avantage car la carte est alors exemple de détails inutiles non pertinents. L&rsquo;idée d&rsquo;utilité apparaît ainsi naturellement. L&rsquo;utilité d&rsquo;une modélisation se mesure en fonction du but visé. Il y a des modélisations faites pour aider à comprendre, elles doivent être relativement simples, et d&rsquo;autres pour aider à agir, elles peuvent accepter une plus grande complexité. »</em></p>



<p class="wp-block-paragraph">Nous ajouterons, en guise d&rsquo;avertissement, la citation d&rsquo;un auteur inconnu.</p>



<p class="wp-block-paragraph"><em> » L&rsquo;esprit limite sa perception des choses à ce qu&rsquo;il peut concevoir et il appelle cela la Réalité. »</em></p>



<h4 class="wp-block-heading">1.6.2&nbsp;&nbsp;&nbsp; UNE DEMARCHE DE MODELISATION</h4>



<p class="wp-block-paragraph"><strong>1-</strong>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; <strong>Isoler le système à étudier</strong>.</p>



<p class="wp-block-paragraph">Pour cela il faut parfaitement définir les limites du système, puis remplacer son environnement par un ensemble de relations d&rsquo;entrées-sorties ayant des relations d&rsquo;équivalentes</p>



<p class="wp-block-paragraph"><strong>2- Effectuer une décomposition en sous systèmes </strong>couplés par des relations. Pour chaque sous système, ilfaudra les isoler parfaitement. Les frontières de découpage passent en général par des points de rupture des caractéristiques que l&rsquo;on observe (Structures matérielles discontinues dans l&rsquo;espace, dans le temps, au niveau des fonctionnalités.) De la finesse du découpage dépend la connaissance recherchée plus ou moins poussée du système à étudier.</p>



<p class="wp-block-paragraph">3- <strong>Etablir un modèle de connaissance ou de comportement</strong> pour chaque sous-système.</p>



<h4 class="wp-block-heading">1.6.3&nbsp;BOUCLAGE, REGULATION, ET ASSERVISSEMENT.</h4>



<p class="wp-block-paragraph">Un bouclage apparaît chaque fois au cours d’une opération. Un système prend en compte l’observation de son état pour la modifier</p>



<p class="wp-block-paragraph"><em><u>Exemples de système bouclé :</u></em></p>



<ul class="wp-block-list"><li>Régulation de température d’un fer à repasser pendant l’opération de repassage. (système automatique)</li></ul>



<ul class="wp-block-list"><li>Automobile et pilote réalisant une opération de conduite(système non automatique)</li></ul>



<p class="wp-block-paragraph">Le bouclage est nécessaire dans les opérations où :</p>



<ul class="wp-block-list"><li><strong>La précision </strong>mise en jeu est importante. L’observation de la grandeur à asservir permet deconstater un <strong>écart</strong> ou une dérive quelconque par rapport à la référence (ou consigne) souhaitée et donc de réagir en conséquence. Dans une solution sans bouclage, on est amené à faire a priori des hypothèses quant au comportement de certains éléments, toute variation par rapport au comportement de certains composants présupposé entraîne irrémédiablement un défaut sur le résultat final. On remarque ici que la précision du résultat final dépend essentiellement de la précision avec laquelle est faite l’observation.</li></ul>



<ul class="wp-block-list"><li><strong>Des perturbations </strong>interviennent en cours d’opérations modifiant ainsi l’état du système. Parexemple, l’ouverture de la porte d’un four thermostaté entraîne une déperdition de chaleur et donc une baisse de température que l’on compense en augmentant la puissance de chauffage. Dans le cas d’un système bouclé, l’observation de la grandeur de sortie rendra compte de l’apparition d’une perturbation.</li></ul>



<ul class="wp-block-list"><li><strong>Le comportement </strong>du système est mal connu ou variable.</li></ul>



<ul class="wp-block-list"><li><strong>La stabilité </strong>est en cause. Soit que l’on souhaite stabiliser u système naturellement divergent,soit l’on souhaite améliorer un comportement dynamique insatisfaisant. (Attention, le bouclage d’un système naturellement stable le déstabilise toujours un peu)</li></ul>



<p class="wp-block-paragraph">Ces différents besoins peuvent naturellement coexister dans certaines opérations. Parmi les classes de systèmes asservis on distingue :</p>



<ul class="wp-block-list"><li>Les régulations qui sont des systèmes destinés à maintenir une ou plusieurs grandeurs physiques à des valeurs fixées constantes, même en présence de perturbations.</li></ul>



<p class="wp-block-paragraph">Exemples : régulation de vitesse (régulateur de Watt), pilote automatique d’avion (maintient d’attitude et d’altitude malgré les perturbations)</p>



<ul class="wp-block-list"><li>Les asservissements qui sont des systèmes destinés à faire suivre une loi généralement non fixée à l’avance à une ou plusieurs grandeurs physiques. Dans ce cas, l’asservissement a pour mission d’assurer une recopie la plus fidèle possible quelles que soient les lois de variations fixées. Les asservissements font souvent intervenir les organes mécaniques. On parle alors de servomécanismes.</li></ul>



<p class="wp-block-paragraph">Par exemple, un pilote automatique d’un engin mobile ou une commande numérique de machine outil réalisent des asservissements.</p>



<h4 class="wp-block-heading">1.6.4&nbsp;&nbsp;&nbsp;&nbsp; SCHEMA FONCTIONNEL D&rsquo;UN SYSTEME DE COMMANDE OU ASSERVI.</h4>



<p class="wp-block-paragraph">On relie entre eux les éléments représentant les différents organes d&rsquo;un système de commande de façon à caractériser leurs rapports fonctionnels au sein de l&rsquo;ensemble. Le schéma fonctionnel ci-dessous illustre la structure de base d&rsquo;un système de commande élémentaire en boucle fermée (à retour).</p>


<div class="wp-block-image">
<figure class="aligncenter size-large"><img decoding="async" src="https://telechargercours.com/wp-content/uploads/2021/07/SCHEMA-FONCTIONNEL-D-UN-SYSTEME-DE-COMMANDE-OU-ASSERVI.png" alt="" class="wp-image-18408"/></figure>
</div>


<p class="wp-block-paragraph">Il est nécessaire de bien comprendre que les flèches de la boucle fermée qui relient les éléments les uns aux autres, représentent la circulation de l&rsquo;énergie nécessaire à l&rsquo;asservissement (à la commande), ou l&rsquo;information, et non la principale source d&rsquo;énergie du système.</p>



<p class="wp-block-paragraph">Par exemple, la principale source d&rsquo;énergie d&rsquo;un four à thermostat peut être classique et provenir de la combustion du charbon et du fuel. Mais ce n&rsquo;est pas cette source d&rsquo;énergie qui apparaît dans la boucle d&rsquo;asservissement (fermée) du système.</p>



<h5 class="wp-block-heading"><strong>1.6.4.1&nbsp; Terminologie</strong></h5>



<p class="wp-block-paragraph">Il est important de bien se rappeler les termes dont on se sert dans les schémas fonctionnels en boucles fermées.</p>



<p class="wp-block-paragraph">On utilise des lettres minuscules pour représenter les variables d&rsquo;entrées et de sorties de chaques élément ainsi que pour désigner chaque élément g1, g2 et h. Ces quantités sont fonction du temps sauf indication contraire.</p>



<p class="wp-block-paragraph">On utilise les lettres majuscules pour désigner les transformées de Laplace développées dans un prochain chapitre.</p>



<h5 class="wp-block-heading"><strong>1.6.4.2&nbsp; Définitions</strong></h5>



<ul class="wp-block-list"><li>On appelle <strong>appareil ou processus</strong> g2 le système, le sous-système ou l&rsquo;opération placé(e) sous asservissement, au moyen d&rsquo;un moyen d&rsquo;un dispositif de contrôle.</li></ul>



<ul class="wp-block-list"><li>On appelle <strong>signal de sortie réglée</strong> c la grandeur produite par l&rsquo;appareil sous asservissement au moyen d&rsquo;un dispositif de contrôle.</li><li>On appelle <strong>chaîne d&rsquo;action</strong> l&rsquo;ensemble des éléments placés entre le point de sommation et la sortie</li></ul>



<p class="wp-block-paragraph">réglée.</p>



<ul class="wp-block-list"><li>On appelle <strong>organes de commande</strong> g1 les éléments de la chaîne d&rsquo;action qui engendrent le signal de commande u et m fourni par l&rsquo;appareil.</li></ul>



<p class="wp-block-paragraph">Les organes de commande sont en général des régulateurs, des correcteurs, des égaliseurs et/ou des amplificateurs.</p>



<ul class="wp-block-list"><li>On appelle <strong>signal de commande</strong> u (ou grandeur réglée m) le signal de sortie délivré par les organes de commandes g1 à l&rsquo;appareil g2.</li></ul>



<ul class="wp-block-list"><li>On appelle chaîne de retour l&rsquo;ensemble des éléments situés entre la sortie réglée c et le point de sommation.</li></ul>



<ul class="wp-block-list"><li>On appelle <strong>organe de retour</strong> h les éléments de la chaîne de retour établissant une relation entre la sotie réglée c et le signal de retour b.</li></ul>



<p class="wp-block-paragraph">Les organes de retour sont en général des capteurs de mesure de la sortie réglée c, des correcteurs et/u des régulateurs.</p>



<ul class="wp-block-list"><li>On appelle <strong>signal de référence ou valeur de consigne</strong> r un signal externe fourni au système asservi, en général sur le sommateur, pour obtenir une réponse déterminée de l&rsquo;appareil sous contrôle. Sa valeur correspond au fonctionnement optimal ou désiré de l&rsquo;appareil.</li></ul>



<ul class="wp-block-list"><li>On appelle <strong>signal de retour primaire</strong> b une fonction de l sortie réglée c fournie par la chaîne de retour, que l&rsquo;on somme algébriquement avec la référence r pour obtenir un signal d&rsquo;erreur e.</li></ul>



<p class="wp-block-paragraph">Ce signal n&rsquo;existe pas en boucle ouverte.</p>



<ul class="wp-block-list"><li>On parle de <strong>retour positif</strong> lorsque le sommateur est un additionneur, c&rsquo;est à dire e = r+b et de <strong>retour</strong> <strong>négatif </strong>ou contre réaction lorsque le sommateur est un soustracteur soit e = r-b.</li></ul>



<h4 class="wp-block-heading">1.6.5&nbsp;&nbsp;&nbsp;&nbsp; STRUCTURE D&rsquo;UN SYSTEME ASSERVI OU BOUCLE</h4>



<p class="wp-block-paragraph">Trois fonctions essentielles gèrent la réalisation d&rsquo;un système asservi ou bouclé :</p>



<ol class="wp-block-list" type="1"><li>L’<strong>OBSERVATION</strong></li></ol>



<ul class="wp-block-list"><li>La <strong>REFLEXION</strong> (dans le cadre d’un opérateur humain) ou <strong>TRAITEMENT ALGORITHMIQUE</strong> (par un système électronique ou informatique)</li><li>L’<strong>ACTION</strong>.</li></ul>


<div class="wp-block-image">
<figure class="aligncenter size-large"><img decoding="async" src="https://telechargercours.com/wp-content/uploads/2021/07/STRUCTURE-DUN-SYSTEME-ASSERVI-OU-BOUCLE.png" alt="" class="wp-image-18410"/></figure>
</div>


<p class="wp-block-paragraph">Suivant les réalisations, ces trois opérations sont exécutées simultanément ou successivement.</p>



<p class="wp-block-paragraph">Explicitons ces trois opérations à propos d’un exemple. Prenons pour tâche de réaliser le remplissage d’une cuve dont on souhaite maintenir le niveau à une hauteur de consigne donnée. Dans ce cas, les trois opérations sont les suivantes :</p>



<ol class="wp-block-list" type="1"><li><strong>Observation </strong>du niveau d’eau dans la cuve,</li></ol>



<ul class="wp-block-list"><li><strong>Comparaison </strong>avec le niveau de consigne</li></ul>



<ul class="wp-block-list"><li><strong>Action </strong>sur le robinet (ouverture ou fermeture)</li></ul>



<p class="wp-block-paragraph"><strong>Représentation générale d’un système automatique sous forme de schéma fonctionnel.</strong> Ce schéma fonctionnel est la représentation générique d’un système automatique bouclé.</p>


<div class="wp-block-image">
<figure class="aligncenter size-large"><img decoding="async" src="https://telechargercours.com/wp-content/uploads/2021/07/Representation-generale-d-un-systeme-automatique-sous-forme-de-schema-fonctionnel.png" alt="" class="wp-image-18413"/></figure>
</div>


<p class="wp-block-paragraph"><strong>régulateur </strong>(comparateur et correcteur) élabore l’ordre de commande de l’actionneur à partir de la consigne et dela mesure. C’est l’organe intelligent. Dans la pratique, le régulateur est en général réalisé par un dispositif électronique ou par un calculateur numérique.</p>



<p class="wp-block-paragraph"><strong>L’actionneur </strong>ou l’organe d’action apporte, en général, la puissance nécessaire à la réalisation de latâche : c’est l’organe musclé (vanne, moteur).</p>



<p class="wp-block-paragraph"><strong>Le système dynamique </strong>évolue suivant les lois physiques (thermodynamique, mécanique, mécanique desfluides, chimique de réaction, etc.) qui lui sont propres. La sortie ou grandeur réglée est, en générale, une grandeur physique que l’on considère comme importante dans la tâche à réaliser. De plus, cette sortie peut fluctuer en fonction de perturbations extérieures souvent imprévisibles.</p>



<p class="wp-block-paragraph"><strong>Le capteur </strong>délivre à partir de la sortie une grandeur caractérisant l’observation. Sa principale qualité estla précision dont dépendra la précision du système global. Il doit fournir une information quasi instantanée par rapport aux évolutions temporelles de la sortie.</p>



<h4 class="wp-block-heading">1.6.6&nbsp;&nbsp;&nbsp;&nbsp; QUALITES D&rsquo;UN ASSERVISSEMENT.</h4>



<p class="wp-block-paragraph">L&rsquo;étude d&rsquo;un système asservi consiste à déterminer sa réponse y(t) à une entrée fonction du temps x(t) donnée.</p>


<div class="wp-block-image">
<figure class="aligncenter size-full"><img decoding="async" src="https://telechargercours.com/wp-content/uploads/2021/07/fonction-du-temps-x.png" alt="" class="wp-image-21419"/></figure>
</div>


<p class="wp-block-paragraph">Pour caractériser les performances du système, on se place le plus souvent dans les conditions suivantes:</p>



<p class="wp-block-paragraph">L&rsquo;entrée x(t) est l&rsquo;une des fonctions du temps précisées ci-dessous :</p>


<div class="wp-block-image">
<figure class="aligncenter size-full"><img decoding="async" src="https://telechargercours.com/wp-content/uploads/2021/07/L-entree-xt-est-l-une-des-fonctions-du-temps-precisees-ci-dessous.png" alt="" class="wp-image-21420"/></figure>
</div>


<p class="wp-block-paragraph">Pour tenir compte des discontinuités, on distingue l&rsquo;instant t = 0- et l&rsquo;instant 0+.</p>



<p class="wp-block-paragraph">La sortie y(t) et ses dérivées par rapport au temps sont supposées nulles pour les instants : t&lt;O- : On dira que le système est initialement au repos.</p>



<p class="wp-block-paragraph">Dans ces conditions, la réponse y(t) comporte en général un <strong>régime transitoire</strong> et un <strong>régime établi ou</strong> <strong>régime permanent </strong>(obtenu pour t suffisamment grand). Voir figure ci-dessous.</p>


<div class="wp-block-image">
<figure class="aligncenter size-full"><img decoding="async" src="https://telechargercours.com/wp-content/uploads/2021/07/reponse-indicielle.png" alt="" class="wp-image-21421"/></figure>
</div>


<p class="wp-block-paragraph">Exemple de réponse indicielle (réponse à un échelon d&rsquo;un système initialement au repos)</p>



<p class="wp-block-paragraph"><strong>Le régime transitoire </strong>est caractérisé essentiellement par :</p>



<ul class="wp-block-list"><li>Le dépassement D : D = AM / HA, en général, le dépassement est exprimé en pourcentage.</li></ul>



<ul class="wp-block-list"><li>Le temps de réponse tr (durée du régime transitoire). Par exemple, pour un critère à 5%, le temps de réponse à 5% est le temps au-delà duquel l&rsquo;écart relatif entre y(t) et ye (régime établi) est inférieur à 5%.</li></ul>



<p class="wp-block-paragraph">En général pour les servomécanismes, il est souhaité un régime transitoire de courte durée et à faible dépassement</p>



<h5 class="wp-block-heading"><strong>1.6.6.1&nbsp; L’aspect dynamique &#8211; L’aspect statique</strong></h5>



<p class="wp-block-paragraph"><strong>Dans l&rsquo;analyse des systèmes asservis, nous distinguons donc l&rsquo;aspect statique de l&rsquo;aspect dynamique.</strong></p>



<p class="wp-block-paragraph"><strong>L’aspect statique </strong>concerne l’étude des systèmes asservis en régime permanent (entrée fixe). On définitl’erreur statique comme étant la différence entre la sortie demandée (consigne) et la sortie réalisée lorsque que le régime d’équilibre est établi (c’est à dire pour t grand). Au cours de la conception des systèmes asservis, on s’efforcera en général d’annuler cette erreur.</p>



<p class="wp-block-paragraph"><strong>L’aspect dynamique</strong>, essentiellement en automatique, s’étudie par les notions de précision dynamique,de rapidité et de stabilité. (il est à noter qu’à l’heure d’aujourd’hui, la précision dynamique des systèmes automatisés n’est pas au programme de C.P.G.E.). Il s’intéresse au comportement transitoire de la sortie, soit suite à une variation de la consigne, soit suite à l’apparition d’une perturbation dans la chaîne fonctionnelle.</p>



<h4 class="wp-block-heading">1.6.7&nbsp;&nbsp;&nbsp;&nbsp; LES PERFORMANCES D’UN ASSERVISSEMENT</h4>



<h5 class="wp-block-heading"><strong>1.6.7.1&nbsp; Les quatre critères de performance</strong></h5>



<ul class="wp-block-list"><li><strong>La stabilité </strong>:<strong>&nbsp;&nbsp; </strong>absence de divergence entre la valeur de sortie et la valeur visée.</li></ul>



<p class="wp-block-paragraph">La stabilité est la performance nécessaire des servomécanismes. Sans stabilité pas d’étude. Elle se</p>



<p class="wp-block-paragraph">caractérise par son degré, qui peut être quantifié par le dépassement, ou l’amortissement, ou par la marge de gain, ou la marge de phase.</p>



<p class="wp-block-paragraph">Le réglage du degré de stabilité d’un asservissement est souvent la première étude à mener lors d’une analyse ou dune conception d’un système automatisé. C’est en agissant sur le régulateur et plus précisément sur le correcteur que les réglables du degré de stabilité sont effectués.</p>



<p class="wp-block-paragraph">Remarque : Le choix d’un degré de stabilité influe forcement sur la rapidité</p>



<p class="wp-block-paragraph"><strong><u>L’étude en régime permanent permet de caractériser :</u></strong></p>



<p class="wp-block-paragraph">&nbsp;&nbsp;&nbsp;&nbsp; <strong>La précision statique </strong>(t tend vers l’infini): mesurée par l&rsquo;écart entre la valeur de sortie et la valeurvisée.</p>



<p class="wp-block-paragraph"><strong><u>L’étude du régime transitoire permet de caractériser</u></strong><strong> </strong>:</p>



<ul class="wp-block-list"><li><strong>La rapidité </strong>: mesurée par le temps de réponse au bout duquel la valeur de sortie reste dans unefourchette exprimée en pourcentage (souvent 5 %) centrée sur la valeur visée. Le temps de réaction d&rsquo;un processus est étroitement lié à son inertie. (très liée au degré de stabilité)</li></ul>



<ul class="wp-block-list"><li><strong>L&rsquo;amortissement </strong>: caractérisé par le rapport entre les amplitudes successives des oscillations de lavaleur de sortie (très lié au degré de stabilité)</li></ul>



<ul class="wp-block-list"><li><strong>précision dynamique </strong>(régime transitoire) (hors programme en C.P.G.E.)</li></ul>



<h5 class="wp-block-heading">1.6.7.1.1 L&rsquo;asservissement idéal</h5>



<p class="wp-block-paragraph">En manière de systèmes de commande comme ailleurs il n&rsquo;y a pas de critère absolu, il n&rsquo;y a que des critères relatifs à des problèmes précis. C&rsquo;est en ayant présent à l&rsquo;esprit cette réserve fondamentale qu&rsquo;il convient de lire ce qui suit.</p>



<p class="wp-block-paragraph">On recherche une bonne stabilité et une bonne précision, mais le régime transitoire doit être rapide et bien amorti.</p>


<div class="wp-block-image">
<figure class="aligncenter size-full"><img decoding="async" src="https://telechargercours.com/wp-content/uploads/2021/07/le-regime-transitoire.png" alt="" class="wp-image-21422"/></figure>
</div>


<p class="wp-block-paragraph">Si le régime transitoire était nul, quel plaisir. On donne un ordre à un système et Hop !!???, il est immédiatement réalisé. Sûrement trop facile et donc ennuyeux.</p>



<p class="wp-block-paragraph">Nous verrons par la suite, pourquoi le fait de ne pas savoir aujourd’hui, fabriquer et concrétiser un dérivateur pure nous empêche d’obtenir instantanément notre demande (la consigne), mais heureusement, cela ne nous empêche pas de rêver.</p>



<h6 class="wp-block-heading">1.6.7.1.2 Dilemme rapidité-amortissement <strong>ou degré de stabilité</strong>.</h6>



<p class="wp-block-paragraph">Dilemme stabilité-précision</p>



<p class="wp-block-paragraph">Ces critères de performance ne sont pas toujours compatibles et nous conduisent souvent à des compromis qui font le plaisir de l’automatisme. La solution unique n’existant pas, il est donc nécessaire de mettre en place des critères d’évaluation et de choix de solutions.</p>



<p class="wp-block-paragraph">Par exemple en mécanique un processus rapide est léger, il a ainsi faible inertie, il risque donc d&rsquo;être peu amorti voire instable.</p>



<p class="wp-block-paragraph">D&rsquo;autre part si on veut améliorer la précision on raidit l&rsquo;asservissement mais on risque de tomber alors sur le phénomène de pompage mais si on cherche à être tranquille du côté de la stabilité on risque de réaliser un asservissement mou et peu précis.</p>



<h6 class="wp-block-heading">1.6.7.1.3 Stabilité</h6>



<p class="wp-block-paragraph">La stabilité traduit la propriété de convergence temporelle asymptotique vers un état d’équilibre. Un système stable peut en effet présenter une sortie divergente soit en raison du comportement dynamique intrinsèque de système commandé, soit en raison d’un bouclage. Ce comportement instable est intolérable pour un système asservi. Il sera donc souvent exiger un comportement transitoire correctement amorti (le correctement est difficile à définir).</p>


<div class="wp-block-image">
<figure class="aligncenter size-full"><img decoding="async" src="https://telechargercours.com/wp-content/uploads/2021/07/Screenshot-26.png" alt="" class="wp-image-21425"/></figure>
</div>


<p class="wp-block-paragraph">la rapidité d’un système est quantifiée par le temps de réponse affecté d’un critère sous égal à 5%. tr5% est déterminé à partir de l’instant pour lequel la valeur instantanée de la sortie s(t) est à une distance inférieure à 5% de la valeur finale (ici K). (voir figure ci-dessous)</p>


<div class="wp-block-image">
<figure class="aligncenter size-full"><img decoding="async" src="https://telechargercours.com/wp-content/uploads/2021/07/amortissement-ou-degre-de-stabilite-2.png" alt="" class="wp-image-21428"/></figure>
</div>


<h6 class="wp-block-heading">1.6.7.1.6 L’amortissement ou degré de stabilité</h6>



<p class="wp-block-paragraph">L’amortissement permet de quantifier le degré de stabilité d’un système. Plus il est grand, plus le système est amorti.</p>



<p class="wp-block-paragraph">Il est à remarquer que plus le système est amorti moins il est rapide ; et si le système est trop peu amorti sont temps de réponse à 5% grandi. Il est donc nécessaire, comme toujours de trouver le bon compromis.</p>



<p class="wp-block-paragraph">Le dépassement est lui aussi important et peu être en fonction des cas (cahier de charges) autorisé ou non. (là encore les compromis sont de rigueur et les critères de choix doivent être clairement identifiés)</p>


<div class="wp-block-image">
<figure class="aligncenter size-full"><img decoding="async" src="https://telechargercours.com/wp-content/uploads/2021/07/amortissement-ou-degre-de-stabilite.png" alt="" class="wp-image-21429"/></figure>
</div>


<h6 class="wp-block-heading">1.6.7.1.7 La qualité d’un système automatique</h6>



<p class="wp-block-paragraph"><strong>Le cahier des charges de tout système bouclé s&rsquo;énonce en trois points : Stabilité, Rapidité de réponse, Précision. La qualité du système dépend donc de ses trois points.</strong></p>



<p class="wp-block-paragraph"><strong><u>Conclusion :</u></strong></p>



<p class="wp-block-paragraph">La structure bouclée, la possibilité de trouver des lois de commande convenables et même parfois assez sophistiquées, permettent de créer automatiquement la commande appropriée. Les contraintes du cahier des charges sont nombreuses et contradictoires ; s&rsquo;y ajoutent parfois des exigences nouvelles : Économie d&rsquo;énergie, économie du matériel, qu&rsquo;il faut prendre en compte (une voiture qui démarre vite use prématurément ses pneus). La synthèse d&rsquo;une commande automatique résulte, avant tout, de compromis.</p>



<p class="has-text-align-center wp-block-paragraph"><strong><a href="https://www.mediafire.com/file/bzz02qvznyl7kxg/Telechargercours.com+SI-CO01-AAS-AC_unlocked.pdf/file" target="_blank" rel="noreferrer noopener">Telecharger</a></strong></p>
]]></content:encoded>
					
		
		
			</item>
		<item>
		<title>Sciences Indusrielles Systèmes linéaires continus invariants Performance des systèmes asservis</title>
		<link>https://telechargercours.com/sciences-indusrielles-systemes-lineaires-performance-des-systemes/</link>
		
		<dc:creator><![CDATA[Cours Zilla]]></dc:creator>
		<pubDate>Mon, 05 Jul 2021 14:05:45 +0000</pubDate>
				<category><![CDATA[Science & Technologie]]></category>
		<guid isPermaLink="false">https://telechargercours.com/?p=18326</guid>

					<description><![CDATA[Sciences Indusrielles Systèmes linéaires continus invariants Performance des systèmes asservis SYSTEMES LINEAIRES CONTINUS INVARIANTS PERFORMANCES DES SYSTEMES ASSERVIS 1 Stabilité des systèmes asservis 1.1 Notion de stabilité La stabilité est communément reconnue comme étant associée à la notion d’équilibre : Prenons les deux positions d’équilibre d’un pendule : En équilibre instable En équilibre stable A [&#8230;]]]></description>
										<content:encoded><![CDATA[<p style="text-align: center;"><strong>Sciences Indusrielles</strong></p>
<p style="text-align: center;"><strong>Systèmes linéaires continus invariants</strong></p>
<h4 style="text-align: center;">Performance des systèmes asservis</h4>
<p style="text-align: center;"><strong>SYSTEMES LINEAIRES CONTINUS INVARIANTS</strong></p>
<p style="text-align: center;"><em>PERFORMANCES DES</em></p>
<p style="text-align: center;"><em>SYSTEMES ASSERVIS</em></p>
<h2><strong>1</strong> <strong>Stabilité des systèmes asservis</strong></h2>
<h3><strong>1.1</strong> <strong>Notion de stabilité</strong></h3>
<p>La stabilité est communément reconnue comme étant associée à la notion d’équilibre :</p>
<p><strong><em>Prenons les deux positions d’équilibre d’un pendule :</em></strong></p>
<p><img decoding="async" class="wp-image-18327" src="https://telechargercours.com/wp-content/uploads/2021/07/word-image-213.png"></p>
<p><strong><em>En équilibre</em></strong></p>
<p><strong><em>instable</em></strong></p>
<p><strong><em>En équilibre stable</em></strong></p>
<p>A gauche si on écarte le pendule de sa position d’équilibre, il finira par la reprendre.</p>
<p>A droite, si on écarte le pendule de sa position d’équilibre, il s’en écarte définitivement.</p>
<p><strong><em>On peut élargir cette notion :</em></strong></p>
<p>Etudions le mouvement d’une bille reposant sur différentes formes.</p>
<table>
<tbody>
<tr>
<td>
<p><strong><em>Equilibre instable</em></strong></p>
</td>
<td>&nbsp;</td>
<td>
<p><strong><em>Equilibre stable</em></strong></p>
</td>
<td>&nbsp;</td>
<td>
<p><strong><em>Equilibre indifférent</em></strong></p>
</td>
</tr>
<tr>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
</tr>
<tr>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
</tr>
<tr>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
</tr>
</tbody>
</table>
<p><img decoding="async" class="wp-image-18328" src="https://telechargercours.com/wp-content/uploads/2021/07/word-image-214.png"> <img decoding="async" class="wp-image-18329" src="https://telechargercours.com/wp-content/uploads/2021/07/word-image-215.png"> <img decoding="async" class="wp-image-18330" src="https://telechargercours.com/wp-content/uploads/2021/07/word-image-216.png"></p>
<p><strong><em>Equilibres multiples stable</em></strong></p>
<p><strong><em>et instable suivant le</em></strong></p>
<p><strong><em>domaine d’évolution</em></strong></p>
<p><strong><em>Il faut donc étudier la qualité de la stabilité d’un système asservis de façon plus précise.</em></strong></p>
<p>Page 1 <strong>Emmanuel FARGES</strong> Ó EduKlub S.A.</p>
<p>Tous droits de l’auteur des œuvres réservés. Sauf autorisation, la reproduction ainsi que toute utilisation des œuvres autre que la consultation individuelle et privée sont interdites.</p>
<p></p>
<p><strong>Sciences Indusrielles</strong></p>
<p><strong>Systèmes linéaires continus invariants</strong></p>
<p>Performance des systèmes asservis</p>
<h3><strong>1.2 Aspect mathématique</strong></h3>
<p>Un système est stable si à toute entrée bornée, le système répond par une sortie bornée</p>
<p>Un système est stable si sa réponse libre (équation différentielle sans second membre) tend vers 0 lorsque le temps tend vers l’infini.</p>
<p>On a vu que toute fonction de transfert et donc en particulier la fo nction de transfert globale d’un système pouvait se mettre après décomposition en éléments simples sous la forme :</p>
<table>
<tbody>
<tr>
<td rowspan="2">
<p><em>H </em>(<em> p</em>) = å</p>
</td>
<td>
<p><em>c<sub>i</sub></em></p>
</td>
<td rowspan="2">
<p><em>avec p<sub>i</sub> </em>Î£ les pôles de la fonctions de transfert globale du système.</p>
</td>
<td>&nbsp;</td>
</tr>
<tr>
<td rowspan="2">
<p><em>p </em>&#8211;<em> p<sub>i</sub></em></p>
</td>
<td>&nbsp;</td>
</tr>
<tr>
<td>
<p><em>i</em></p>
</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
</tr>
</tbody>
</table>
<p>Or après transformée inverse de Laplace, on s’aperçoit que :</p>
<ul>
<li>Les pôles nuls engendrent des solutions de la forme <em>t</em> <em><sup>n</sup></em> (non borné)</li>
<li>Les pôles réels engendrent des solutions de la forme <em>e</em><sup>(</sup><em><sup>pi t</sup></em> <sup>)</sup> (<em>borné si p<sub>i</sub></em> &lt; 0)</li>
<li>Les pôles complexes qui sont nécessairement conjugués engendrent des solutions de la</li>
</ul>
<table>
<tbody>
<tr>
<td>
<p>(<sup>Re</sup>(<em><sup>p</sup><sub>i</sub></em> )<em><sup>t</sup></em> )</p>
</td>
<td rowspan="2">
<p>sin (<em>w</em> <em><sub>i</sub></em> <em>t</em> +<em>j<sub>i</sub></em> ) (<em>borné si</em> Re ( <em>p<sub>i</sub></em> ) &lt; 0)</p>
</td>
<td>&nbsp;</td>
</tr>
<tr>
<td>
<p>forme : <em>e</em></p>
</td>
<td>&nbsp;</td>
</tr>
</tbody>
</table>
<p><strong><em>La condition générale de stabilité est donc que les pôles de la fonction de transfert globale du système soient à partie réelle strictement négative.</em></strong></p>
<p>Ce critère est une condition de stabilité mais ne permet pas de quantifier la qualité de cette éventuelle stabilité, ce que nous ferons plus tard.</p>
<h3><strong>1.3 Critère algébrique de ROUTH</strong></h3>
<p>Ce critère algébrique permet de savoir si un système est stable ou non sans en déterminer la qualité. Il permet de savoir si les zéros d’un polynôme sont à parties réelles négatives ou non. Pour la stabilité on s’intéresse aux pôles de la fonction de transfert, donc aux zéros du</p>
<p><em><sup>N </sup></em><sup>(</sup><em><sup> p</sup></em><sup>)</sup></p>
<p>polynôme au dénominateur : <em>D</em> ( <em>p</em>) <em>avec H</em> ( <em>p</em>) =</p>
<p>Posons : <em>D</em>( <em>p</em>) = <em>b<sub>n</sub></em> <em>p<sup>n</sup></em> + &#8230;&#8230;.. + <em>b</em><sub>1</sub> <em>p</em> + <em>b</em><sub>0</sub></p>
<p>Les zéros de D(p) donc les pôles de la fonction de transfert sont à parties réelles strictement négatives si deux conditions :</p>
<p><strong><em>Condition 1 : Tous les coefficients </em></strong><em>b<sub>i</sub></em><strong><em> sont positifs.</em></strong></p>
<p><strong><em>Condition 2 : Tous les coefficients de la première colonne du tableau ci-dessous doivent être positif.</em></strong></p>
<p>Le tableau se constitue de la façon suivante :</p>
<p>Les deux premières lignes se remplissent avec les coefficients du polynôme D(p).</p>
<p>Les lignes suivantes se déduisent des deux immédiatement précédentes, de la façon suivante :</p>
<p>Page 2 <strong>Emmanuel FARGES</strong> Ó EduKlub S.A.</p>
<p>Tous droits de l’auteur des œuvres réservés. Sauf autorisation, la reproduction ainsi que toute utilisation des œuvres autre que la consultation individuelle et privée sont interdites.</p>
<p></p>
<p><strong>Sciences Indusrielles</strong></p>
<p><strong>Systèmes linéaires continus invariants</strong></p>
<p>Performance des systèmes asservis</p>
<table>
<tbody>
<tr>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td colspan="2">
<p>p<sup>n</sup></p>
</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>
<p>b<sub>n</sub></p>
</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>
<p>bn-2</p>
</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>
<p>bn-4</p>
</td>
<td>&nbsp;</td>
</tr>
<tr>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>
<p><sub>p</sub>n-1</p>
</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>
<p>bn-1</p>
</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>
<p>bn-3</p>
</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>
<p>bn-5</p>
</td>
<td>&nbsp;</td>
</tr>
<tr>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>
<p><sub>P</sub>n-2</p>
</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td colspan="2" rowspan="3">
<p><em>c </em>=</p>
</td>
<td>
<p><em>b b</em></p>
</td>
<td colspan="3">
<p>&#8211;<em> b b</em></p>
</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td rowspan="3">
<p><em>d </em>=</p>
</td>
<td>
<p><em>b b</em></p>
</td>
<td colspan="2">
<p>&#8211;<em> b b</em></p>
</td>
<td>&nbsp;</td>
<td>
<p>etc</p>
</td>
<td>&nbsp;</td>
</tr>
<tr>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>
<p><em>n</em>-1<em> n</em>&#8211; 2</p>
</td>
<td colspan="2">
<p><em>n n</em>-3</p>
</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>
<p><em>n</em>-1<em> n </em>-4</p>
</td>
<td>
<p><em>n n</em>-5</p>
</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
</tr>
<tr>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td colspan="4" rowspan="2">
<p><em><sup>b</sup>n </em>-1</p>
</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td colspan="3" rowspan="2">
<p><em><sup>b</sup>n</em>-1</p>
</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
</tr>
<tr>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
</tr>
<tr>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td colspan="4">
<p><em><sup>cb</sup>n </em>-3 <sup>&#8211;</sup><em> <sup>db</sup>n</em>-1</p>
</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>
<p>…</p>
</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>
<p>…</p>
</td>
<td>&nbsp;</td>
</tr>
<tr>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>
<p><em>c</em></p>
</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
</tr>
<tr>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>
<p>p<sup>1</sup>=p</p>
</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>
<p>…</p>
</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>
<p>…</p>
</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>
<p>…</p>
</td>
<td>&nbsp;</td>
</tr>
<tr>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>
<p>p<sup>0</sup>=1</p>
</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>
<p>…</p>
</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>
<p>…</p>
</td>
<td>&nbsp;</td>
</tr>
</tbody>
</table>
<p><img decoding="async" class="wp-image-18331" src="https://telechargercours.com/wp-content/uploads/2021/07/word-image-217.png"></p>
<p><strong><em>1<sup>ère</sup> colonne</em></strong></p>
<p><strong><em>Remarque : </em></strong>si les coefficients n’existent pas, on y met des zéros et on ne va pas plus loin.</p>
<p><strong><em>Remarque : </em></strong>Si un zéro apparaît dans la première colonne, on ne peut plus calculer les autres lignes. On s’intéresse donc aux zéros du polynôme (p+a)D(p) qui a les mêmes zéros que D(p) avec –a en plus. Donc si a&gt;0, cela ne change rien a la stabilité de notre système et cela nous permet d’effectuer les calculs (en général on prend a=1).</p>
<h3><strong>1.4</strong> <strong>Application à un deuxième ordre</strong></h3>
<p>On prend un système représenté par un deuxième ordre à retour unitaire dont le schéma bloc est donné ci-dessous :</p>
<table>
<tbody>
<tr>
<td colspan="2">
<p><strong><em>E</em></strong></p>
</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td colspan="2">
<p><em>K</em></p>
</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>
<p><strong><em>S</em></strong></p>
</td>
<td>&nbsp;</td>
</tr>
<tr>
<td rowspan="2">
<p>+</p>
</td>
<td rowspan="2">&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
</tr>
<tr>
<td colspan="2" rowspan="3">
<p>1+</p>
</td>
<td rowspan="2">
<p>2<em>x</em></p>
</td>
<td>&nbsp;</td>
<td rowspan="3">
<p><em>p </em>+</p>
</td>
<td rowspan="2">
<p><em>p</em><sup>2</sup></p>
</td>
<td colspan="2" rowspan="2">&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
</tr>
<tr>
<td colspan="2" rowspan="3">
<p>&#8211;</p>
</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
</tr>
<tr>
<td rowspan="2">
<p><em>w</em></p>
</td>
<td>&nbsp;</td>
<td colspan="3" rowspan="2">
<p><em>w </em><sup>2</sup></p>
</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
</tr>
<tr>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
</tr>
<tr>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>
<p>0</p>
</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>
<p>0</p>
</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
</tr>
<tr>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
</tr>
</tbody>
</table>
<p><img decoding="async" class="wp-image-18332" src="https://telechargercours.com/wp-content/uploads/2021/07/word-image-218.png"> <img decoding="async" class="wp-image-18333" src="https://telechargercours.com/wp-content/uploads/2021/07/word-image-219.png"></p>
<table>
<tbody>
<tr>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>
<p><em>K</em></p>
</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
</tr>
<tr>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td colspan="4">
<p>1 +</p>
</td>
<td colspan="2">
<p>2<em>x</em></p>
</td>
<td>&nbsp;</td>
<td colspan="2">
<p><em><sub>p</sub> </em><sub>+</sub><em> p</em><sup>2</sup></p>
</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td colspan="2" rowspan="2">
<p><em><sup>K</sup> BF</em></p>
</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
</tr>
<tr>
<td colspan="3" rowspan="2">
<p><em>FTBF </em>=</p>
</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td colspan="2">
<p><em>w</em><sub>0</sub></p>
</td>
<td>&nbsp;</td>
<td colspan="2">
<p><em>w</em><sub>0</sub><sup>2</sup></p>
</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td rowspan="2">
<p>=</p>
</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
</tr>
<tr>
<td colspan="2" rowspan="4">
<p>1+</p>
</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td rowspan="3">
<p><em>K</em></p>
</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td colspan="2">&nbsp;</td>
<td colspan="3" rowspan="4">
<p>1+</p>
</td>
<td colspan="3" rowspan="4">
<p>2<em>x<sub>BF</sub></em></p>
</td>
<td rowspan="4">
<p><em>p </em>+<em> <sup>p</sup></em></p>
</td>
<td rowspan="2">
<p>2</p>
</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
</tr>
<tr>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
</tr>
<tr>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
</tr>
<tr>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td rowspan="2">
<p>2<em>x</em></p>
</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td colspan="2" rowspan="2">
<p><em>p</em><sup>2</sup></p>
</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td colspan="3" rowspan="2">
<p><em>w</em><sub>0</sub><sup>2</sup><em><sub>BF</sub></em></p>
</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
</tr>
<tr>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td colspan="5" rowspan="2">
<p>1+</p>
</td>
<td>&nbsp;</td>
<td rowspan="2">
<p><em>p </em>+</p>
</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td colspan="2">
<p><em><sup>w</sup></em>0<em>BF</em></p>
</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
</tr>
<tr>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td colspan="2" rowspan="2">
<p><em>w</em><sub>0</sub><sup>2</sup></p>
</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
</tr>
<tr>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td colspan="2">
<p><em>w</em><sub>0</sub></p>
</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
</tr>
<tr>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td colspan="2">&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td colspan="2" rowspan="2">
<p><em>K</em></p>
</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td colspan="2">&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td rowspan="2">
<p><em>x</em></p>
</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td colspan="3">&nbsp;</td>
<td>&nbsp;</td>
</tr>
<tr>
<td>
<p>avec</p>
</td>
<td>
<p><em>K</em></p>
</td>
<td rowspan="2">
<p><em>BF</em></p>
</td>
<td colspan="2">
<p>=</p>
</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>
<p><em>x</em></p>
</td>
<td colspan="3" rowspan="2">
<p><em>BF</em></p>
</td>
<td colspan="2">
<p>=</p>
</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>
<p><em>w</em></p>
</td>
<td>
<p>= <em>w</em></p>
</td>
<td>
<p>1+ <em>K</em></p>
</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
</tr>
<tr>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td colspan="4" rowspan="2">
<p>1+ <em>K</em></p>
</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td colspan="2" rowspan="2">
<p>1+ <em>K</em></p>
</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>
<p>0 <em>BF</em></p>
</td>
<td>
<p>0</p>
</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
</tr>
<tr>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
</tr>
</tbody>
</table>
<p><img decoding="async" class="wp-image-18334" src="https://telechargercours.com/wp-content/uploads/2021/07/word-image-220.png"> <img decoding="async" class="wp-image-18335" src="https://telechargercours.com/wp-content/uploads/2021/07/word-image-221.png"> <img decoding="async" class="wp-image-18336" src="https://telechargercours.com/wp-content/uploads/2021/07/word-image-222.png"> <img decoding="async" class="wp-image-18337" src="https://telechargercours.com/wp-content/uploads/2021/07/word-image-223.png"> <img decoding="async" class="wp-image-18338" src="https://telechargercours.com/wp-content/uploads/2021/07/word-image-224.png"> <img decoding="async" class="wp-image-18339" src="https://telechargercours.com/wp-content/uploads/2021/07/word-image-225.png"> <img decoding="async" class="wp-image-18340" src="https://telechargercours.com/wp-content/uploads/2021/07/word-image-226.png"> <img decoding="async" class="wp-image-18341" src="https://telechargercours.com/wp-content/uploads/2021/07/word-image-227.png"></p>
<p>Pôles de la fonction de transfert en boucle fermée (FTBF) :</p>
<table>
<tbody>
<tr>
<td rowspan="2">
<p>Si <em>x<sub>BF</sub></em></p>
</td>
<td rowspan="2">
<p>&lt;1</p>
</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
</tr>
<tr>
<td>
<p>les pôles sont :</p>
</td>
<td>
<p><sup>&#8211;</sup><em><sup>x</sup>BF <sup>w</sup></em>0<em> BF</em></p>
</td>
<td>
<p>±<em> jw </em><sub>0</sub><em><sub>BF</sub></em></p>
</td>
<td>&nbsp;</td>
<td colspan="3">
<p>1- <em>x<sub>BF</sub></em><sup>2</sup></p>
</td>
<td colspan="2">
<p>Re &lt; 0</p>
</td>
<td>&nbsp;</td>
</tr>
<tr>
<td rowspan="2">
<p>Si <em>x<sub>BF</sub></em></p>
</td>
<td rowspan="2">
<p>&gt; 1</p>
</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td colspan="2">&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
</tr>
<tr>
<td>
<p>les pôles sont :</p>
</td>
<td>
<p><sup>&#8211;</sup><em><sup>x</sup>BF <sup>w</sup> </em>0<em> BF</em></p>
</td>
<td>
<p><sup>±</sup><em><sup>w</sup></em> 0 <em>BF</em></p>
</td>
<td colspan="4">
<p><em>x<sub>BF</sub></em><sup>2</sup> -1</p>
</td>
<td>
<p>&lt; 0</p>
</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
</tr>
</tbody>
</table>
<p><img decoding="async" class="wp-image-18342" src="https://telechargercours.com/wp-content/uploads/2021/07/word-image-228.png"></p>
<p><strong><em>Conclusion : </em></strong>Un deuxième ordre est toujours stable (encore une fois sans notion de qualité de cette stabilité)</p>
<p>Page 3 <strong>Emmanuel FARGES</strong> Ó EduKlub S.A.</p>
<p>Tous droits de l’auteur des œuvres réservés. Sauf autorisation, la reproduction ainsi que toute utilisation des œuvres autre que la consultation individuelle et privée sont interdites.</p>
<p></p>
<p><strong>Sciences Indusrielles</strong></p>
<p><strong>Systèmes linéaires continus invariants</strong></p>
<p>Performance des systèmes asservis</p>
<h3><strong>1.5</strong> <strong>Stabilité des systèmes bouclés</strong></h3>
<p>On va chercher à quantifier la qualité de cette stabilité en définissant des marges vis-à-vis de la stabilité limite.</p>
<p><strong><em>On considère le système bouclé classique :</em></strong></p>
<table>
<tbody>
<tr>
<td rowspan="2">
<p><strong><em>E</em></strong></p>
</td>
<td>&nbsp;</td>
<td rowspan="2">
<p><strong><em>S</em></strong></p>
</td>
<td>&nbsp;</td>
</tr>
<tr>
<td>&nbsp;</td>
<td>&nbsp;</td>
</tr>
</tbody>
</table>
<p><img decoding="async" class="wp-image-18343" src="https://telechargercours.com/wp-content/uploads/2021/07/word-image-229.png"></p>
<ul>
<li><img decoding="async" class="wp-image-18344" src="https://telechargercours.com/wp-content/uploads/2021/07/word-image-230.png"><strong><em> A(p)</em></strong></li>
</ul>
<p><img decoding="async" class="wp-image-18345" src="https://telechargercours.com/wp-content/uploads/2021/07/word-image-231.png"></p>
<p>&#8211;</p>
<p><strong><em>B(p)</em></strong></p>
<table>
<tbody>
<tr>
<td rowspan="2">
<p><em>FTBO </em>=<em> A</em>(<em> p </em>)<em> B</em>(<em> p</em>)</p>
</td>
<td rowspan="2">
<p>et</p>
</td>
<td rowspan="2">
<p><em>FTBF </em>=</p>
</td>
<td>
<p><em>A</em>(<em> p</em>)</p>
</td>
<td rowspan="2">
<p>=</p>
</td>
<td>
<p><em>A</em>(<em> p</em>)</p>
</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
</tr>
<tr>
<td rowspan="2">
<p>1+ <em>A</em>( <em>p</em> ) <em>B</em>( <em>p</em>)</p>
</td>
<td colspan="2" rowspan="2">
<p>1+ <em>FTBO</em></p>
</td>
<td>&nbsp;</td>
</tr>
<tr>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
</tr>
<tr>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
</tr>
</tbody>
</table>
<p>Faire l’étude de la stabilité de ce système revient à faire l’étude des solutions du polynôme :1 + <em>FTBO</em> = 0 soit <em>FTBO</em> = -1</p>
<p><strong><em>Donc l’étude de la stabilité se fait dans ce cas sur la FTBO</em></strong></p>
<p><strong><em>Le point -1 est appelé point critique</em></strong></p>
<p><strong><em>Condition pratique de stabilité : </em></strong>Le but est de s’écarter le plus possible du point critique pour</p>
<p>(<sup>&#8211;</sup><em> <sup>jp</sup> </em>) <sup>ì</sup><em><sup>H</sup></em> <em>dB</em> <sup>=</sup> <sup>0</sup><em><sup>dB</sup></em></p>
<p><sub>lequel : &#8211; 1 = </sub><em><sub>e</sub></em> <sub>. Soit í</sub></p>
<p><strong><em>Marge de Gain :Valeur courante :10dB</em></strong></p>
<p>On lit la marge de gain sur les diagrammes de Bode de la <strong><em>FTBO</em></strong></p>
<p><em>H <sub>dB</sub> </em>(<em>w</em>)</p>
<p><img decoding="async" class="wp-image-18346" src="https://telechargercours.com/wp-content/uploads/2021/07/word-image-232.png"></p>
<p><strong><em>MG &lt; 0</em></strong></p>
<p><em>w</em></p>
<p><em>j </em>(<em>w </em>)</p>
<p><em>w</em></p>
<p><strong>-180°</strong></p>
<p><strong><em>Point critique</em></strong></p>
<p><strong><em>en phase</em></strong></p>
<p><strong><em>On se place au point critique sur le diagramme de phase </em></strong><em>j</em> (<em>w</em> ) = -180° <strong><em>et on lit la</em></strong></p>
<p><strong><em>marge de gain sur le diagramme de gain.</em></strong></p>
<p><strong><em>La marge de gain est l’écart entre 0dB et le gain au point critique en phase.</em></strong></p>
<ul>
<li><strong><em>Si on amplifie au point critique, le système est instable : MG&lt;0.</em></strong></li>
<li><strong><em>Si on atténue au point critique , le système est stable avec une marge MG&gt;0</em></strong></li>
</ul>
<p>Page 4 <strong>Emmanuel FARGES</strong> Ó EduKlub S.A.</p>
<p>Tous droits de l’auteur des œuvres réservés. Sauf autorisation, la reproduction ainsi que toute utilisation des œuvres autre que la consultation individuelle et privée sont interdites.</p>
<p></p>
<p><strong>Sciences Indusrielles</strong></p>
<p><strong>Systèmes linéaires continus invariants</strong></p>
<p>Performance des systèmes asservis</p>
<p><strong><em>Marge de Phase :Valeur courante 45°</em></strong></p>
<p>On lit la marge de phase sur les diagrammes de Bode de la <strong><em>FTBO</em></strong></p>
<p><img decoding="async" class="wp-image-18347" src="https://telechargercours.com/wp-content/uploads/2021/07/word-image-233.png"></p>
<table>
<tbody>
<tr>
<td rowspan="2">
<p><em><sup>H</sup> dB</em></p>
</td>
<td rowspan="2">
<p>(<em>w</em>)</p>
</td>
<td>
<p><strong><em>Point critique</em></strong></p>
</td>
<td>&nbsp;</td>
</tr>
<tr>
<td rowspan="2">
<p><strong><em>en gain 0 dB</em></strong></p>
</td>
<td>&nbsp;</td>
</tr>
<tr>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
</tr>
</tbody>
</table>
<p><em>w</em></p>
<p><em>j </em>(<em>w </em>)</p>
<p><em>w</em></p>
<p><em>Mj </em>&lt; 0</p>
<p><strong>-180°</strong></p>
<p><strong><em>On se place au point critique en gain (0 dB) et on lit la marge de phase sur le diagramme de phase La marge de phase </em></strong><em>Mj</em><strong><em> est</em></strong></p>
<p><img decoding="async" class="wp-image-18348" src="https://telechargercours.com/wp-content/uploads/2021/07/word-image-234.png"></p>
<p><strong><em>l’écart en -180° et la phase au point critique en gain.</em></strong></p>
<ul>
<li><strong><em>Si la phase au point critique en gain est inférieure à -180°, alors le système est instable et </em></strong><em>Mj</em> &lt; 0</li>
<li><strong><em>Si la phase au point critique en gain est supérieure à -180°, le système est stable avec une marge de phase </em></strong><em>Mj</em> &gt; 0</li>
</ul>
<p><strong><em>Marge de phase et de gain sur un diagramme de Black :</em></strong></p>
<p>Elles sont définies de façons identiques mais se lisent différemment, comme suit :</p>
<p><img decoding="async" class="wp-image-18349" src="https://telechargercours.com/wp-content/uploads/2021/07/word-image-235.png"></p>
<p><strong><em>Système instable</em></strong> <strong><em>Point critique</em></strong></p>
<p><strong><em>MG&lt;0</em></strong></p>
<p><em>Mj </em>&gt; 0</p>
<p><em>Mj </em>&lt; 0</p>
<p><strong><em>MG&gt;0</em></strong></p>
<p><strong><em>Système stable</em></strong></p>
<p>Page 5 <strong>Emmanuel FARGES</strong> Ó EduKlub S.A.</p>
<p>Tous droits de l’auteur des œuvres réservés. Sauf autorisation, la reproduction ainsi que toute utilisation des œuvres autre que la consultation individuelle et privée sont interdites.</p>
<p></p>
<p><strong>Sciences</strong> <strong>Indusrielles</strong></p>
<p><strong>Systèmes</strong> <strong>linéaires</strong> <strong>continus</strong> <strong>invariants</strong></p>
<p>Performance des systèmes asservis</p>
<h3><strong>1.</strong> <strong>6</strong> <strong>Influence</strong> <strong>du</strong> <strong>gain</strong> <strong>sur</strong> <strong>la</strong> <strong>stabilité</strong> <strong>des</strong> <strong>systèmes</strong> <strong>bouclés</strong></h3>
<p>On considère le système bouclé classique avec un correcteur proportionnel K dans la boucle ouverte suivant :</p>
<p><img decoding="async" class="wp-image-18350" src="https://telechargercours.com/wp-content/uploads/2021/07/word-image-236.png"></p>
<table>
<tbody>
<tr>
<td>
<p><strong><em>E</em></strong></p>
</td>
<td rowspan="2">
<p>+</p>
</td>
<td>&nbsp;</td>
<td>
<p><strong><em>S</em></strong></p>
</td>
<td>&nbsp;</td>
</tr>
<tr>
<td rowspan="2">
<p>Sans gain dans la boucle ouverte</p>
</td>
<td rowspan="2">
<p><strong><em>K</em></strong></p>
</td>
<td rowspan="2">
<p><strong><em>A(p)</em></strong></p>
</td>
<td>&nbsp;</td>
</tr>
<tr>
<td>&nbsp;</td>
<td>&nbsp;</td>
</tr>
<tr>
<td>
<p>la FTBO de ce système vaut :</p>
</td>
<td>&nbsp;</td>
<td rowspan="2">
<p>&#8211;</p>
</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
</tr>
<tr>
<td rowspan="2">
<p>FTBO = A(p) B(p)</p>
</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
</tr>
<tr>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
</tr>
</tbody>
</table>
<p><img decoding="async" class="wp-image-18351" src="https://telechargercours.com/wp-content/uploads/2021/07/word-image-237.png"></p>
<p>Avec gain dans la boucle ouverte, la</p>
<p>FTBO de ce système devient :</p>
<p>FTBO = K A(p) B(p)</p>
<p><strong><em>20LogK</em></strong></p>
<p><img decoding="async" class="wp-image-18352" src="https://telechargercours.com/wp-content/uploads/2021/07/word-image-238.png"><strong><em> B(p)</em></strong></p>
<table>
<tbody>
<tr>
<td>
<p><em>H </em>(<em>w</em>)</p>
</td>
<td>
<p><strong><em>Avec gain K</em></strong></p>
</td>
</tr>
</tbody>
</table>
<p>Si on augmente le gain en boucle ouverte , la phase de la FTBO reste inchangée, alors que le gain est multiplié par K. Cela veut dire que la courbe de phase dans Bode est la même avec ou sans gain alors que la courbe de gain avec gain se déduit de celle sans gain par translation de 20LogK vers le haut :</p>
<p>On passe donc d’un système stable puisque la marge de gain MG est positive à un système instable puisque la marge MG peut devenir négative avec le correcteur proportionnel (K suffisamment grand) dans la boucle ouverte.</p>
<p><strong><em>Sans gain K</em></strong></p>
<p><strong>-180°</strong></p>
<p><em>dB</em></p>
<p><strong><em>MG</em></strong> <strong><em>&lt;</em></strong> <strong><em>0</em></strong></p>
<p><em>w</em></p>
<p><strong><em>MG</em></strong> <strong><em>&gt;</em></strong> <strong><em>0</em></strong></p>
<p><em>j</em> ( <em>w</em> )</p>
<p><em>w</em></p>
<p><strong><em>Point</em></strong> <strong><em>critique</em></strong></p>
<p><strong><em>en</em></strong> <strong><em>phase</em></strong></p>
<table>
<tbody>
<tr>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>
<p><strong><em>E</em></strong></p>
</td>
<td rowspan="2">
<p>+</p>
</td>
<td>&nbsp;</td>
<td colspan="2" rowspan="2">
<p>1</p>
</td>
<td>&nbsp;</td>
</tr>
<tr>
<td rowspan="2">
<p>(</p>
</td>
<td rowspan="2">
<p>)</p>
</td>
<td rowspan="3">
<p>dans la</p>
</td>
<td rowspan="2">
<p>&#8211;</p>
</td>
<td>&nbsp;</td>
</tr>
<tr>
<td>&nbsp;</td>
<td>
<p><em>p</em></p>
</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
</tr>
<tr>
<td>
<p>Sans intégration <em>a</em> = 0</p>
</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td rowspan="2">
<p><em>a</em></p>
</td>
<td>&nbsp;</td>
</tr>
<tr>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
</tr>
</tbody>
</table>
<p>boucle ouverte la FTBO de ce système vaut :FTBO = A(p) B(p)</p>
<p><strong><em>S</em></strong></p>
<p><strong><em>A(p)</em></strong></p>
<p><strong><em>B(p)</em></strong></p>
<p>Page 6 <strong>Emmanuel</strong> <strong>FARGES</strong> Ó EduKlub S.A.</p>
<p>Tous droits de l’auteur des œuvres réservés. Sauf autorisation, la reproduction ainsi que toute utilisation des œuvres autre que la consultation individuelle et privée sont interdites.</p>
<p><img fetchpriority="high" decoding="async" class="alignnone wp-image-18353" title="Performance des systèmes asservis" src="https://telechargercours.com/wp-content/uploads/2021/07/word-image-239.png" alt="Performance des systèmes asservis" width="848" height="272"></p>
<p></p>
<p><strong>Sciences Indusrielles</strong></p>
<p><strong>Systèmes linéaires continus invariants</strong></p>
<p>Performance des systèmes asservis</p>
<p>Avec intégration dans la boucle ouverte, la FTBO de ce système</p>
<p><sup>1</sup></p>
<p>devient : <em>FTBO</em> = <em>A </em>(<em> p </em>)<em> B</em>(<em> p</em>)</p>
<p>L’influence de l’action intégrale est très simple sur les courbes de phase des diagrammes de Bode. En effet la phase d’un intégrateur est constante et vaut -90°. Donc les courbes de Bode sont translatées de -90° vers le bas à chaque fois que l’on rajoute un intégrateur.</p>
<p><img decoding="async" class="wp-image-18354" src="https://telechargercours.com/wp-content/uploads/2021/07/word-image-240.png"></p>
<p><strong><em>Aucune intégration </em></strong> <img decoding="async" class="wp-image-18355" src="https://telechargercours.com/wp-content/uploads/2021/07/word-image-241.png"></p>
<p><strong><em>Une intégration </em></strong> <img decoding="async" class="wp-image-18356" src="https://telechargercours.com/wp-content/uploads/2021/07/word-image-242.png"></p>
<p><strong><em>Seconde intégration</em></strong></p>
<p><strong><em>Point critique en phase avec une double intégration déplacé vers la gauche, c’est à dire vers les zones d’amplification au gain statique a MG ® -20LogK</em></strong></p>
<p><strong><em>Le système est déstabilisé</em></strong></p>
<table>
<tbody>
<tr>
<td>
<p>(</p>
</td>
<td>
<p>)</p>
</td>
<td rowspan="2">
<p><strong><em>Point critique</em></strong></p>
</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
</tr>
<tr>
<td>
<p><em>j w</em></p>
</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
</tr>
<tr>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>
<p><strong><em>en phase sans</em></strong></p>
</td>
<td rowspan="2">
<p><em>w</em></p>
</td>
<td>&nbsp;</td>
</tr>
<tr>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>
<p><strong><em>intégration</em></strong></p>
</td>
<td>&nbsp;</td>
</tr>
</tbody>
</table>
<p><strong>-180°</strong></p>
<p><strong><em>Point critique en phase avec une</em></strong></p>
<p><strong><em>intégration déplacé vers la</em></strong></p>
<p><strong><em>gauche, c’est à dire vers les</em></strong></p>
<p><strong><em>zones d’amplification :</em></strong></p>
<p><strong><em>Mauvais pour la stabilité</em></strong></p>
<p><strong><em>Conclusion :</em></strong> <strong><em>L’action intégrale dégrade la stabilité des systèmes</em></strong></p>
<h3><strong>2</strong> <strong>Précision des systèmes asservis.</strong></h3>
<h3><strong>2.1</strong> <strong>Ecart – Erreur &#8211; Précision</strong></h3>
<p>La précision caractérise l’écart entre la consigne et la valeur atteinte.</p>
<table>
<tbody>
<tr>
<td>
<p><em>e </em>=</p>
</td>
<td>
<p>lim</p>
</td>
<td>
<p><em>e</em>(<em>t </em>) &#8211;<em> s</em>(<em>t</em>) = lim</p>
</td>
<td>
<p><em>p </em>[<em>E</em>(<em> p</em>) &#8211;<em> S </em>(<em> p</em>)]</p>
</td>
<td colspan="3">
<p>d’après le théorème de la valeur finale.</p>
</td>
<td>&nbsp;</td>
</tr>
<tr>
<td>&nbsp;</td>
<td>
<p><em>t </em>® ¥</p>
</td>
<td>
<p><em>p </em>® 0</p>
</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
</tr>
<tr>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td rowspan="2">&nbsp;</td>
<td>&nbsp;</td>
</tr>
<tr>
<td rowspan="2">
<p><em>e </em>=</p>
</td>
<td rowspan="2">
<p>lim</p>
</td>
<td rowspan="2">
<p><em>pE</em>(<em> p</em>)[1 &#8211;<em> H </em>(<em> p</em>)]</p>
</td>
<td rowspan="2">
<p><em>avec H </em>(<em> p</em>) =</p>
</td>
<td rowspan="2">
<p><em>FTBF </em>=</p>
</td>
<td>
<p><em>S </em>(<em> p</em>)</p>
</td>
<td>&nbsp;</td>
</tr>
<tr>
<td rowspan="2">
<p><em>E </em>(<em> p</em>)</p>
</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
</tr>
<tr>
<td>&nbsp;</td>
<td>
<p><em>p </em>® 0</p>
</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
</tr>
<tr>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
</tr>
</tbody>
</table>
<p>Considérons un système à retour unitaire pour regarder la précision d’un système.</p>
<table>
<tbody>
<tr>
<td rowspan="2">
<p><em>A</em>(<em> p</em>) =<em> FTBO </em>=</p>
</td>
<td>
<p><em><sup>K</sup> BO</em></p>
</td>
<td>&nbsp;</td>
<td>
<p><em>N </em>(<em> p</em>)</p>
</td>
<td rowspan="2">
<p><em>avec</em></p>
</td>
<td>&nbsp;</td>
</tr>
<tr>
<td rowspan="2">
<p><em>p<sup>a</sup></em></p>
</td>
<td>&nbsp;</td>
<td rowspan="2">
<p><em>D </em>(<em> p</em>)</p>
</td>
<td>&nbsp;</td>
</tr>
<tr>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
</tr>
</tbody>
</table>
<p><em>pE </em>(<em> p</em>)</p>
<p>Donc : <em>e</em> = lim</p>
<p><em>p </em>®0 1+<em> FTBO</em></p>
<table>
<tbody>
<tr>
<td>
<p><strong><em>E</em></strong><sub>+</sub></p>
</td>
<td>&nbsp;</td>
<td>
<p><em>e</em></p>
</td>
<td rowspan="2">
<p><strong><em>A(p)</em></strong></p>
</td>
<td>&nbsp;</td>
</tr>
<tr>
<td>&nbsp;</td>
<td rowspan="2">
<p>&#8211;</p>
</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
</tr>
<tr>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
</tr>
</tbody>
</table>
<p><img decoding="async" class="wp-image-18357" src="https://telechargercours.com/wp-content/uploads/2021/07/word-image-243.png"></p>
<p><em>N </em>(0) =<em> D</em>(0) = 0</p>
<p><strong><em>S</em></strong></p>
<p>Page 7 <strong>Emmanuel FARGES</strong> Ó EduKlub S.A.</p>
<p>Tous droits de l’auteur des œuvres réservés. Sauf autorisation, la reproduction ainsi que toute utilisation des œuvres autre que la consultation individuelle et privée sont interdites.</p>
<p></p>
<p><strong>Sciences Indusrielles</strong></p>
<p><strong>Systèmes linéaires continus invariants</strong></p>
<p>Performance des systèmes asservis</p>
<h3><strong>2.2</strong> <strong>Précision statique</strong></h3>
<p>L’erreur statique notée <em>e<sub>s</sub></em> est l’erreur en régime établie pour une entrée de type</p>
<p><em><sup>A</sup></em></p>
<p>échelon : <em>E</em> ( <em>p</em>) =</p>
<table>
<tbody>
<tr>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td rowspan="3">
<p><em>A</em></p>
</td>
<td colspan="2">
<p>ì</p>
</td>
<td>&nbsp;</td>
<td>
<p><em>A</em></p>
</td>
<td rowspan="3">
<p>si <em>a</em> = 0:pasd&rsquo;intégrationenboucleouverte</p>
</td>
<td>&nbsp;</td>
</tr>
<tr>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td colspan="2" rowspan="2">
<p>ï</p>
</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
</tr>
<tr>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
</tr>
<tr>
<td rowspan="2">
<p>Donc :</p>
</td>
<td rowspan="2">
<p><em>e<sub>s</sub></em></p>
</td>
<td rowspan="2">
<p>= lim</p>
</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td rowspan="2">
<p>= <sub>í</sub></p>
</td>
<td rowspan="2">
<p>1</p>
</td>
<td rowspan="2">
<p>+ <em>K<sub>BO</sub></em></p>
</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
</tr>
<tr>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td rowspan="2">
<p><em><sup>K</sup>BO</em></p>
</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
</tr>
<tr>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>
<p><em>p </em>®0</p>
</td>
<td rowspan="2">
<p>1</p>
</td>
<td rowspan="2">
<p>+</p>
</td>
<td colspan="2">
<p>ï</p>
</td>
<td>&nbsp;</td>
<td colspan="2" rowspan="2">
<p>si <em>a</em> ³1:aumoinsunintégrateurenboucleouverte</p>
</td>
<td>&nbsp;</td>
</tr>
<tr>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>
<p><em>a</em></p>
</td>
<td colspan="3">
<p>î<sup>0</sup></p>
</td>
<td>&nbsp;</td>
</tr>
<tr>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>
<p><em>p</em></p>
</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
</tr>
</tbody>
</table>
<p><strong><em>Conclusion :</em></strong></p>
<ul>
<li><strong><em>L’erreur statique d’un système bouclé est nulle si on a au moins un intégrateur en boucle ouverte.</em></strong></li>
<li><strong><em>L’erreur statique est d’autant plus faible que le gain en boucle ouverte K<sub>BO</sub> est grand quand il n’y a pas d’intégrateur en boucle ouverte.</em></strong></li>
</ul>
<h3><strong>2.3</strong> <strong>Précision dynamique</strong></h3>
<p>L’erreur en vitesse ou précision dynamique ou encore erreur de traînage <em>e<sub>s</sub></em> est l’erreur en</p>
<p>régime établi pour une entrée rampe du type <em>E</em> ( <em>p</em>) = <em><sup>V</sup></em> <em>p</em><sup>2</sup></p>
<table>
<tbody>
<tr>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td colspan="2">
<p>ì</p>
</td>
<td>&nbsp;</td>
<td colspan="3">
<p>¥ si a=0pasd&rsquo;intégrateurenboucleouverte</p>
</td>
<td>&nbsp;</td>
</tr>
<tr>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td colspan="2" rowspan="2">
<p><em>V</em></p>
</td>
<td colspan="2">
<p>ï</p>
</td>
<td>&nbsp;</td>
<td colspan="3" rowspan="2">
<p><em>V</em></p>
</td>
<td>&nbsp;</td>
</tr>
<tr>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td colspan="2">
<p>ï</p>
</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
</tr>
<tr>
<td rowspan="2">
<p>Donc :</p>
</td>
<td rowspan="2">
<p><em>e<sub>t</sub></em></p>
</td>
<td rowspan="2">
<p>= lim</p>
</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td rowspan="2">
<p>= í</p>
</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td rowspan="2">
<p>si a=1unint égrateur enboucleouverte</p>
</td>
<td>&nbsp;</td>
</tr>
<tr>
<td>&nbsp;</td>
<td rowspan="2">
<p><em><sup>K</sup>BO</em></p>
</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td rowspan="2">
<p><em>K</em></p>
</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
</tr>
<tr>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>
<p><em>p</em>®0</p>
</td>
<td rowspan="2">
<p><em>p </em>+</p>
</td>
<td colspan="2" rowspan="2">
<p>ï</p>
</td>
<td>&nbsp;</td>
<td colspan="2" rowspan="2">
<p><em>BO</em></p>
</td>
<td>&nbsp;</td>
</tr>
<tr>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>
<p><em>a</em>-1</p>
</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
</tr>
<tr>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>
<p><em>p</em></p>
</td>
<td colspan="2">
<p>ï</p>
</td>
<td colspan="4" rowspan="2">
<p>si a ³ 2 au moinsdeuxintégrateursenboucleouverte</p>
</td>
<td>&nbsp;</td>
</tr>
<tr>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td colspan="2">
<p>î<sup>0</sup></p>
</td>
<td>&nbsp;</td>
</tr>
</tbody>
</table>
<h3><strong>2.4</strong> <strong>Résumé</strong></h3>
<p>On peut généraliser la notion de précision en fonction du nombre d’intégrateur dans la boucle ouverte sous forme d’un tableau :</p>
<table>
<tbody>
<tr>
<td>&nbsp;</td>
<td colspan="2">
<p><strong><em>Entrée</em></strong></p>
</td>
<td>&nbsp;</td>
<td colspan="2">
<p><strong><em>Précision, erreur</em></strong></p>
</td>
<td colspan="3">
<p><strong><em>0 intégrateur</em></strong></p>
</td>
<td colspan="3">
<p><strong><em>1 intégrateur</em></strong></p>
</td>
<td>&nbsp;</td>
<td colspan="3">
<p><strong><em>2 intégrateur</em></strong></p>
</td>
<td>
<p><strong><em>3 intégrateur</em></strong></p>
</td>
<td>&nbsp;</td>
</tr>
<tr>
<td>&nbsp;</td>
<td>
<p><strong><em>Echelon : A u(t)</em></strong></p>
</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>
<p><strong><em>Erreur statique</em></strong></p>
</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td colspan="2">
<p><em>A</em></p>
</td>
<td colspan="2">
<p><strong><em>0</em></strong></p>
</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td colspan="2">
<p><strong><em>0</em></strong></p>
</td>
<td>&nbsp;</td>
<td>
<p><strong><em>0</em></strong></p>
</td>
<td>&nbsp;</td>
</tr>
<tr>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
</tr>
<tr>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td colspan="3">
<p>1+ <em>K</em> <em><sub>BO</sub></em></p>
</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
</tr>
<tr>
<td>&nbsp;</td>
<td>
<p><strong><em>Rampe : V t u(t)</em></strong></p>
</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>
<p><strong><em>Erreur en vitesse</em></strong></p>
</td>
<td>&nbsp;</td>
<td colspan="2">
<p><strong><em>¥</em></strong></p>
</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td colspan="2">
<p><em>V</em></p>
</td>
<td>&nbsp;</td>
<td colspan="2">
<p><strong><em>0</em></strong></p>
</td>
<td>&nbsp;</td>
<td>
<p><strong><em>0</em></strong></p>
</td>
<td>&nbsp;</td>
</tr>
<tr>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>
<p><em><sup>K</sup> BO</em></p>
</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
</tr>
<tr>
<td>&nbsp;</td>
<td>
<p><strong><em>Parabole :</em></strong></p>
</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>
<p><strong><em>Erreur en</em></strong></p>
</td>
<td>&nbsp;</td>
<td colspan="2">
<p><strong><em>¥</em></strong></p>
</td>
<td>&nbsp;</td>
<td colspan="2">
<p><strong><em>¥</em></strong></p>
</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td colspan="2">
<p><em>g</em></p>
</td>
<td>
<p><strong><em>0</em></strong></p>
</td>
<td>&nbsp;</td>
</tr>
<tr>
<td>&nbsp;</td>
<td rowspan="2">
<p><em>gt </em><sup>2</sup><em>u </em>(<em>t </em>) <strong><em>²</em></strong></p>
</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>
<p><strong><em>accélération</em></strong></p>
</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td rowspan="2">&nbsp;</td>
<td rowspan="2">
<p><em><sup>K</sup> BO</em></p>
</td>
<td rowspan="2">&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
</tr>
<tr>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
</tr>
</tbody>
</table>
<h3><strong>2.5</strong> <strong>Influence des perturbations</strong></h3>
<p>Soit le système perturbé représenté par le schéma bloc ci-dessous :</p>
<p>Page 8 <strong>Emmanuel FARGES</strong> Ó EduKlub S.A.</p>
<p>Tous droits de l’auteur des œuvres réservés. Sauf autorisation, la reproduction ainsi que toute utilisation des œuvres autre que la consultation individuelle et privée sont interdites.</p>
<p></p>
<p><strong>Sciences Indusrielles</strong></p>
<p><strong>Systèmes linéaires continus invariants</strong></p>
<p>Performance des systèmes asservis</p>
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<p><strong><em>E</em></strong></p>
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<p><strong><em>F<sub>1</sub>(p)</em></strong></p>
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<p><strong><em>F<sub>2</sub>(p)</em></strong></p>
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<p><img decoding="async" class="wp-image-18358" src="https://telechargercours.com/wp-content/uploads/2021/07/word-image-244.png"></p>
<p><strong><em>E est l’entrée du système, c’est à dire la consigne</em></strong></p>
<p><strong><em>S est la sortie du système, c’est à dire la réponse à l’entrée E</em></strong></p>
<p><strong><em>P est une seconde entrée du système appelée perturbation car elle est subie, c’est à dire qu’on ne la maîtrise pas : le vent , la houle, les courant dans le cas d’un pilote automatique de bateau chargé de maintenir le cap de l’embarcation.</em></strong></p>
<ul>
<li><strong><em>: est l’erreur du système puisque c’est la différence entre l’entrée E (ce que l’on veut faire) et la sortie S (ce que l’on a fait)</em></strong></li>
</ul>
<p>Plaçons nous dans le cas général pour les fonctions de transfert F<sub>1</sub> et F<sub>2</sub> :</p>
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<p>1</p>
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<p><em>N</em></p>
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<p>( <em>p</em>)</p>
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<p>ì<em>N</em></p>
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<p>( <em>p</em> )polynomeen p avec <em>N</em></p>
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<p>(0) = 1</p>
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<p>C’est à dire : <em>F</em> ( <em>p</em>) =</p>
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<p>1</p>
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<p>avec</p>
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<p>í <sup>1</sup></p>
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<p>1</p>
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<p><em>D</em><sub>1</sub> (<em> p</em>)</p>
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<p>( <em>p</em> )polynomeen p avec <em>D</em><sub>1</sub> (0) = 1</p>
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<p>1</p>
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<p><em><sub>p</sub><sup>a</sup></em>1</p>
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<p>î <em><sup>D</sup></em>1</p>
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<p>Et : <em>F</em> ( <em>p</em>) =</p>
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<p>2</p>
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<p><em>N</em></p>
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<p>2</p>
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<p>( <em>p</em>)</p>
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<p>avec</p>
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<p>ì<em>N</em></p>
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<p>2</p>
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<p>( <em>p</em>)polynomeenpavec <em>N</em></p>
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<p>2</p>
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<p>(0)</p>
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<p>= 1</p>
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<p>í</p>
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<p><em>D </em><sub>2</sub> (<em>p </em>)</p>
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<p>( <em>p</em>)polynomeenpavec <em>D</em><sub>2</sub></p>
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<p>2</p>
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<p><em><sub>p</sub><sup>a</sup></em>2</p>
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<p>î <em><sup>D</sup></em>2</p>
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<p>(0)</p>
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<p>=1</p>
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<p>Il existe deux types d’erreur sur ces systèmes puisque l’on a deux types d’entrées différentes.</p>
<p>On peut donc décomposer l’erreur en deux :</p>
<p>Une erreur vis-à-vis de l’entrée, c’est à dire la consigne (à perturbation nulle) : <em>e</em><sub>1</sub></p>
<p>Une erreur vis-à-vis de la perturbation (à consigne nulle) : <em>e</em><sub>2</sub></p>
<p>On a alors : <em>e</em> = <em>e</em><sub>1</sub> + <em>e</em><sub>2</sub></p>
<p><strong><em>Calculons </em></strong><em><sup>e</sup></em>1<strong><em> :</em></strong></p>
<p>On peut refaire le schéma bloc avec une perturbation nulle :</p>
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<p><em>e</em><sub>1</sub> = <em>E</em> &#8211; <em>S</em></p>
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<p><strong><em>E</em></strong></p>
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<p>+</p>
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<p><em>e</em><sub>1</sub></p>
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<p><strong><em>S</em></strong></p>
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<p><strong><em>F<sub>1</sub>(p) F<sub>2</sub>(p)</em></strong></p>
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<p><em>S </em>=<em> F</em><sub>1</sub><em> F</em><sub>2</sub><em> e</em><sub>1</sub></p>
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<p>Donc</p>
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<p><em>e</em><sub>1</sub> =</p>
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<p><em>E</em></p>
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<p>1 + <em>F</em><sub>1</sub> <em>F</em><sub>2</sub></p>
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<p><strong><em>Calculons </em></strong><em>e</em><sub>2</sub></p>
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<p><strong><em>:</em></strong></p>
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<p><strong><em>P</em></strong></p>
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<p><img decoding="async" class="wp-image-18359" src="https://telechargercours.com/wp-content/uploads/2021/07/word-image-245.png"> <img decoding="async" class="wp-image-18360" src="https://telechargercours.com/wp-content/uploads/2021/07/word-image-246.png"> <img decoding="async" class="wp-image-18361" src="https://telechargercours.com/wp-content/uploads/2021/07/word-image-247.png"></p>
<p>On peut refaire le schéma bloc avec une entrée nulle :</p>
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<p>+</p>
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<p><em>e</em><sub>2</sub></p>
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<p><strong><em>S</em></strong></p>
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<p><strong><em>F<sub>1</sub>(p)</em></strong></p>
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<p><strong><em>F<sub>2</sub>(p)</em></strong></p>
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<p><em>e</em><sub>2</sub> = &#8211;<em>S</em></p>
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<p><em>S </em>=<em> F</em><sub>2</sub> [</p>
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<p><em>P </em>+<em> F</em><sub>1</sub><em>e</em></p>
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<p>2 ]</p>
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<p>Page 9</p>
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<p><strong>Emmanuel FARGES</strong></p>
</td>
<td>&nbsp;</td>
<td colspan="2">
<p>Ó EduKlub S.A.</p>
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<p><img decoding="async" class="wp-image-18362" src="https://telechargercours.com/wp-content/uploads/2021/07/word-image-248.png"> <img decoding="async" class="wp-image-18363" src="https://telechargercours.com/wp-content/uploads/2021/07/word-image-249.png"></p>
<p>Tous droits de l’auteur des œuvres réservés. Sauf autorisation, la reproduction ainsi que toute utilisation des œuvres autre que la consultation individuelle et privée sont interdites.</p>
<p></p>
<p><strong>Sciences Indusrielles</strong></p>
<p><strong>Systèmes linéaires continus invariants</strong></p>
<p>Performance des systèmes asservis</p>
<p>D’où : <em>e</em><sub>2</sub> = &#8211; <em><sup>F</sup></em><sup>2</sup><em><sup> P</sup></em></p>
<p>1+ <em>F</em><sub>1</sub> <em>F</em><sub>2</sub></p>
<p>Les erreurs vis-à-vis de l’entrée ayant déjà été abordées, on ne s’intéresse ici qu’à l’erreur statique vis-à-vis de la perturbation.</p>
<p>On soumet donc le système à une perturbation échelon (erreur statique), et on regarde quelle</p>
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<p>est l’erreur générée : <em>e<sub>s</sub></em> <sub>2</sub></p>
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<p>= lim <em>e</em></p>
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<p><sub>2</sub> (<em>t</em> ) = lim <em>pe</em></p>
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<p><sub>2</sub> ( <em>p</em>)</p>
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<p><em>t </em>®¥</p>
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<p><em>p</em>®0</p>
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<p><em><sup>e</sup> s </em>2</p>
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<p>= lim <em>pe</em><sub>2</sub> ( <em>p</em>) = lim <em>p</em></p>
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<p><em>P </em>(<em>p </em>)<em>F</em><sub>2</sub> (<em> p</em>)</p>
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<p>avec</p>
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<p><em>P </em>(<em> p</em>) =</p>
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<p><em>P</em><sub>0</sub></p>
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<p>1+ <em>F</em><sub>1</sub> ( <em>p</em>)<em>F</em><sub>2</sub> ( <em>p</em> )</p>
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<p><em>p</em></p>
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<p><em>p </em>®0</p>
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<p><em>p</em>®0</p>
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<td colspan="3">&nbsp;</td>
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<td colspan="2">&nbsp;</td>
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<td colspan="2">&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
</tr>
<tr>
<td rowspan="3">
<p><em><sup>e</sup>s </em>2</p>
</td>
<td rowspan="3">
<p>= lim</p>
</td>
<td>
<p><em>P K</em></p>
</td>
<td rowspan="2">
<p>2</p>
</td>
<td>
<p><em><sub>p</sub></em><sup>&#8211;</sup><em><sup>a</sup></em> 2</p>
</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td rowspan="3">
<p>= lim</p>
</td>
<td>&nbsp;</td>
<td colspan="3">
<p><em>P K</em></p>
</td>
<td rowspan="2">
<p>2</p>
</td>
<td>
<p><em><sub>p</sub><sup>a</sup></em>1</p>
</td>
<td>&nbsp;</td>
<td rowspan="3">
<p>= lim</p>
</td>
<td colspan="2">
<p><em>P K</em></p>
</td>
<td rowspan="2">
<p>2</p>
</td>
<td colspan="3">
<p><em><sub>p</sub><sup>a</sup></em>1</p>
</td>
<td>&nbsp;</td>
</tr>
<tr>
<td>
<p>0</p>
</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>
<p>0</p>
</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>
<p>0</p>
</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
</tr>
<tr>
<td>&nbsp;</td>
<td>&nbsp;</td>
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<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td colspan="4" rowspan="2">
<p><sub>+</sub> <em><sub>p</sub></em><sup>(</sup><em><sup>a</sup></em> 1 <sup>+</sup><em><sup>a</sup></em>2 <sup>)</sup></p>
</td>
<td colspan="6" rowspan="2">
<p><em>K </em><sub>1</sub><em> K</em><sub>2</sub></p>
</td>
<td>&nbsp;</td>
</tr>
<tr>
<td>&nbsp;</td>
<td colspan="6">
<p><em>p</em>®0 <sub>1+</sub><em> <sub>K K p</sub></em><sup>-(</sup><em><sup>a</sup></em>1 <sup>+</sup><em><sup>a</sup></em>2 <sup>)</sup></p>
</td>
<td>
<p><em>p </em>® 0</p>
</td>
<td>&nbsp;</td>
<td>
<p><em>K K</em></p>
</td>
<td rowspan="2">
<p>2</p>
</td>
<td>
<p><em>p</em>®0</p>
</td>
<td>&nbsp;</td>
</tr>
<tr>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>
<p>1</p>
</td>
<td>
<p>2</p>
</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td colspan="2">
<p>1</p>
</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
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<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
</tr>
</tbody>
</table>
<p>Donc si on veut annuler l’erreur statique vis-à-vis d’une perturbation (ce qui est très intéressant étant donnée qu’en général on ne maîtrise pas cette perturbation) : il faut <em>a</em><sub>1</sub> ³ 1</p>
<p><strong><em>Conclusion : Pour annuler l’erreur statique vis-à-vis d’une perturbation, il faut placer au moins un intégrateur avant celle-ci.</em></strong></p>
<p>Page 10 <strong>Emmanuel FARGES</strong> Ó EduKlub S.A.</p>
<p>Tous droits de l’auteur des œuvres réservés. Sauf autorisation, la reproduction ainsi que toute utilisation des œuvres autre que la consultation individuelle et privée sont interdites.</p>


<p class="has-text-align-center wp-block-paragraph"><strong><a href="https://www.mediafire.com/file/kvnxb6b4ez2fxpx/Systèmes+linéaires+continus+invariants+Performance+des+systèmes+asservis+SI-CO09-AAS-EF.pdf/file" target="_blank" rel="noreferrer noopener">Telecharger</a></strong></p>
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		<item>
		<title>Systèmes linéaires continus invariants ANALYSE FREQUENTIELLE partie 3</title>
		<link>https://telechargercours.com/systemes-lineaires-continus-invariants-analyse-frequentielle-partie-3/</link>
		
		<dc:creator><![CDATA[Cours Zilla]]></dc:creator>
		<pubDate>Mon, 05 Jul 2021 12:41:47 +0000</pubDate>
				<category><![CDATA[Science & Technologie]]></category>
		<guid isPermaLink="false">https://telechargercours.com/?p=18256</guid>

					<description><![CDATA[Sciences Indusrielles Systèmes linéaires continus invariants ANALYSE FREQUENTIELLE SYSTEMES LINEAIRES CONTINUS INVARIANTS ANALYSE FREQUENTIELLE 3 Système du deuxième ordre. 3.1 Définition d’un système du deuxième ordre Un système du second ordre a son compo rtement régi par une équation différentielle du deuxième ordre de la forme : &#160; &#160; &#160; &#160; &#160; &#160; &#160; &#160; [&#8230;]]]></description>
										<content:encoded><![CDATA[<p style="text-align: center;"><strong>Sciences Indusrielles</strong></p>
<h4 style="text-align: center;"><strong>Systèmes linéaires continus invariants</strong></h4>
<p style="text-align: center;">ANALYSE FREQUENTIELLE</p>
<p style="text-align: center;"><strong>SYSTEMES LINEAIRES CONTINUS</strong></p>
<p style="text-align: center;"><strong>INVARIANTS</strong></p>
<p style="text-align: center;"><em>ANALYSE FREQUENTIELLE</em></p>
<h2><strong>3</strong> <strong>Système du deuxième ordre.</strong></h2>
<h3><strong>3.1</strong> <strong>Définition d’un système du deuxième ordre</strong></h3>
<p><strong>Un système du second ordre a son compo rtement régi par une équation différentielle du deuxième ordre de la forme :</strong></p>
<table>
<tbody>
<tr>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
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<td>&nbsp;</td>
<td>
<p>ì</p>
</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>
<p><em>e </em>=<em> e </em>(<em>t</em>) entréedusystème</p>
</td>
<td>&nbsp;</td>
</tr>
<tr>
<td>&nbsp;</td>
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<td>
<p>ï</p>
</td>
<td colspan="3" rowspan="2">
<p><em>s </em>=<em> s</em>(<em>t</em>) réponsedusystèmeàl&rsquo;entréee(t)</p>
</td>
<td>&nbsp;</td>
</tr>
<tr>
<td colspan="2" rowspan="3">
<p>1</p>
</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td rowspan="3">
<p><em>d</em></p>
</td>
<td rowspan="2">
<p>2</p>
</td>
<td rowspan="3">
<p><em>s</em></p>
</td>
<td>&nbsp;</td>
<td colspan="3" rowspan="3">
<p>2<em>x</em> <em>ds</em></p>
</td>
<td>&nbsp;</td>
<td rowspan="2">
<p>ï</p>
</td>
<td>&nbsp;</td>
</tr>
<tr>
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<td>&nbsp;</td>
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<p>+</p>
</td>
<td rowspan="3">
<p>+ <em>s</em> = <em>Ke</em></p>
</td>
<td>
<p>ï</p>
</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td rowspan="3">
<p>K:gainstatique(&gt;0)</p>
</td>
<td>&nbsp;</td>
</tr>
<tr>
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<td>&nbsp;</td>
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<p>í</p>
</td>
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<td>&nbsp;</td>
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</tr>
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<td>&nbsp;</td>
<td colspan="3" rowspan="2">
<p><em>w</em><sub>0</sub><sup>2</sup></p>
</td>
<td colspan="3" rowspan="2">
<p><em>dt </em><sup>2</sup></p>
</td>
<td rowspan="2">
<p><em>w</em><sub>0</sub></p>
</td>
<td>&nbsp;</td>
<td rowspan="2">
<p><em>dt</em></p>
</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
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</tr>
<tr>
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<td>&nbsp;</td>
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<p><sup>ï</sup><em>x</em> :coefficient d&rsquo;amortissement (sans dimension) (&gt;0)</p>
</td>
<td>&nbsp;</td>
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<td>
<p>ï</p>
</td>
<td rowspan="2">
<p><em>w</em></p>
</td>
<td>&nbsp;</td>
<td rowspan="2">
<p>pulsationpropredusystème(enrads<sup>-1</sup> )</p>
</td>
<td>&nbsp;</td>
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<p>ï</p>
</td>
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<p>0</p>
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<p>î</p>
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</tr>
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<td colspan="4">
<p><strong>&nbsp;</strong></p>
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<td colspan="9">
<p><strong>&nbsp;</strong></p>
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</table>
<h3><strong>3.2 Fonction de transfert globale d’un deuxième ordre</strong></h3>
<p>On applique la transformée de Laplace à l’ensemble de l’équation différentielle ci-dessus, avec des conditions initiales nulles :</p>
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<p>Donc :</p>
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<td colspan="3">
<p>1 <em>d</em> <sup>2</sup> <em>s</em></p>
</td>
<td rowspan="3">
<p>+</p>
</td>
<td colspan="4">
<p>2<em>x</em> <em>ds</em></p>
</td>
<td rowspan="3">
<p>+ <em>s</em> = <em>Ke</em> transforméedeLaplace</p>
</td>
<td>
<p><em>p</em><sup>2</sup></p>
</td>
<td rowspan="3">
<p><em>S </em>(<em> p</em>) +</p>
</td>
<td>
<p>2<em>x</em></p>
</td>
<td rowspan="3">
<p><em>pS</em>(<em> p</em>) +<em> S </em>(<em> p</em>) =<em> KE </em>(<em> p</em>)</p>
</td>
<td>&nbsp;</td>
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<td>&nbsp;</td>
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<td>&nbsp;</td>
<td colspan="3" rowspan="2">
<p><em>w</em><sup>2</sup><em> dt </em><sup>2</sup></p>
</td>
<td rowspan="2">
<p><em>w</em></p>
</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td rowspan="2">
<p><em>dt</em></p>
</td>
<td>&nbsp;</td>
<td rowspan="2">
<p><em>w</em></p>
</td>
<td>&nbsp;</td>
</tr>
<tr>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td rowspan="2">
<p>0</p>
</td>
<td>&nbsp;</td>
<td colspan="2">
<p>uuuuuuuuuuuuuuuuuuuuuuuuru <em><sub>w</sub></em> 2</p>
</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
</tr>
<tr>
<td colspan="2">
<p>0</p>
</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>
<p>0</p>
</td>
<td>&nbsp;</td>
<td>
<p>0</p>
</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
</tr>
</tbody>
</table>
<p>On peut alors présenter le rapport de la sortie S(p) sur l’entrée E(p), c’est à dire la fonction de</p>
<table>
<tbody>
<tr>
<td rowspan="2">
<p>transfert globale du système :</p>
</td>
<td rowspan="2">
<p><em>H </em>(<em> p</em>) =</p>
</td>
<td>
<p><em>S</em>(<em> p</em>)</p>
</td>
<td rowspan="2">
<p>=</p>
</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td colspan="2">
<p><em>K</em></p>
</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
</tr>
<tr>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td rowspan="2">
<p>2<em>x</em></p>
</td>
<td>&nbsp;</td>
<td colspan="2" rowspan="2">
<p><em>p</em><sup>2</sup></p>
</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
</tr>
<tr>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td colspan="2">
<p><em>E </em>(<em> p</em>)</p>
</td>
<td rowspan="2">
<p>1+</p>
</td>
<td rowspan="2">
<p><em>p </em>+</p>
</td>
<td rowspan="2">&nbsp;</td>
<td rowspan="2">&nbsp;</td>
<td>&nbsp;</td>
</tr>
<tr>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td rowspan="2">
<p><em>w</em><sub>0</sub></p>
</td>
<td colspan="2">&nbsp;</td>
<td>&nbsp;</td>
</tr>
<tr>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>
<p><em>w</em><sub>0</sub></p>
</td>
<td>
<p>2</p>
</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
</tr>
</tbody>
</table>
<p><strong><em>On notera :</em></strong></p>
<ul>
<li>Le gain statique vaut : K=H(0)</li>
<li>Pour identifier les caractéristiques d’un système du deuxième ordre (c’est à dire</li>
</ul>
<p><em>K </em>,<em> x </em>et<em> w</em><sub>0</sub> ), on veillera bien à présenter la fonction de transfert globale H(p) avec le coefficient en p<sup>0</sup> du polynôme au dénominateur égal à 1. Ainsi le numérateur peut être identifié au gain statique K, coefficient en p<sup>2</sup> du polynôme au dénominateur peut être identifié à l’inverse de la pulsation propre<em>w</em><sub>0</sub> et on déduit l’amortissement du</p>
<p>coefficient en p<sup>1</sup> : <sup>2</sup><em><sup>x</sup></em> .</p>
<p><em>w</em><sub>0</sub></p>
<p>Page 1 <strong>Emmanuel FARGES</strong> Ó EduKlub S.A.</p>
<p>Tous droits de l’auteur des œuvres réservés. Sauf autorisation, la reproduction ainsi que toute utilisation des œuvres autre que la consultation individuelle et privée sont interdites.</p>
<p></p>
<p><strong>Sciences Indusrielles</strong></p>
<p><strong>Systèmes linéaires continus invariants</strong></p>
<p>ANALYSE FREQUENTIELLE</p>
<h3><strong>3.3</strong> <strong>Gain et phase réels d’un deuxième ordre</strong></h3>
<table>
<tbody>
<tr>
<td rowspan="2">
<p><em>H </em>(<em> jw</em>) =</p>
</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td colspan="2">
<p><em>K</em></p>
</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td rowspan="2">
<p>=</p>
</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td colspan="2">
<p><em>K</em></p>
</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
</tr>
<tr>
<td colspan="2">&nbsp;</td>
<td>&nbsp;</td>
<td rowspan="3">
<p>(<em> jw </em>)</p>
</td>
<td>
<p>2</p>
</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td rowspan="2">
<p>2</p>
</td>
<td rowspan="2">
<p>ö</p>
</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
</tr>
<tr>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td rowspan="2">
<p>2<em>x</em></p>
</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td colspan="3">
<p>æ</p>
</td>
<td rowspan="3">
<p>&#8211;</p>
</td>
<td rowspan="2">
<p><em>w</em></p>
</td>
<td rowspan="3">
<p>+ <em>j</em></p>
</td>
<td rowspan="2">
<p>2<em>xw</em></p>
</td>
<td rowspan="2">&nbsp;</td>
<td>&nbsp;</td>
</tr>
<tr>
<td colspan="2" rowspan="2">
<p>1+</p>
</td>
<td rowspan="2">
<p><em>jw </em>+</p>
</td>
<td>&nbsp;</td>
<td colspan="3" rowspan="2">
<p>ç<sup>1</sup></p>
</td>
<td>&nbsp;</td>
<td rowspan="2">
<p>÷</p>
</td>
<td>&nbsp;</td>
</tr>
<tr>
<td rowspan="2">
<p><em>w</em><sub>0</sub></p>
</td>
<td rowspan="2">
<p><em>w</em><sub>0</sub> <sup>2</sup></p>
</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>
<p>2</p>
</td>
<td colspan="2" rowspan="2">
<p><em>w</em><sub>0</sub></p>
</td>
<td>&nbsp;</td>
</tr>
<tr>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td colspan="3">
<p>è</p>
</td>
<td>&nbsp;</td>
<td colspan="2">
<p><em>w </em><sub>0</sub></p>
</td>
<td>
<p>ø</p>
</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
</tr>
</tbody>
</table>
<p><strong><em>Calcul du gain :</em></strong></p>
<table>
<tbody>
<tr>
<td>&nbsp;</td>
<td colspan="3" rowspan="5">
<p><em>H </em>(<em> jw</em>)</p>
</td>
<td>&nbsp;</td>
<td colspan="2" rowspan="5">
<p>=</p>
</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td rowspan="2">
<p><em>K</em></p>
</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td colspan="9" rowspan="5">
<p>soit en dB :</p>
</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td colspan="10" rowspan="7">
<p><em>H </em>(<em> jw </em>)</p>
</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td colspan="6" rowspan="7">
<p><em><sub>dB</sub></em> = 20<em>Log</em></p>
</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td colspan="2" rowspan="7">
<p><em>H </em>(<em> jw</em>)</p>
</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
</tr>
<tr>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
</tr>
<tr>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td rowspan="4">&nbsp;</td>
<td rowspan="4">&nbsp;</td>
<td rowspan="4">&nbsp;</td>
<td rowspan="5">&nbsp;</td>
<td rowspan="4">&nbsp;</td>
<td rowspan="5">&nbsp;</td>
<td rowspan="4">&nbsp;</td>
<td rowspan="5">&nbsp;</td>
<td rowspan="4">&nbsp;</td>
<td rowspan="4">&nbsp;</td>
<td rowspan="5">&nbsp;</td>
<td rowspan="4">&nbsp;</td>
<td rowspan="5">&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
</tr>
<tr>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td rowspan="2">&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td rowspan="2">&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
</tr>
<tr>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td rowspan="3">
<p>2</p>
</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
</tr>
<tr>
<td colspan="9" rowspan="3">
<p>æ</p>
</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td rowspan="3">
<p><em>w</em><sup>2</sup></p>
</td>
<td colspan="6" rowspan="3">
<p>4<em>x</em> <sup>2</sup><em>w</em> <sup>2</sup></p>
</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
</tr>
<tr>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
</tr>
<tr>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>
<p>ö</p>
</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
</tr>
<tr>
<td colspan="9" rowspan="2">
<p>ç</p>
</td>
<td colspan="2" rowspan="2">
<p>1-</p>
</td>
<td>&nbsp;</td>
<td rowspan="2">
<p>÷ <sup>+</sup></p>
</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
</tr>
<tr>
<td>
<p><em>w </em><sup>2</sup></p>
</td>
<td colspan="2">
<p><em>w</em><sup>2</sup></p>
</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
</tr>
<tr>
<td colspan="9">
<p>è</p>
</td>
<td colspan="3">
<p>0</p>
</td>
<td>
<p>ø</p>
</td>
<td colspan="2">
<p>0</p>
</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
</tr>
<tr>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td colspan="5">&nbsp;</td>
<td>
<p>æ</p>
</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td colspan="5">
<p><em><sub>w</sub></em> 2 <sub>ö</sub><sup>2</sup></p>
</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td colspan="12">
<p>4<em>x</em> <sup>2</sup> <em>w</em><sup>2</sup></p>
</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
</tr>
<tr>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td colspan="4" rowspan="2">
<p><em>H </em>(<em> jw</em>)</p>
</td>
<td colspan="3" rowspan="2">
<p><em>dB</em></p>
</td>
<td colspan="11" rowspan="2">
<p>= 20<em>LogK</em> &#8211; 20<em>Log</em> <sub>ç</sub>1</p>
</td>
<td rowspan="2">
<p>&#8211;</p>
</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td rowspan="2">
<p>÷</p>
</td>
<td colspan="3" rowspan="2">
<p>+</p>
</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
</tr>
<tr>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td colspan="3">
<p>2</p>
</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td colspan="2">
<p>2</p>
</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
</tr>
<tr>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td colspan="6">
<p>è</p>
</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td colspan="4">
<p><em>w</em><sub>0</sub></p>
</td>
<td>
<p>ø</p>
</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td colspan="4">
<p><em>w</em><sub>0</sub></p>
</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
</tr>
<tr>
<td colspan="14">
<p><strong><em>Existence d’une résonance :</em></strong></p>
</td>
<td colspan="5">&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
</tr>
<tr>
<td colspan="34">
<p>Il existe une résonance si il existe une pulsation <em>w<sub>r</sub></em></p>
</td>
<td>&nbsp;</td>
<td colspan="26">
<p>dite pulsation de résonance pour laquelle</p>
</td>
<td>&nbsp;</td>
</tr>
<tr>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td colspan="4">&nbsp;</td>
<td colspan="2">&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td colspan="8">&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
</tr>
<tr>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td colspan="3">
<p>æ</p>
</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td colspan="6">
<p><em>w</em> <sup>2</sup> ö</p>
</td>
<td>
<p>2</p>
</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td colspan="9">
<p>4<em>x</em> <sup>2</sup> <em>w</em><sup>2</sup></p>
</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
</tr>
<tr>
<td colspan="3" rowspan="2">
<p>le gain</p>
</td>
<td>&nbsp;</td>
<td colspan="6" rowspan="2">
<p><em>H </em>(<em> jw</em>)</p>
</td>
<td colspan="11" rowspan="2">
<p><em><sub>dB</sub></em> = 20<em>LogK</em> &#8211; 20<em>Log</em></p>
</td>
<td colspan="4" rowspan="2">
<p>ç<sup>1</sup></p>
</td>
<td rowspan="2">
<p>&#8211;</p>
</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td colspan="2">&nbsp;</td>
<td>&nbsp;</td>
<td rowspan="2">
<p>÷</p>
</td>
<td colspan="3" rowspan="2">
<p>+</p>
</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td colspan="14" rowspan="2">
<p>présente un maximum.</p>
</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
</tr>
<tr>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td colspan="2">
<p>2</p>
</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>
<p>2</p>
</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
</tr>
<tr>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td colspan="3" rowspan="2">
<p>è</p>
</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td rowspan="2">&nbsp;</td>
<td colspan="5" rowspan="2">
<p><em>w</em><sub>0</sub></p>
</td>
<td rowspan="2">
<p>ø</p>
</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td colspan="2" rowspan="3">
<p>2</p>
</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td colspan="4" rowspan="2">
<p><em>w</em><sub>0</sub></p>
</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
</tr>
<tr>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
</tr>
<tr>
<td colspan="25" rowspan="3">
<p>Le gain est maximum si la fonction de <em>w</em> :</p>
</td>
<td>
<p>æ</p>
</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td colspan="4">
<p><em>w</em><sup>2</sup> ö</p>
</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td colspan="9">
<p>4<em>x</em> <sup>2</sup> <em>w</em><sup>2</sup></p>
</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
</tr>
<tr>
<td rowspan="2">
<p>ç</p>
</td>
<td colspan="5" rowspan="2">
<p>1-</p>
</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td rowspan="2">
<p>÷</p>
</td>
<td colspan="7" rowspan="2">
<p>+</p>
</td>
<td colspan="3">&nbsp;</td>
<td>&nbsp;</td>
<td colspan="12" rowspan="2">
<p>est minimale :</p>
</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
</tr>
<tr>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>
<p>2</p>
</td>
<td>&nbsp;</td>
<td colspan="3">
<p>2</p>
</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
</tr>
<tr>
<td colspan="32">
<p>è</p>
</td>
<td colspan="2">
<p><em>w</em><sub>0</sub></p>
</td>
<td colspan="9">
<p>ø</p>
</td>
<td colspan="18">
<p><em>w</em><sub>0</sub></p>
</td>
<td>&nbsp;</td>
</tr>
<tr>
<td colspan="19" rowspan="3">
<p>Etude de cette fonction : Posons <em>x</em> =</p>
</td>
<td colspan="6">
<p><em>w</em><sup>2</sup></p>
</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td colspan="6">
<p>æ</p>
</td>
<td colspan="6">
<p><em>w</em> <sup>2</sup> ö</p>
</td>
<td>
<p>2</p>
</td>
<td colspan="2">
<p>4<em>x</em> <sup>2</sup> <em>w</em><sup>2</sup></p>
</td>
<td>&nbsp;</td>
</tr>
<tr>
<td>&nbsp;</td>
<td colspan="2">&nbsp;</td>
<td colspan="24" rowspan="2">
<p>, alors la fonction :</p>
</td>
<td colspan="6" rowspan="2">
<p><em>f </em>(<em>w </em>) = <sub>ç</sub> 1-</p>
</td>
<td colspan="5">&nbsp;</td>
<td rowspan="2">
<p>÷</p>
</td>
<td rowspan="2">
<p>+</p>
</td>
<td>&nbsp;</td>
<td rowspan="2">
<p>est</p>
</td>
<td>&nbsp;</td>
</tr>
<tr>
<td>&nbsp;</td>
<td colspan="2">
<p>2</p>
</td>
<td colspan="5">
<p>2</p>
</td>
<td>
<p>2</p>
</td>
<td>&nbsp;</td>
</tr>
<tr>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td colspan="2">
<p><em>w</em><sub>0</sub></p>
</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td colspan="6">
<p>è</p>
</td>
<td colspan="5">
<p><em>w </em><sub>0</sub></p>
</td>
<td>
<p>ø</p>
</td>
<td>&nbsp;</td>
<td>
<p><em>w</em><sub>0</sub></p>
</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
</tr>
<tr>
<td colspan="43">
<p>minimale si la fonction : <em>g</em>( <em>x</em>) = (1- <em>x</em>)<sup>2</sup> + 4<em>x</em> <sup>2</sup> <em>x</em> = 1+ (4<em>x</em><sup>2</sup></p>
</td>
<td colspan="17">
<p>&#8211; 2)<em>x</em> + <em>x</em><sup>2</sup> est minimale .</p>
</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
</tr>
</tbody>
</table>
<p><img decoding="async" class="wp-image-18257" src="https://telechargercours.com/wp-content/uploads/2021/07/word-image-147.png"> <img decoding="async" class="wp-image-18258" src="https://telechargercours.com/wp-content/uploads/2021/07/word-image-148.png"></p>
<p><em>g</em>¢(<em> x</em>) = 4<em>x </em><sup>2</sup> &#8211; 2 + 2<em>x </em>, donc on a un extremum pour<em> x </em>telque<em> g </em>¢(<em>x</em>) = 0 , c’est à dire pour</p>
<ol>
<li>= 1- 2<em>x</em> <sup>2</sup> .</li>
</ol>
<table>
<tbody>
<tr>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
</tr>
<tr>
<td>
<p>Donc la pulsation pour laquelle on a résonance est :</p>
</td>
<td>
<p><em>w</em> = <em>w</em></p>
</td>
<td rowspan="2">
<p>0</p>
</td>
<td colspan="2">
<p>1 &#8211; 2<em>x</em> <sup>2</sup></p>
</td>
<td>&nbsp;</td>
</tr>
<tr>
<td>&nbsp;</td>
<td>
<p><em>r</em></p>
</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
</tr>
</tbody>
</table>
<p><img decoding="async" class="wp-image-18259" src="https://telechargercours.com/wp-content/uploads/2021/07/word-image-149.png"></p>
<p><strong><em>Remarque : </em></strong>On voit donc bien que la résonance ne peut exister que dans certains cas d’amortissement (terme sous la racine positif).</p>
<p><img decoding="async" class="wp-image-18260" src="https://telechargercours.com/wp-content/uploads/2021/07/word-image-150.png"></p>
<p><strong><em>On a résonance pour des valeurs d’amortissement </em></strong>: <em>x</em> &lt; <sup>2</sup> 2</p>
<p><strong><em>Quantification de la résonance : facteur de surtension Q</em></strong></p>
<p>Le facteur de surtension exprimé en dB est la différence entre la valeur du pic de résonance et le gain statique : <em>Q<sub>dB</sub></em> = <img decoding="async" class="wp-image-18261" src="https://telechargercours.com/wp-content/uploads/2021/07/word-image-151.png"> <img decoding="async" class="wp-image-18262" src="https://telechargercours.com/wp-content/uploads/2021/07/word-image-152.png"><em> H </em>(<em> jw<sub>r</sub> </em>) <img decoding="async" class="wp-image-18263" src="https://telechargercours.com/wp-content/uploads/2021/07/word-image-153.png"> <img decoding="async" class="wp-image-18264" src="https://telechargercours.com/wp-content/uploads/2021/07/word-image-154.png"><em><sub>dB </sub></em>&#8211; 20<em>LogK</em></p>
<p><img decoding="async" class="wp-image-18265" src="https://telechargercours.com/wp-content/uploads/2021/07/word-image-155.png"></p>
<table>
<tbody>
<tr>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>
<p>æ</p>
</td>
<td>&nbsp;</td>
<td>
<p><em>w</em><sup>2</sup></p>
</td>
<td>
<p>ö</p>
</td>
<td>
<p>2</p>
</td>
<td>
<p>4<em>x</em> <sup>2</sup><em>w</em> <sup>2</sup></p>
</td>
<td>&nbsp;</td>
</tr>
<tr>
<td rowspan="2">
<p><em>Q</em></p>
</td>
<td rowspan="2">
<p>=</p>
</td>
<td rowspan="2">
<p><em>H </em>(<em> jw</em></p>
</td>
<td rowspan="2">
<p>)</p>
</td>
<td>&nbsp;</td>
<td rowspan="2">
<p>&#8211; 20<em>LogK</em> = -20<em>Log</em></p>
</td>
<td rowspan="2">
<p>1</p>
</td>
<td rowspan="2">
<p>&#8211;</p>
</td>
<td>
<p><em>r</em></p>
</td>
<td>&nbsp;</td>
<td rowspan="2">
<p>+</p>
</td>
<td>
<p><em>r</em></p>
</td>
<td>&nbsp;</td>
</tr>
<tr>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
</tr>
<tr>
<td>
<p><em>dB</em></p>
</td>
<td>&nbsp;</td>
<td>
<p><em>r</em></p>
</td>
<td>&nbsp;</td>
<td rowspan="2">
<p><em>dB</em></p>
</td>
<td>&nbsp;</td>
<td rowspan="2">
<p>ç</p>
</td>
<td>&nbsp;</td>
<td rowspan="2">
<p>2</p>
</td>
<td rowspan="2">
<p>÷</p>
</td>
<td>&nbsp;</td>
<td rowspan="2">
<p>2</p>
</td>
<td>&nbsp;</td>
</tr>
<tr>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
</tr>
<tr>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>
<p>è</p>
</td>
<td>&nbsp;</td>
<td>
<p><em>w</em><sub>0</sub></p>
</td>
<td>
<p>ø</p>
</td>
<td>&nbsp;</td>
<td>
<p><em>w</em><sub>0</sub></p>
</td>
<td>&nbsp;</td>
</tr>
</tbody>
</table>
<p><img decoding="async" class="wp-image-18266" src="https://telechargercours.com/wp-content/uploads/2021/07/word-image-156.png"></p>
<table>
<tbody>
<tr>
<td>
<p><em>Q<sub>dB</sub> </em>= -20<em>Log </em>(1 &#8211; (1- 2<em>x </em><sup>2</sup> ))<sup>2</sup></p>
</td>
<td>
<p>+ 4<em>x</em> <sup>2</sup> (1- 2<em>x</em> <sup>2</sup> )</p>
</td>
</tr>
</tbody>
</table>
<table>
<tbody>
<tr>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td rowspan="2">
<p>(</p>
</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td rowspan="2">
<p>)</p>
</td>
<td>&nbsp;</td>
</tr>
<tr>
<td>
<p><em>Q </em>= -20<em>Log</em></p>
</td>
<td colspan="4">
<p>4<em>x</em> <sup>4</sup> + 4<em>x</em> <sup>2</sup> -8<em>x</em> <sup>4</sup> = -20<em>Log</em> 4<em>x</em> <sup>2</sup> &#8211; 4<em>x</em> <sup>4</sup> = &#8211; 20<em>Log</em></p>
</td>
<td>
<p>2<em>x</em></p>
</td>
<td colspan="2">
<p>1- <em>x</em> <sup>2</sup></p>
</td>
<td>&nbsp;</td>
</tr>
<tr>
<td>
<p><em>dB</em></p>
</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
</tr>
<tr>
<td>
<p>Page 2</p>
</td>
<td>&nbsp;</td>
<td colspan="3">
<p><strong>Emmanuel FARGES</strong></p>
</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td colspan="2">
<p>Ó EduKlub S.A.</p>
</td>
<td>&nbsp;</td>
</tr>
</tbody>
</table>
<p><img decoding="async" class="wp-image-18267" src="https://telechargercours.com/wp-content/uploads/2021/07/word-image-157.png"></p>
<p>Tous droits de l’auteur des œuvres réservés. Sauf autorisation, la reproduction ainsi que toute utilisation des œuvres autre que la consultation individuelle et privée sont interdites.</p>
<p></p>
<p><strong>Sciences Indusrielles</strong></p>
<p><strong>Systèmes linéaires continus invariants</strong></p>
<p>ANALYSE FREQUENTIELLE</p>
<table>
<tbody>
<tr>
<td colspan="5" rowspan="3">
<p>D’où le résultat final :</p>
</td>
<td rowspan="3">
<p><em><sup>Q</sup>dB</em></p>
</td>
<td rowspan="3">
<p>= 20<em>Log</em></p>
</td>
<td>&nbsp;</td>
<td colspan="2">
<p>1</p>
</td>
<td>&nbsp;</td>
<td rowspan="3">
<p>Or Q est tel que</p>
</td>
<td rowspan="3">
<p><em><sup>Q</sup>dB</em></p>
</td>
<td rowspan="3">
<p>= 20<em>L o g Q</em></p>
</td>
<td rowspan="3">
<p>, donc :</p>
</td>
<td>&nbsp;</td>
</tr>
<tr>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
</tr>
<tr>
<td rowspan="2">
<p>(2<em>x</em></p>
</td>
<td colspan="3" rowspan="2">
<p>1 &#8211;<em>x</em> <sup>2</sup> )</p>
</td>
<td>&nbsp;</td>
</tr>
<tr>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
</tr>
<tr>
<td rowspan="2">
<p><em>Q </em>=</p>
</td>
<td colspan="2">
<p>1</p>
</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
</tr>
<tr>
<td>
<p>(2 <em>x</em></p>
</td>
<td>
<p>1- <em>x</em> <sup>2</sup></p>
</td>
<td>
<p>)</p>
</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
</tr>
</tbody>
</table>
<p><img decoding="async" class="wp-image-18268" src="https://telechargercours.com/wp-content/uploads/2021/07/word-image-158.png"> <img decoding="async" class="wp-image-18269" src="https://telechargercours.com/wp-content/uploads/2021/07/word-image-159.png"></p>
<p><strong><em>Calcul de la phase :</em></strong></p>
<ul>
<li>(<em>w</em> ) = <em>Arg</em> [ <em>H</em> ( <em>jw</em>)]</li>
</ul>
<table>
<tbody>
<tr>
<td colspan="42">
<p><em>K</em></p>
</td>
<td colspan="15">
<p><em>K</em></p>
</td>
<td colspan="5">
<p>é æ</p>
</td>
<td colspan="11">
<p><em>w</em> <sup>2</sup> ö</p>
</td>
<td colspan="27">
<p>2<em>xw</em> <sup>ù</sup></p>
</td>
<td>&nbsp;</td>
</tr>
<tr>
<td colspan="2" rowspan="2">
<p><em>H </em>(<em> jw</em>) =</p>
</td>
<td colspan="27">&nbsp;</td>
<td>&nbsp;</td>
<td rowspan="2">
<p>=</p>
</td>
<td rowspan="2">&nbsp;</td>
<td colspan="25">&nbsp;</td>
<td colspan="5" rowspan="2">
<p>ê ç<sup>1</sup> <sup>&#8211;</sup></p>
</td>
<td colspan="4">&nbsp;</td>
<td>&nbsp;</td>
<td colspan="5" rowspan="2">
<p><sub>÷</sub> &#8211; <em>j</em></p>
</td>
<td colspan="3">&nbsp;</td>
<td colspan="10" rowspan="2">
<p><sub>ú</sub> , donc :</p>
</td>
<td colspan="15">&nbsp;</td>
<td>&nbsp;</td>
</tr>
<tr>
<td colspan="2">
<p>æ</p>
</td>
<td colspan="15">
<p><em>w</em> <sup>2</sup> ö</p>
</td>
<td colspan="11">
<p>2<em>xw</em></p>
</td>
<td colspan="9">
<p>2</p>
</td>
<td colspan="2" rowspan="2">
<p>ö</p>
</td>
<td colspan="9">
<p>2</p>
</td>
<td colspan="4">
<p>2</p>
</td>
<td>
<p>2</p>
</td>
<td colspan="2">
<p><em>w</em></p>
</td>
<td colspan="3">
<p>2</p>
</td>
<td colspan="3">
<p><em>w</em></p>
</td>
<td colspan="15">&nbsp;</td>
<td>&nbsp;</td>
</tr>
<tr>
<td>&nbsp;</td>
<td colspan="2" rowspan="3">
<p>ç<sup>1</sup></p>
</td>
<td rowspan="3">
<p>&#8211;</p>
</td>
<td colspan="7">&nbsp;</td>
<td colspan="4" rowspan="3">
<p>÷ <sup>+</sup></p>
</td>
<td colspan="3" rowspan="3">
<p><em>j</em></p>
</td>
<td colspan="10">&nbsp;</td>
<td colspan="2">&nbsp;</td>
<td colspan="7" rowspan="3">
<p><sup>æ</sup>1-</p>
</td>
<td colspan="2" rowspan="2">
<p><em>w</em></p>
</td>
<td colspan="2">&nbsp;</td>
<td>&nbsp;</td>
<td colspan="2" rowspan="3">
<p>+</p>
</td>
<td rowspan="3">&nbsp;</td>
<td rowspan="2">&nbsp;</td>
<td colspan="5">
<p>4<em>x</em></p>
</td>
<td colspan="3">
<p><em>w</em></p>
</td>
<td rowspan="2">&nbsp;</td>
<td colspan="3" rowspan="2">
<p>ë è</p>
</td>
<td colspan="7">
<p>0</p>
</td>
<td colspan="6">
<p>ø</p>
</td>
<td colspan="2">
<p>0</p>
</td>
<td rowspan="2">
<p>û</p>
</td>
<td rowspan="2">&nbsp;</td>
<td rowspan="2">&nbsp;</td>
<td colspan="22">&nbsp;</td>
<td>&nbsp;</td>
</tr>
<tr>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td colspan="4" rowspan="2">
<p>2</p>
</td>
<td colspan="2">&nbsp;</td>
<td colspan="12">&nbsp;</td>
<td colspan="3">&nbsp;</td>
<td colspan="2">&nbsp;</td>
<td colspan="8">&nbsp;</td>
<td colspan="15">&nbsp;</td>
<td colspan="22">&nbsp;</td>
<td>&nbsp;</td>
</tr>
<tr>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td colspan="2">&nbsp;</td>
<td colspan="2">&nbsp;</td>
<td colspan="10" rowspan="2">
<p><em>w</em><sub>0</sub></p>
</td>
<td colspan="2" rowspan="2">
<p><em>w</em></p>
</td>
<td colspan="2">&nbsp;</td>
<td colspan="2" rowspan="2">
<p>÷</p>
</td>
<td>&nbsp;</td>
<td colspan="2">&nbsp;</td>
<td colspan="4" rowspan="2">
<p><em>w</em></p>
</td>
<td colspan="47">&nbsp;</td>
<td>&nbsp;</td>
</tr>
<tr>
<td colspan="4">
<p>è</p>
</td>
<td colspan="5">
<p><em>w</em><sub>0</sub></p>
</td>
<td colspan="11">
<p>ø</p>
</td>
<td colspan="7">
<p>ç</p>
</td>
<td colspan="2">
<p>2</p>
</td>
<td colspan="6">&nbsp;</td>
<td colspan="47">
<p>2</p>
</td>
<td>&nbsp;</td>
</tr>
<tr>
<td colspan="36">
<p>è</p>
</td>
<td colspan="5">
<p>0</p>
</td>
<td colspan="12">
<p>ø</p>
</td>
<td colspan="47">
<p>0</p>
</td>
<td>&nbsp;</td>
</tr>
<tr>
<td colspan="36">
<p>é æ</p>
</td>
<td colspan="11">
<p><em>w</em><sup>2</sup> ö</p>
</td>
<td colspan="53">
<p>2<em>xw</em> ù</p>
</td>
<td>&nbsp;</td>
</tr>
<tr>
<td colspan="34" rowspan="2">
<p><em>j</em> (<em>w</em> ) = <em>Arg</em> [<em>H</em>( <em>jw</em>) ] = <em>Arg</em> ê <sub>ç</sub> 1</p>
</td>
<td rowspan="2">
<p>&#8211;</p>
</td>
<td rowspan="2">&nbsp;</td>
<td colspan="4">&nbsp;</td>
<td>&nbsp;</td>
<td colspan="5" rowspan="2">
<p><sub>÷</sub> &#8211; <em>j</em></p>
</td>
<td colspan="8">&nbsp;</td>
<td>&nbsp;</td>
<td colspan="2" rowspan="2">
<p>ú</p>
</td>
<td colspan="43">&nbsp;</td>
<td>&nbsp;</td>
</tr>
<tr>
<td colspan="4">
<p>2</p>
</td>
<td>&nbsp;</td>
<td colspan="2">&nbsp;</td>
<td colspan="5" rowspan="2">
<p><em>w</em><sub>0</sub></p>
</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
</tr>
<tr>
<td colspan="36">
<p>ë è</p>
</td>
<td colspan="5">
<p><em>w</em><sub>0</sub></p>
</td>
<td colspan="7">
<p>ø</p>
</td>
<td colspan="47">
<p>û</p>
</td>
<td>&nbsp;</td>
</tr>
<tr>
<td>&nbsp;</td>
<td colspan="3">
<p>ææ</p>
</td>
<td>&nbsp;</td>
<td colspan="2" rowspan="3">
<p>&#8211;</p>
</td>
<td>&nbsp;</td>
<td colspan="5">
<p><em>w </em><sup>2</sup></p>
</td>
<td colspan="2">
<p>ö</p>
</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td colspan="7" rowspan="3">
<p>&#8211; <em>j</em></p>
</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td colspan="7">
<p>2<em>xw</em> ö</p>
</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td colspan="5">
<p>ææ</p>
</td>
<td colspan="4" rowspan="3">
<p>1 &#8211;</p>
</td>
<td colspan="3">
<p><em>w</em><sup>2</sup></p>
</td>
<td colspan="2">
<p>ö</p>
</td>
<td colspan="2" rowspan="3">
<p>&#8211; <em>j</em></p>
</td>
<td colspan="9">
<p>2<em>xw</em> ö</p>
</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td colspan="6">
<p>æ æ</p>
</td>
<td colspan="3" rowspan="3">
<p>1-</p>
</td>
<td colspan="4">
<p><em>w</em><sup>2</sup></p>
</td>
<td>
<p>ö</p>
</td>
<td colspan="2" rowspan="3">
<p>&#8211; <em>j</em></p>
</td>
<td colspan="5">
<p>2<em>xw</em> ö</p>
</td>
<td>&nbsp;</td>
</tr>
<tr>
<td>&nbsp;</td>
<td colspan="4" rowspan="2">
<p>Im ç 1</p>
</td>
<td>&nbsp;</td>
<td colspan="6">&nbsp;</td>
<td colspan="3">&nbsp;</td>
<td colspan="2">&nbsp;</td>
<td colspan="5">&nbsp;</td>
<td colspan="2" rowspan="2">
<p>÷</p>
</td>
<td colspan="11">&nbsp;</td>
<td colspan="7" rowspan="2">
<p>Re ç</p>
</td>
<td colspan="2">&nbsp;</td>
<td colspan="3">&nbsp;</td>
<td colspan="2">&nbsp;</td>
<td>&nbsp;</td>
<td colspan="6">&nbsp;</td>
<td colspan="2" rowspan="2">
<p>÷</p>
</td>
<td colspan="6">&nbsp;</td>
<td colspan="5" rowspan="2">
<p>Im ç</p>
</td>
<td>&nbsp;</td>
<td colspan="4">&nbsp;</td>
<td>&nbsp;</td>
<td colspan="4">&nbsp;</td>
<td rowspan="2">
<p>÷</p>
</td>
<td>&nbsp;</td>
</tr>
<tr>
<td>&nbsp;</td>
<td colspan="2">&nbsp;</td>
<td colspan="4" rowspan="2">
<p>2</p>
</td>
<td colspan="2" rowspan="2">
<p>÷</p>
</td>
<td colspan="2">&nbsp;</td>
<td colspan="7">&nbsp;</td>
<td colspan="11">&nbsp;</td>
<td colspan="2" rowspan="2">
<p>ç</p>
</td>
<td>&nbsp;</td>
<td colspan="2" rowspan="2">
<p>2</p>
</td>
<td colspan="2" rowspan="2">
<p>÷</p>
</td>
<td colspan="7">&nbsp;</td>
<td colspan="6">&nbsp;</td>
<td rowspan="2">
<p>ç</p>
</td>
<td colspan="2">&nbsp;</td>
<td rowspan="2">
<p>2</p>
</td>
<td>&nbsp;</td>
<td rowspan="2">
<p>÷</p>
</td>
<td colspan="4">&nbsp;</td>
<td>&nbsp;</td>
</tr>
<tr>
<td colspan="9">
<p>ç<sup>ç</sup></p>
</td>
<td colspan="33">
<p>÷</p>
</td>
<td colspan="3">
<p>ç</p>
</td>
<td colspan="5">&nbsp;</td>
<td colspan="17">
<p>÷</p>
</td>
<td colspan="5">
<p>ç</p>
</td>
<td colspan="5">&nbsp;</td>
<td>&nbsp;</td>
<td colspan="7">
<p>÷</p>
</td>
<td>&nbsp;</td>
</tr>
<tr>
<td rowspan="2">
<p>tan <em>j</em> =</p>
</td>
<td colspan="3">
<p>èè</p>
</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td colspan="5">
<p><em>w</em><sub>0</sub></p>
</td>
<td colspan="2">
<p>ø</p>
</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td colspan="7">
<p><em><sup>w</sup></em>0 ø</p>
</td>
<td colspan="11" rowspan="2">
<p>;cos<em>j</em> =</p>
</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td colspan="3">
<p>è</p>
</td>
<td colspan="2">
<p>è</p>
</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td colspan="3">
<p><em>w</em><sub>0</sub></p>
</td>
<td colspan="2">
<p>ø</p>
</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td colspan="7">
<p><em><sup>w</sup></em>0 ø</p>
</td>
<td colspan="5" rowspan="2">
<p>;sin <em>j</em> =</p>
</td>
<td>&nbsp;</td>
<td colspan="5">
<p>è</p>
</td>
<td>
<p>è</p>
</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td colspan="4">
<p><em>w </em><sub>0</sub></p>
</td>
<td>
<p>ø</p>
</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td colspan="5">
<p><em><sup>w</sup></em>0 ø</p>
</td>
<td>&nbsp;</td>
</tr>
<tr>
<td colspan="3" rowspan="2">
<p>æ æ</p>
</td>
<td>&nbsp;</td>
<td colspan="2" rowspan="4">
<p>&#8211;</p>
</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td colspan="4" rowspan="2">
<p><em>w</em><sup>2</sup></p>
</td>
<td colspan="2" rowspan="2">
<p>ö</p>
</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td colspan="8" rowspan="4">
<p>&#8211; <em>j</em></p>
</td>
<td>&nbsp;</td>
<td colspan="7" rowspan="2">
<p>2<em>xw</em> ö</p>
</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td colspan="3" rowspan="2">
<p>æ</p>
</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td colspan="3" rowspan="4">
<p>&#8211;</p>
</td>
<td colspan="2" rowspan="2">
<p><em>w </em><sup>2</sup></p>
</td>
<td colspan="2" rowspan="2">
<p>ö</p>
</td>
<td colspan="3" rowspan="4">
<p>&#8211; <em>j</em></p>
</td>
<td colspan="6" rowspan="2">
<p>2<em>xw</em></p>
</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td rowspan="2">
<p>æ</p>
</td>
<td colspan="4" rowspan="4">
<p>1-</p>
</td>
<td>&nbsp;</td>
<td colspan="3" rowspan="2">
<p><em>w</em><sup>2</sup></p>
</td>
<td colspan="2" rowspan="2">
<p>ö</p>
</td>
<td colspan="2" rowspan="4">
<p>&#8211; <em>j</em></p>
</td>
<td colspan="4" rowspan="2">
<p>2<em>xw</em></p>
</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
</tr>
<tr>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td colspan="3">&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td colspan="2">&nbsp;</td>
<td>&nbsp;</td>
</tr>
<tr>
<td>&nbsp;</td>
<td colspan="4" rowspan="2">
<p>Re ç 1</p>
</td>
<td>&nbsp;</td>
<td colspan="6">&nbsp;</td>
<td colspan="3">&nbsp;</td>
<td>&nbsp;</td>
<td colspan="5">&nbsp;</td>
<td colspan="2" rowspan="2">
<p>÷</p>
</td>
<td colspan="15">&nbsp;</td>
<td colspan="2">&nbsp;</td>
<td colspan="3" rowspan="2">
<p>1</p>
</td>
<td colspan="3">&nbsp;</td>
<td>&nbsp;</td>
<td colspan="5">&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td colspan="11">&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td colspan="2">&nbsp;</td>
<td colspan="3">&nbsp;</td>
<td colspan="3">&nbsp;</td>
<td>&nbsp;</td>
<td colspan="2">&nbsp;</td>
<td>&nbsp;</td>
</tr>
<tr>
<td>&nbsp;</td>
<td colspan="2">&nbsp;</td>
<td colspan="4" rowspan="2">
<p><em>w</em><sup>2</sup></p>
</td>
<td colspan="2" rowspan="2">
<p>÷</p>
</td>
<td colspan="2">&nbsp;</td>
<td>&nbsp;</td>
<td colspan="4" rowspan="2">
<p><em>w</em></p>
</td>
<td>&nbsp;</td>
<td colspan="15">&nbsp;</td>
<td colspan="2">&nbsp;</td>
<td colspan="2" rowspan="2">
<p><em>w </em><sup>2</sup></p>
</td>
<td colspan="2" rowspan="2">
<p>÷</p>
</td>
<td>&nbsp;</td>
<td colspan="3" rowspan="2">
<p><em>w</em></p>
</td>
<td colspan="2">&nbsp;</td>
<td>&nbsp;</td>
<td colspan="11">&nbsp;</td>
<td rowspan="2">
<p>ç</p>
</td>
<td>&nbsp;</td>
<td colspan="3" rowspan="2">
<p><em>w </em><sup>2</sup></p>
</td>
<td colspan="2" rowspan="2">
<p>÷</p>
</td>
<td colspan="2" rowspan="2">
<p><em>w</em></p>
</td>
<td colspan="2">&nbsp;</td>
<td colspan="2">&nbsp;</td>
<td>&nbsp;</td>
</tr>
<tr>
<td colspan="9">
<p>ç <sup>ç</sup></p>
</td>
<td colspan="11">&nbsp;</td>
<td colspan="18">
<p>÷</p>
</td>
<td colspan="8">
<p>ç</p>
</td>
<td colspan="4">&nbsp;</td>
<td rowspan="2">
<p>0</p>
</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td colspan="11">&nbsp;</td>
<td colspan="5">&nbsp;</td>
<td colspan="2">&nbsp;</td>
<td rowspan="2">
<p>0</p>
</td>
<td>&nbsp;</td>
<td colspan="2">&nbsp;</td>
<td>&nbsp;</td>
</tr>
<tr>
<td colspan="9">
<p>è è</p>
</td>
<td colspan="4">
<p>0</p>
</td>
<td colspan="17">
<p>ø</p>
</td>
<td colspan="18">
<p>0 ø</p>
</td>
<td colspan="9">
<p>è</p>
</td>
<td>
<p>0</p>
</td>
<td colspan="9">
<p>ø</p>
</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td colspan="11">&nbsp;</td>
<td colspan="7">
<p>è</p>
</td>
<td colspan="2">
<p>0</p>
</td>
<td colspan="6">
<p>ø</p>
</td>
<td>&nbsp;</td>
<td colspan="2">&nbsp;</td>
<td>&nbsp;</td>
</tr>
<tr>
<td colspan="100">&nbsp;</td>
<td>&nbsp;</td>
</tr>
<tr>
<td>&nbsp;</td>
<td colspan="43">
<p>&#8211; 2<em>xw</em></p>
</td>
<td colspan="5" rowspan="2">
<p><sup>æ</sup>1-</p>
</td>
<td colspan="7">
<p><em>w </em><sup>2</sup></p>
</td>
<td rowspan="2">
<p>ö</p>
</td>
<td colspan="5">&nbsp;</td>
<td colspan="7">&nbsp;</td>
<td>&nbsp;</td>
<td colspan="20">
<p>-2<em>xw</em></p>
</td>
<td>&nbsp;</td>
<td colspan="9">&nbsp;</td>
<td>&nbsp;</td>
</tr>
<tr>
<td>&nbsp;</td>
<td colspan="8">&nbsp;</td>
<td colspan="11">&nbsp;</td>
<td colspan="24">&nbsp;</td>
<td colspan="7">&nbsp;</td>
<td colspan="5">&nbsp;</td>
<td colspan="7">&nbsp;</td>
<td>&nbsp;</td>
<td colspan="7">&nbsp;</td>
<td colspan="4">&nbsp;</td>
<td colspan="3">&nbsp;</td>
<td colspan="6">&nbsp;</td>
<td>&nbsp;</td>
<td colspan="9">&nbsp;</td>
<td>&nbsp;</td>
</tr>
<tr>
<td>&nbsp;</td>
<td colspan="10">&nbsp;</td>
<td colspan="4" rowspan="2">
<p><em>w</em><sub>0</sub></p>
</td>
<td colspan="29">&nbsp;</td>
<td colspan="4">
<p>ç</p>
</td>
<td colspan="8">
<p>2</p>
</td>
<td colspan="6">
<p>÷</p>
</td>
<td colspan="7">&nbsp;</td>
<td>&nbsp;</td>
<td colspan="8">&nbsp;</td>
<td colspan="4" rowspan="2">
<p><em>w</em><sub>0</sub></p>
</td>
<td colspan="8">&nbsp;</td>
<td colspan="10">&nbsp;</td>
<td>&nbsp;</td>
</tr>
<tr>
<td rowspan="2">
<p>Donc :</p>
</td>
<td colspan="4" rowspan="2">
<p>tan <em>j</em> =</p>
</td>
<td>&nbsp;</td>
<td colspan="5">&nbsp;</td>
<td colspan="7">&nbsp;</td>
<td colspan="2">&nbsp;</td>
<td colspan="11" rowspan="2">
<p>;cos<em>j</em> =</p>
</td>
<td colspan="9">&nbsp;</td>
<td colspan="4">
<p>è</p>
</td>
<td colspan="8">
<p><em>w</em><sub>0</sub></p>
</td>
<td colspan="6">
<p>ø</p>
</td>
<td colspan="10" rowspan="2">
<p>;sin<em>j</em> =</p>
</td>
<td colspan="6">&nbsp;</td>
<td colspan="8">&nbsp;</td>
<td>&nbsp;</td>
<td colspan="9">&nbsp;</td>
<td>&nbsp;</td>
</tr>
<tr>
<td colspan="2" rowspan="3">
<p>æ</p>
</td>
<td colspan="7">&nbsp;</td>
<td colspan="3" rowspan="3">
<p><em>w</em></p>
</td>
<td colspan="2" rowspan="2">
<p>2</p>
</td>
<td colspan="3" rowspan="3">
<p>ö</p>
</td>
<td colspan="2">&nbsp;</td>
<td colspan="9">&nbsp;</td>
<td colspan="4">&nbsp;</td>
<td colspan="8">&nbsp;</td>
<td colspan="6">&nbsp;</td>
<td colspan="9">&nbsp;</td>
<td colspan="9">&nbsp;</td>
<td>&nbsp;</td>
<td colspan="9">&nbsp;</td>
<td>&nbsp;</td>
</tr>
<tr>
<td>&nbsp;</td>
<td colspan="4">&nbsp;</td>
<td colspan="7">&nbsp;</td>
<td colspan="22">&nbsp;</td>
<td>&nbsp;</td>
<td colspan="3" rowspan="2">
<p>2</p>
</td>
<td colspan="8">
<p>2</p>
</td>
<td colspan="2">&nbsp;</td>
<td colspan="2" rowspan="2">
<p>2</p>
</td>
<td>&nbsp;</td>
<td rowspan="2">
<p>2</p>
</td>
<td colspan="7">&nbsp;</td>
<td>&nbsp;</td>
<td colspan="3">&nbsp;</td>
<td colspan="3" rowspan="4">
<p><sup>æ</sup>1 &#8211;</p>
</td>
<td colspan="2">&nbsp;</td>
<td rowspan="2">
<p>2</p>
</td>
<td>&nbsp;</td>
<td colspan="3" rowspan="4">
<p>ö <sub>+</sub></p>
</td>
<td colspan="3">&nbsp;</td>
<td colspan="4" rowspan="2">
<p>2 2</p>
</td>
<td colspan="10">&nbsp;</td>
<td>&nbsp;</td>
</tr>
<tr>
<td>&nbsp;</td>
<td colspan="4">&nbsp;</td>
<td colspan="7">&nbsp;</td>
<td colspan="2">&nbsp;</td>
<td colspan="22">&nbsp;</td>
<td>&nbsp;</td>
<td colspan="8">&nbsp;</td>
<td colspan="2">&nbsp;</td>
<td>&nbsp;</td>
<td colspan="7">&nbsp;</td>
<td>&nbsp;</td>
<td colspan="3">&nbsp;</td>
<td colspan="2">&nbsp;</td>
<td>&nbsp;</td>
<td colspan="3">&nbsp;</td>
<td colspan="10">&nbsp;</td>
<td>&nbsp;</td>
</tr>
<tr>
<td>&nbsp;</td>
<td colspan="4">&nbsp;</td>
<td colspan="2" rowspan="2">
<p>ç</p>
</td>
<td colspan="6" rowspan="2">
<p>1-</p>
</td>
<td>&nbsp;</td>
<td colspan="6">&nbsp;</td>
<td colspan="2" rowspan="2">
<p>÷</p>
</td>
<td colspan="17">
<p>æ</p>
</td>
<td colspan="5" rowspan="3">
<p>1-</p>
</td>
<td colspan="4">
<p><em>w</em></p>
</td>
<td colspan="4">
<p>ö</p>
</td>
<td colspan="4" rowspan="3">
<p>+</p>
</td>
<td colspan="6">
<p>4<em>x w</em></p>
</td>
<td colspan="7">&nbsp;</td>
<td>&nbsp;</td>
<td colspan="3">&nbsp;</td>
<td colspan="4">
<p><em>w</em></p>
</td>
<td colspan="7">
<p>4<em>x w</em></p>
</td>
<td>&nbsp;</td>
<td colspan="9">&nbsp;</td>
<td>&nbsp;</td>
</tr>
<tr>
<td>&nbsp;</td>
<td colspan="4">&nbsp;</td>
<td colspan="7">
<p>2</p>
</td>
<td colspan="17">&nbsp;</td>
<td colspan="4">&nbsp;</td>
<td colspan="4">&nbsp;</td>
<td colspan="6">&nbsp;</td>
<td colspan="7">&nbsp;</td>
<td>&nbsp;</td>
<td colspan="3">&nbsp;</td>
<td colspan="2">&nbsp;</td>
<td rowspan="2">
<p>2</p>
</td>
<td>&nbsp;</td>
<td colspan="7">&nbsp;</td>
<td>&nbsp;</td>
<td colspan="9">&nbsp;</td>
<td>&nbsp;</td>
</tr>
<tr>
<td>&nbsp;</td>
<td colspan="4">&nbsp;</td>
<td colspan="2" rowspan="2">
<p>è</p>
</td>
<td colspan="7">&nbsp;</td>
<td colspan="10" rowspan="2">
<p><em><sup>w</sup></em>0 ø</p>
</td>
<td colspan="12">&nbsp;</td>
<td colspan="3" rowspan="2">
<p>ç</p>
</td>
<td colspan="4" rowspan="2">
<p><em>w </em><sup>2</sup></p>
</td>
<td colspan="2" rowspan="2">
<p>÷</p>
</td>
<td colspan="2">&nbsp;</td>
<td>&nbsp;</td>
<td colspan="3" rowspan="2">
<p><em>w</em><sup>2</sup></p>
</td>
<td colspan="2">&nbsp;</td>
<td colspan="7">&nbsp;</td>
<td>&nbsp;</td>
<td colspan="8">
<p>ç</p>
</td>
<td colspan="2">
<p>÷</p>
</td>
<td colspan="9">
<p>2</p>
</td>
<td colspan="10">&nbsp;</td>
<td>&nbsp;</td>
</tr>
<tr>
<td>&nbsp;</td>
<td colspan="4">&nbsp;</td>
<td colspan="7">&nbsp;</td>
<td colspan="12">&nbsp;</td>
<td colspan="5">&nbsp;</td>
<td colspan="6">&nbsp;</td>
<td>&nbsp;</td>
<td colspan="2">&nbsp;</td>
<td colspan="7">&nbsp;</td>
<td>&nbsp;</td>
<td colspan="6">
<p>è</p>
</td>
<td colspan="3">
<p><em>w</em><sub>0</sub></p>
</td>
<td colspan="2">
<p>ø</p>
</td>
<td colspan="9">
<p><em>w</em><sub>0</sub></p>
</td>
<td colspan="10">&nbsp;</td>
<td>&nbsp;</td>
</tr>
<tr>
<td>&nbsp;</td>
<td colspan="43">
<p>è</p>
</td>
<td colspan="4">
<p>0</p>
</td>
<td colspan="8">
<p>ø</p>
</td>
<td colspan="6">
<p>0</p>
</td>
<td colspan="7">&nbsp;</td>
<td>&nbsp;</td>
<td colspan="11">&nbsp;</td>
<td colspan="9">&nbsp;</td>
<td colspan="10">&nbsp;</td>
<td>&nbsp;</td>
</tr>
<tr>
<td>&nbsp;</td>
<td colspan="11">
<p>é</p>
</td>
<td colspan="13">
<p>2<em>xw</em></p>
</td>
<td colspan="75">
<p>ù</p>
</td>
<td>&nbsp;</td>
</tr>
<tr>
<td>&nbsp;</td>
<td colspan="4">&nbsp;</td>
<td colspan="4" rowspan="2">
<p>ê</p>
</td>
<td colspan="3">&nbsp;</td>
<td colspan="9">&nbsp;</td>
<td colspan="4">&nbsp;</td>
<td colspan="5" rowspan="2">
<p>ú</p>
</td>
<td colspan="70">&nbsp;</td>
<td>&nbsp;</td>
</tr>
<tr>
<td>&nbsp;</td>
<td colspan="4">&nbsp;</td>
<td colspan="5">&nbsp;</td>
<td colspan="5" rowspan="2">
<p><em>w</em><sub>0</sub></p>
</td>
<td colspan="6">&nbsp;</td>
<td colspan="70">&nbsp;</td>
<td>&nbsp;</td>
</tr>
<tr>
<td colspan="9" rowspan="2">
<p><em>j</em> (<em>w</em>) = &#8211; arctan <sup>ê</sup></p>
</td>
<td>&nbsp;</td>
<td colspan="4">&nbsp;</td>
<td colspan="8">&nbsp;</td>
<td colspan="3" rowspan="2">
<p>ú</p>
</td>
<td colspan="70">&nbsp;</td>
<td>&nbsp;</td>
</tr>
<tr>
<td colspan="7">&nbsp;</td>
<td colspan="8" rowspan="2">
<p><em>w</em><sup>2</sup></p>
</td>
<td colspan="3">&nbsp;</td>
<td colspan="70">&nbsp;</td>
<td>&nbsp;</td>
</tr>
<tr>
<td>&nbsp;</td>
<td colspan="12">
<p>ê <sub>æ</sub></p>
</td>
<td colspan="2" rowspan="3">
<p>&#8211;</p>
</td>
<td>&nbsp;</td>
<td colspan="76">
<p>ö<sup>ú</sup></p>
</td>
<td>&nbsp;</td>
</tr>
<tr>
<td>&nbsp;</td>
<td colspan="4">&nbsp;</td>
<td colspan="4" rowspan="2">
<p>ê</p>
</td>
<td colspan="4" rowspan="2">
<p>ç <sup>1</sup></p>
</td>
<td>&nbsp;</td>
<td colspan="7">&nbsp;</td>
<td colspan="2">&nbsp;</td>
<td colspan="5" rowspan="2">
<p>÷ú</p>
</td>
<td colspan="70">&nbsp;</td>
<td>&nbsp;</td>
</tr>
<tr>
<td>&nbsp;</td>
<td colspan="4">&nbsp;</td>
<td colspan="10">
<p>2</p>
</td>
<td colspan="70">&nbsp;</td>
<td>&nbsp;</td>
</tr>
<tr>
<td>&nbsp;</td>
<td colspan="8">
<p>ê</p>
</td>
<td colspan="3" rowspan="2">
<p>è</p>
</td>
<td colspan="13">
<p><em>w</em><sub>0</sub></p>
</td>
<td colspan="75">
<p>ú</p>
</td>
<td>&nbsp;</td>
</tr>
<tr>
<td>&nbsp;</td>
<td colspan="8">
<p>ë</p>
</td>
<td colspan="88">
<p>øû</p>
</td>
<td>&nbsp;</td>
</tr>
</tbody>
</table>
<p><img decoding="async" class="wp-image-18270" src="https://telechargercours.com/wp-content/uploads/2021/07/word-image-160.png"></p>
<h3><strong>3.4</strong> <strong>Asymptotes des diagrammes de Bode</strong></h3>
<p>Deux méthodes sont possibles pour déterminer les asymptotes des diagrammes de Bode : Soit on détermine les équivalents en 0 et en +¥ du gain réel et de la phase réelle, soit on cherche des équivalents directement sur la fonction de transfert.</p>
<p><strong><em>1<sup>ère</sup> méthode : équivalents à partir des fonctions de gain et de phase réels.</em></strong></p>
<p><strong><em>Asymptote sur le diagramme de gain :</em></strong></p>
<table>
<tbody>
<tr>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
</tr>
<tr>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>
<p>æ</p>
</td>
<td>&nbsp;</td>
<td>
<p><em>w </em><sup>2</sup></p>
</td>
<td>
<p>ö</p>
</td>
<td>
<p>2</p>
</td>
<td colspan="2">
<p>4<em>x</em> <sup>2</sup> <em>w</em><sup>2</sup></p>
</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
</tr>
<tr>
<td>&nbsp;</td>
<td rowspan="2">
<p><em>H </em>(<em> jw</em>)</p>
</td>
<td rowspan="2">
<p><em><sub>dB</sub></em> = 20<em>LogK</em> &#8211; 20<em>Log</em></p>
</td>
<td rowspan="2">
<p>ç<sup>1</sup></p>
</td>
<td rowspan="2">
<p>&#8211;</p>
</td>
<td>&nbsp;</td>
<td rowspan="2">
<p>÷</p>
</td>
<td rowspan="2">
<p>+</p>
</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td colspan="2" rowspan="2">
<p>est la fonction du gain (en dB) réelle</p>
</td>
<td>&nbsp;</td>
</tr>
<tr>
<td>&nbsp;</td>
<td>
<p>2</p>
</td>
<td>
<p>2</p>
</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
</tr>
<tr>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td rowspan="2">
<p>è</p>
</td>
<td>&nbsp;</td>
<td rowspan="2">
<p><em>w</em><sub>0</sub></p>
</td>
<td rowspan="2">
<p>ø</p>
</td>
<td>&nbsp;</td>
<td colspan="2" rowspan="2">
<p><em>w</em><sub>0</sub></p>
</td>
<td>&nbsp;</td>
<td rowspan="3">
<p>Ó EduKlub S.A.</p>
</td>
<td>&nbsp;</td>
</tr>
<tr>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
</tr>
<tr>
<td colspan="3">
<p>Page 3</p>
</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td colspan="6">
<p><strong>Emmanuel FARGES</strong></p>
</td>
<td>&nbsp;</td>
</tr>
</tbody>
</table>
<p><img decoding="async" class="wp-image-18271" src="https://telechargercours.com/wp-content/uploads/2021/07/word-image-161.png"></p>
<p>Tous droits de l’auteur des œuvres réservés. Sauf autorisation, la reproduction ainsi que toute utilisation des œuvres autre que la consultation individuelle et privée sont interdites.</p>
<p></p>
<p><strong>Sciences Indusrielles</strong></p>
<p><strong>Systèmes linéaires continus invariants</strong></p>
<p>ANALYSE FREQUENTIELLE</p>
<table>
<tbody>
<tr>
<td colspan="4">
<p>Ø <em>équivalent lorsque</em> <em>w</em> ® 0 <em>: on peut négliger les termes en</em> <em>w</em> <sup>2</sup> et <em>w</em><sup>4</sup></p>
</td>
<td>
<p>devant1.Donc :</p>
</td>
</tr>
<tr>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
</tr>
<tr>
<td>&nbsp;</td>
<td>
<p><em>H </em>(<em> jw</em>)</p>
</td>
<td>
<p>; 20<em>LogK</em></p>
</td>
<td>
<p>, c’est une droite horizontale d’ordonnée</p>
</td>
<td>
<p>20<em>LogK</em> dB.</p>
</td>
</tr>
<tr>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>
<p>{</p>
</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
</tr>
<tr>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>
<p><em><sup>dB</sup></em> <em>w</em> ® 0</p>
</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
</tr>
</tbody>
</table>
<ul>
<li><em>équivalent lorsque w </em>® ¥<em> : on peut négliger le terme en</em></li>
</ul>
<p><img decoding="async" class="wp-image-18272" src="https://telechargercours.com/wp-content/uploads/2021/07/word-image-162.png"></p>
<table>
<tbody>
<tr>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td colspan="2" rowspan="3">
<p><em>H </em>(<em> jw</em>)</p>
</td>
<td>&nbsp;</td>
<td rowspan="2">&nbsp;</td>
<td>&nbsp;</td>
<td rowspan="3">
<p>;</p>
</td>
<td colspan="2" rowspan="3">
<p>20<em>LogK</em> &#8211; 20<em>Log</em></p>
</td>
<td colspan="2" rowspan="2">
<p><em>w</em><sup>4</sup></p>
</td>
<td rowspan="3">&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
</tr>
<tr>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
</tr>
<tr>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
</tr>
<tr>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>
<p>{</p>
</td>
<td>&nbsp;</td>
<td colspan="3">
<p>4</p>
</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
</tr>
<tr>
<td colspan="6">
<p>1et <em>w</em> <sup>2</sup> devant celuien <em>w</em><sup>4</sup> .Donc :</p>
</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td colspan="2">
<p><em><sup>dB</sup></em> <em>w</em> ® ¥</p>
</td>
<td>&nbsp;</td>
<td colspan="3">
<p><em>w</em><sub>0</sub></p>
</td>
<td>
<p>,</p>
</td>
<td>&nbsp;</td>
</tr>
<tr>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td colspan="2">
<p>;</p>
</td>
<td colspan="5">
<p>20<em>LogK</em> + 20<em>Logw</em><sup>2</sup> &#8211; 20<em>Logw</em><sup>2</sup></p>
</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
</tr>
<tr>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td colspan="2">
<p>{</p>
</td>
<td>&nbsp;</td>
<td colspan="2">
<p>0</p>
</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
</tr>
<tr>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td colspan="2">
<p><em>w</em> ® ¥</p>
</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
</tr>
<tr>
<td colspan="16">
<p>soit, mis en forme différemment pour faire clairement apparaître la pente de la droite :</p>
</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
</tr>
<tr>
<td>&nbsp;</td>
<td>
<p><em>H </em>(<em> jw</em>)</p>
</td>
<td colspan="4">
<p>; 20 <em>LogK</em> + 40<em>Logw</em></p>
</td>
<td rowspan="2">
<p>0</p>
</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>
<p>-40</p>
</td>
<td colspan="2">
<p><em>Logw</em></p>
</td>
<td colspan="2">
<p>c’est une</p>
</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
</tr>
<tr>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>
<p>{</p>
</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>
<p>{</p>
</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
</tr>
<tr>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>
<p><em><sup>dB</sup></em> <em>w</em> ® ¥</p>
</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td colspan="8">
<p>pente d e l a d r o iteasymptote</p>
</td>
<td colspan="2" rowspan="2">
<p><em>Logw</em><sub>0</sub> ) .</p>
</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
</tr>
<tr>
<td colspan="3">
<p>droite de pente -40 <em><sup>dB</sup></em></p>
</td>
<td rowspan="2">
<p><em>décade</em></p>
</td>
<td colspan="8">
<p>passant par le point (<em>w</em> = 1;</p>
</td>
<td colspan="2">
<p>20<em>LogK</em> + 40</p>
</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
</tr>
<tr>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
</tr>
</tbody>
</table>
<p><img decoding="async" class="wp-image-18273" src="https://telechargercours.com/wp-content/uploads/2021/07/word-image-163.png"></p>
<p><strong><em>Intersection entre les deux asymptotes :</em></strong></p>
<p>On cherche la pulsation dite <strong><em>pulsation propre du système</em></strong> notée <em>w</em> <em><sub>p</sub></em> pour laquelle les deux asymptotes ont des valeurs égales. L’équation permettant de déterminer <em>w</em> <em><sub>p</sub></em> est donc :</p>
<p>20<em>LogK</em> = 20<em>LogK</em> + 40<em>Logw</em> <sub>0</sub> &#8211; 40<em>Logw</em> <em><sub>p</sub></em> .</p>
<p>Soit 40<em>Logw</em> <em><sub>p</sub></em> = 40<em>Logw</em><sub>0</sub> ; d’où :<em>w</em> <em><sub>p</sub></em> = <em>w</em><sub>0</sub></p>
<p><strong><em>Asymptote sur le diagramme de phase :</em></strong></p>
<p>La fonction de phase réelle est définie par</p>
<table>
<tbody>
<tr>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td colspan="3">
<p>&#8211; 2<em>xw</em></p>
</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td colspan="3" rowspan="2">
<p><sup>æ</sup>1-</p>
</td>
<td>
<p><em>w </em><sup>2</sup></p>
</td>
<td rowspan="2">
<p>ö</p>
</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td colspan="6">
<p>-2<em>xw</em></p>
</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
</tr>
<tr>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
</tr>
<tr>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td colspan="2" rowspan="2">
<p><em>w</em><sub>0</sub></p>
</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>
<p>ç</p>
</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>
<p>2</p>
</td>
<td>
<p>÷</p>
</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td colspan="5" rowspan="2">
<p><em>w</em><sub>0</sub></p>
</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
</tr>
<tr>
<td rowspan="2">
<p>tan <em>j</em> =</p>
</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td colspan="6" rowspan="2">
<p>;cos<em>j</em> =</p>
</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>
<p>è</p>
</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>
<p><em>w</em><sub>0</sub></p>
</td>
<td>
<p>ø</p>
</td>
<td rowspan="2">
<p>;sin<em>j</em> =</p>
</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
</tr>
<tr>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td rowspan="2">
<p>2</p>
</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
</tr>
<tr>
<td colspan="2" rowspan="2">
<p>æ</p>
</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td rowspan="2">
<p><em>w</em></p>
</td>
<td rowspan="2">
<p>ö</p>
</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td rowspan="2">
<p>2</p>
</td>
<td>&nbsp;</td>
<td>
<p>2</p>
</td>
<td rowspan="2">
<p>2 2</p>
</td>
<td colspan="3" rowspan="4">
<p><sup>æ</sup>1 &#8211;</p>
</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td rowspan="2">
<p>2</p>
</td>
<td>&nbsp;</td>
<td rowspan="3">
<p>ö</p>
</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td rowspan="2">
<p>2 2</p>
</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
</tr>
<tr>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
</tr>
<tr>
<td colspan="2" rowspan="2">
<p>ç</p>
</td>
<td colspan="2" rowspan="2">
<p>1-</p>
</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td rowspan="2">
<p>÷</p>
</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td colspan="3">
<p>æ</p>
</td>
<td rowspan="3">
<p>1</p>
</td>
<td rowspan="3">
<p>&#8211;</p>
</td>
<td>
<p><em>w</em></p>
</td>
<td>&nbsp;</td>
<td>
<p>ö</p>
</td>
<td rowspan="3">
<p>+</p>
</td>
<td>
<p>4<em>x w</em></p>
</td>
<td colspan="2">
<p><em>w</em></p>
</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td rowspan="2">
<p>+</p>
</td>
<td>&nbsp;</td>
<td>
<p>4<em>x w</em></p>
</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
</tr>
<tr>
<td>&nbsp;</td>
<td>
<p>2</p>
</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td colspan="2">&nbsp;</td>
<td rowspan="2">
<p>2</p>
</td>
<td>&nbsp;</td>
<td rowspan="2">
<p>÷</p>
</td>
<td>&nbsp;</td>
<td rowspan="2">
<p>2</p>
</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
</tr>
<tr>
<td colspan="2" rowspan="2">
<p>è</p>
</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td colspan="4" rowspan="2">
<p><em><sup>w</sup></em>0 ø</p>
</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td colspan="3" rowspan="2">
<p>ç</p>
</td>
<td colspan="2" rowspan="2">
<p><em>w </em><sup>2</sup></p>
</td>
<td rowspan="2">
<p>÷</p>
</td>
<td rowspan="2">
<p><em>w</em><sup>2</sup></p>
</td>
<td colspan="3">
<p>ç</p>
</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
</tr>
<tr>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td colspan="3">
<p>è</p>
</td>
<td colspan="4">
<p><em>w</em><sub>0</sub></p>
</td>
<td>
<p>ø</p>
</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>
<p><em>w</em><sub>0</sub></p>
</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
</tr>
<tr>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td colspan="3">
<p>è</p>
</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>
<p>0</p>
</td>
<td>
<p>ø</p>
</td>
<td>&nbsp;</td>
<td>
<p>0</p>
</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
</tr>
<tr>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td colspan="2">
<p>é</p>
</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td colspan="3">
<p>2<em>xw</em></p>
</td>
<td colspan="2">
<p>ù</p>
</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
</tr>
<tr>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td colspan="2" rowspan="2">
<p>ê</p>
</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td colspan="2" rowspan="2">
<p>ú</p>
</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
</tr>
<tr>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td colspan="3" rowspan="2">
<p><em>w</em><sub>0</sub></p>
</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
</tr>
<tr>
<td colspan="6" rowspan="2">
<p><em>j</em> (<em>w</em>) = &#8211; arctan</p>
</td>
<td colspan="2" rowspan="2">
<p>ê</p>
</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td rowspan="2">
<p>ú</p>
</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
</tr>
<tr>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td colspan="2" rowspan="2">
<p><em>w</em><sup>2</sup></p>
</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
</tr>
<tr>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td colspan="3">
<p>ê <sub>æ</sub></p>
</td>
<td>&nbsp;</td>
<td rowspan="3">
<p>&#8211;</p>
</td>
<td colspan="2">
<p>ö<sup>ú</sup></p>
</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
</tr>
<tr>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td colspan="2" rowspan="2">
<p>ê</p>
</td>
<td colspan="2" rowspan="2">
<p>ç <sup>1</sup></p>
</td>
<td colspan="2">&nbsp;</td>
<td colspan="2" rowspan="2">
<p>÷ú</p>
</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
</tr>
<tr>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td colspan="2">
<p>2</p>
</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
</tr>
<tr>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td colspan="2">
<p>ê</p>
</td>
<td rowspan="2">
<p>è</p>
</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td colspan="2">
<p><em>w</em><sub>0</sub></p>
</td>
<td colspan="2">
<p>ú</p>
</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
</tr>
<tr>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td colspan="2">
<p>ë</p>
</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td colspan="2">
<p>øû</p>
</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
</tr>
<tr>
<td colspan="23">
<p>Ø <em>équivalent lorsque</em> <em>w</em> ® 0 <em>:</em> tan <em>j</em> ® 0</p>
</td>
<td colspan="5">
<p>; cos<em>j</em> ® 1</p>
</td>
<td colspan="2">
<p>;</p>
</td>
<td>&nbsp;</td>
<td colspan="4">
<p>sin <em>j</em> ® 0 .Donc,</p>
</td>
<td>&nbsp;</td>
</tr>
<tr>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>
<p>{</p>
</td>
<td colspan="3">
<p>{</p>
</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>
<p>{</p>
</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
</tr>
<tr>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>
<p><em>w</em>®0</p>
</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>
<p><em>w</em> ®0</p>
</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td colspan="2">
<p><em>w</em>®0</p>
</td>
<td>&nbsp;</td>
</tr>
<tr>
<td colspan="5">
<p><em>j</em> (<em>w</em>)</p>
</td>
<td>&nbsp;</td>
<td colspan="2">
<p>»</p>
</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td colspan="16">
<p>0° on a une asymptote horizontale à 0°</p>
</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
</tr>
<tr>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td colspan="2">
<p>{</p>
</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
</tr>
</tbody>
</table>
<p><img decoding="async" class="wp-image-18274" src="https://telechargercours.com/wp-content/uploads/2021/07/word-image-164.png"></p>
<p><em>w</em> ® 0</p>
<p>Page 4 <strong>Emmanuel FARGES</strong> Ó EduKlub S.A.</p>
<p>Tous droits de l’auteur des œuvres réservés. Sauf autorisation, la reproduction ainsi que toute utilisation des œuvres autre que la consultation individuelle et privée sont interdites.</p>
<p></p>
<p><strong>Sciences Indusrielles</strong></p>
<p><strong>Systèmes linéaires continus invariants</strong></p>
<p>ANALYSE FREQUENTIELLE</p>
<p>Ø <em>équivalent lorsque</em> <em>w</em> ® ¥ <em>:</em> tan <em>j</em> ® 0 ; cos<em>j</em> ® &#8211; 1 ; sin<em>j</em> ® 0<sup>&#8211;</sup> .Donc,</p>
<p>{ { {</p>
<p><em>w</em>®¥ <em>w </em>®¥ <em>w</em>®¥</p>
<p><em>j</em>(<em>w</em> ) » &#8211;<em>p</em> (- 180°) (car le cosinus tend vers -1 et le sinus vers 0 par valeurs</p>
<p>{</p>
<ul>
<li>® ¥</li>
</ul>
<p>négatives) on a une asymptote horizontale à -180°</p>
<table>
<tbody>
<tr>
<td>&nbsp;</td>
<td>
<p>æ</p>
</td>
<td>
<p>ö</p>
</td>
<td>&nbsp;</td>
<td rowspan="3">
<p>(</p>
</td>
<td>&nbsp;</td>
<td rowspan="2">
<p>= &#8211; arctan (+¥) = &#8211;</p>
</td>
<td>
<p><em>p</em></p>
</td>
<td rowspan="2">
<p>(- 90°)</p>
</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
</tr>
<tr>
<td>
<p>Comme</p>
</td>
<td>
<p><em>j</em> ç <em>w</em></p>
</td>
<td>
<p>÷</p>
</td>
<td>
<p>= <em>j</em></p>
</td>
<td>
<p><em>w</em></p>
</td>
<td rowspan="2">
<p>2</p>
</td>
<td>
<p>, on relie les deux asymptotes qui</p>
</td>
<td>&nbsp;</td>
</tr>
<tr>
<td>&nbsp;</td>
<td>
<p>è</p>
</td>
<td>
<p><em>p </em><sub>ø</sub></p>
</td>
<td>&nbsp;</td>
<td>
<p>0 <sup>)</sup></p>
</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
</tr>
</tbody>
</table>
<p>sont normalement parallèles (puisque toutes les deux horizontales) au niveau de la pulsation propre du système <em>w</em> <em><sub>p</sub></em> = <em>w</em><sub>0</sub> .</p>
<p><strong><em>2<sup>ème</sup> méthode : équivalents à partir de la fonction de transfert :</em></strong></p>
<p><em>A</em></p>
<p>On cherche des équivalents de la fonction de transfert de la forme</p>
<p><em>p<sup>n</sup></em></p>
<p><strong><em>Fonction de transfert équivalente pour </em></strong><em>w</em> ® 0<strong><em> , c’est à dire pour </em></strong><em>p</em> ® 0<strong><em> (puisque </em></strong><em>p</em> = <em>jw</em><strong><em> )</em></strong></p>
<table>
<tbody>
<tr>
<td rowspan="2">
<p><em>H </em>(<em> p</em>) =</p>
</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td colspan="2">
<p><em>K</em></p>
</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td rowspan="2">
<p>équivalent à</p>
</td>
<td>
<p><em>K</em></p>
</td>
<td rowspan="2">
<p>soit</p>
</td>
<td colspan="2">
<p><em>K</em></p>
</td>
<td>&nbsp;</td>
</tr>
<tr>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td rowspan="2">
<p>2<em>x</em></p>
</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td colspan="2" rowspan="2">
<p><em>p</em><sup>2</sup></p>
</td>
<td>&nbsp;</td>
<td rowspan="2">
<p>1</p>
</td>
<td rowspan="2">
<p>0</p>
</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
</tr>
<tr>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td colspan="2">
<p>14243</p>
</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
</tr>
<tr>
<td>&nbsp;</td>
<td colspan="2" rowspan="2">
<p>1+</p>
</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td rowspan="2">
<p><em>p </em>+</p>
</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td rowspan="2">
<p>quand p ® 0</p>
</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td colspan="2">
<p><em>p</em></p>
</td>
<td>&nbsp;</td>
</tr>
<tr>
<td>&nbsp;</td>
<td colspan="2">&nbsp;</td>
<td rowspan="2">
<p><em>w</em></p>
</td>
<td>
<p>2</p>
</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
</tr>
<tr>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>
<p><em>w</em></p>
</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td rowspan="2">
<p>0</p>
</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
</tr>
<tr>
<td>&nbsp;</td>
<td colspan="3">
<p>0</p>
</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
</tr>
</tbody>
</table>
<ul>
<li>On retrouve donc l’asymptote horizontale pour <em>w</em> ® 0 à 20<em>LogK</em> pour la courbe de gain en dB (d’après les résultats démontrés paragraphe 1.3.3)</li>
<li>On retrouve donc l’asymptote horizontale pour <em>w</em> ® 0 à 0° pour la courbe de phase (d’après les résultats démontrés paragraphe 1.3.3)</li>
</ul>
<p><strong><em>Fonction de transfert équivalente pour </em></strong><em>w</em> ® ¥<strong><em> , c’est à dire pour </em></strong><em>p</em> ® ¥<strong><em> (puisque </em></strong><em>p</em> = <em>jw</em><strong><em> )</em></strong></p>
<table>
<tbody>
<tr>
<td rowspan="2">
<p><em>H </em>(<em> p</em>) =</p>
</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td colspan="3">
<p><em>K</em></p>
</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td rowspan="2">
<p>équivalentà</p>
</td>
<td colspan="3">
<p><em>K w</em><sub>0</sub><sup>2</sup></p>
</td>
<td>&nbsp;</td>
</tr>
<tr>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td rowspan="2">
<p>2</p>
</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td rowspan="2">
<p>2</p>
</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
</tr>
<tr>
<td colspan="2" rowspan="3">
<p>1+</p>
</td>
<td rowspan="2">
<p>2<em>x</em></p>
</td>
<td>&nbsp;</td>
<td rowspan="3">
<p><em>p </em>+</p>
</td>
<td rowspan="2">
<p><em>p</em></p>
</td>
<td colspan="2">
<p>14243</p>
</td>
<td rowspan="2">
<p><em>p</em></p>
</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
</tr>
<tr>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td rowspan="2">&nbsp;</td>
<td rowspan="2">
<p>quand p ® ¥</p>
</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
</tr>
<tr>
<td rowspan="2">
<p><em>w </em><sub>0</sub></p>
</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>
<p>2</p>
</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
</tr>
<tr>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td colspan="2">
<p><em>w</em><sub>0</sub></p>
</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
</tr>
</tbody>
</table>
<ul>
<li>On retrouve donc l’asymptote de pente -40dB/décade pour <em>w</em> ® ¥ à</li>
</ul>
<p>20<em>LogKw</em><sub>0</sub><sup>2</sup> &#8211; (20 ´ 2) <em>Logw</em> = 20<em>LogK</em> + 40<em>Logw</em><sub>0</sub> &#8211; 40<em>Logw</em> pour la courbe de gain en dB (d’après les résultats démontrés paragraphe 1.3.3)</p>
<p>ü On retrouve donc l’asymptote horizontale pour <em>w</em> ® ¥ à &#8211;<em>p</em> (- 180°) pour la courbe de phase (d’après les résultats démontrés paragraphe 1.3.3)</p>
<p>Page 5 <strong>Emmanuel FARGES</strong> Ó EduKlub S.A.</p>
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<p></p>
<p><strong>Sciences Indusrielles</strong></p>
<p><strong>Systèmes linéaires continus invariants</strong></p>
<p>ANALYSE FREQUENTIELLE</p>
<p><strong><em>Graphiquement cela donne les diagrammes asymptotiques schématiques suivants :</em></strong></p>
<p><img decoding="async" class="alignnone wp-image-18275" title="Systèmes linéaires continus invariants" src="https://telechargercours.com/wp-content/uploads/2021/07/word-image-165.png" alt="Systèmes linéaires continus invariants" width="1247" height="463"></p>
<table>
<tbody>
<tr>
<td>
<p>20<em>LogK</em></p>
</td>
<td rowspan="2">
<p><em>H</em></p>
</td>
<td rowspan="3">
<p><em>dB</em></p>
</td>
<td rowspan="2">
<p>(<em>w</em>)</p>
</td>
<td colspan="2" rowspan="2">
<p><strong><em>Pente : </em></strong>-40 <em>dB</em> / <em>décade</em></p>
</td>
<td>&nbsp;</td>
</tr>
<tr>
<td>&nbsp;</td>
<td>&nbsp;</td>
</tr>
<tr>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
</tr>
<tr>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>
<p><strong><em>Ecart de 40dB</em></strong></p>
</td>
<td>&nbsp;</td>
</tr>
<tr>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>
<p><em>w</em></p>
</td>
<td>&nbsp;</td>
</tr>
<tr>
<td>&nbsp;</td>
<td colspan="3">
<p><em>w</em> <em><sub>p</sub></em> = <em>w</em><sub>0</sub></p>
</td>
<td>
<p><strong><em>1 décade</em></strong></p>
</td>
<td>
<p>10<em>w</em><sub>0</sub></p>
</td>
<td>&nbsp;</td>
</tr>
</tbody>
</table>
<ul>
<li>(<em>w</em> )</li>
</ul>
<p><img decoding="async" class="wp-image-18276" src="https://telechargercours.com/wp-content/uploads/2021/07/word-image-166.png"></p>
<table>
<tbody>
<tr>
<td>
<p><strong>0°</strong></p>
</td>
<td rowspan="2">
<p><em>w <sub>p</sub></em></p>
</td>
<td rowspan="2">
<p>= <em>w</em><sub>0</sub></p>
</td>
<td>&nbsp;</td>
</tr>
<tr>
<td>&nbsp;</td>
<td>&nbsp;</td>
</tr>
</tbody>
</table>
<p><em>w</em></p>
<p><strong>-180°</strong></p>
<h3><strong>3.5</strong> <strong>Diagrammes de Bode d’un deuxième ordre</strong></h3>
<p><em>H <sub>dB</sub> </em>(<em>w</em>)</p>
<p><strong><em>Asymptote horizontale à</em></strong></p>
<p><strong><em>20LogK</em></strong></p>
<p><strong><em>1 décade</em></strong></p>
<p><strong><em>Asymptote de pente -40 dB /décade et d’équation</em></strong></p>
<p>20<em>LogK</em> + 40<em>Logw</em> <sub>0</sub> &#8211; 40<em>Logw</em></p>
<p><strong><em>Gain du deuxième ordre (K=10 , x =0,1 et w<sub>0</sub> = 10rads<sup>-1</sup>) Avec résonance : </em></strong><em>x</em> = 0,1 Þ <em>Q<sub>dB</sub></em> ; 14<em>dB</em></p>
<p><img decoding="async" class="wp-image-18277" src="https://telechargercours.com/wp-content/uploads/2021/07/word-image-167.png"></p>
<p><strong><em>Gain du deuxième ordre (K=10 , x =0,5 et w<sub>0</sub> = 10rads<sup>-1</sup>).Avec résonance : </em></strong><em>x</em> = 0,5 Þ <em>Q<sub>dB</sub></em> =1,25<em>dB</em></p>
<p><strong><em>Gain du deuxième ordre (K=10 , x =0,9 et w<sub>0</sub> = 10rads<sup>-1</sup>) sans résonance</em></strong></p>
<p><strong><em>Gain du deuxième ordre (K=10 , x =2 et w<sub>0</sub> = 10rads<sup>-1</sup>) sans résonance</em></strong></p>
<p><em>w </em>(<em>rad s</em><sup>-1</sup> )</p>
<p><strong><em>Pulsation propre </em></strong><em>w</em><sub>0</sub></p>
<p>Page 6 <strong>Emmanuel FARGES</strong> Ó EduKlub S.A.</p>
<p>Tous droits de l’auteur des œuvres réservés. Sauf autorisation, la reproduction ainsi que toute utilisation des œuvres autre que la consultation individuelle et privée sont interdites.</p>
<p></p>
<p><strong>Sciences Indusrielles</strong></p>
<p><strong>Systèmes linéaires continus invariants</strong></p>
<p>ANALYSE FREQUENTIELLE</p>
<p><em>j</em> (<em>w</em> )<strong><em> ici</em></strong></p>
<p><strong><em>en degré</em></strong></p>
<p><strong><em>Asymptote horizontale à 0°</em></strong></p>
<p><em><sup>p</sup></em></p>
<p><em>j</em> (<em>w</em><sub>0</sub> ) = -arctan ¥ = &#8211; <strong><em>= -90°</em></strong></p>
<p><strong><em>-180°</em></strong></p>
<p><strong><em>Phase du deuxième ordre (K=10 , x =0,1 et w<sub>0</sub> = 10rads<sup>-1</sup>)</em></strong></p>
<p><img decoding="async" class="wp-image-18278" src="https://telechargercours.com/wp-content/uploads/2021/07/word-image-168.png"></p>
<p><strong><em>Phase du deuxième ordre (K=10 ,</em></strong></p>
<ul>
<li><strong><em>=0,5 et w<sub>0</sub> = 10rads<sup>-1</sup>).</em></strong></li>
</ul>
<p><strong><em>Phase du deuxième ordre (K=10 ,</em></strong></p>
<ul>
<li><strong><em>=0,9 et w<sub>0</sub> = 10rads<sup>-1</sup>)</em></strong></li>
</ul>
<p><strong><em>Phase du deuxième ordre</em></strong></p>
<p><strong><em>(K=10 , x =2 et w<sub>0</sub> = 10rads<sup>-1</sup>)</em></strong></p>
<p><em>w </em>(<em>rad s</em><sup>-1</sup> )</p>
<p><strong><em>Asymptote horizontale à -180°</em></strong></p>
<ol>
<li><strong>Construction des courbes de Bode asymptotiques d’un système de fonction de transfert quelconque.</strong></li>
</ol>
<h3><strong>4.1</strong> <strong>Diagrammes de Bode de l’inverse d’une fonction de transfert</strong></h3>
<table>
<tbody>
<tr>
<td rowspan="3">
<p>Considérons un système de fonction de transfert <em>H</em> ( <em>p</em> ) =</p>
</td>
<td>
<p>1</p>
</td>
<td rowspan="3">
<p>avec</p>
</td>
<td colspan="2" rowspan="3">
<p><em>H </em><sub>1</sub>(<em> p</em>) une fonction de</p>
</td>
<td>&nbsp;</td>
</tr>
<tr>
<td>&nbsp;</td>
<td>&nbsp;</td>
</tr>
<tr>
<td rowspan="2">
<p><em>H </em><sub>1</sub> (<em> p </em>)</p>
</td>
<td>&nbsp;</td>
</tr>
<tr>
<td rowspan="2">
<p>transfert dont on connaît les diagrammes de Bode.</p>
</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
</tr>
<tr>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
</tr>
<tr>
<td colspan="5">
<p>Peut on déduire les diagrammes de Bode de <em>H</em> ( <em>p</em>) connaissant ceux de <em>H</em> <sub>1</sub>( <em>p</em>) ?</p>
</td>
<td>&nbsp;</td>
</tr>
<tr>
<td colspan="3">
<p>On pose <em>j</em> (<em>w</em> ) et <em>H</em> <em><sub>dB</sub></em> (<em>w</em>) respectivement la phase et le gain en dB de</p>
</td>
<td colspan="2">
<p><em>H </em>(<em> p</em>)</p>
</td>
<td>&nbsp;</td>
</tr>
<tr>
<td colspan="4">
<p>On pose <em>j</em><sub>1</sub> (<em>w</em> ) et <em>H</em><sub>1</sub><em><sub>dB</sub></em> (<em>w</em>) respectivement la phase et le gain en dB de</p>
</td>
<td>
<p><em>H </em><sub>1</sub>(<em> p</em>)</p>
</td>
<td>&nbsp;</td>
</tr>
<tr>
<td colspan="4">
<p>On cherche donc à trouver une relation entre <em>j</em> (<em>w</em> ) et <em>j</em><sub>1</sub> (<em>w</em> ) , puis entre</p>
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<p><em>H <sub>dB</sub> </em>(<em>w</em>) et<em> H</em><sub>1</sub><em><sub>dB</sub> </em>(<em>w</em>)</p>
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<p><em><sup>jj </sup></em><sup>(</sup><em><sup>w </sup></em><sup>)</sup></p>
<p>On peut écrire le complexe <em>H</em> ( <em>jw</em>) sous sa forme exponentielle : <em>H</em> ( <em>jw</em> ) = <em>H</em> ( <em>jw</em>) <em>e</em></p>
<p><em><sup>jj </sup></em><sup>(</sup><em><sup>w </sup></em><sup>)</sup></p>
<p>On peut écrire le complexe <em>H</em> <sub>1</sub>( <em>jw</em>) sous sa forme exponentielle : <em>H</em><sub>1</sub> ( <em>jw</em> ) = <em>H</em><sub>1</sub> ( <em>jw</em>) <em>e</em> <sup>1</sup></p>
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<p><em>H </em>(<em> jw</em>)</p>
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<p><em><sub>e</sub> jj </em>(<em>w </em>) <sub>=</sub></p>
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<p><em><sub>e</sub></em><sup>&#8211;</sup> <em><sup>jj</sup></em> 1 (<em><sup>w</sup></em> ) <sub>.</sub></p>
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<p><em>jj</em><sub>1</sub> (<em>w </em>)</p>
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<p><em>H</em><sub>1</sub> (<em> jw</em>)</p>
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<p><em>H</em><sub>1</sub> (<em> jw</em>)</p>
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<p>D’où :</p>
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<p><em>H </em>(<em> jw</em>)</p>
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<p><em>j</em> (<em>w</em>) = &#8211;<em>j</em><sub>1</sub> (<em>w</em> ) soit encore :</p>
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<p><em>H </em><sub>1</sub> (<em> jw</em>)</p>
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<p><em>H <sub>dB</sub> </em>(<em>w </em>) = 20<em>Log</em></p>
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<p>æ</p>
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<p>ö</p>
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<p>= &#8211;<em>H</em><sub>1</sub><em><sub>db</sub></em> (<em>w</em> )</p>
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<p><em>H</em>(<em> jw</em>)</p>
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<p>= 20<em>Log</em> ç</p>
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<p>÷ = -20<em>Log</em></p>
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<p><em>H</em><sub>1</sub> (<em> jw </em>)</p>
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<p>ç</p>
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<p><em>H</em></p>
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<p>( <em>jw</em>)</p>
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<p>÷</p>
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<p>è</p>
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<p>1</p>
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<p>ø</p>
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<p>Ó EduKlub S.A.</p>
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<p>Page 7</p>
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<p><strong>Emmanuel FARGES</strong></p>
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<p>Tous droits de l’auteur des œuvres réservés. Sauf autorisation, la reproduction ainsi que toute utilisation des œuvres autre que la consultation individuelle et privée sont interdites.</p>
<p></p>
<p><strong>Sciences Indusrielles</strong></p>
<p><strong>Systèmes linéaires continus invariants</strong></p>
<p>ANALYSE FREQUENTIELLE</p>
<p><sub>Finalement :</sub> <sup>ì</sup>ï<sub>í</sub><em><sup>H</sup></em> <em>dB</em> (<em><sup>w</sup></em> ) <sup>= &#8211;</sup><em><sup>H</sup></em>1 <em>dB</em> (<em><sup>w</sup></em> )</p>
<p>ï<sub>î</sub> <em>j </em>(<em>w </em>) = &#8211;<em>j</em><sub>1</sub> (<em>w </em>)</p>
<p><strong><em>Ce qui revient graphiquement à effectuer la symétrie de la courbe de gain par rapport à l’axe 0dB et la symétrie de la courbe de phase par rapport à l’axe 0°.</em></strong></p>
<p><strong><em>Application : Voit TD</em></strong></p>
<h3><strong>4.2</strong> <strong>Diagrammes de Bode d’un produit de fonctions de transfert</strong></h3>
<p>Diagrammes de Bode d’un produit de deux fonctions de transfert :</p>
<p>Considérons un système de fonction de transfert <em>H</em> ( <em>p</em>) = <em>H</em><sub>1</sub> ( <em>p</em>)<em>H</em> <sub>2</sub> ( <em>p</em>) avec <em>H</em> <sub>1</sub>( <em>p</em>) et <em>H </em><sub>2</sub> (<em> p</em>) deux fonctions de transfert dont on connaît les diagrammes de Bode.</p>
<p>On pose <em>j</em> (<em>w</em> ) et <em>H</em> <em><sub>dB</sub></em> (<em>w</em>) respectivement la phase et le gain en dB de <em>H</em> ( <em>p</em>)</p>
<p>On pose <em>j</em><sub>1</sub> (<em>w</em> ) et <em>H</em><sub>1</sub><em><sub>dB</sub></em> (<em>w</em>) respectivement la phase et le gain en dB de <em>H</em> <sub>1</sub>( <em>p</em>)</p>
<p>On pose <em>j</em> <sub>2</sub> (<em>w</em>) et <em>H</em> <sub>2</sub> <em><sub>dB</sub></em> (<em>w</em>) respectivement la phase et le gain en dB de <em>H</em> <sub>2</sub> ( <em>p</em>)</p>
<p>Peut on déduire les diagrammes de Bode de <em>H</em> ( <em>p</em>) connaissant ceux de <em>H</em> <sub>1</sub>( <em>p</em>) et <em>H</em> <sub>2</sub> ( <em>p</em>) ?</p>
<p><em><sup>jj </sup></em><sup>(</sup><em><sup>w </sup></em><sup>)</sup></p>
<p>On peut écrire le complexe <em>H</em> ( <em>jw</em>) sous sa forme exponentielle : <em>H</em> ( <em>jw</em> ) = <em>H</em> ( <em>jw</em>) <em>e</em></p>
<p><em><sup>jj </sup></em><sup>(</sup><em><sup>w </sup></em><sup>)</sup></p>
<p><em><sup>jj </sup></em><sup>(</sup><em><sup>w </sup></em><sup>)</sup></p>
<p>On peut écrire le complexe <em>H</em> <sub>2</sub> ( <em>jw</em>) sous sa forme exponentielle : <em>H</em> <sub>2</sub> ( <em>jw</em> ) = <em>H</em> <sub>2</sub> ( <em>jw</em>) <em>e</em> <sup>2</sup></p>
<p><em>H </em>(<em> jw</em>) =<em> H</em><sub>1</sub> (<em> jw </em>)<em>H </em><sub>2</sub> (<em> jw</em>) soit :</p>
<p><img decoding="async" class="wp-image-18279" src="https://telechargercours.com/wp-content/uploads/2021/07/word-image-169.png"> <img decoding="async" class="wp-image-18280" src="https://telechargercours.com/wp-content/uploads/2021/07/word-image-170.png"><em><sub> H </sub></em><sub>(</sub><em><sub> jw </sub></em><sub>)</sub> <img decoding="async" class="wp-image-18281" src="https://telechargercours.com/wp-content/uploads/2021/07/word-image-171.png"> <img decoding="async" class="wp-image-18282" src="https://telechargercours.com/wp-content/uploads/2021/07/word-image-172.png"><em><sub> e </sub><sup>jj</sup> </em>(<em><sup>w</sup> </em>) <sub>= </sub> <img decoding="async" class="wp-image-18283" src="https://telechargercours.com/wp-content/uploads/2021/07/word-image-173.png"> <img decoding="async" class="wp-image-18284" src="https://telechargercours.com/wp-content/uploads/2021/07/word-image-174.png"><em><sub> H</sub></em><sub>1</sub> <sub>(</sub><em><sub> jw </sub></em><sub>)</sub> <img decoding="async" class="wp-image-18285" src="https://telechargercours.com/wp-content/uploads/2021/07/word-image-175.png"> <img decoding="async" class="wp-image-18286" src="https://telechargercours.com/wp-content/uploads/2021/07/word-image-176.png"><em><sub> e </sub><sup>jj</sup></em>1 (<em><sup>w</sup> </em>) <img decoding="async" class="wp-image-18287" src="https://telechargercours.com/wp-content/uploads/2021/07/word-image-177.png"> <img decoding="async" class="wp-image-18288" src="https://telechargercours.com/wp-content/uploads/2021/07/word-image-178.png"><em><sub> H </sub></em><sub>2</sub> <sub>(</sub><em><sub> jw </sub></em><sub>)</sub> <img decoding="async" class="wp-image-18289" src="https://telechargercours.com/wp-content/uploads/2021/07/word-image-179.png"> <img decoding="async" class="wp-image-18290" src="https://telechargercours.com/wp-content/uploads/2021/07/word-image-180.png"><em><sub> e </sub><sup>jj</sup> </em>2 (<em><sup>w</sup> </em>)</p>
<p><em>j </em>é<em>j </em>(<em>w </em>) +<em>j </em>(<em>w </em>)ù</p>
<p>= <em>H</em><sub>1</sub> ( <em>jw</em> ) <em>H</em> <sub>2</sub> ( <em>jw</em> ) <em>e</em> <sup>ë</sup> <sup>1</sup> <sup>2</sup> <sup>û</sup></p>
<p>Donc : <img decoding="async" class="wp-image-18291" src="https://telechargercours.com/wp-content/uploads/2021/07/word-image-181.png"> <img decoding="async" class="wp-image-18292" src="https://telechargercours.com/wp-content/uploads/2021/07/word-image-182.png"><em> H </em>(<em> jw </em>) <img decoding="async" class="wp-image-18293" src="https://telechargercours.com/wp-content/uploads/2021/07/word-image-183.png"> <img decoding="async" class="wp-image-18294" src="https://telechargercours.com/wp-content/uploads/2021/07/word-image-184.png"> = <img decoding="async" class="wp-image-18295" src="https://telechargercours.com/wp-content/uploads/2021/07/word-image-185.png"> <img decoding="async" class="wp-image-18296" src="https://telechargercours.com/wp-content/uploads/2021/07/word-image-186.png"><em> H</em><sub>1</sub> (<em> jw </em>) <img decoding="async" class="wp-image-18297" src="https://telechargercours.com/wp-content/uploads/2021/07/word-image-187.png"> <img decoding="async" class="wp-image-18298" src="https://telechargercours.com/wp-content/uploads/2021/07/word-image-188.png"> <img decoding="async" class="wp-image-18299" src="https://telechargercours.com/wp-content/uploads/2021/07/word-image-189.png"> <img decoding="async" class="wp-image-18300" src="https://telechargercours.com/wp-content/uploads/2021/07/word-image-190.png"><em> H </em><sub>2</sub> (<em> jw </em>) <img decoding="async" class="wp-image-18301" src="https://telechargercours.com/wp-content/uploads/2021/07/word-image-191.png"> <img decoding="async" class="wp-image-18302" src="https://telechargercours.com/wp-content/uploads/2021/07/word-image-192.png"> et <em>j</em> (<em>w</em>) = <em>j</em><sub>1</sub> (<em>w</em>)+ <em>j</em> <sub>2</sub> (<em>w</em>) , soit :</p>
<p><em>H <sub>dB</sub> </em>(<em>w </em>) = 20<em>Log </em> <img decoding="async" class="wp-image-18303" src="https://telechargercours.com/wp-content/uploads/2021/07/word-image-193.png"> <img decoding="async" class="wp-image-18304" src="https://telechargercours.com/wp-content/uploads/2021/07/word-image-194.png"><em> H </em>(<em> jw </em>) <img decoding="async" class="wp-image-18305" src="https://telechargercours.com/wp-content/uploads/2021/07/word-image-195.png"> <img decoding="async" class="wp-image-18306" src="https://telechargercours.com/wp-content/uploads/2021/07/word-image-196.png"> = 20<em>Log</em> <sup>é</sup><sub>ë</sub> <img decoding="async" class="wp-image-18307" src="https://telechargercours.com/wp-content/uploads/2021/07/word-image-197.png"> <img decoding="async" class="wp-image-18308" src="https://telechargercours.com/wp-content/uploads/2021/07/word-image-198.png"><em> H</em><sub>1</sub> (<em> jw </em>) <img decoding="async" class="wp-image-18309" src="https://telechargercours.com/wp-content/uploads/2021/07/word-image-199.png"> <img decoding="async" class="wp-image-18310" src="https://telechargercours.com/wp-content/uploads/2021/07/word-image-200.png"> <img decoding="async" class="wp-image-18311" src="https://telechargercours.com/wp-content/uploads/2021/07/word-image-201.png"> <img decoding="async" class="wp-image-18312" src="https://telechargercours.com/wp-content/uploads/2021/07/word-image-202.png"><em> H </em><sub>2</sub> (<em> jw </em>) <img decoding="async" class="wp-image-18313" src="https://telechargercours.com/wp-content/uploads/2021/07/word-image-203.png"> <img decoding="async" class="wp-image-18314" src="https://telechargercours.com/wp-content/uploads/2021/07/word-image-204.png"><sup>ù</sup><sub>û</sub></p>
<ul>
<li>20<em>Log </em> <img decoding="async" class="wp-image-18315" src="https://telechargercours.com/wp-content/uploads/2021/07/word-image-205.png"> <img decoding="async" class="wp-image-18316" src="https://telechargercours.com/wp-content/uploads/2021/07/word-image-206.png"><em> H</em><sub>1</sub> (<em> jw </em>) <img decoding="async" class="wp-image-18317" src="https://telechargercours.com/wp-content/uploads/2021/07/word-image-207.png"> <img decoding="async" class="wp-image-18318" src="https://telechargercours.com/wp-content/uploads/2021/07/word-image-208.png"> + 20<em>Log </em> <img decoding="async" class="wp-image-18319" src="https://telechargercours.com/wp-content/uploads/2021/07/word-image-209.png"> <img decoding="async" class="wp-image-18320" src="https://telechargercours.com/wp-content/uploads/2021/07/word-image-210.png"><em> H</em><sub>2</sub> (<em> jw </em>) <img decoding="async" class="wp-image-18321" src="https://telechargercours.com/wp-content/uploads/2021/07/word-image-211.png"> <img decoding="async" class="wp-image-18322" src="https://telechargercours.com/wp-content/uploads/2021/07/word-image-212.png"> = <em>H</em><sub>1</sub> <em><sub>dB</sub></em> (<em>w</em> ) + <em>H</em><sub>2</sub> <em><sub>dB</sub></em> (<em>w</em> )</li>
</ul>
<table>
<tbody>
<tr>
<td rowspan="2">
<p>Finalement :</p>
</td>
<td>
<p>ì<em>H</em></p>
</td>
<td rowspan="2">
<p><em>dB </em>(</p>
</td>
<td>
<p><em>w</em></p>
</td>
<td rowspan="2">
<p>)</p>
</td>
<td>
<p>= <em>H</em></p>
</td>
<td>
<p><em>w</em></p>
</td>
<td>
<p><em>H</em></p>
</td>
<td rowspan="2">
<p>2 <em>dB</em> (</p>
</td>
<td>
<p><em>w</em></p>
</td>
<td rowspan="2">
<p>)</p>
</td>
<td>&nbsp;</td>
</tr>
<tr>
<td>
<p>í<sup>ï</sup></p>
</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>
<p>1 <em>dB</em> ( )</p>
</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
</tr>
<tr>
<td>&nbsp;</td>
<td colspan="8">
<p>ï <em>j</em> (<em>w</em> ) = <em>j</em><sub>1</sub> (<em>w</em> ) +<em>j</em> <sub>2</sub> (<em>w</em> )</p>
</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
</tr>
<tr>
<td>&nbsp;</td>
<td>
<p>î</p>
</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
</tr>
</tbody>
</table>
<p><strong><em>Ce qui revient graphiquement à « sommer » les deux courbes de gain en dB et les deux courbes de phase.</em></strong></p>
<p><strong><em>Application : Voir TD</em></strong></p>
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<p></p>
<p><strong>Sciences Indusrielles</strong></p>
<p><strong>Systèmes linéaires continus invariants</strong></p>
<p>ANALYSE FREQUENTIELLE</p>
<h3><strong>4. 3</strong> <strong>Diagrammes de Bode d’un système de fonction de transfert quelconque</strong></h3>
<p>Le but est d’écrire cette fonction de transfert quelconque sous forme d’un produit de fonction de transfert connues (1<sup>er</sup> ordre, 2<sup>ème</sup> ordre, intégrateur ou dérivateur simple ou multiple :</p>
<p><em>K</em></p>
<p><em>H </em>(<em> p</em>) =<em> <sub>pa</sub> </em>) pour pouvoir ensuite « sommer » les courbes connues.</p>
<p>On obtiendra ainsi un maximum <strong>d’équivalents successifs intermédiaires</strong> pour les diagrammes de Bode asymptotiques qui s’approche donc ainsi un peu plus du diagramme réel</p>
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<p class="has-text-align-center wp-block-paragraph"><strong><a href="https://www.mediafire.com/file/h9g6ncvvqtedxlx/Telechargercours.com++Systèmes+linéaires+continus+invariants+ANALYSE+FREQUENTIELLE+partie+3+SI-CO08-AAS-EF.pdf/file" target="_blank" rel="noreferrer noopener">Telecharger</a></strong></p>
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		<title>Systèmes linéaires continus invariants ANALYSE FREQUENTIELLE (Partie 1 &amp; 2)</title>
		<link>https://telechargercours.com/systemes-lineaires-continus-invariants-analyse-frequentielle-1-2/</link>
		
		<dc:creator><![CDATA[Cours Zilla]]></dc:creator>
		<pubDate>Sun, 04 Jul 2021 17:39:32 +0000</pubDate>
				<category><![CDATA[Science & Technologie]]></category>
		<guid isPermaLink="false">https://telechargercours.com/?p=18107</guid>

					<description><![CDATA[Sciences Indusrielles Systèmes linéaires continus invariants ANALYSE FREQUENTIELLE SYSTEMES LINEAIRES CONTINUS INVARIANTS ANALYSE FREQUENTIELLE (Partie 1 &#38; 2) L’étude détaillée se limite aux systèmes de bases, c’est à dire aux systèmes du premier ordre et du second ordre. En effet l’étude des autres systèmes se construit à partir des résultats issus des premiers et seconds [&#8230;]]]></description>
										<content:encoded><![CDATA[<p style="text-align: center;"><strong>Sciences Indusrielles</strong></p>
<p style="text-align: center;"><strong>Systèmes linéaires continus invariants</strong></p>
<p style="text-align: center;">ANALYSE FREQUENTIELLE</p>
<p style="text-align: center;"><strong>SYSTEMES LINEAIRES CONTINUS INVARIANTS</strong></p>
<p style="text-align: center;"><em>ANALYSE FREQUENTIELLE (Partie 1 &amp; 2)</em></p>
<p>L’étude détaillée se limite aux systèmes de bases, c’est à dire aux systèmes du premier ordre et du second ordre. En effet l’étude des autres systèmes se construit à partir des résultats issus des premiers et seconds ordres.</p>
<h2><strong>1</strong> <strong>Généralités &#8211; Rappels</strong></h2>
<h3><strong>1.1</strong> <strong>Fonctions de transfert</strong></h3>
<p><strong><em>Considérons le schéma bloc d’un asservissement :</em></strong></p>
<table>
<tbody>
<tr>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
</tr>
<tr>
<td>&nbsp;</td>
<td rowspan="2">&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
</tr>
<tr>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
</tr>
<tr>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
</tr>
</tbody>
</table>
<p><em><img decoding="async" class="wp-image-18108" src="https://telechargercours.com/wp-content/uploads/2021/07/word-image.png"></em></p>
<p><strong><em>Définition des fonctions sur le schéma bloc usuel :</em></strong></p>
<p><em>E </em>(<em> p</em>) est l’entrée du système ou encore la consigne donnée au système.</p>
<p><em>S </em>(<em> p</em>) est la sortie du système ou encore la réponse système.</p>
<p><em>S </em>¢(<em> p</em>) est l’image de la sortie.</p>
<ul>
<li>( <em>p</em>) est l’erreur (ou écart) constatée entre l’entrée <em>E</em> ( <em>p</em>) et l’image de la sortie <em>S</em> ¢( <em>p</em>) .</li>
</ul>
<p><strong><em>Définition de la FTBF et de la FTBO :</em></strong></p>
<ul>
<li>Fonction de Transfert en Boucle Fermée (abrégé sous la forme FTBF) :</li>
</ul>
<p><em>S </em>(<em> p</em>)</p>
<p><em>E</em>(<em> p</em>)</p>
<ul>
<li>Fonction de Transfert en Boucle Ouverte (abrégé sous la forme FTBO) :</li>
</ul>
<p><em><sub>FTBO</sub> </em><sub>=</sub><em> S</em>¢(<em> p</em>)</p>
<ul>
<li>( <em>p</em>)</li>
</ul>
<p><strong><em>Calcul de la FTBF et de la FTBO dans le cas de la boucle d’asservissement classique représentée ci-dessus :</em></strong></p>
<p><em>FTBO </em>=<em> <sup>S</sup></em><sup>¢(</sup><em> <sup>p</sup></em><sup>)</sup> =<em> A</em>(<em> p</em>)<em>B</em>(<em> p</em>)</p>
<ul>
<li>( <em>p</em>)</li>
</ul>
<p>Page 1 <strong>Emmanuel FARGES</strong> Ó EduKlub S.A.</p>
<p>Tous droits de l’auteur des œuvres réservés. Sauf autorisation, la reproduction ainsi que toute utilisation des œuvres autre que la consultation individuelle et privée sont interdites.</p>
<p><a id="post-18107-page2"></a><strong>Sciences Indusrielles</strong></p>
<p><strong>Systèmes linéaires continus invariants</strong></p>
<p>ANALYSE FREQUENTIELLE</p>
<p><em>S</em>(<em> p</em>) =<em> A</em>(<em> p </em>)<em>e </em>(<em> p</em>)</p>
<ul>
<li><em>A</em>(<em> p</em>)[<em>E</em>(<em> p</em>) &#8211;<em> S </em>¢(<em> p</em>)]</li>
<li><em>A</em>(<em> p</em>)[<em>E</em>(<em> p</em>) &#8211;<em> B </em>(<em> p</em>)<em> S</em>(<em> p</em>)]</li>
</ul>
<table>
<tbody>
<tr>
<td>
<p>[</p>
</td>
<td rowspan="2">
<p>+ <em>A</em>( <em>p</em>) <em>B</em>( <em>p</em>)</p>
</td>
<td>
<p>]</p>
</td>
<td colspan="4" rowspan="2">
<p>= <em>A</em> ( <em>p</em> ) <em>E</em>( <em>p</em>)</p>
</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
</tr>
<tr>
<td>
<p><em>S</em>(<em> p</em>) 1</p>
</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
</tr>
<tr>
<td colspan="4" rowspan="3">
<p>Soit la fonction de transfert en boucle fermée : <em>FTBF</em> = <em>H</em> ( <em>p</em>) =</p>
</td>
<td rowspan="2">
<p><em>S </em>(<em> p</em>)</p>
</td>
<td rowspan="2">&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
</tr>
<tr>
<td rowspan="2">
<p><em>H </em>(<em> p</em>) =</p>
</td>
<td>&nbsp;</td>
<td>
<p><em>A </em>(<em> p</em>)</p>
</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
</tr>
<tr>
<td colspan="2" rowspan="2">
<p><em>E </em>(<em> p</em>)</p>
</td>
<td colspan="3" rowspan="2">
<p>1+ <em>A</em> ( <em>p</em> ) <em>B</em>( <em>p</em>)</p>
</td>
<td>&nbsp;</td>
</tr>
<tr>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
</tr>
<tr>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
</tr>
</tbody>
</table>
<p><em>A </em>(<em> p</em>)</p>
<p>Ce qui peut encore se présenter sous la forme : <em>FTBF</em> = <em>H</em> ( <em>p</em>) =</p>
<p>1+ <em>FTBO</em></p>
<p>Chaque fonction de transfert étant un rapport polynomial, peut donc s’écrire :</p>
<p><em>K</em></p>
<p><em>H </em>(<em> p</em>) =<em> <sub>pa</sub></em></p>
<table>
<tbody>
<tr>
<td colspan="3" rowspan="2">
<p><em>m </em>æ</p>
</td>
<td rowspan="3">
<p><em>p</em></p>
</td>
<td>&nbsp;</td>
<td colspan="2" rowspan="2">
<p>ö</p>
</td>
<td>&nbsp;</td>
<td colspan="3">
<p>ì<em>a</em> :classedusystème</p>
</td>
<td>&nbsp;</td>
</tr>
<tr>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td rowspan="2">
<p>ï</p>
</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
</tr>
<tr>
<td rowspan="3">
<p>Õ</p>
</td>
<td rowspan="3">
<p>ç1-</p>
</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td colspan="2" rowspan="3">
<p>÷</p>
</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
</tr>
<tr>
<td rowspan="2">&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td colspan="3" rowspan="2">
<p><sub>ï</sub><em>K </em>:gainstatiquedusystème</p>
</td>
<td>&nbsp;</td>
</tr>
<tr>
<td rowspan="2">
<p><em>z</em></p>
</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
</tr>
<tr>
<td>&nbsp;</td>
<td>
<p>ç</p>
</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td colspan="2">
<p>÷</p>
</td>
<td>&nbsp;</td>
<td colspan="3" rowspan="2">
<p>ï<em>n</em> :ordredusystème</p>
</td>
<td>&nbsp;</td>
</tr>
<tr>
<td colspan="3">
<p><em>i </em>=1è</p>
</td>
<td>
<p><em>i</em></p>
</td>
<td>&nbsp;</td>
<td colspan="2">
<p>ø</p>
</td>
<td rowspan="3">
<p>où :</p>
</td>
<td>&nbsp;</td>
</tr>
<tr>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td rowspan="2">
<p>í</p>
</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
</tr>
<tr>
<td rowspan="2">
<p><em>n</em></p>
</td>
<td rowspan="2">
<p>æ</p>
</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td colspan="2" rowspan="2">
<p>ö</p>
</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
</tr>
<tr>
<td>&nbsp;</td>
<td colspan="2" rowspan="2">
<p><em>p</em></p>
</td>
<td>&nbsp;</td>
<td>
<p>ï <em>p</em></p>
</td>
<td rowspan="3">
<p><em>j</em></p>
</td>
<td>
<p>:(zérosdudénominateur)polesdelafonctiondetransfert</p>
</td>
<td>&nbsp;</td>
</tr>
<tr>
<td rowspan="2">
<p>Õ</p>
</td>
<td rowspan="2">
<p><sup>ç</sup> 1-</p>
</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td rowspan="2">
<p>÷</p>
</td>
<td>&nbsp;</td>
<td rowspan="2">
<p>ï</p>
</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
</tr>
<tr>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
</tr>
<tr>
<td>&nbsp;</td>
<td>
<p>ç</p>
</td>
<td>&nbsp;</td>
<td>
<p><em>p</em></p>
</td>
<td>&nbsp;</td>
<td colspan="2">
<p>÷</p>
</td>
<td>&nbsp;</td>
<td colspan="3">
<p>ï <em>z</em> :(zérosdunumérateur)zérosdelafonctiondetransfert</p>
</td>
<td>&nbsp;</td>
</tr>
<tr>
<td colspan="3">
<p><em>j </em>= 1<sub>è</sub></p>
</td>
<td>&nbsp;</td>
<td colspan="3">
<p><em>j </em><sub>ø</sub></p>
</td>
<td>&nbsp;</td>
<td>
<p>î <em>i</em></p>
</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
</tr>
</tbody>
</table>
<h3><strong>1.3 Analyse fréquentielle</strong></h3>
<h4><strong>1.3.1 Réponse harmonique</strong></h4>
<p>On étudie la réponse du système soumis à une entrée harmonique, c’est à dire sinusoïdale :</p>
<p><em>e</em>(<em>t</em>) =<em> E </em><sub>0</sub> sin (<em>w t </em>) .</p>
<p>On démontre qu’une des caractéristiques des systèmes linéaires continus invariants est qu’ils présentent une réponse en régime établi (solution particulière de l’équation différentielle) de la même forme que l’entrée, c’est à dire sinusoïdale mais déphasée et d’amplitude</p>
<p>différente : <em>s</em>(<em>t</em>) = <em>S</em><sub>0</sub> sin (<em>w</em> <em>t</em> + <em>j</em>)</p>
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<p></p>
<p><strong>Sciences Indusrielles</strong></p>
<p><strong>Systèmes linéaires continus invariants</strong></p>
<p>ANALYSE FREQUENTIELLE</p>
<p><img loading="lazy" decoding="async" class="alignnone wp-image-18109" title="ANALYSE FREQUENTIELLE" src="https://telechargercours.com/wp-content/uploads/2021/07/word-image-1.png" alt="ANALYSE FREQUENTIELLE" width="1076" height="694"></p>
<p><strong><em>Signal d’entrée :</em></strong></p>
<p>E<sub>0 </sub> <img decoding="async" class="wp-image-18110" src="https://telechargercours.com/wp-content/uploads/2021/07/word-image-2.png"></p>
<p><em>e</em>(<em>t</em>) =<em> E </em><sub>0</sub> sin (<em>w t </em>)</p>
<p><strong><em>Signal de sortie :</em></strong></p>
<p><em>s </em>(<em>t</em>) =<em> S </em><sub>0</sub> sin (<em>w t </em>+<em>j </em>)</p>
<p>S<sub>0 </sub> <img decoding="async" class="wp-image-18111" src="https://telechargercours.com/wp-content/uploads/2021/07/word-image-3.png"></p>
<p><strong><em>Déphasage </em></strong><em>j</em></p>
<p>On étudie alors le rapport d’amplitude sortie sur entrée <em><sup>S</sup></em><sup>0</sup> et le déphasage<em> j </em>en fonction de<em> E</em><sub>0</sub></p>
<p>la pulsation du signal d’entrée <em>w</em> .</p>
<p>On montre que le rapport des amplitudes est égal au module de la fonction de transfert avec <em>p </em>=<em> jw </em>et que le déphasage est égal à l’argument de la fonction de transfert avec<em> p </em>=<em> jw </em>.</p>
<p><strong><em>Conclusion : </em>L’étude fréque ntielle d’un système de fonction de transfert </strong><em>H</em> ( <em>p</em>)<strong> consiste donc à éffectuer </strong><em>p</em> = <em>jw</em><strong> dans l’expression de la fonction de transfert et à étudier le complexe </strong><em>H</em> ( <em>jw</em>)<strong> souvent par son module (alors exprimé en décibels :</strong></p>
<p><img decoding="async" class="wp-image-18112" src="https://telechargercours.com/wp-content/uploads/2021/07/word-image-4.png"> <img decoding="async" class="wp-image-18113" src="https://telechargercours.com/wp-content/uploads/2021/07/word-image-5.png"><em> H </em>(<em> jw </em>) <img decoding="async" class="wp-image-18114" src="https://telechargercours.com/wp-content/uploads/2021/07/word-image-6.png"> <img decoding="async" class="wp-image-18115" src="https://telechargercours.com/wp-content/uploads/2021/07/word-image-7.png"><em><sub>dB </sub></em>= 20<em>Log </em> <img decoding="async" class="wp-image-18116" src="https://telechargercours.com/wp-content/uploads/2021/07/word-image-8.png"> <img decoding="async" class="wp-image-18117" src="https://telechargercours.com/wp-content/uploads/2021/07/word-image-9.png"><em> H </em>(<em> jw</em>) <img decoding="async" class="wp-image-18118" src="https://telechargercours.com/wp-content/uploads/2021/07/word-image-10.png"> <img decoding="async" class="wp-image-18119" src="https://telechargercours.com/wp-content/uploads/2021/07/word-image-11.png"><strong> ) et son argument ( </strong><em>j</em> (<em>w</em> ) = <em>Arg</em> [ <em>H</em> ( <em>jw</em>)]<strong>) ou bien par sa partie réelle et sa partie imaginaire.</strong></p>
<h4><strong>1.3.2 Représentation fonctionnelle</strong></h4>
<p><strong><em>Représentation dans le plan de Black : Diagramme de Black :</em></strong></p>
<p>On représente le gain exprimé en décibels en fonction de la phase en degré pour une même pulsation <em>w</em> . On obtient donc une courbe paramétrée par <em>w</em> .</p>
<p>Cette représentation est très utilisée pour concevoir la correction d’un système. Elle est aussi assez souvent rencontrée dans les sujets de SI en Classe Préparatoire aux Grandes Ecoles.</p>
<p><img decoding="async" class="wp-image-18120" src="https://telechargercours.com/wp-content/uploads/2021/07/word-image-12.png"></p>
<p><img decoding="async" class="wp-image-18121" src="https://telechargercours.com/wp-content/uploads/2021/07/word-image-13.png"> <img decoding="async" class="wp-image-18122" src="https://telechargercours.com/wp-content/uploads/2021/07/word-image-14.png"><em> H </em>(<em> jw </em>) <img decoding="async" class="wp-image-18123" src="https://telechargercours.com/wp-content/uploads/2021/07/word-image-15.png"> <img decoding="async" class="wp-image-18124" src="https://telechargercours.com/wp-content/uploads/2021/07/word-image-16.png"><em><sub>dB </sub></em>=<em> H <sub>dB</sub> </em>(<em>w</em>)</p>
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<p></p>
<p><strong>Sciences Indusrielles</strong></p>
<p><strong>Systèmes linéaires continus invariants</strong></p>
<p>ANALYSE FREQUENTIELLE</p>
<p><img decoding="async" class="wp-image-18125" src="https://telechargercours.com/wp-content/uploads/2021/07/word-image-17.png"></p>
<p><em>H <sub>dB</sub> </em>(<em>w</em><sub>1</sub>)</p>
<p><em>H <sub>dB</sub> </em>(<em>w</em><sub>2</sub> )</p>
<p><em>j </em>(<em>w</em><sub>2</sub> ) <em>j </em>(<em>w</em><sub>1</sub> ) <em>j </em>(<em>w </em>)</p>
<p><strong><em>Représentation dans le plan de Bode : Diagrammes de Bode</em></strong></p>
<p>On représente le gain du système en fonction de la pulsation <em>w</em> sur une courbe avec une échelle logarithmique en abscisse (voir exemple ci-dessous).</p>
<p>On représente la phase du système en fonction de la pulsation <em>w</em> sur une autre courbe avec toujours l’échelle logarithmique pour les pulsations <em>w</em> (voir exemple ci-dessous).</p>
<p><img decoding="async" class="wp-image-18126" src="https://telechargercours.com/wp-content/uploads/2021/07/word-image-18.png"></p>
<p><em>H <sub>dB</sub></em>(<em>w</em>)</p>
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<p><strong>Sciences Indusrielles</strong></p>
<p><strong>Systèmes linéaires continus invariants</strong></p>
<p>ANALYSE FREQUENTIELLE</p>
<p><img decoding="async" class="wp-image-18127" src="https://telechargercours.com/wp-content/uploads/2021/07/word-image-19.png"></p>
<ol>
<li><strong>dB</strong></li>
</ol>
<p><strong>0 dB</strong></p>
<ul>
<li>(<em>w</em> )</li>
</ul>
<p><strong>0°</strong></p>
<p><strong><em>Droite de pente -20dB/décade</em></strong></p>
<p><strong><em>correspondant à lafonction de</em></strong></p>
<p><strong><em>transfert : </em></strong><sup>1</sup><strong><em> ,c’est à dire au</em></strong></p>
<p><em>p</em></p>
<p><strong><em>gain : </em></strong><em>H</em> <em><sub>dB</sub></em> (<em>w</em> ) = -20<em>Logw</em></p>
<p><em>w</em> (<em>rad s</em><sup>-1</sup> )</p>
<p><strong>1 décade</strong> <strong>1 décade</strong></p>
<p><strong><em>Droite horizontale à -90°</em></strong></p>
<p><strong><em>correspondant à la phase de</em></strong></p>
<p><sup>1</sup></p>
<p><strong><em>la fonction de transfert</em></strong> <strong><em>,</em></strong></p>
<p><strong><em>c’est à dire </em></strong><em>j</em> (<em>w</em> ) = -90°</p>
<p><strong>-90°</strong></p>
<p><em>w</em> (<em>rad s</em><sup>-1</sup> )</p>
<p><strong><em>Représentation dans le plan de Nyquist :</em></strong></p>
<p>On trace la courbe paramétrée par la pulsation <em>w</em> avec en abscisse, la partie réelle du complexe <em>H</em> ( <em>jw</em>) et en ordonnée la partie imaginaire du complexe <em>H</em> ( <em>jw</em>) .</p>
<p>Page 5 <strong>Emmanuel FARGES</strong> Ó EduKlub S.A.</p>
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<p></p>
<p><strong>Sciences Indusrielles</strong></p>
<p><strong>Systèmes linéaires continus invariants</strong></p>
<p>ANALYSE FREQUENTIELLE</p>
<p>Ce type d’étude dans le plan dit de Nyquist n’est pas très fréquemment rencontré dans les sujets de concours aux écoles d’ingénieurs.</p>
<p>Im é<sub>ë</sub> <em>H</em> ( <em>jw</em> )ù<sub>û</sub></p>
<p><img decoding="async" class="wp-image-18128" src="https://telechargercours.com/wp-content/uploads/2021/07/word-image-20.png"></p>
<p>Re é<sub>ë</sub> <em>H</em> ( <em>jw</em> )ù<sub>û</sub></p>
<h4><strong>1.3.3 Asymptote dans les diagrammes de Bode</strong></h4>
<p>Rechercher des asymptotes pour les courbes de Bode revient à chercher des équivalents pour</p>
<p><em><sup>K</sup></em></p>
<p>les fonctions de transfert sous la forme : <em>H</em> ( <em>p</em> ) = .</p>
<p><strong><em>En effet les systèmes qui répondent à cette forme de fonction de transfert ( </em></strong><em><sup>n</sup></em> <sup>=</sup> <sup>0</sup><strong><em> : H(p) modélise un gain pur, </em></strong><em><sup>n</sup></em> <sup>Î</sup> <sup>¥</sup> <sup>*</sup> <sup>+</sup><strong><em> : intégrateurs, </em></strong><em><sup>n</sup></em> <sup>Î¥</sup> <sup>*</sup> <sup>&#8211;</sup><strong><em> : dérivateurs), ont des diagrammes de Bode réels qui sont des droites.</em></strong></p>
<p>On pose <em>p</em> = <em>jw</em> et on s’intéresse au module et à l’argument du complexe <em>H</em> ( <em>jw</em>)</p>
<table>
<tbody>
<tr>
<td>&nbsp;</td>
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<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td colspan="6" rowspan="3">
<p><em>K </em>æ 1 ö<em><sup>n</sup></em></p>
</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>
<p>æ</p>
</td>
<td colspan="2">
<p><em>p</em> ö<em><sup>n</sup></em></p>
</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
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<td>
<p><em>p</em></p>
</td>
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</tr>
<tr>
<td>&nbsp;</td>
<td>&nbsp;</td>
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<td rowspan="2">
<p><em>K</em></p>
</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td rowspan="2">
<p><em>K</em></p>
</td>
<td>&nbsp;</td>
<td rowspan="4">
<p>(- <em>j</em> )<em><sup>n</sup></em></p>
</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td colspan="2" rowspan="2">
<p><em>K</em></p>
</td>
<td>&nbsp;</td>
<td rowspan="2">
<p><sub>ç</sub> &#8211; <em>j</em></p>
</td>
<td>&nbsp;</td>
<td rowspan="2">
<p>÷</p>
</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td colspan="2" rowspan="2">
<p><em>K</em></p>
</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td rowspan="2">
<p>&#8211; <em>jn</em></p>
</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
</tr>
<tr>
<td>&nbsp;</td>
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</tr>
<tr>
<td colspan="3" rowspan="2">
<p><em>H </em>(<em> jw</em>) =</p>
</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td colspan="2" rowspan="2">
<p>=</p>
</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td rowspan="2">
<p>ç</p>
</td>
<td>&nbsp;</td>
<td rowspan="2">
<p>÷ <sup>=</sup></p>
</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td rowspan="2">
<p>=</p>
</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td rowspan="2">
<p>ç <em>e</em></p>
</td>
<td colspan="2" rowspan="2">
<p>2 <sub>÷</sub></p>
</td>
<td rowspan="2">
<p>=</p>
</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td rowspan="2">
<p><em>e</em></p>
</td>
<td>&nbsp;</td>
<td rowspan="2">
<p>2</p>
</td>
<td>&nbsp;</td>
</tr>
<tr>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td colspan="2" rowspan="2">
<p><em>n</em></p>
</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td rowspan="2">
<p><em>n</em></p>
</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td rowspan="2">
<p><em>n</em></p>
</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td rowspan="2">
<p><em>n</em></p>
</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
</tr>
<tr>
<td colspan="4" rowspan="2">
<p>(</p>
</td>
<td>&nbsp;</td>
<td rowspan="2">
<p>)</p>
</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td colspan="5">
<p><em>n </em><sub>è</sub><em> j </em><sub>ø</sub></p>
</td>
<td rowspan="2">
<p>( )</p>
</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td rowspan="2">
<p>(</p>
</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td rowspan="2">
<p>ç</p>
</td>
<td colspan="2" rowspan="2">
<p>÷</p>
</td>
<td>&nbsp;</td>
<td rowspan="2">
<p>(</p>
</td>
<td>&nbsp;</td>
<td rowspan="2">
<p>)</p>
</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
</tr>
<tr>
<td rowspan="2">
<p><em>jw</em></p>
</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td colspan="5">
<p>( )</p>
</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td rowspan="2">
<p><em>w</em></p>
</td>
<td colspan="2">
<p>)</p>
</td>
<td>&nbsp;</td>
<td rowspan="2">
<p><em>w</em></p>
</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
</tr>
<tr>
<td>&nbsp;</td>
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<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>
<p><em>w</em></p>
</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>
<p><em>w</em></p>
</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td rowspan="2">
<p>è</p>
</td>
<td colspan="2" rowspan="2">
<p>ø</p>
</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
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<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
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</tr>
<tr>
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</tr>
<tr>
<td colspan="34">
<p>On vient donc de mettre en forme <em>H</em> ( <em>jw</em>) de manière à faire apparaître son</p>
</td>
<td>&nbsp;</td>
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<td>&nbsp;</td>
</tr>
<tr>
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</tr>
<tr>
<td rowspan="3">
<p>module :</p>
</td>
<td>&nbsp;</td>
<td colspan="6" rowspan="3">
<p><em>H </em>(<em> jw</em>)</p>
</td>
<td>&nbsp;</td>
<td colspan="2" rowspan="3">
<p>=</p>
</td>
<td rowspan="2">
<p><em>K</em></p>
</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td colspan="22" rowspan="3">
<p>et son argument, c’est à dire la phase de la fonction de</p>
</td>
<td>&nbsp;</td>
</tr>
<tr>
<td>&nbsp;</td>
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<td>&nbsp;</td>
<td>&nbsp;</td>
</tr>
<tr>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td colspan="2" rowspan="2">
<p><em>w<sup>n</sup></em></p>
</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
</tr>
<tr>
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</tr>
</tbody>
</table>
<p><em>p</em></p>
<p>transfert : <em>j</em> (<em>w</em>) = &#8211;<em>n</em> <sub>2</sub> . Le module en dB est donc : <em>H </em>(<em> jw </em>)<em> <sub>dB</sub> </em>= 20<em>LogK </em>&#8211; 20<em> nLogw </em>,</p>
<p>c’est à dire une <strong>droite de pente</strong> -20<em>n</em> <em><sup>dB</sup></em> <em><sub>décade</sub></em> <strong>passant par le point</strong> (<em>w</em> = 1, 20<em>LogK</em> )</p>
<p><img decoding="async" class="wp-image-18129" src="https://telechargercours.com/wp-content/uploads/2021/07/word-image-21.png"></p>
<p>En effet l’échelle de pulsation en abscisse étant logarithmique, une droite y a pour équation :</p>
<p><em>y </em>=<em> a L og w </em>+<em> b </em>.</p>
<p>Page 6 <strong>Emmanuel FARGES</strong> Ó EduKlub S.A.</p>
<p>Tous droits de l’auteur des œuvres réservés. Sauf autorisation, la reproduction ainsi que toute utilisation des œuvres autre que la consultation individuelle et privée sont interdites.</p>
<p></p>
<p><strong>Sciences Indusrielles</strong></p>
<p><strong>Systèmes linéaires continus invariants</strong></p>
<p>ANALYSE FREQUENTIELLE</p>
<p><strong><em>On obtient donc les diagrammes de Bode suivants pour</em></strong></p>
<p><em>H <sub>dB</sub> </em>(<em>w</em>)</p>
<p><strong>20logK</strong></p>
<p><strong>20LogK-40n (2décade s d’écart)</strong></p>
<p><em>j </em>(<em>w </em>)</p>
<p><strong>0</strong></p>
<ul>
<li><em>n <sup>p</sup></em></li>
</ul>
<p>2</p>
<p><em><sup>K</sup></em></p>
<p><em>H </em>(<em> p </em>) =</p>
<p><strong><em>Droite d’équation :</em></strong></p>
<p>20<em>LogK</em> &#8211; 20<em>nLogw</em> <strong><em>:</em></strong></p>
<p><strong><em>Droite de pente -20n dB /décade passant par le point </em></strong>(<em>w</em> = 1;20<em>LogK</em> )</p>
<p><em>w </em>(<em>rad s</em><sup>-1</sup> )</p>
<p><em>w </em>(<em>rad s</em><sup>-1</sup> )</p>
<h4><strong>1.3.4 Notions importantes</strong></h4>
<h5><strong>1.3.4.1 Pulsation au gain unité</strong></h5>
<p><strong><em>Définition :</em></strong> La pulsation au gain unité est la pulsation pour laquelle le gain du système (c’est à dire le module de sa fonction de transfert) vaut 0 dB. Cette pulsation au</p>
<p>gain unité est notée <em>w<sub>c</sub></em><sub>0</sub></p>
<p><img decoding="async" class="wp-image-18130" src="https://telechargercours.com/wp-content/uploads/2021/07/word-image-22.png"> <img decoding="async" class="wp-image-18131" src="https://telechargercours.com/wp-content/uploads/2021/07/word-image-23.png"><em> H </em>(<em> jw <sub>c</sub> </em><sub>0</sub> ) <img decoding="async" class="wp-image-18132" src="https://telechargercours.com/wp-content/uploads/2021/07/word-image-24.png"> <img decoding="async" class="wp-image-18133" src="https://telechargercours.com/wp-content/uploads/2021/07/word-image-25.png"><em><sub> dB </sub></em>= 0 d B Û <img decoding="async" class="wp-image-18134" src="https://telechargercours.com/wp-content/uploads/2021/07/word-image-26.png"> <img decoding="async" class="wp-image-18135" src="https://telechargercours.com/wp-content/uploads/2021/07/word-image-27.png"><em> H </em>(<em> jw<sub>c</sub> </em><sub>0</sub> ) <img decoding="async" class="wp-image-18136" src="https://telechargercours.com/wp-content/uploads/2021/07/word-image-28.png"> <img decoding="async" class="wp-image-18137" src="https://telechargercours.com/wp-content/uploads/2021/07/word-image-29.png"> =1</p>
<ul>
<li>Graphiquement, <em>w<sub>c</sub></em><sub>0</sub> est la pulsation où le diagramme de Bode en gain coupe les abcisses (0 dB) :</li>
</ul>
<p><strong>Exemple :</strong></p>
<p><img decoding="async" class="wp-image-18138" src="https://telechargercours.com/wp-content/uploads/2021/07/word-image-30.png"></p>
<p><em>H <sub>dB</sub> </em>(<em>w</em>)</p>
<p><strong><em>Pulsation au gain unité ou</em></strong></p>
<p><strong><em>pulsation de coupure à</em></strong></p>
<p><strong><em>0dB : ici </em></strong><em>w<sub>c</sub></em><sub>0</sub> =1000 <em>rad s</em><sup>-1</sup></p>
<p><strong><em>Diagramme de Bode en gain du</em></strong></p>
<p><strong><em>système de fonction de transfert :</em></strong></p>
<p>10</p>
<p><em>H </em>(<em> p</em>) =</p>
<p>1+ 0,01<em>p</em></p>
<p><em>w </em>(<em>rad s</em><sup>-1</sup> )</p>
<p>Page 7 <strong>Emmanuel FARGES</strong> Ó EduKlub S.A.</p>
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<p></p>
<p><strong>Sciences Indusrielles</strong></p>
<p><strong>Systèmes linéaires continus invariants</strong></p>
<p>ANALYSE FREQUENTIELLE</p>
<h5><strong>1.3.4.2 Bande passante</strong></h5>
<p><strong><em>Définition :</em></strong> La bande passante est définie à partir de ou des pulsations de coupure à un</p>
<p>certain nombre de dB notés <em>l</em> , on la note : <em>w<sub>c</sub></em> <sub>&#8211;</sub> <em><sub>ldB</sub></em></p>
<p><img decoding="async" class="wp-image-18139" src="https://telechargercours.com/wp-content/uploads/2021/07/word-image-31.png"></p>
<table>
<tbody>
<tr>
<td>&nbsp;</td>
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<td>
<p>ç<em><sup>l</sup></em></p>
</td>
<td>&nbsp;</td>
<td>
<p>÷</p>
</td>
<td>&nbsp;</td>
</tr>
<tr>
<td>&nbsp;</td>
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<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td rowspan="2">
<p><em>H </em>(0)</p>
</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>
<p>æ</p>
</td>
<td rowspan="2">
<p>20</p>
</td>
<td>
<p>ö</p>
</td>
<td>&nbsp;</td>
</tr>
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<td>
<p>ç</p>
</td>
<td>
<p>÷</p>
</td>
<td>&nbsp;</td>
</tr>
<tr>
<td>&nbsp;</td>
<td rowspan="4">
<p><em>H </em>(0)</p>
</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td rowspan="4">
<p><em>dB </em><sup>&#8211;</sup></p>
</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td rowspan="4">
<p><em>H </em>(<em> jw <sub>c</sub> </em><sub>&#8211;</sub><em><sub>ldB</sub> </em>)</p>
</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td rowspan="4">
<p><em><sub>dB</sub></em> = <em>l</em> dB Û</p>
</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td rowspan="3">
<p>= 10<sup>è</sup></p>
</td>
<td>&nbsp;</td>
<td>
<p>ø</p>
</td>
<td>&nbsp;</td>
</tr>
<tr>
<td>&nbsp;</td>
<td>&nbsp;</td>
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<td rowspan="3">&nbsp;</td>
<td>&nbsp;</td>
<td rowspan="3">&nbsp;</td>
<td>&nbsp;</td>
<td colspan="4" rowspan="4">
<p><em>H </em>(<em> jw<sub>c</sub> </em><sub>&#8211;</sub><em><sub>ldB</sub> </em>)</p>
</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
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<tr>
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<tr>
<td colspan="6">
<p>14243</p>
</td>
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</tbody>
</table>
<ol>
<li><em>LogK</em>
<ul>
<li>Graphiquement, <em>w<sub>c</sub></em> <sub>&#8211;</sub> <em><sub>ldB</sub></em> est la pulsation à partir de laquelle le diagramme de Bode</li>
</ul>
</li>
</ol>
<p>en gain est en-dessous de <img decoding="async" class="wp-image-18140" src="https://telechargercours.com/wp-content/uploads/2021/07/word-image-32.png"> <img decoding="async" class="wp-image-18141" src="https://telechargercours.com/wp-content/uploads/2021/07/word-image-33.png"><em> H </em>(0) <img decoding="async" class="wp-image-18142" src="https://telechargercours.com/wp-content/uploads/2021/07/word-image-34.png"> <img decoding="async" class="wp-image-18143" src="https://telechargercours.com/wp-content/uploads/2021/07/word-image-35.png"><em><sub>dB </sub></em>&#8211;<em> l </em>= 20<em>LogK </em>&#8211;<em> l </em>:</p>
<p><strong>Exemple :</strong></p>
<p><img decoding="async" class="wp-image-18144" src="https://telechargercours.com/wp-content/uploads/2021/07/word-image-36.png"></p>
<p><em>H <sub>dB</sub> </em>(<em>w</em>)</p>
<p><strong><em>20LogK &#8211; 3dB</em></strong></p>
<p><strong><em>Diagramme de Bode en gain du</em></strong></p>
<p><strong><em>système de fonction de transfert :</em></strong></p>
<p>10</p>
<p><em>H </em>(<em> p</em>) =</p>
<p>1+ 0,01<em>p</em></p>
<p><strong><em>20LogK</em></strong></p>
<p><strong><em>Bande passante (ici à -3dB) ou pulsation de coupure à -3dB : ici</em></strong></p>
<p><em>w<sub>c</sub></em> <sub>&#8211;</sub> <sub>3</sub><em><sub>dB</sub></em> = 100 <em>r a d s</em><sup>-1</sup></p>
<p><em>w </em>(<em>rad s</em><sup>-1</sup> )</p>
<h5><strong>1.3.4.3 Résonnance-Surtension</strong></h5>
<p><strong><em>Définition :</em></strong> On parle de phénomène de résonance dès que la courbe de gain du diagramme de Bode présente un extremum (maximum avec tangente nulle : autre que la tangente nulle de l’asymptote horizontale 20Log K pour <em>w</em> ® 0 ).</p>
<ul>
<li>La pulsation pour laquelle existe ce maximum est appelée pulsation de résonance et est notée : <em>w<sub>r</sub></em></li>
<li>La surtension notée <strong>Q</strong> (<strong>Q<sub>dB</sub></strong> <strong>lorsqu’elle est exprimée en dB</strong>) quantifie le pic de résonance. En dB c’est l’écart entre le pic de résonance et le gain statique en dB.</li>
</ul>
<table>
<tbody>
<tr>
<td rowspan="2">
<p><em>Q<sub>dB</sub> </em>=</p>
</td>
<td rowspan="2">
<p><em>H </em>(<em>j<sup>w</sup> <sub>r</sub> </em>)</p>
</td>
<td rowspan="2">
<p>&#8211;</p>
</td>
<td>&nbsp;</td>
<td rowspan="2">&nbsp;</td>
<td>&nbsp;</td>
<td rowspan="2">
<p><em>H </em>(0)</p>
</td>
<td>&nbsp;</td>
<td rowspan="2">&nbsp;</td>
<td>&nbsp;</td>
<td rowspan="3">
<p><em>dB</em></p>
</td>
<td rowspan="2">
<p>=</p>
</td>
<td rowspan="2">
<p><em>H </em>(<em>j<sup>w</sup><sub>r</sub> </em>)</p>
</td>
<td rowspan="2">
<p>&#8211; 20<em>LogK</em></p>
</td>
<td>&nbsp;</td>
</tr>
<tr>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
</tr>
<tr>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td colspan="7" rowspan="2">
<p><em>dB</em></p>
</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td rowspan="2">
<p><em>dB</em></p>
</td>
<td>&nbsp;</td>
</tr>
<tr>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
</tr>
</tbody>
</table>
<p>Page 8 <strong>Emmanuel FARGES</strong> Ó EduKlub S.A.</p>
<p>Tous droits de l’auteur des œuvres réservés. Sauf autorisation, la reproduction ainsi que toute utilisation des œuvres autre que la consultation individuelle et privée sont interdites.</p>
<p></p>
<p><strong>Sciences Indusrielles</strong></p>
<p><strong>Systèmes linéaires continus invariants</strong></p>
<p>ANALYSE FREQUENTIELLE</p>
<p><strong>Exemple :</strong></p>
<p><img decoding="async" class="wp-image-18145" src="https://telechargercours.com/wp-content/uploads/2021/07/word-image-37.png"></p>
<p><em>H <sub>dB</sub> </em>(<em>w</em>)</p>
<table>
<tbody>
<tr>
<td rowspan="2">
<p><strong><em>Pic de résonance</em></strong></p>
</td>
<td>
<p><strong><em>Coefficient de</em></strong></p>
</td>
<td>&nbsp;</td>
</tr>
<tr>
<td rowspan="2">
<p><strong><em>surtension Q<sub>dB</sub></em></strong></p>
</td>
<td>&nbsp;</td>
</tr>
<tr>
<td>&nbsp;</td>
<td>&nbsp;</td>
</tr>
</tbody>
</table>
<p><strong><em>20LogK</em></strong></p>
<p><em>w </em>(<em>rad s</em><sup>-1</sup> )</p>
<p><strong><em>Pulsation de résonance </em></strong><em>w<sub>r</sub></em></p>
<h2><strong>2</strong> <strong>Système du premier ordre.</strong></h2>
<h3><strong>2.1</strong> <strong>Définition d’un système du premier ordre</strong></h3>
<p><strong>Un système du premier ordre a son comportement régi par une équation différentielle du premier ordre de la forme :</strong></p>
<table>
<tbody>
<tr>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>
<p>ì</p>
</td>
<td>
<p><em>e </em>=<em> e</em>(<em>t </em>) entréedusystème</p>
</td>
<td>&nbsp;</td>
</tr>
<tr>
<td>&nbsp;</td>
<td rowspan="2">
<p><em>ds</em></p>
</td>
<td>&nbsp;</td>
<td>
<p>ï</p>
</td>
<td rowspan="2">
<p><em>s </em>=<em> s </em>(<em>t </em>) réponsedusystèmeàl&rsquo;entréee(t)</p>
</td>
<td>&nbsp;</td>
</tr>
<tr>
<td rowspan="3">
<p><em>s </em>+<em>t</em></p>
</td>
<td rowspan="3">
<p>= <em>Ke</em></p>
</td>
<td>
<p>ï</p>
</td>
<td>&nbsp;</td>
</tr>
<tr>
<td>&nbsp;</td>
<td rowspan="2">
<p>í</p>
</td>
<td rowspan="3">
<p>K:gainstatique(&gt;0)</p>
</td>
<td>&nbsp;</td>
</tr>
<tr>
<td rowspan="2">
<p><em>dt</em></p>
</td>
<td>&nbsp;</td>
</tr>
<tr>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>
<p>ï</p>
</td>
<td>&nbsp;</td>
</tr>
</tbody>
</table>
<p><sup>ï</sup><sub>î</sub><em>t</em> :constantedetemps(&gt;0,homogèneà untemps(seconde (s))</p>
<h3><strong>2.2</strong> <strong>Fonction de transfert globale d’un premier ordre</strong></h3>
<p>On applique la transformée de Laplace à l’ensemble de l’équation différentielle ci-dessus, avec des conditions initiales nulles :</p>
<p><strong><em>Rappel :</em></strong> La transformée de Laplace d’une fonction f(t) est une fonction de p, notée par</p>
<p>convention avec la lettre majuscule de la fonction du temps transformée : F(p)</p>
<p>Donc : <em>s</em> +<em>t</em> <em><sup>ds</sup></em> = <em>Ke</em> transforméedeLaplace <em>S </em>+<em> t pS </em>=<em> KE</em></p>
<p><em><sub>dt</sub></em> uuuuuuuuuuuuuuuuuuuuuuuuuuru</p>
<p>On peut alors présenter le rapport de la sortie S(p) sur l’entrée E(p), c’est à dire la fonction de</p>
<table>
<tbody>
<tr>
<td rowspan="2">
<p>transfert globale du système :</p>
</td>
<td rowspan="2">
<p><em>H </em>(<em> p</em>) =</p>
</td>
<td>
<p><em>S</em>(<em> p </em>)</p>
</td>
<td rowspan="2">
<p>=</p>
</td>
<td>
<p><em>K</em></p>
</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
</tr>
<tr>
<td>&nbsp;</td>
<td colspan="2" rowspan="2">
<p>1+<em>t</em> <em>p</em></p>
</td>
<td>&nbsp;</td>
</tr>
<tr>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td colspan="2">
<p><em>E</em>(<em> p</em>)</p>
</td>
<td>&nbsp;</td>
</tr>
</tbody>
</table>
<p>Page 9 <strong>Emmanuel FARGES</strong> Ó EduKlub S.A.</p>
<p>Tous droits de l’auteur des œuvres réservés. Sauf autorisation, la reproduction ainsi que toute utilisation des œuvres autre que la consultation individuelle et privée sont interdites.</p>
<p></p>
<p><strong>Sciences Indusrielles</strong></p>
<p><strong>Systèmes linéaires continus invariants</strong></p>
<p>ANALYSE FREQUENTIELLE</p>
<p><strong><em>On note :</em></strong></p>
<ul>
<li>Le gain statique vaut : K=H(0)</li>
<li>Pour identifier les caractéristiques d’un système du premier ordre (c’est à dire <em>K </em>et<em> t </em>), on veillera bien à présenter la fonction de transfert globale H(p) avec le</li>
</ul>
<p>coefficient en p<sup>0</sup> du polynôme au dénominateur égal à 1. Ainsi le numérateur peut être identifié au gain statique K et le coefficient en p<sup>1</sup> du polynôme au dénominateur peut être identifié à la constante de temps <em>t</em></p>
<h3><strong>2.3</strong> <strong>Gain et phase réels d’un premier ordre</strong></h3>
<table>
<tbody>
<tr>
<td rowspan="2">
<p><em>H </em>(<em> jw</em>) =</p>
</td>
<td>
<p><em>K</em></p>
</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
</tr>
<tr>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
</tr>
<tr>
<td colspan="2">
<p>1+ <em>jtw</em></p>
</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
</tr>
<tr>
<td colspan="2">
<p><strong><em>Calcul du gain :</em></strong></p>
</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
</tr>
<tr>
<td rowspan="2">
<p><em>H </em>(<em> jw</em>) =</p>
</td>
<td>
<p><em>K</em></p>
</td>
<td rowspan="2">
<p>soit en dB : <em>H</em> ( <em>jw</em> ) <em><sub>dB</sub></em> = 20<em>Log H</em> ( <em>jw</em>)</p>
</td>
<td>&nbsp;</td>
</tr>
<tr>
<td>
<p>1 +<em>t</em> <sup>2</sup> <em>w</em><sup>2</sup></p>
</td>
<td>&nbsp;</td>
</tr>
</tbody>
</table>
<p><img decoding="async" class="wp-image-18146" src="https://telechargercours.com/wp-content/uploads/2021/07/word-image-38.png"></p>
<p><img decoding="async" class="wp-image-18147" src="https://telechargercours.com/wp-content/uploads/2021/07/word-image-39.png"> <img decoding="async" class="wp-image-18148" src="https://telechargercours.com/wp-content/uploads/2021/07/word-image-40.png"><em> H </em>(<em> jw</em>) <img decoding="async" class="wp-image-18149" src="https://telechargercours.com/wp-content/uploads/2021/07/word-image-41.png"> <img decoding="async" class="wp-image-18150" src="https://telechargercours.com/wp-content/uploads/2021/07/word-image-42.png"><em><sub>dB </sub></em>= 20<em>LogK </em>&#8211; 20<em> Log </em> <img decoding="async" class="wp-image-18151" src="https://telechargercours.com/wp-content/uploads/2021/07/word-image-43.png"> <img decoding="async" class="wp-image-18152" src="https://telechargercours.com/wp-content/uploads/2021/07/word-image-44.png"> 1 +<em>t</em> <sup>2</sup> <em>w</em><sup>2</sup></p>
<p><img decoding="async" class="wp-image-18153" src="https://telechargercours.com/wp-content/uploads/2021/07/word-image-45.png"></p>
<p><strong><em>Calcul de la phase :</em></strong></p>
<ul>
<li>(<em>w </em>) =<em> Arg </em>[<em> H </em>(<em> jw</em>)]</li>
</ul>
<table>
<tbody>
<tr>
<td rowspan="2">
<p><em>H </em>(<em> jw</em>) =</p>
</td>
<td>&nbsp;</td>
<td>
<p><em>K</em></p>
</td>
<td rowspan="2">
<p>=</p>
</td>
<td>&nbsp;</td>
<td colspan="3">
<p><em>K</em>(1-<em> jtw</em>)</p>
</td>
<td>&nbsp;</td>
<td rowspan="2">
<p>=</p>
</td>
<td>&nbsp;</td>
<td>
<p><em>K</em></p>
</td>
<td rowspan="2">
<p>1 &#8211; <em>jtw</em></p>
</td>
<td rowspan="3">
<p>)</p>
</td>
<td rowspan="2">
<p>, donc :</p>
</td>
<td>&nbsp;</td>
</tr>
<tr>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td rowspan="2">
<p>2 2</p>
</td>
<td>&nbsp;</td>
</tr>
<tr>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td rowspan="2">
<p>(</p>
</td>
<td>&nbsp;</td>
<td rowspan="2">
<p>)(</p>
</td>
<td>&nbsp;</td>
<td rowspan="2">
<p>)</p>
</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>
<p>(</p>
</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
</tr>
<tr>
<td>&nbsp;</td>
<td colspan="2" rowspan="2">
<p>1 + <em>jtw</em></p>
</td>
<td>&nbsp;</td>
<td rowspan="2">
<p>+<em> jtw</em></p>
</td>
<td rowspan="2">
<p>&#8211;<em> jtw</em></p>
</td>
<td colspan="2" rowspan="2">
<p>1</p>
</td>
<td rowspan="2">
<p>+<em>t w</em></p>
</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
</tr>
<tr>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>
<p>1</p>
</td>
<td>
<p>1</p>
</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
</tr>
</tbody>
</table>
<ul>
<li>(<em>w </em>) =<em> Arg </em>[<em> H</em>(<em> jw</em>) ] =<em> Arg </em>[1-<em> jtw</em>]</li>
</ul>
<table>
<tbody>
<tr>
<td>&nbsp;</td>
<td>
<p>(</p>
</td>
<td colspan="2" rowspan="2">
<p>&#8211;<em> jtw</em></p>
</td>
<td colspan="2">
<p>)</p>
</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td colspan="3">
<p>(</p>
</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>
<p>)</p>
</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td colspan="5">
<p>(</p>
</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>
<p>)</p>
</td>
<td>&nbsp;</td>
</tr>
<tr>
<td rowspan="3">
<p>tan <em>j</em> =</p>
</td>
<td>
<p>Im 1</p>
</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td colspan="2" rowspan="3">
<p>;cos<em>j</em> =</p>
</td>
<td colspan="6">
<p>Re 1- <em>jtw</em></p>
</td>
<td>&nbsp;</td>
<td colspan="3" rowspan="3">
<p>;sin<em>j</em> =</p>
</td>
<td colspan="11">
<p>Im 1- <em>jtw</em></p>
</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
</tr>
<tr>
<td rowspan="2">
<p>Re 1</p>
</td>
<td colspan="2" rowspan="2">
<p>&#8211;<em> jtw</em></p>
</td>
<td colspan="2" rowspan="3">
<p>)</p>
</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td colspan="2" rowspan="2">
<p>1 &#8211; <em>jtw</em></p>
</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td rowspan="2">&nbsp;</td>
<td rowspan="2">&nbsp;</td>
<td>&nbsp;</td>
<td rowspan="2">&nbsp;</td>
<td>&nbsp;</td>
<td colspan="3" rowspan="2">
<p>1- <em>jtw</em></p>
</td>
<td>&nbsp;</td>
<td rowspan="2">&nbsp;</td>
<td>&nbsp;</td>
<td rowspan="2">&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
</tr>
<tr>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
</tr>
<tr>
<td>&nbsp;</td>
<td>
<p>(</p>
</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
</tr>
<tr>
<td rowspan="2">
<p>Donc :</p>
</td>
<td colspan="2">&nbsp;</td>
<td colspan="2">&nbsp;</td>
<td colspan="2">&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td colspan="2">&nbsp;</td>
<td>&nbsp;</td>
<td colspan="4">&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td colspan="5">&nbsp;</td>
<td>&nbsp;</td>
<td colspan="5" rowspan="2">&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
</tr>
<tr>
<td colspan="2">
<p>tan <em>j</em> =</p>
</td>
<td colspan="2">
<p>&#8211;<em>tw</em></p>
</td>
<td colspan="2">
<p>;cos<em>j</em> =</p>
</td>
<td>&nbsp;</td>
<td colspan="3">
<p>1</p>
</td>
<td>&nbsp;</td>
<td colspan="4">
<p>;sin<em>j</em> =</p>
</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td colspan="6">
<p>&#8211;<em>tw</em></p>
</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
</tr>
<tr>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td colspan="2">
<p>1</p>
</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td colspan="4">
<p>1+ <em>t</em> <sup>2</sup><em>w</em> <sup>2</sup></p>
</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td colspan="8">
<p>1+<em>t</em> <sup>2</sup><em>w</em><sup>2</sup></p>
</td>
<td colspan="5">&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
</tr>
</tbody>
</table>
<p><img decoding="async" class="wp-image-18155" src="https://telechargercours.com/wp-content/uploads/2021/07/word-image-47.png"> <img decoding="async" class="wp-image-18157" src="https://telechargercours.com/wp-content/uploads/2021/07/word-image-49.png"></p>
<ul>
<li>(<em>w</em>) = &#8211; arctan [<em>tw</em>]</li>
</ul>
<h3><strong>2.4</strong> <strong>Asymptotes des diagrammes de Bode</strong></h3>
<p>Deux méthodes sont possibles pour déterminer les asymptotes des diagrammes de Bode : Soit on détermine les équivalents en 0 et en +¥ du gain réel et de la phase réelle, soit on cherche des équivalents directement sur la fonction de transfert.</p>
<p><strong><em>1<sup>ère</sup> méthode : équivalents à partir des fonctions de gain et de phase réels.</em></strong></p>
<p><strong><em>Asymptote sur le diagramme de gain :</em></strong></p>
<p><img decoding="async" class="wp-image-18158" src="https://telechargercours.com/wp-content/uploads/2021/07/word-image-50.png"> <img decoding="async" class="wp-image-18160" src="https://telechargercours.com/wp-content/uploads/2021/07/word-image-52.png"> <img decoding="async" class="wp-image-18162" src="https://telechargercours.com/wp-content/uploads/2021/07/word-image-54.png"><em> H </em>(<em> jw</em>) <img decoding="async" class="wp-image-18164" src="https://telechargercours.com/wp-content/uploads/2021/07/word-image-56.png"> <img decoding="async" class="wp-image-18167" src="https://telechargercours.com/wp-content/uploads/2021/07/word-image-59.png"><em><sub>dB </sub></em>= 20<em>LogK </em>&#8211; 20<em> Log </em> <img decoding="async" class="wp-image-18168" src="https://telechargercours.com/wp-content/uploads/2021/07/word-image-60.png"> <img decoding="async" class="wp-image-18170" src="https://telechargercours.com/wp-content/uploads/2021/07/word-image-62.png"> 1 +<em>t</em> <sup>2</sup> <em>w</em><sup>2</sup> est la fonction du ga in (en dB) réelle</p>
<p><img decoding="async" class="wp-image-18171" src="https://telechargercours.com/wp-content/uploads/2021/07/word-image-63.png"></p>
<ul>
<li><em>équivalent lorsque w </em>® 0<em> : on peut négliger le terme en t </em><sup>2</sup><em>w</em><sup>2</sup> devant1.Donc :</li>
</ul>
<table>
<tbody>
<tr>
<td>&nbsp;</td>
<td rowspan="2">
<p><em>H </em>(<em> jw</em>)</p>
</td>
<td>&nbsp;</td>
<td rowspan="2">
<p>; 20<em>LogK</em></p>
</td>
<td colspan="2" rowspan="2">
<p>, c’est une droite horizontale d’ordonnée 20<em>LogK</em> dB.</p>
</td>
<td>&nbsp;</td>
</tr>
<tr>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
</tr>
<tr>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>
<p>{</p>
</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
</tr>
<tr>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>
<p><em><sup>dB</sup></em> <em>w</em> ® 0</p>
</td>
<td>&nbsp;</td>
<td rowspan="2">
<p>Ó EduKlub S.A.</p>
</td>
<td>&nbsp;</td>
</tr>
<tr>
<td colspan="2">
<p>Page 10</p>
</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>
<p><strong>Emmanuel FARGES</strong></p>
</td>
<td>&nbsp;</td>
</tr>
</tbody>
</table>
<p>Tous droits de l’auteur des œuvres réservés. Sauf autorisation, la reproduction ainsi que toute utilisation des œuvres autre que la consultation individuelle et privée sont interdites.</p>
<p></p>
<p><strong>Sciences Indusrielles</strong></p>
<p><strong>Systèmes linéaires continus invariants</strong></p>
<p>ANALYSE FREQUENTIELLE</p>
<ul>
<li><em>équivalent lorsque w </em>® ¥<em> : on peut négliger le terme en </em>1devant<em> t </em><sup>2</sup><em>w </em><sup>2</sup> .Donc :</li>
</ul>
<table>
<tbody>
<tr>
<td>
<p><em>H </em>(<em> jw</em>)</p>
</td>
<td>
<p>; 20<em>LogK</em> &#8211; 20<em>Logtw</em> , soit, mis en forme différemment pour faire</p>
</td>
</tr>
<tr>
<td>&nbsp;</td>
<td>
<p>{</p>
</td>
</tr>
<tr>
<td>&nbsp;</td>
<td>
<p><em><sup>dB </sup>w </em>® ¥</p>
</td>
</tr>
</tbody>
</table>
<p>clairement apparaître la pente de la droite :</p>
<table>
<tbody>
<tr>
<td>&nbsp;</td>
<td rowspan="2">
<p><em>H </em>(<em> jw</em>)</p>
</td>
<td>&nbsp;</td>
<td colspan="3" rowspan="2">
<p>; 20<em>LogK</em> &#8211; 20 <em>Logt</em></p>
</td>
<td rowspan="2">
<p>-20</p>
</td>
<td rowspan="2">
<p><em>Logw</em></p>
</td>
<td rowspan="2">
<p>c’est une</p>
</td>
<td>&nbsp;</td>
</tr>
<tr>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
</tr>
<tr>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>
<p>{</p>
</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>
<p>{</p>
</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
</tr>
<tr>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>
<p><em><sup>dB </sup>w </em>® ¥</p>
</td>
<td>&nbsp;</td>
<td colspan="3">
<p>pentedeladroiteasymp to te</p>
</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
</tr>
<tr>
<td colspan="4">
<p>droite de pente -20 <em><sup>dB</sup></em></p>
</td>
<td rowspan="2">
<p><em>décade</em></p>
</td>
<td>
<p>passant par le point</p>
</td>
<td>
<p>(<em>w</em> = 1;</p>
</td>
<td colspan="2">
<p>20<em>LogK</em> &#8211; 20<em>Logt</em> ).</p>
</td>
<td>&nbsp;</td>
</tr>
<tr>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
</tr>
</tbody>
</table>
<p><img decoding="async" class="wp-image-18173" src="https://telechargercours.com/wp-content/uploads/2021/07/word-image-65.png"></p>
<p><strong><em>Intersection entre les deux asymptotes :</em></strong></p>
<p>On cherche la pulsation dite <strong><em>pulsation propre du système</em></strong> notée <em>w</em> <em><sub>p</sub></em> pour laquelle les deux asymptotes ont des valeurs égales. L’équation permettant de déterminer <em>w</em> <em><sub>p</sub></em> est donc :</p>
<p>20<em>LogK</em> = 20<em>LogK</em> &#8211; 20<em>Logt</em> &#8211; 20<em>Logw</em> <em><sub>p</sub></em> .</p>
<p>1</p>
<p>Soit 20<em>Logw</em> <em><sub>p</sub></em> = -20<em>Logt</em> ; d’où :<em>w</em> <em><sub>p</sub></em> = <em><sub>t</sub></em></p>
<p><strong><em>Explication sur la pente de la droite et la notion de décade :</em></strong></p>
<p>Un décade est une plage fréquentielle contenue entre une pulsation <em>w</em><sub>0</sub> et 10 fois sa valeur : 10<em>w</em><sub>0</sub> . Donc les deux points qui correspondent à ces pulsations sur l’asymptote sont : (<em>w</em><sub>0</sub> ; 20<em>LogK</em> &#8211; 20<em>Logt</em> &#8211; 20<em>Logw</em><sub>0</sub> )et</p>
<table>
<tbody>
<tr>
<td>
<p>æ</p>
</td>
<td>&nbsp;</td>
<td>
<p>ö</p>
</td>
<td>&nbsp;</td>
</tr>
<tr>
<td>
<p>ç</p>
</td>
<td rowspan="2">
<p>-20<em>Log</em> (10<em>w</em><sub>0</sub> )</p>
</td>
<td>
<p>÷</p>
</td>
<td>&nbsp;</td>
</tr>
<tr>
<td>
<p>ç 10<em>w</em><sub>0</sub> ;20<em>LogK</em> &#8211; 20<em>Logt</em></p>
</td>
<td>
<p>÷. Ils y a donc un écart de -20 dB entre</p>
</td>
<td>&nbsp;</td>
</tr>
<tr>
<td>
<p>ç</p>
</td>
<td>
<p>1442443</p>
</td>
<td>
<p>÷</p>
</td>
<td>&nbsp;</td>
</tr>
<tr>
<td>
<p>ç</p>
</td>
<td>
<p>= &#8211; 20<em>Logw</em> &#8211; 20</p>
</td>
<td>
<p>÷</p>
</td>
<td>&nbsp;</td>
</tr>
<tr>
<td>
<p>è</p>
</td>
<td>
<p>0</p>
</td>
<td>
<p>ø</p>
</td>
<td>&nbsp;</td>
</tr>
</tbody>
</table>
<p><img decoding="async" class="wp-image-18175" src="https://telechargercours.com/wp-content/uploads/2021/07/word-image-67.png"></p>
<p>les deux points séparées d’une décade, d’où la pente de -20 <em><sup>dB</sup></em> <em><sub>décade</sub></em> .</p>
<p><strong><em>Asymptote sur le diagramme de phase :</em></strong></p>
<table>
<tbody>
<tr>
<td>
<p>La fonction de phase réelle est définie par</p>
</td>
<td>
<p>tan <em>j</em> =</p>
</td>
<td>
<p>&#8211;<em>tw</em></p>
</td>
<td>
<p>;cos<em>j</em> =</p>
</td>
<td colspan="2">
<p>1</p>
</td>
<td>&nbsp;</td>
<td>
<p>;sin<em>j</em> =</p>
</td>
<td>&nbsp;</td>
<td colspan="2">
<p>&#8211;<em>tw</em></p>
</td>
<td>&nbsp;</td>
</tr>
<tr>
<td>&nbsp;</td>
<td colspan="2">
<p>1</p>
</td>
<td>&nbsp;</td>
<td colspan="2">
<p>1+ <em>t</em> <sup>2</sup><em>w</em> <sup>2</sup></p>
</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td colspan="4">
<p>1+<em>t</em> <sup>2</sup><em>w</em><sup>2</sup></p>
</td>
</tr>
</tbody>
</table>
<p><img decoding="async" class="wp-image-18177" src="https://telechargercours.com/wp-content/uploads/2021/07/word-image-69.png"> <img decoding="async" class="wp-image-18178" src="https://telechargercours.com/wp-content/uploads/2021/07/word-image-70.png"></p>
<ul>
<li>(<em>w</em>) = &#8211; arctan [<em>tw</em>]</li>
<li><em>équivalent lorsque w </em>® 0<em> : j </em>(<em>w </em>) » 0°.Donc, on a une asymptote horizontale à 0°</li>
</ul>
<table>
<tbody>
<tr>
<td rowspan="3">
<p>Ø</p>
</td>
<td rowspan="3">
<p><em>équivalent lorsque w </em>® ¥</p>
</td>
<td>
<p>(</p>
</td>
<td>
<p>)</p>
</td>
<td>&nbsp;</td>
<td rowspan="2">
<p><em>p</em></p>
</td>
<td>
<p>(</p>
</td>
<td>&nbsp;</td>
<td>
<p>)</p>
</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
</tr>
<tr>
<td rowspan="2">
<p><em>: j w</em></p>
</td>
<td>&nbsp;</td>
<td rowspan="2">
<p>» &#8211;</p>
</td>
<td>&nbsp;</td>
<td rowspan="2">
<p>&#8211; 90°</p>
</td>
<td>&nbsp;</td>
<td rowspan="2">
<p>.Donc, on a une asymptote</p>
</td>
<td>&nbsp;</td>
</tr>
<tr>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
</tr>
<tr>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td colspan="2">
<p>2</p>
</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
</tr>
</tbody>
</table>
<p>horizontale à -90°</p>
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<p>Tous droits de l’auteur des œuvres réservés. Sauf autorisation, la reproduction ainsi que toute utilisation des œuvres autre que la consultation individuelle et privée sont interdites.</p>
<p></p>
<p><strong>Sciences Indusrielles</strong></p>
<p><strong>Systèmes linéaires continus invariants</strong></p>
<p>ANALYSE FREQUENTIELLE</p>
<table>
<tbody>
<tr>
<td rowspan="2">
<p>Comme</p>
</td>
<td>
<p>æ</p>
</td>
<td>
<p>ö</p>
</td>
<td>
<p>æ</p>
</td>
<td>
<p>1</p>
</td>
<td>
<p>ö</p>
</td>
<td rowspan="2">
<p>= -arctan (1) = &#8211;</p>
</td>
<td>
<p><em>p</em></p>
</td>
<td rowspan="2">
<p>(- 45°)</p>
</td>
<td rowspan="2">
<p>, on relie les deux asymptotes qui sont</p>
</td>
<td>&nbsp;</td>
</tr>
<tr>
<td>
<p><em>j</em> ç <em>w</em></p>
</td>
<td>
<p>÷</p>
</td>
<td>
<p>= <em>j</em> <sub>ç</sub></p>
</td>
<td>&nbsp;</td>
<td>
<p>÷</p>
</td>
<td rowspan="2">
<p>4</p>
</td>
<td>&nbsp;</td>
</tr>
<tr>
<td>&nbsp;</td>
<td>
<p>è</p>
</td>
<td>
<p><em>p </em><sub>ø</sub></p>
</td>
<td colspan="2">
<p>è <em>t</em></p>
</td>
<td>
<p>ø</p>
</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
</tr>
</tbody>
</table>
<p>normalement parallèles (puisque toutes les deux horizontales) au niveau de la pulsation propre</p>
<p>1</p>
<p>du système <em>w</em> <em><sub>p</sub></em> = <em><sub>t</sub></em> .</p>
<p><strong><em>2<sup>ème</sup> méthode : équivalents à partir de la fonction de transfert :</em></strong></p>
<p><em>A</em></p>
<p>On cherche des équivalents de la fonction de transfert de la forme</p>
<p><em>p<sup>n</sup></em></p>
<p><strong><em>Fonction de transfert équivalente pour </em></strong><em>w</em> ® 0<strong><em> , c’est à dire pour </em></strong><em>p</em> ® 0<strong><em> (puisque </em></strong><em>p</em> = <em>jw</em><strong><em> )</em></strong></p>
<table>
<tbody>
<tr>
<td rowspan="2">
<p><em>H </em>(<em> p </em>) =</p>
</td>
<td>
<p><em>K</em></p>
</td>
<td rowspan="2">
<p>équivalentà</p>
</td>
<td>
<p><em>K</em></p>
</td>
<td rowspan="2">
<p>soit</p>
</td>
<td colspan="3">
<p><em>K</em></p>
</td>
<td>&nbsp;</td>
</tr>
<tr>
<td>&nbsp;</td>
<td rowspan="2">
<p>1</p>
</td>
<td>&nbsp;</td>
<td rowspan="2">
<p>0</p>
</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
</tr>
<tr>
<td colspan="2">
<p>1 +<em>t</em> <em>p</em></p>
</td>
<td>
<p>14243</p>
</td>
<td>&nbsp;</td>
<td rowspan="2">
<p><em>p</em></p>
</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
</tr>
<tr>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td rowspan="2">
<p>quand p ® 0</p>
</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
</tr>
<tr>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
</tr>
<tr>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
</tr>
</tbody>
</table>
<ul>
<li>On retrouve donc l’asymptote horizontale pour <em>w</em> ® 0 à 20<em>LogK</em> pour la courbe de gain en dB (d’après les résultats démontrés paragraphe 1.3.3)</li>
<li>On retrouve donc l’asymptote horizontale pour <em>w</em> ® 0 à 0° pour la courbe de phase (d’après les résultats démontrés paragraphe 1.3.3)</li>
</ul>
<p><strong><em>Fonction de transfert équivalente pour </em></strong><em>w</em> ® ¥<strong><em> , c’est à dire pour </em></strong><em>p</em> ® ¥<strong><em> (puisque </em></strong><em>p</em> = <em>jw</em><strong><em> )</em></strong></p>
<table>
<tbody>
<tr>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td rowspan="2">
<p><em>K</em></p>
</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td rowspan="2">
<p><em>K</em></p>
</td>
<td>&nbsp;</td>
<td colspan="2">
<p><em>K</em></p>
</td>
<td>&nbsp;</td>
</tr>
<tr>
<td rowspan="2">
<p><em>H </em>(<em> p</em>) =</p>
</td>
<td>&nbsp;</td>
<td rowspan="2">
<p>équivalentà</p>
</td>
<td>&nbsp;</td>
<td rowspan="2">
<p>soit</p>
</td>
<td>
<p><em>t</em></p>
</td>
<td rowspan="2">&nbsp;</td>
<td>&nbsp;</td>
</tr>
<tr>
<td>&nbsp;</td>
<td rowspan="2">
<p>+<em>t p</em></p>
</td>
<td colspan="2" rowspan="2">
<p><em>t p</em></p>
</td>
<td rowspan="2">
<p>1</p>
</td>
<td>&nbsp;</td>
</tr>
<tr>
<td colspan="2">
<p>1</p>
</td>
<td>
<p>14243</p>
</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
</tr>
<tr>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td rowspan="2">
<p>quand p ® ¥</p>
</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td colspan="2">
<p><em>p</em></p>
</td>
<td>&nbsp;</td>
</tr>
<tr>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
</tr>
<tr>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
</tr>
</tbody>
</table>
<p><img decoding="async" class="wp-image-18180" src="https://telechargercours.com/wp-content/uploads/2021/07/word-image-72.png"></p>
<ul>
<li>On retrouve donc l’asymptote de pente -20dB/décade pour <em>w</em> ® ¥ à</li>
</ul>
<p>20<em>Log</em> <em><sup>K</sup><sub>t</sub></em> &#8211; (20 ´ 1)<em>Logw</em> = 20<em>LogK</em> &#8211; 20<em>Logt</em> &#8211; 20<em>Logw</em> pour la courbe de gain en dB (d’après les résultats démontrés paragraphe 1.3.3)</p>
<p>ü On retrouve donc l’asymptote horizontale pour <em>w</em> ® ¥ à &#8211; <em><sup>p</sup></em> (-90°) pour la courbe 2</p>
<p>de phase (d’après les résultats démontrés paragraphe 1.3.3)</p>
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<p></p>
<p><strong>Sciences Indusrielles</strong></p>
<p><strong>Systèmes linéaires continus invariants</strong></p>
<p>ANALYSE FREQUENTIELLE</p>
<p><strong><em>Graphiquement cela donne les diagrammes asymptotiques schématiques suivants :</em></strong></p>
<p><img decoding="async" class="wp-image-18181" src="https://telechargercours.com/wp-content/uploads/2021/07/word-image-73.png"></p>
<table>
<tbody>
<tr>
<td>
<p>20<em>LogK</em></p>
</td>
<td rowspan="2">
<p><em><sup>H</sup> dB</em></p>
</td>
<td rowspan="2">
<p>(<em>w</em>)</p>
</td>
<td>&nbsp;</td>
</tr>
<tr>
<td>&nbsp;</td>
<td>&nbsp;</td>
</tr>
</tbody>
</table>
<p>Pente : -20 <em>dB</em> / <em>décade</em></p>
<p><em>w</em></p>
<table>
<tbody>
<tr>
<td rowspan="2">
<p><strong>0°</strong></p>
</td>
<td>&nbsp;</td>
<td rowspan="2">
<p><em>w <sub>p</sub> </em>=</p>
</td>
<td>
<p>1</p>
</td>
<td>&nbsp;</td>
</tr>
<tr>
<td>
<p><em>j w</em></p>
</td>
<td>
<p><em>t</em></p>
</td>
<td>&nbsp;</td>
</tr>
<tr>
<td>&nbsp;</td>
<td>
<p>(</p>
</td>
<td>
<p>)</p>
</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
</tr>
</tbody>
</table>
<p><em>w</em></p>
<p><strong>-90°</strong></p>
<h3><strong>2.5</strong> <strong>Diagrammes de Bode d’un premier ordre</strong></h3>
<p><img decoding="async" class="wp-image-18184" src="https://telechargercours.com/wp-content/uploads/2021/07/word-image-76.png"></p>
<p><strong><em>Asymptote horizontale</em></strong></p>
<p><strong><em>à 20LogK</em></strong></p>
<p><em>H <sub>dB</sub> </em>(<em>w</em>)</p>
<p><strong><em>20 &#8211; 3= 17dB</em></strong></p>
<p><strong><em>Courbe de Gain en</em></strong></p>
<p><strong><em>dB réelle.</em></strong></p>
<p><strong><em>Pulsation de coupure à -3dB ou bande</em></strong></p>
<p>1</p>
<p><strong><em>passante à -3dB : </em></strong><em>w</em> <em><sub>c</sub></em> <sub>-3</sub><em><sub>dB</sub></em> = <em>w</em> <em><sub>p</sub></em> = <em><sub>t</sub></em></p>
<p><strong><em>Asymptote de pente : -20 dB/décade, d’équation :</em></strong></p>
<p><em><sup>w</sup>c </em>0</p>
<p>20<em>LogK</em> &#8211; 20<em>Logt</em> &#8211; 20<em>Logw</em></p>
<table>
<tbody>
<tr>
<td rowspan="2">
<p><em>w</em></p>
</td>
<td rowspan="3">
<p><em>c </em>-3<em>dB</em></p>
</td>
<td rowspan="2">
<p>= <em>w</em></p>
</td>
<td rowspan="3">
<p><em>p</em></p>
</td>
<td rowspan="2">
<p>= <sup>1</sup></p>
</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
</tr>
<tr>
<td rowspan="2">
<p>1 décade</p>
</td>
<td>&nbsp;</td>
</tr>
<tr>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>
<p><em>t</em></p>
</td>
<td>&nbsp;</td>
</tr>
</tbody>
</table>
<p><em>w </em>(<em>r a d s</em><sup>-1</sup> )</p>
<ul>
<li>(<em>w</em> )</li>
</ul>
<p>Page 13 <strong>Emmanuel FARGES</strong> Ó EduKlub S.A.</p>
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<p></p>
<p><strong>Sciences Indusrielles</strong></p>
<p><strong>Systèmes linéaires continus invariants</strong></p>
<p>ANALYSE FREQUENTIELLE</p>
<p><img decoding="async" class="wp-image-18187" src="https://telechargercours.com/wp-content/uploads/2021/07/word-image-79.png"></p>
<p><strong><em>Asymptote à </em></strong>0°</p>
<ul>
<li><sup>æ</sup><sub>ç</sub> <em>w<sub>p</sub></em> = <em><sub>t</sub></em><sup>1</sup> <sup>ö</sup><sub>÷</sub> = &#8211; arctan (1) = &#8211; 45°</li>
</ul>
<p>èø</p>
<p><strong><em>Courbe réelle de phase</em></strong></p>
<p><strong><em>Asymptote à </em></strong>&#8211; <em><sup>p</sup></em></p>
<p>2</p>
<p><strong><em>-90°</em></strong></p>
<p><em>w </em>(<em>rad s</em><sup>-1</sup> )</p>
<p><strong><em>Justification de la bande passante à -3dB :</em></strong></p>
<p>La pulsation de coupure à -3dB ou bande passante à -3dB est telle que :</p>
<p>20 <em>LogK</em> &#8211; <img decoding="async" class="wp-image-18189" src="https://telechargercours.com/wp-content/uploads/2021/07/word-image-81.png"> <img decoding="async" class="wp-image-18191" src="https://telechargercours.com/wp-content/uploads/2021/07/word-image-83.png"><em> H </em>(<em> jw<sub>c</sub></em><sub>-3</sub><em><sub>dB</sub> </em>) <img decoding="async" class="wp-image-18192" src="https://telechargercours.com/wp-content/uploads/2021/07/word-image-84.png"> <img decoding="async" class="wp-image-18194" src="https://telechargercours.com/wp-content/uploads/2021/07/word-image-86.png"><em><sub>dB </sub></em>= 3<em>dB </em>, soit pour un premier ordre :</p>
<p><img decoding="async" class="wp-image-18195" src="https://telechargercours.com/wp-content/uploads/2021/07/word-image-87.png"></p>
<p>20<em>LogK</em> &#8211; 20<em>LogK</em> + 20<em>Log </em> <img decoding="async" class="wp-image-18197" src="https://telechargercours.com/wp-content/uploads/2021/07/word-image-89.png"> <img decoding="async" class="wp-image-18199" src="https://telechargercours.com/wp-content/uploads/2021/07/word-image-91.png"> 1+ <em>t</em> <sup>2</sup><em>w<sub>c</sub></em><sup>2</sup><sub>-3</sub><em><sub>dB</sub></em> = 3</p>
<p><img decoding="async" class="wp-image-18201" src="https://telechargercours.com/wp-content/uploads/2021/07/word-image-93.png"></p>
<table>
<tbody>
<tr>
<td colspan="9">
<p>20<em>Log</em> 1 + <em>t</em> <sup>2</sup><em>w<sub>c</sub></em><sup>2</sup><sub>-3</sub><em><sub>dB</sub></em></p>
</td>
<td colspan="2">
<p>=</p>
</td>
<td>&nbsp;</td>
<td>
<p>3</p>
</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
</tr>
<tr>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>
<p>æ</p>
</td>
<td>&nbsp;</td>
<td>
<p>3</p>
</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>
<p>ö</p>
</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>
<p>æ</p>
</td>
<td colspan="2">
<p>6 ö</p>
</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
</tr>
<tr>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td rowspan="2">
<p>ç</p>
</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td rowspan="2">
<p>÷</p>
</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td rowspan="2">
<p>ç</p>
</td>
<td>&nbsp;</td>
<td rowspan="2">
<p>÷</p>
</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
</tr>
<tr>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td colspan="2" rowspan="2">
<p>20</p>
</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td rowspan="2">
<p>20</p>
</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
</tr>
<tr>
<td colspan="4">
<p>2 2</p>
</td>
<td colspan="3" rowspan="2">
<p>=10</p>
</td>
<td>
<p>è</p>
</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>
<p>ø</p>
</td>
<td rowspan="2">
<p>Þ 1</p>
</td>
<td rowspan="2">
<p>+<em>t</em></p>
</td>
<td>
<p>2</p>
</td>
<td>
<p>2</p>
</td>
<td>
<p>è</p>
</td>
<td>
<p>ø</p>
</td>
<td rowspan="2">
<p>; 2</p>
</td>
<td>
<p>2 2</p>
</td>
<td rowspan="2">
<p>= 1</p>
</td>
<td>&nbsp;</td>
</tr>
<tr>
<td colspan="4">
<p><sup>1+</sup><em> <sup>t w</sup>c </em>-3<em>dB</em></p>
</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>
<p><em><sup>w</sup>c</em>&#8211; 3<em> dB</em></p>
</td>
<td>
<p>= 10</p>
</td>
<td>&nbsp;</td>
<td>&nbsp;</td>
<td>
<p><sup>Þ</sup><em><sup>t w</sup>c </em>-3<em>dB</em></p>
</td>
<td>&nbsp;</td>
</tr>
<tr>
<td colspan="2" rowspan="2">
<p>D’où :</p>
</td>
<td rowspan="2">
<p><em>w</em></p>
</td>
<td>&nbsp;</td>
<td rowspan="2">
<p>=</p>
</td>
<td>
<p>1</p>
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<p>= <em>w</em></p>
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<p><em>c </em>-3<em> dB</em></p>
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<p><em>p</em></p>
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<p><em>t</em></p>
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<p><img decoding="async" class="wp-image-18203" src="https://telechargercours.com/wp-content/uploads/2021/07/word-image-95.png"></p>
<p>Page 14 <strong>Emmanuel FARGES</strong> Ó EduKlub S.A.</p>
<p>Tous droits de l’auteur des œuvres réservés. Sauf autorisation, la reproduction ainsi que toute utilisation des œuvres autre que la consultation individuelle et privée sont interdites.</p>


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]]></content:encoded>
					
		
		
			</item>
		<item>
		<title>Systèmes linéaires continus invariants ANALYSE TEMPORELLE (Partie 2)</title>
		<link>https://telechargercours.com/systemes-lineaires-continus-invariants-analyse-temporelle-partie2/</link>
		
		<dc:creator><![CDATA[Cours Zilla]]></dc:creator>
		<pubDate>Sat, 03 Jul 2021 15:08:02 +0000</pubDate>
				<category><![CDATA[Science & Technologie]]></category>
		<guid isPermaLink="false">https://telechargercours.com/?p=18098</guid>

					<description><![CDATA[Systèmes linéaires continus invariants Analyse temporelle SYSTEMES LINEAIRES CONTINUS INVARIANTS ANALYSE TEMPORELLE (Partie 2) 2&#160;&#160;&#160;&#160;&#160;&#160;&#160;&#160;&#160;&#160;&#160; Système du deuxième ordre. 2.1&#160;&#160;&#160;&#160;&#160;&#160; Définition d’un système du deuxième ordre Un système du second ordre a son comportement régi par une équation différentielle du deuxième ordre de la forme : &#160; &#160; &#160; &#160; &#160; &#160; &#160; &#160; &#160; [&#8230;]]]></description>
										<content:encoded><![CDATA[
<p class="has-text-align-center wp-block-paragraph"><strong>Systèmes linéaires continus invariants</strong></p>



<p class="has-text-align-center wp-block-paragraph">Analyse temporelle</p>



<p class="has-text-align-center wp-block-paragraph"><strong>SYSTEMES LINEAIRES CONTINUS INVARIANTS</strong></p>



<p class="has-text-align-center wp-block-paragraph"><em>ANALYSE TEMPORELLE (Partie 2)</em></p>



<h2 class="wp-block-heading"><strong>2</strong>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; <strong>Système du deuxième ordre.</strong></h2>



<h3 class="wp-block-heading"><strong>2.1</strong>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; <strong>Définition d’un système du deuxième ordre</strong></h3>



<p class="wp-block-paragraph"><strong>Un système du second ordre a son comportement régi par une équation différentielle du deuxième ordre de la forme :</strong></p>



<figure class="wp-block-table"><table><tbody><tr><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>ì</td><td>&nbsp;</td><td>&nbsp;</td><td><em>e </em>=<em> e </em>(<em>t</em>)&nbsp;&nbsp; entréedusystème</td><td>&nbsp;</td></tr><tr><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>ï</td><td><em>s </em>=<em> s</em>(<em>t</em>)&nbsp;&nbsp; réponsedusystèmeàl&rsquo;entréee(t)</td><td>&nbsp;</td></tr><tr><td>1</td><td>&nbsp;</td><td><em>d</em></td><td>2</td><td><em>s</em></td><td>&nbsp;</td><td>2<em>x</em> <em>ds</em></td><td>&nbsp;</td><td>ï</td><td>&nbsp;</td></tr><tr><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td></tr><tr><td>&nbsp;</td><td>&nbsp;</td><td>+</td><td>+ <em>s</em> = <em>Ke</em></td><td>ï</td><td>&nbsp;</td><td>&nbsp;</td><td>K:gainstatique(&gt;0)</td><td>&nbsp;</td></tr><tr><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>í</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td></tr><tr><td>&nbsp;</td><td><em>w</em><sub>0</sub><sup>2</sup></td><td>&nbsp;</td><td><em>dt </em><sup>2</sup></td><td><em>w</em><sub>0</sub><em>&nbsp; dt</em></td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td></tr><tr><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td><sup>ï</sup><em>x</em> :coefficient d&rsquo;amortissement (sans dimension) (&gt;0)</td><td>&nbsp;</td></tr><tr><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>ï</td><td><em>w</em></td><td>&nbsp;</td><td>pulsationpropredusystème(enrads<sup>-1</sup> )</td><td>&nbsp;</td></tr><tr><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>ï</td><td>0</td><td>&nbsp;</td></tr><tr><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>î</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td></tr></tbody></table></figure>



<h3 class="wp-block-heading"><strong>2.2</strong>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; <strong>Fonction de transfert globale d’un deuxième ordre</strong></h3>



<p class="wp-block-paragraph">On applique la transformée de Laplace à l’ensemble de l’équation différentielle ci-dessus, avec des conditions initiales nulles :</p>



<figure class="wp-block-table"><table><tbody><tr><td>Donc :</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td></tr><tr><td>&nbsp;</td><td>1&nbsp; <em>d</em> <sup>2</sup> <em>s</em></td><td>+</td><td>2<em>x</em> <em>ds</em></td><td>+ <em>s</em> = <em>Ke</em>&nbsp;&nbsp; transforméedeLaplace</td><td><em>p</em><sup>2</sup></td><td><em>S </em>(<em> p</em>)+</td><td>2<em>x</em></td><td><em>pS</em>(<em> p</em>)+<em> S </em>(<em> p</em>)=<em> KE </em>(<em> p</em>)</td><td>&nbsp;</td></tr><tr><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td></tr><tr><td>&nbsp;</td><td><em>w</em><sup>2</sup><em>&nbsp; dt </em><sup>2</sup></td><td><em>w</em></td><td>&nbsp;</td><td>&nbsp;</td><td><em>dt</em></td><td>&nbsp;</td><td><em>w</em></td><td>&nbsp;</td></tr><tr><td>&nbsp;</td><td>&nbsp;</td><td>0</td><td>&nbsp;</td><td>uuuuuuuuuuuuuuuuuuuuuuuuru <em><sub>w</sub></em> 2</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td></tr><tr><td>0</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>0</td><td>&nbsp;</td><td>0</td><td>&nbsp;</td><td>&nbsp;</td></tr></tbody></table></figure>



<p class="wp-block-paragraph">On peut alors présenter le rapport de la sortie S(p) sur l’entrée E(p), c’est à dire la fonction de</p>



<figure class="wp-block-table"><table><tbody><tr><td>transfert globale du système :</td><td><em>H </em>(<em> p</em>)=</td><td><em>S</em>(<em> p</em>)</td><td>=</td><td>&nbsp;</td><td>&nbsp;</td><td><em>K</em></td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td></tr><tr><td>&nbsp;</td><td>&nbsp;</td><td>2<em>x</em></td><td>&nbsp;</td><td><em>p</em><sup>2</sup></td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td></tr><tr><td>&nbsp;</td><td>&nbsp;</td><td><em>E </em>(<em> p</em>)</td><td>1+</td><td><em>p </em>+</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td></tr><tr><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td><em>w</em><sub>0</sub></td><td>&nbsp;</td><td>&nbsp;</td></tr><tr><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td><em>w</em><sub>0</sub></td><td>2</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td></tr></tbody></table></figure>



<p class="wp-block-paragraph"><strong><em><u>On notera :</u></em></strong></p>



<ul class="wp-block-list"><li>Le gain statique vaut : K=H(0)</li></ul>



<ul class="wp-block-list"><li>Pour identifier les caractéristiques d’un système du deuxième ordre (c’est à dire</li></ul>



<p class="wp-block-paragraph"><em>K </em>,<em>x</em><em> </em>et<em>w</em><sub>0</sub><em> </em>), on veillera bien à présenter la fonction de transfert globale H(p) avec le<em> </em>coefficient en p<sup>0</sup> du polynôme au dénominateur égal à 1. Ainsi le numérateur peut être identifié au gain statique K, coefficient en p<sup>2</sup> du polynôme au dénominateur peut être</p>



<p class="wp-block-paragraph">identifié à la pulsation propre <em>w</em>0 et on déduit l’amortissement du coefficient en p<sup>1</sup> : <sup>2</sup><em><sup>x</sup></em> . <em>w</em><sub>0</sub></p>



<h3 class="wp-block-heading"><strong>2.3 Réponse à un échelon (réponse dite indicielle)</strong></h3>



<h4 class="wp-block-heading"><strong>2.3.1 Réponse dans le domaine fréquentielle</strong></h4>



<p class="wp-block-paragraph">On soumet le système à une entrée échelon de taille A. L’entrée est donc la fonction du temps : <strong>e(t)= A u(t)</strong>.</p>



<p class="wp-block-paragraph">Page 1&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; <strong>Emmanuel FARGES</strong>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; Ó EduKlub S.A.</p>



<p class="wp-block-paragraph">Tous droits de l’auteur des œuvres réservés. Sauf autorisation, la reproduction ainsi que toute utilisation des œuvres autre que la consultation individuelle et privée sont interdites.</p>



<p class="wp-block-paragraph"><strong>Systèmes linéaires continus invariants</strong></p>



<p class="wp-block-paragraph">Analyse temporelle</p>



<figure class="wp-block-table"><table><tbody><tr><td>&nbsp;</td><td>&nbsp;</td><td>e (t)</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td></tr><tr><td>&nbsp;</td><td>A</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>Temps</td><td>t</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td></tr><tr><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td></tr><tr><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td></tr><tr><td>&nbsp;</td><td>t=0</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td></tr><tr><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td></tr><tr><td>e(t)=A u(t) donc&nbsp; <em>E</em> ( <em>p</em>) =</td><td><em>A</em></td><td>or <em>H</em> ( <em>p</em>) =</td><td><em>S</em>(<em> p</em>)</td><td>=</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td><em>K</em></td><td>&nbsp;</td><td>&nbsp;</td><td>Þ <em>S</em> ( <em>p</em>) =</td><td>&nbsp;</td><td>&nbsp;</td><td><em>K A</em></td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td></tr><tr><td><em>p</em></td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>2<em>x</em></td><td>&nbsp;</td><td><em>p</em><sup>2</sup></td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>2<em>x</em></td><td>&nbsp;</td><td>&nbsp;</td><td><em>p</em><sup>2</sup></td><td>ù</td><td>&nbsp;</td></tr><tr><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td><em>E </em>(<em> p</em>)</td><td>1</td><td>+</td><td><em>p </em>+</td><td>é</td><td>+</td><td>&nbsp;</td><td>+</td><td>&nbsp;</td></tr><tr><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td><em>p </em><sub>ê</sub>1</td><td>&nbsp;</td><td><em>p</em></td><td>&nbsp;</td><td>ú</td><td>&nbsp;</td></tr><tr><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td><em>w</em><sub>0</sub></td><td>2</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td></tr><tr><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td><em>w</em><sub>0</sub></td><td>2</td><td>&nbsp;</td></tr><tr><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td><em>w</em><sub>0</sub></td><td>ë</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td><em>w</em><sub>0</sub></td><td>û</td><td>&nbsp;</td></tr></tbody></table></figure>



<p class="wp-block-paragraph">On obtient donc très rapidement la solution (réponse à une entrée) dans le domaine fréquentielle (solution en p).</p>



<p class="wp-block-paragraph">Le problème est qu’il faut désormais effectuer la décomposition en éléments simples.</p>



<p class="wp-block-paragraph">La décomposition en éléments simples dépend des racines du dénominateur :</p>



<figure class="wp-block-table"><table><tbody><tr><td>é</td><td>2<em>x</em></td><td>&nbsp;</td><td><em>p</em></td><td>2</td><td>ù</td><td>qui admet 0 comme racine et pour le polynôme du second degré, on a soit</td><td>&nbsp;</td></tr><tr><td><em>p&nbsp; </em>1+</td><td><em>p </em>+</td><td>&nbsp;</td><td>ú</td><td>&nbsp;</td></tr><tr><td>&nbsp;</td><td>&nbsp;</td><td>2</td><td>&nbsp;</td></tr><tr><td>ê</td><td><em>w</em><sub>0</sub></td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td></tr><tr><td>ë</td><td><em>w</em><sub>0</sub></td><td>û</td><td>&nbsp;</td><td>&nbsp;</td></tr></tbody></table></figure>



<p class="wp-block-paragraph">deux racines réelles (discriminant &gt;0) soit deux racines complexes conjuguées (discriminant &lt;0).</p>



<p class="wp-block-paragraph">On va donc procéder à une première décomposition en éléments simples, puis on discutera de la valeur des racines du polynôme du second degré pour poursuivre la décomposition.</p>



<h4 class="wp-block-heading"><strong>2.3.2 Première décomposition en éléments simples</strong></h4>



<p class="wp-block-paragraph">La première décomposition en éléments simples revient à chercher dans notre cas particulier à mettre S(p) sous la forme :</p>



<figure class="wp-block-table"><table><tbody><tr><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td><em>K A</em></td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td><em>KA</em><em>w</em><sub>0</sub><sup>2</sup></td><td>&nbsp;</td><td>é <em><sub>a</sub></em></td><td>&nbsp;</td><td>&nbsp;</td><td><em>bp </em>+<em> c</em></td><td>ù</td><td>&nbsp;</td></tr><tr><td><em>S</em>(<em> p</em>)=</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>=</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>= <em>KA</em> <sub>ê</sub></td><td>&nbsp;</td><td>+</td><td>&nbsp;</td><td>&nbsp;</td><td><sub>ú</sub> avec a, b et c</td><td>&nbsp;</td></tr><tr><td>&nbsp;</td><td>&nbsp;</td><td>2<em>x</em></td><td>&nbsp;</td><td>&nbsp;</td><td><em>p</em><sup>2</sup></td><td>ù</td><td><em>p </em>é<em> p </em><sup>2</sup></td><td>+ 2<em>xw</em> <sub>0</sub> <em>p</em> + <em>w</em><sub>0</sub><sup>2</sup></td><td>ù</td><td>&nbsp;</td><td><em>p </em><sup>2</sup></td><td>2</td><td>&nbsp;</td></tr><tr><td>é</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>ë <em><sup>p</sup></em></td><td>&nbsp;</td><td>+ 2<em>xw</em> <sub>0</sub> <em>p</em> +<em>w</em><sub>0</sub></td><td>û</td><td>&nbsp;</td></tr><tr><td>&nbsp;</td><td><em>p </em><sub>ê</sub>1</td><td>+</td><td>&nbsp;</td><td><em>p</em></td><td>+</td><td>&nbsp;</td><td>ú</td><td>ë</td><td>&nbsp;</td><td>û</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td></tr><tr><td><em>w</em><sub>0</sub></td><td>2</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td></tr><tr><td>ë</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td><em>w</em><sub>0</sub></td><td>û</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td></tr></tbody></table></figure>



<p class="wp-block-paragraph">trois coefficients à déterminer.</p>



<p class="wp-block-paragraph">Employons désormais la méthode la plus rapide pour déterminer les coefficients :</p>



<figure class="wp-block-table"><table><tbody><tr><td>&nbsp;</td><td><em>KAa </em>=</td><td><em>pS</em>(<em> p</em>)</td><td>=</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td><em>KA</em></td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>= <em>KA</em> Þ <em>a</em> = 1</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td></tr><tr><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td></tr><tr><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>2</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td></tr><tr><td>&nbsp;</td><td>&nbsp;</td><td>123</td><td>1+</td><td>&nbsp;</td><td>2<em>x</em></td><td><em>p </em>+</td><td>&nbsp;</td><td><em>p</em></td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td></tr><tr><td>&nbsp;</td><td>&nbsp;</td><td>pour p=0</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td></tr><tr><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td><em>w</em></td><td>2</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td></tr><tr><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td><em>w</em></td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td></tr><tr><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>1442443</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td></tr><tr><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>0</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>0</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td></tr><tr><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>pourp=0</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td></tr><tr><td>Ensuite on identifie :</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td></tr><tr><td>&nbsp;</td><td>&nbsp;</td><td><em>w</em><sub>0</sub><sup>2</sup></td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>=</td><td>1</td><td>+</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td><em>bp </em>+<em> c</em></td><td>=</td><td><em>p</em><sup>2</sup><em> </em>+2<em>xw</em><em> </em><sub>0</sub><em> p </em>+<em>w</em><sub>0</sub><em> </em><sup>2</sup><em>&nbsp; </em>+<em> bp</em><sup>2</sup><em> </em>+<em>cp</em></td><td>&nbsp;</td></tr><tr><td>&nbsp;</td><td><em>p </em>é<em>p</em><sup>2</sup><em> </em>+2<em>xw</em></td><td><em>p </em>+<em>w</em><em> </em><sup>2</sup></td><td>ù</td><td>&nbsp;</td><td><em>p</em></td><td>&nbsp;</td><td><em>p </em><sup>2</sup><em> </em>+2<em>xw</em></td><td><em>p </em>+<em>w</em><em> </em><sup>2</sup></td><td><em>p </em>é<em> p </em><sup>2</sup></td><td>+ 2<em>xw</em></td><td>0</td><td><em>p </em>+<em>w</em></td><td>2 <sub>ù</sub></td><td>&nbsp;</td></tr><tr><td>ë</td><td>0</td><td>&nbsp;</td><td>0</td><td>û</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>0</td><td>0</td><td>&nbsp;</td><td>ë</td><td>&nbsp;</td><td>0</td><td>û</td><td>&nbsp;</td></tr></tbody></table></figure>



<ul class="wp-block-list"><li><em>p</em><sup>2</sup><em> </em>(<em>b </em>+1)+(2<em>xw</em><em> </em><sub>0</sub><em> </em>+<em> c</em>)<em> p </em>+<em>w</em><sub>0</sub><sup>2</sup><sub></sub></li></ul>



<p class="wp-block-paragraph"><em>p </em>é<sub>ë</sub><em> p </em><sup>2</sup><em> </em>+2<em>xw</em><em> </em><sub>0</sub><em> p </em>+<em>w</em><sub>0</sub><sup>2</sup><em> </em>ù<sub>û</sub></p>



<p class="wp-block-paragraph">On trouve donc le système :</p>



<p class="wp-block-paragraph">ì&nbsp; <em>b</em> +1 = 0&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; ì&nbsp; <em>b</em> = -1</p>



<p class="wp-block-paragraph">í&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; Þ í&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; , on obtient donc une première décomposition en éléments</p>



<p class="wp-block-paragraph"><sub>î</sub> 2<em>xw</em><sub>0</sub> + <em>c</em> = 0&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; <sub>î</sub><em>c</em> = -2<em>xw</em><sub>0</sub></p>



<p class="wp-block-paragraph">simples :</p>



<p class="wp-block-paragraph">Page 2&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; <strong>Emmanuel FARGES</strong>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; Ó EduKlub S.A.</p>



<p class="wp-block-paragraph">Tous droits de l’auteur des œuvres réservés. Sauf autorisation, la reproduction ainsi que toute utilisation des œuvres autre que la consultation individuelle et privée sont interdites.</p>



<p class="wp-block-paragraph"><strong>Sciences Indusrielles</strong></p>



<p class="wp-block-paragraph"><strong>Systèmes linéaires continus invariants</strong></p>



<p class="wp-block-paragraph">Analyse temporelle</p>



<figure class="wp-block-table"><table><tbody><tr><td>é</td><td>1</td><td>&nbsp;</td><td><em>p </em>+2<em>xw</em><sub>0</sub></td><td>ù</td><td>&nbsp;</td><td>&nbsp;</td></tr><tr><td><em>S</em>(<em> p</em>)=<em> KA </em><sub>ê</sub></td><td>&nbsp;</td><td>&#8211;</td><td><em>p </em><sup>2</sup><em> </em>+2<em>xw</em><em> </em><sub>0</sub><em> p </em>+<em>w</em><sub>0</sub><sup>2</sup></td><td>ú</td><td>&nbsp;</td><td>&nbsp;</td></tr><tr><td>ë <em><sup>p</sup></em></td><td>&nbsp;</td><td>û</td><td>&nbsp;</td><td>&nbsp;</td></tr><tr><td>&nbsp;</td><td>&nbsp;</td><td><strong>2.3.3 Seconde décomposition en éléments simples</strong></td><td>&nbsp;</td></tr></tbody></table></figure>



<figure class="wp-block-table"><table><tbody><tr><td>La suite de la décomposition dépend du signe du discriminant de&nbsp; <em>p</em> <sup>2</sup> + 2<em>xw</em> <em>p</em> + <em>w</em> <sup>2</sup></td></tr><tr><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>0</td><td>0</td></tr><tr><td>D = 4<em>x</em> <sup>2</sup> <em>w</em></td><td><sup>2</sup> &#8211; 4<em>w</em></td><td><sup>2</sup> = 4<em>w</em> <sup>2</sup></td><td>(<em>x</em> <sup>2</sup></td><td>-1)</td><td>&nbsp;</td></tr><tr><td>0</td><td>0</td><td>0</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td></tr></tbody></table></figure>



<p class="wp-block-paragraph">Donc trois cas se présente à nous</p>



<ul class="wp-block-list"><li><em>x</em> &gt; 1&nbsp; systèmetrèsamorti:2racinesréelles</li></ul>



<figure class="wp-block-table"><table><tbody><tr><td>ï</td><td><em>x</em> = 1&nbsp; caslimite:1racinedouble</td><td>&nbsp;</td></tr><tr><td>í</td><td>&nbsp;</td></tr><tr><td>ï</td><td>systèmefaiblementamorti:2racinescomplexesconjuguées</td><td>&nbsp;</td></tr><tr><td><sub>î</sub><em>x</em> &lt; 1</td><td>&nbsp;</td></tr></tbody></table></figure>



<h5 class="wp-block-heading"><strong>2.3.3.1 Deuxième ordre fortement amorti</strong></h5>



<p class="wp-block-paragraph">On a donc <em>x</em> &gt; 1, <em>p</em> <sup>2</sup> + 2<em>xw</em><sub>0</sub> <em>p</em> + <em>w</em><sub>0</sub><sup>2</sup> admet donc deux racines réelles :</p>



<figure class="wp-block-table"><table><tbody><tr><td><sup>ì</sup><em>l</em> = &#8211;<em>xw</em></td><td>0</td><td>+<em>w</em></td><td>0</td><td><em>x </em><sup>2</sup></td><td>-1</td><td>&nbsp;</td><td>&nbsp;</td></tr><tr><td>ï</td><td>1</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>, on remarque d’ailleurs que ces deux racines sont négatives (donc le</td><td>&nbsp;</td></tr><tr><td>í</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td></tr><tr><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&#8211;<em>w</em></td><td>&nbsp;</td><td><em>x </em><sup>2</sup></td><td>&nbsp;</td><td>&nbsp;</td></tr><tr><td>ï<em>l</em> = &#8211;<em>xw</em></td><td>0</td><td>0</td><td>-1</td><td>&nbsp;</td></tr><tr><td>î</td><td>2</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td></tr></tbody></table></figure>



<p class="wp-block-paragraph">système est stable) et homogè ne à des s<sup>-1</sup>.</p>



<p class="wp-block-paragraph">On poursuit donc la décomposition en éléments simples sous la forme :</p>



<figure class="wp-block-table"><table><tbody><tr><td>é 1</td><td>&nbsp;</td><td><em>p </em>+2<em>xw</em><sub>0</sub></td><td>&nbsp;</td><td>&nbsp;</td><td>ù</td><td>é</td><td>1</td><td>&nbsp;</td><td><em>a</em></td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td><em>b</em></td><td>&nbsp;</td><td>ù</td><td>&nbsp;</td></tr><tr><td><em>S</em>(<em> p</em>)=<em> KA </em><sub>ê</sub></td><td>&nbsp;</td><td>&#8211;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>ú</td><td>= <em>KA</em> ê</td><td>&nbsp;</td><td>+</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>+</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>ú</td><td>&nbsp;</td></tr><tr><td><em>p</em></td><td><em>p </em><sup>2</sup><em> </em>+2<em>xw</em><em> p </em>+<em>w</em></td><td>2</td><td>&nbsp;</td><td><em>p </em>+<em>xw</em><em> </em>&#8211;<em>w</em></td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td></tr><tr><td>ë</td><td>&nbsp;</td><td>û</td><td>ê <em><sup>p</sup></em></td><td>&nbsp;</td><td>0</td><td><em>x</em> <sup>2</sup> -1&nbsp;&nbsp; <em>p</em> + <em>xw</em> + <em>w</em></td><td>0</td><td><em>x</em> <sup>2</sup> -1 ú</td><td>&nbsp;</td></tr><tr><td>&nbsp;</td><td>&nbsp;</td><td>0</td><td>0</td><td>&nbsp;</td><td>ë</td><td>&nbsp;</td><td>&nbsp;</td><td>0</td><td>&nbsp;</td><td>&nbsp;</td><td>0</td><td>û</td><td>&nbsp;</td></tr></tbody></table></figure>



<p class="wp-block-paragraph">Par identification on doit donc avoir :</p>



<p class="wp-block-paragraph">-( <em>p</em> + 2<em>xw</em><sub>0</sub> ) = <em>a</em>( <em>p</em> + <em>xw</em><sub>0</sub> +<em>w</em><sub>0 </sub><img loading="lazy" decoding="async" src="" width="7" height="10"><img loading="lazy" decoding="async" src="" width="5" height="23"><em>x </em><sup>2</sup><em> </em>-1)<em> </em>+<em> </em><em>b</em>(<em> </em><em>p</em><em> </em>+<em> xw</em><sub>0</sub><em> </em>&#8211;<em>w </em><sub>0 </sub><img loading="lazy" decoding="async" src="" width="6" height="10"><img loading="lazy" decoding="async" src="" width="6" height="23"><em>x </em><sup>2</sup><em> </em>-1)</p>



<p class="wp-block-paragraph">= (<em>a</em> + <em>b</em>) <em>p</em> + <em>xw</em><sub>0</sub> (<em>a</em> + <em>b</em>) +<em>w</em><sub>0 </sub><img loading="lazy" decoding="async" src="" width="7" height="9"><em>&nbsp;x </em><sup>2</sup><em> </em>&#8211;<em> </em>1(<em>a</em><em> </em>&#8211;<em> </em><em>b</em>)</p>



<p class="wp-block-paragraph">C’est à dire le système :</p>



<figure class="wp-block-table"><table><tbody><tr><td>ì</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td><em>a </em>+<em> b </em>= -1</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td></tr><tr><td>ï</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td></tr><tr><td>í</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td></tr><tr><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>-1 = -2<em>xw</em><sub>0</sub></td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td></tr><tr><td>ï<em>xw</em><sub>0</sub> (<em>b</em> + <em>a</em>) + (<em>a</em> &#8211; <em>b</em>)<em>w</em><sub>0</sub>&nbsp;&nbsp; <em>x</em> <sup>2</sup></td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td></tr><tr><td>î</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td></tr><tr><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>ì</td><td><em>x</em> &#8211;</td><td><em>x</em> <sup>2</sup> -1</td><td>&nbsp;</td></tr><tr><td>ì</td><td><em>a </em>+<em> b </em>= -1</td><td>ì&nbsp; <em>a</em> = -1 &#8211; <em>b</em></td><td>ï <em>b</em> =</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td></tr><tr><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td></tr><tr><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>2</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td></tr><tr><td>ï</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>ï</td><td>&nbsp;</td><td>2&nbsp; <em>x</em> -1</td><td>&nbsp;</td></tr><tr><td>ï</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td><em>x</em></td><td>&nbsp;</td><td>&nbsp;</td></tr><tr><td>Þ <sub>í</sub></td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>Þ</td><td>í</td><td>+ 2<em>b</em> =</td><td>&nbsp;</td><td>&nbsp;</td><td>Þ <sub>í</sub></td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td></tr><tr><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td></tr><tr><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td></tr><tr><td>ï(<em>a</em> &#8211; <em>b</em>)&nbsp; <em>x</em></td><td>2</td><td>-1 = &#8211;<em>x</em></td><td>1</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>2</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td></tr><tr><td>&nbsp;</td><td>2</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td></tr><tr><td>&nbsp;</td><td>ï</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>ï</td><td>&nbsp;</td><td><em>x</em> +</td><td>&nbsp;</td><td><em>x</em>&nbsp; -1</td><td>&nbsp;</td></tr><tr><td>î</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td><em>x</em> -1</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td></tr><tr><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>î</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td><sub>ï</sub><em>a </em>= &#8211;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td></tr><tr><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td></tr><tr><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>2</td><td>&nbsp;</td><td><em>x</em> <sup>2</sup> -1</td><td>&nbsp;</td></tr><tr><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>ï</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td></tr><tr><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>î</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td></tr></tbody></table></figure>



<p class="wp-block-paragraph">On trouve donc la décomposition en éléments simples dans le cas d’un système fortement amorti :</p>



<p class="wp-block-paragraph">Analyse temporelle</p>



<p class="wp-block-paragraph"><strong><em><u>Réponse temporelle :</u></em></strong></p>



<p class="wp-block-paragraph">On peut désormais appliquer la transformée inverse de Laplace, puisque l’on sait inverser tous les termes (même si ils sont très lourds !!!) :</p>



<figure class="wp-block-table"><table><tbody><tr><td>&nbsp;</td><td>é</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>ù</td><td>&nbsp;</td><td>&nbsp;</td></tr><tr><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>æ</td><td>2</td><td>ö</td><td>&nbsp;</td><td>&nbsp;</td><td>æ</td><td>&nbsp;</td><td>2</td><td>ö</td><td>&nbsp;</td><td>&nbsp;</td></tr><tr><td></td><td></td><td></td><td></td><td></td><td></td><td></td><td></td><td></td><td></td><td></td><td></td><td></td><td></td><td></td><td></td></tr></tbody></table></figure>



<p class="wp-block-paragraph">On a donc une <strong><u>pente nulle à l’origine</u></strong> . C’est un des critères qui permet de ne pas confondre la réponse d’un deuxième ordre fortement amorti avec la réponse d’un premier ordre</p>



<p class="wp-block-paragraph"><strong><em><u>Précision statique , ou erreur statique:</u></em></strong></p>



<p class="wp-block-paragraph">Les exponentielles de la solution ci-dessus sont décroissantes, on voit donc que : lim <em>s</em>(<em>t</em> ) = <em>KA</em> .</p>



<p class="wp-block-paragraph"><em>t</em>®¥</p>



<p class="wp-block-paragraph">Donc <em>e</em><em><sub>s</sub></em>&nbsp; = lim[<em>e</em>(<em>t</em>) &#8211; <em>s</em>(<em>t</em> )] = <em>A</em> (1 &#8211; <em>K</em> ) , soit le même résultat que pour un premier ordre.</p>



<p class="wp-block-paragraph"><em>t</em>®¥</p>



<p class="wp-block-paragraph"><strong>C’est à dire que le système est précis ( </strong><em>e</em><em><sub>S</sub></em><strong> </strong>=<strong> </strong>0<strong> ) si le gain de la fonction de transfert globale du système est unitaire.</strong></p>



<p class="wp-block-paragraph">Néanmoins il n’est pas nécessaire de calculer la solution dans le domaine temporelle (d’ailleurs il faut toujours l’éviter vu la lourdeur des calculs) pour trouver l’erreur statique. Il suffit pour cela d’utiliser le théorème de la valeur finale :</p>



<figure class="wp-block-table"><table><tbody><tr><td><em>e</em><em><sub>s</sub></em></td><td>= lim [<em>e</em>(<em>t</em> ) &#8211; <em>s</em>(<em>t</em>)] = lim <em>p</em> [<em>E</em>( <em>p</em>) &#8211; <em>S</em>( <em>p</em>)] = lim <em>pE</em>( <em>p</em>) [1- <em>H</em> ( <em>p</em>)]</td><td>&nbsp;</td></tr><tr><td>&nbsp;</td><td><em>t </em>®¥</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td><em>p </em>®0</td><td>&nbsp;</td><td>&nbsp;</td><td><em>p </em>®0</td><td>&nbsp;</td></tr><tr><td>&nbsp;</td><td>&nbsp;</td><td>é</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>ù</td><td>&nbsp;</td><td>&nbsp;</td></tr><tr><td>&nbsp;</td><td>&nbsp;</td><td>ê</td><td>&nbsp;</td><td>&nbsp;</td><td><em>K</em></td><td>&nbsp;</td><td>ú</td><td>&nbsp;</td><td>&nbsp;</td></tr><tr><td>&nbsp;</td><td>= lim <em>A</em> <sup>ê</sup>1-</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>ú</td><td>= <em>A</em>(1- <em>K</em> )</td><td>&nbsp;</td></tr><tr><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td></tr><tr><td>&nbsp;</td><td>&nbsp;</td><td>2<em>x</em></td><td>&nbsp;</td><td>&nbsp;</td><td><em>p</em><sup>2</sup></td><td>&nbsp;</td><td>&nbsp;</td></tr><tr><td>&nbsp;</td><td><em>p </em>®0</td><td>ê</td><td>1+</td><td>&nbsp;</td><td><em>p </em>+</td><td>ú</td><td>&nbsp;</td><td>&nbsp;</td></tr><tr><td>&nbsp;</td><td>&nbsp;</td><td>ê</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>ú</td><td>&nbsp;</td><td>&nbsp;</td></tr><tr><td>&nbsp;</td><td>&nbsp;</td><td><em>w </em><sub>0</sub></td><td>&nbsp;</td><td>2</td><td>&nbsp;</td><td>&nbsp;</td></tr><tr><td>&nbsp;</td><td>&nbsp;</td><td>ë</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td><em>w</em><sub>0</sub></td><td>û</td><td>&nbsp;</td><td>&nbsp;</td></tr></tbody></table></figure>



<p class="wp-block-paragraph">Page 4&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; <strong>Emmanuel FARGES</strong>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; Ó EduKlub S.A.</p>



<p class="wp-block-paragraph">Tous droits de l’auteur des œuvres réservés. Sauf autorisation, la reproduction ainsi que toute utilisation des œuvres autre que la consultation individuelle et privée sont interdites.</p>



<p class="wp-block-paragraph"><strong>Sciences Indusrielles</strong></p>



<p class="wp-block-paragraph"><strong>Systèmes linéaires continus invariants</strong></p>



<p class="wp-block-paragraph">Analyse temporelle</p>



<p class="wp-block-paragraph"><strong><em><u>Courbe réponse :</u></em></strong></p>



<p class="wp-block-paragraph"><strong><em>Pas de dépassement</em></strong></p>



<p class="wp-block-paragraph"><strong>KA</strong></p>



<p class="wp-block-paragraph"><strong><em><u>Réponse du système :</u></em></strong></p>



<p class="wp-block-paragraph"><em>H </em>(<em> p </em>)=&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; <sup>10</sup>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; <sub>(</sub><em>K</em> = 10,<em>x</em> = 2,<em>w</em><sub>0</sub> &nbsp;= 0,1 <em>rads</em><sup>&#8211;</sup><sup>1</sup> <sub>)</sub></p>



<p class="wp-block-paragraph">1+ 40 <em>p</em> +100 <em>p</em><sub>2</sub></p>



<p class="wp-block-paragraph"><strong><em>à un échelon de taille A=1</em></strong></p>



<p class="wp-block-paragraph"><strong>Pente nulle à l’origine</strong></p>



<p class="wp-block-paragraph"><strong><em><u>Notion de pôles dominants :</u></em></strong></p>



<figure class="wp-block-table"><table><tbody><tr><td><sup>ì</sup><em> p&nbsp; </em>= &#8211;<em>xw</em></td><td>0</td><td>+<em>w</em></td><td>0</td><td><em>x </em><sup>2</sup></td><td>-1</td><td>&nbsp;</td><td>&nbsp;</td></tr><tr><td>ï</td><td>1</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td></tr><tr><td>Les deux solutions du polynôme du second degré : <sub>í</sub></td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>sont les pôles de</td><td>&nbsp;</td></tr><tr><td>&nbsp;</td><td>&nbsp;</td><td>= &#8211;<em>xw</em></td><td>&nbsp;</td><td>&#8211;<em>w</em></td><td>&nbsp;</td><td><em>x </em><sup>2</sup></td><td>&nbsp;</td><td>&nbsp;</td></tr><tr><td>ï <em>p</em></td><td>2</td><td>0</td><td>0</td><td>-1</td><td>&nbsp;</td></tr><tr><td>î</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td></tr></tbody></table></figure>



<p class="wp-block-paragraph">la fonction de la fonction de transfert</p>



<p class="wp-block-paragraph">Réécrivons la solution en t en fonction des pôles de la fonction de transfert :</p>



<figure class="wp-block-table"><table><tbody><tr><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td></tr><tr><td>é</td><td><em>x</em> +&nbsp; <em>x</em> <sup>2</sup> -1&nbsp;&nbsp; <em>p t</em></td><td>&nbsp;</td><td><em>x</em> &#8211;&nbsp; <em>x</em> <sup>2</sup> -1&nbsp;&nbsp; <em>p t</em> <sup>ù</sup></td><td>&nbsp;</td></tr><tr><td><em>s</em>(<em>t</em>)=<em> KA </em>ê1-</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>1&nbsp; <sub>+</sub></td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>2</td><td>ú . Si le système est très amorti : <em>x</em> &gt;&gt;1 , alors</td><td>&nbsp;</td></tr><tr><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td></tr><tr><td>ê</td><td>2</td><td>&nbsp;</td><td><em>x</em> <sup>2</sup> &#8211; 1&nbsp; <em><sup>e</sup></em></td><td>&nbsp;</td><td>&nbsp;</td><td>2&nbsp; <em>x</em> <sup>2</sup> -1&nbsp; <em><sup>e</sup></em></td><td>&nbsp;</td><td>ú</td><td>&nbsp;</td></tr><tr><td>ë</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>û</td><td>&nbsp;</td></tr><tr><td>on remarque que</td><td><em>p</em><sub>2</sub><em>&nbsp; </em>&lt;&lt;<em> p</em><sub>1</sub><em> </em>donc l’exponentielle en<em> </em><em><sub>e</sub></em><em> </em><em><sup>p</sup></em><em> </em>2<em> </em><em><sup>t</sup></em><em> </em>(<em> p</em><sub>2</sub><em> </em>est négative) s’amortie beaucoup</td><td>&nbsp;</td></tr><tr><td>plus vite que l’exponentielle en</td><td><em><sub>e</sub></em> <em><sup>p</sup></em>1<em><sup>t</sup></em> ( <em>p</em><sub>1</sub> est négative). <strong>On dit que</strong>&nbsp; <em>p</em><sub>1</sub> <strong>est le pôle dominant</strong></td><td>&nbsp;</td></tr></tbody></table></figure>



<p class="wp-block-paragraph">(en valeur absolu c’est le plus petit) et le système se comporte donc comme un premier ordre</p>



<p class="wp-block-paragraph"><sup>1</sup></p>



<p class="wp-block-paragraph">de constante de temps :<em>t</em><sub>1</sub> = &#8211;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; .</p>



<p class="wp-block-paragraph"><strong>Systèmes linéaires continus invariants</strong></p>



<p class="wp-block-paragraph">Analyse temporelle</p>



<figure class="wp-block-table"><table><tbody><tr><td>&nbsp;</td><td>&nbsp;</td><td>é</td><td>æ</td><td>&#8211;</td><td>&nbsp;</td><td><em>t</em></td><td>öù</td><td>&nbsp;</td><td>&nbsp;</td></tr><tr><td>Soit :</td><td><em>s</em>(<em>t</em>)=<em> KA</em></td><td>ê</td><td>ç</td><td>&nbsp;</td><td>&nbsp;</td><td>÷ú</td><td>, soit la réponse d’un premier ordre à un échelon de taille A.</td><td>&nbsp;</td></tr><tr><td><em>t</em><sub>1</sub></td><td>&nbsp;</td></tr><tr><td>ê<sup>1</sup></td><td><sup>&#8211;</sup><em> e</em><sub>è</sub></td><td>&nbsp;</td><td>øú</td><td>&nbsp;</td></tr><tr><td>&nbsp;</td><td>&nbsp;</td><td>ê</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>ú</td><td>&nbsp;</td><td>&nbsp;</td></tr><tr><td>&nbsp;</td><td>&nbsp;</td><td>ë</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>û</td><td>&nbsp;</td><td>&nbsp;</td></tr></tbody></table></figure>



<p class="wp-block-paragraph"><strong>On peut redémontrer cela directement à partir de la fonction de transfert globale du système :</strong></p>



<figure class="wp-block-table"><table><tbody><tr><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td><em>K</em></td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td><em>K </em><em>w</em><em> </em><sup>2</sup></td><td>&nbsp;</td><td><em>K</em><em>w</em><em> </em><sup>2</sup></td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td></tr><tr><td><em>H </em>(<em> p</em>)=</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>=</td><td>&nbsp;</td><td>0</td><td>=</td><td>0</td><td>&nbsp;</td><td>&nbsp;</td><td>avec&nbsp; <em>p</em><sub>2</sub>&nbsp; et <em>p</em><sub>1</sub> , pôles de la</td><td>&nbsp;</td></tr><tr><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td><em>p</em><sup>2</sup></td><td>&nbsp;</td><td><em>w</em><sub>0</sub></td><td><sup>2</sup> + 2<em>xw</em><sub>0</sub> <em>p</em> + <em>p</em><sup>2</sup></td><td>(<em> p </em>&#8211;<em> p</em><sub>1</sub><em> </em>)(<em>p</em></td><td>&#8211; <em>p</em> <sub>2</sub> )</td><td>&nbsp;</td></tr><tr><td>1+</td><td>2<em>x</em></td><td>&nbsp;</td><td><em>p </em>+</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td></tr><tr><td><em>w</em></td><td>&nbsp;</td><td><em>w </em><sup>2</sup></td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td></tr><tr><td>0</td><td>0</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td></tr></tbody></table></figure>



<ul class="wp-block-list"><li><em>p</em><sub>2</sub><em>&nbsp; </em>&lt;&lt;<em> p</em><sub>1</sub></li></ul>



<p class="wp-block-paragraph">fonction de transfert tel que <sup>ï</sup>í <em>p</em><sub>2</sub> &lt; 0homogèneà <em>s</em><sup>&#8211;</sup><sup>1</sup> .</p>



<figure class="wp-block-table"><table><tbody><tr><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td><sup>ï</sup><em> p </em>&lt;0homogèneà<em> s</em><sup>&#8211;</sup><sup>1</sup></td><td>&nbsp;</td></tr><tr><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>î</td><td>1</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td></tr><tr><td>&nbsp;</td><td>ì <em><sub>p</sub></em></td><td>= &#8211;</td><td>1</td><td>&nbsp;</td><td>avec <em>t</em></td><td>&nbsp;</td><td>constantedetemps</td><td>&nbsp;</td></tr><tr><td>&nbsp;</td><td>&nbsp;</td><td>1</td><td>&nbsp;</td></tr><tr><td>&nbsp;</td><td>ï</td><td>&nbsp;</td><td>1</td><td>&nbsp;</td><td>&nbsp;</td><td><em>t</em><sub>1</sub></td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td></tr><tr><td>On peut donc poser</td><td>ï</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>.</td><td>&nbsp;</td></tr><tr><td>í</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>1</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td></tr><tr><td>&nbsp;</td><td>ï</td><td><em>p</em></td><td>&nbsp;</td><td>= &#8211;</td><td>&nbsp;</td><td>avec <em>t</em></td><td>&nbsp;</td><td>constantedetemps</td><td>&nbsp;</td></tr><tr><td>&nbsp;</td><td>ï</td><td>2</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>2</td><td>&nbsp;</td></tr><tr><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td><em>t </em><sub>2</sub></td><td>&nbsp;</td><td>&nbsp;</td></tr><tr><td>&nbsp;</td><td>î</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td></tr></tbody></table></figure>



<figure class="wp-block-table"><table><tbody><tr><td><em>H </em>(<em> p</em>)=</td><td>&nbsp;</td><td><em>K</em><em>w</em></td><td>2</td><td>&nbsp;</td><td>=</td><td>&nbsp;</td><td>&nbsp;</td><td><em>K </em><em>w</em></td><td>2</td><td>&nbsp;</td><td>&nbsp;</td><td>=</td><td><em>K</em><em>w</em></td><td><sup>2</sup><em>t t</em></td><td>2</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>. On remarque donc que le</td><td>&nbsp;</td></tr><tr><td>&nbsp;</td><td>0</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>0</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>0</td><td>1</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td></tr><tr><td>(</td><td><em>p </em>&#8211;<em> p</em></td><td><em>p </em>&#8211;<em> p</em></td><td>2 )</td><td>æ</td><td>öæ</td><td>&nbsp;</td><td>1</td><td>ö</td><td>1 + <em>t</em></td><td>1</td><td><em>p</em></td><td>&nbsp;</td><td>1+ <em>t</em></td><td>2</td><td><em>p</em></td><td>)</td><td>&nbsp;</td></tr><tr><td>&nbsp;</td><td><sub>1 </sub>)(</td><td>&nbsp;</td><td>&nbsp;</td><td><sub>ç</sub><em> p </em>+</td><td>&nbsp;</td><td>1</td><td>÷ç <em><sup>p</sup></em> <sup>+</sup></td><td>÷</td><td>(</td><td>&nbsp;</td><td>)(</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td></tr><tr><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td><em>t</em><sub>2</sub></td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td></tr><tr><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>è</td><td><em><sup>t</sup></em> 1 øè</td><td>&nbsp;</td><td>ø</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td></tr><tr><td></td><td></td><td></td><td></td><td></td><td></td><td></td><td></td><td></td><td></td><td></td><td></td><td></td><td></td><td></td><td></td><td></td><td></td><td></td><td></td><td></td><td></td><td></td><td></td><td></td></tr></tbody></table></figure>



<p class="wp-block-paragraph">système du second ordre est le produit de deux premiers ordre. Dans l’hypothèse d’un très fort amortissement <em>x</em> &gt;&gt;1 , soit encore <em>p</em><sub>2</sub> &lt;&lt; <em>p</em><sub>1</sub> ou encore sur les constantes de temps <em>t</em> <sub>2</sub> &lt;&lt;<em>t</em><sub>1</sub> . On a donc le produit de deux premiers ordre dont l’un est très lent devant l’autre. Le système se comporte donc bien sûr comme le plus lent, c’est à dire à <em>t</em><sub>1</sub> qui correspond au <strong><em>pôle</em></strong> <strong><em>dominant </em></strong><em>p</em><sub>1</sub></p>



<p class="wp-block-paragraph"><strong><em><u>Réponse du système :</u></em></strong></p>



<p class="wp-block-paragraph"><em>H </em>(<em> p</em>)=&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; <sup>10</sup>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; <sub>(</sub><em>K</em> = 10,<em>x</em> = 5,<em>w</em><sub>0</sub> &nbsp;= 0,1 <em>rads</em><sup>&#8211;</sup><sup>1</sup> <sub>)</sub></p>



<p class="wp-block-paragraph">1+ 100 <em>p</em> +100 <em>p</em><sub>2</sub></p>



<p class="wp-block-paragraph"><strong>K</strong></p>



<p class="wp-block-paragraph"><strong><em>On voit que les critères d’identification au premier ordre de</em></strong></p>



<figure class="wp-block-table"><table><tbody><tr><td><strong><em>constante de temps :</em></strong><em>t</em><strong><em> </em></strong>= &#8211;<sup>1</sup></td><td>=</td><td>1</td><td>; 99 <em>s</em></td><td>&nbsp;</td></tr><tr><td>1</td><td><em>p</em><sub>1</sub></td><td>&nbsp;</td><td><em>w</em><sub>0</sub> (5 &#8211;</td><td>24)</td><td>&nbsp;</td></tr><tr><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td></tr></tbody></table></figure>



<p class="wp-block-paragraph"><strong><em>sont tous très approchés (valeur à </em></strong><em>t</em><strong><em> , pente à l’origine, t</em></strong><strong><em><sub>5%</sub></em></strong><strong><em>)</em></strong></p>



<p class="wp-block-paragraph"><strong><em>Ce système est bien approximé par un premier ordre de</em></strong></p>



<p class="wp-block-paragraph"><em>t</em></p>



<p class="wp-block-paragraph">Page 6&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; <strong>Emmanuel FARGES</strong>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; Ó EduKlub S.A.</p>



<p class="wp-block-paragraph"><strong>Systèmes linéaires continus invariants</strong></p>



<p class="wp-block-paragraph">Analyse temporelle</p>



<h5 class="wp-block-heading"><strong>2.3.3.2 Deuxième ordre limite : amortissement unitaire</strong></h5>



<p class="wp-block-paragraph">L’amortissement est unitaire, donc le discriminant est nul, on a alors une racine double pour</p>



<figure class="wp-block-table"><table><tbody><tr><td>le polynôme</td><td><em>p </em><sup>2</sup><em> </em>+2<em>xw</em><em> p </em>+<em>w</em><em> </em><sup>2</sup></td><td>. On a alors</td><td><em>p</em></td><td>= <em>p</em></td><td>2</td><td>= &#8211;<em>w</em></td><td>. On peut reprendre la</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td></tr><tr><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>0</td><td>&nbsp;</td><td>0</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>1</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>0</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td></tr><tr><td>décomposition en éléments simples avec <em>x</em> = 1 à partir de</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>é 1</td><td>&nbsp;</td><td>&nbsp;</td><td><em>p </em>+2<em>w</em><sub>0</sub></td><td>&nbsp;</td><td>&nbsp;</td><td>ù</td><td>&nbsp;</td></tr><tr><td><em>S</em>(<em> p</em>)=<em> KA </em><sub>ê</sub></td><td>&nbsp;</td><td>&nbsp;</td><td>&#8211;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>ú</td><td>&nbsp;</td></tr><tr><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td></tr><tr><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td><sub>ë</sub><em> p&nbsp;&nbsp; p </em><sup>2</sup><em> </em>+2<em>w</em><em> </em><sub>0</sub><em> p </em>+<em>w</em><sub>0</sub><sup>2</sup></td><td>û</td><td>&nbsp;</td></tr><tr><td>é 1</td><td>&nbsp;</td><td><em>p </em>+2<em>w</em></td><td>&nbsp;</td><td>ù</td><td>&nbsp;</td><td>&nbsp;</td><td>é <sub>1</sub></td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td><em>p </em>+2<em>w</em></td><td>&nbsp;</td><td>ù</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td></tr><tr><td><em>S</em>(<em> p</em>)=<em> KA </em><sub>ê</sub></td><td>&nbsp;</td><td>&#8211;</td><td>&nbsp;</td><td>&nbsp;</td><td>0</td><td>&nbsp;</td><td><sub>ú</sub> = <em>KA</em> ê</td><td>&nbsp;</td><td>&#8211;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>0</td><td>&nbsp;</td><td>&nbsp;</td><td>ú que l’on cherche à décomposer sous</td><td>&nbsp;</td></tr><tr><td><em>p</em></td><td><em>p </em><sup>2</sup><em> </em>+2<em>w</em><em> p </em>+<em>w</em><em> </em><sup>2</sup></td><td>&nbsp;</td><td>&nbsp;</td><td>(</td><td><em>p </em>+<em>w</em></td><td>2</td><td>&nbsp;</td><td>&nbsp;</td></tr><tr><td>ë</td><td>&nbsp;</td><td>û</td><td>&nbsp;</td><td>&nbsp;</td><td>ê <em><sup>p</sup></em></td><td>&nbsp;</td><td>&nbsp;</td><td>ú</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td></tr><tr><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>0</td><td>&nbsp;</td><td>0</td><td>&nbsp;</td><td>&nbsp;</td><td>ë</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>0 )</td><td>&nbsp;</td><td>û</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td></tr><tr><td>la forme :</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td></tr><tr><td>é</td><td>1</td><td>&nbsp;</td><td><em>p </em>+2<em>w</em></td><td>0</td><td>ù</td><td>&nbsp;</td><td>é <sub>1</sub></td><td>æ</td><td>&nbsp;</td><td>&nbsp;</td><td><em>a</em></td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td><em>b</em></td><td>&nbsp;</td><td>öù</td><td>é</td><td>1</td><td>&nbsp;</td><td>&nbsp;</td><td><sup>æ</sup><em> a </em>+<em> b </em>(<em> p </em>+<em>w</em></td><td><sub>)</sub> öù</td><td>&nbsp;</td><td>&nbsp;</td></tr><tr><td><em>S</em>(<em> p</em>)=<em> KA </em>ê</td><td>&nbsp;</td><td>&#8211;</td><td>&nbsp;</td><td>ú = <em>KA</em> ê</td><td>&nbsp;</td><td>&#8211; ç</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>+</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>÷ú = <em>KA</em> ê</td><td>&nbsp;</td><td>&nbsp;</td><td>&#8211; ç</td><td>&nbsp;</td><td>0</td><td>&nbsp;</td><td>÷ú</td><td>&nbsp;</td><td>&nbsp;</td></tr><tr><td>&nbsp;</td><td>(<em> p </em>+<em>w</em><sub>0</sub><em> </em>)<sup>2</sup></td><td>&nbsp;</td><td>(<em> p </em>+<em>w</em><em> </em><sub>0</sub><em> </em>)<sup>2</sup></td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>(<em> p </em>+<em>w</em><sub>0</sub><em> </em>)<sup>2</sup></td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td></tr><tr><td>ê <em><sup>p</sup></em></td><td>&nbsp;</td><td>ú</td><td>&nbsp;</td><td>ê <em>p</em></td><td>ç</td><td>&nbsp;</td><td><em>p </em>+<em>w</em><sub>0</sub><em>&nbsp; </em>÷ú</td><td>ê <em>p</em></td><td>&nbsp;</td><td>&nbsp;</td><td>ç</td><td>&nbsp;</td><td>÷<sub>ú</sub></td><td>&nbsp;</td><td>&nbsp;</td></tr><tr><td>ë</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>û</td><td>&nbsp;</td><td>ë</td><td>&nbsp;</td><td>è</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>øû</td><td>ë</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>è</td><td>&nbsp;</td><td>&nbsp;</td><td>øû</td><td>&nbsp;</td><td>&nbsp;</td></tr></tbody></table></figure>



<p class="wp-block-paragraph">Ce qui nous donne le système à résoudre pour déterminer a et b :</p>



<figure class="wp-block-table"><table><tbody><tr><td>ì&nbsp;&nbsp; <em>b</em> = 1</td><td>Þ</td><td>ì <em>b</em> = 1</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td></tr><tr><td>í</td><td>í</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td></tr><tr><td><sub>î</sub><em> a </em>+<em> b</em><em>w</em><sub>0</sub><em>&nbsp; </em>=</td><td>2<em>w</em><sub>0</sub></td><td><sub>î</sub><em>a </em>=<em>w</em><sub>0</sub></td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td></tr><tr><td>&nbsp;</td><td>é</td><td>1</td><td>æ</td><td>&nbsp;</td><td><em>w</em><sub>0</sub></td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>1</td><td>&nbsp;</td><td>öù</td><td>&nbsp;</td></tr><tr><td>Donc <em>S</em>( <em>p</em>) = <em>KA</em> ê</td><td>&nbsp;</td><td>&#8211; ç</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>+</td><td>&nbsp;</td><td>&nbsp;</td><td>÷ú</td><td>&nbsp;</td></tr><tr><td>&nbsp;</td><td>(</td><td><em>p </em>+<em>w</em></td><td>0 )</td><td>&nbsp;</td><td><em>p </em>+<em>w</em></td><td>&nbsp;</td><td>&nbsp;</td></tr><tr><td>&nbsp;</td><td>ê <em>p</em></td><td>ç</td><td>2</td><td>&nbsp;</td><td>0</td><td>÷<sub>ú</sub></td><td>&nbsp;</td></tr><tr><td>&nbsp;</td><td>ë</td><td>&nbsp;</td><td>è</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>øû</td><td>&nbsp;</td></tr></tbody></table></figure>



<p class="wp-block-paragraph">Or d’après les transformées de Laplace développée dans le cours du même nom :</p>



<figure class="wp-block-table"><table><tbody><tr><td>ì</td><td>&nbsp;</td><td><em>L </em>é<em> e</em><sup>&#8211;</sup><em><sup>at</sup></em><em> u</em>(<em>t</em>)ù</td><td>=</td><td>1</td><td>&nbsp;</td><td>&nbsp;</td></tr><tr><td>ï</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td></tr><tr><td>&nbsp;</td><td>ë</td><td>û</td><td>&nbsp;</td><td>&nbsp;</td><td><em>p </em>+<em> a</em></td><td>&nbsp;</td></tr><tr><td>ï</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td></tr><tr><td>í</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td><em>n</em>!</td><td>&nbsp;</td></tr><tr><td>ï<em>L</em> é<em>t</em> <em><sup>n</sup></em><em>e</em><sup>&#8211;</sup><em><sup>at</sup></em> <em>u</em> (<em>t</em>)ù</td><td>=</td><td>&nbsp;</td><td>&nbsp;</td></tr><tr><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td></tr><tr><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td></tr><tr><td>ï</td><td>ë</td><td>û</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td><em>n</em>+1</td><td>&nbsp;</td></tr><tr><td>î</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>(<em> p </em>+<em> a </em>)</td><td>&nbsp;</td></tr></tbody></table></figure>



<p class="wp-block-paragraph">On trouve donc la réponse du système dans le domaine temporel :</p>



<p class="wp-block-paragraph"><em>s</em>(<em>t</em>)=<em> KA </em>é1-<sup>(</sup>1+<em>w</em><em> t </em><sup>)</sup><em>e</em><sup>&#8211;<em>w</em></sup>0<em> </em><em><sup>t</sup></em><em> </em>ù</p>



<p class="wp-block-paragraph">ë&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; 0&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; û</p>



<p class="wp-block-paragraph">On retrouve encore une fois que l’erreur statique vaut : <em>e</em><em><sub>s</sub></em>&nbsp; = <em>A</em> (1- <em>K</em>)</p>



<p class="wp-block-paragraph">Pente à l’origine :</p>



<figure class="wp-block-table"><table><tbody><tr><td><em>s</em>¢(<em>t </em>)=<em> KA </em>é</td><td>&#8211;<em>w</em></td><td><em><sub>e</sub></em><sup>&#8211;</sup> <em><sup>w</sup></em>0<em><sup>t</sup></em>&nbsp; <sub>+</sub> <em><sub>w</sub></em></td><td>(1 + <em>w</em></td><td><em><sub>t</sub></em> <sub>)<em>e</em></sub><sup>&#8211;<em>w</em></sup>0<em><sup>t</sup></em> <sub>ù</sub></td></tr><tr><td>ë</td><td>0</td><td>0</td><td>0</td><td>û</td></tr></tbody></table></figure>



<p class="wp-block-paragraph">Donc&nbsp;&nbsp;&nbsp;&nbsp; <em>s</em>¢(0)=0</p>



<p class="wp-block-paragraph"><strong>On a aussi une pente nulle à l’origine.</strong></p>



<p class="wp-block-paragraph">Page 7&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; <strong>Emmanuel FARGES</strong>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; Ó EduKlub S.A.</p>



<p class="wp-block-paragraph">Tous droits de l’auteur des œuvres réservés. Sauf autorisation, la reproduction ainsi que toute utilisation des œuvres autre que la consultation individuelle et privée sont interdites.</p>



<p class="wp-block-paragraph"><strong>Sciences Indusrielles</strong></p>



<p class="wp-block-paragraph"><strong>Systèmes linéaires continus invariants</strong></p>



<p class="wp-block-paragraph">Analyse temporelle</p>



<p class="wp-block-paragraph"><strong><em><u>Courbe réponse :</u></em></strong></p>



<figure class="wp-block-table"><table><tbody><tr><td><strong><em>Réponse du système :</em></strong></td><td>&nbsp;</td><td><strong><em>KA</em></strong></td><td>&nbsp;</td></tr><tr><td><em>H </em>(<em> p</em>)=</td><td>&nbsp;</td><td>10</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td></tr><tr><td>1+ 20 <em>p</em> +100 <em>p</em><sup>2</sup></td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td></tr><tr><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td></tr><tr><td></td><td></td><td></td><td></td><td></td><td></td><td></td></tr></tbody></table></figure>



<p class="wp-block-paragraph">(<em>K</em> = 10,<em>x</em> = 1,<em>w</em><sub>0</sub>&nbsp; = 0,1 <em>rads</em><sup>&#8211;</sup><sup>1</sup>)</p>



<ul class="wp-block-list"><li><strong><em>une entrée échelon de taille A=1</em></strong></li></ul>



<p class="wp-block-paragraph"><strong><em>Pente nulle à l’origine</em></strong></p>



<h5 class="wp-block-heading"><strong>2.3.3.3 Deuxième ordre faiblement amorti</strong></h5>



<p class="wp-block-paragraph">C’est le cas le plus intéressant pour un système du second ordre. Le système présente donc un amortissement inférieur à 1 : <em>x</em> &lt; 1</p>



<p class="wp-block-paragraph">Le discriminant est négatif, on a alors deux racines complexes conjuguées pour le polynôme</p>



<figure class="wp-block-table"><table><tbody><tr><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>ì</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td></tr><tr><td>&nbsp;</td><td>2</td><td>2</td><td>&nbsp;</td><td><em>p</em><sub>1</sub><em> </em>= &#8211;<em>xw</em><sub>0</sub><em> </em>+<em> j</em><em>w</em><em> </em><sub>0</sub></td><td>1- <em>x</em></td><td>2</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td></tr><tr><td>&nbsp;</td><td>ï</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td></tr><tr><td><em>p</em></td><td>&nbsp;</td><td>+ 2<em>xw</em><sub>0</sub> <em>p</em> + <em>w</em><sub>0</sub></td><td>. On a alors <sub>í</sub></td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>. On peut reprendre la</td><td>&nbsp;</td></tr><tr><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td></tr><tr><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>ï <em>p</em></td><td>2</td><td>= &#8211;<em>xw</em></td><td>0</td><td>&#8211; <em>j</em><em>w</em></td><td>0</td><td>1 &#8211; <em>x</em> <sup>2</sup>&nbsp; = <em>p</em></td><td>&nbsp;</td></tr><tr><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>î</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>1</td><td>&nbsp;</td><td>&nbsp;</td></tr></tbody></table></figure>



<figure class="wp-block-table"><table><tbody><tr><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>é 1</td><td>&nbsp;</td><td><em>p </em>+2<em>xw</em><sub>0</sub></td><td>ù</td><td>&nbsp;</td></tr><tr><td>décomposition en éléments simples à partir de <em>S</em>( <em>p</em>) = <em>KA</em> <sub>ê</sub></td><td>&nbsp;</td><td>&#8211;</td><td>&nbsp;</td><td>ú</td><td>&nbsp;</td></tr><tr><td>&nbsp;</td><td><em>p </em><sup>2</sup><em> </em>+2<em>xw</em><em> </em><sub>0</sub><em> p </em>+<em>w</em><sub>0</sub><sup>2</sup></td><td>&nbsp;</td></tr><tr><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>ë <em><sup>p</sup></em></td><td>&nbsp;</td><td>û</td><td>&nbsp;</td></tr><tr><td><em>p</em><sup>2</sup><em> </em>+<em> </em>2<em>xw</em></td><td><em>p </em>+<em> </em><em>w</em><em> </em><sup>2</sup></td><td>= ( <em>p</em> &#8211; <em>p</em> )(<em>p</em> &#8211;</td><td><em>p</em></td><td>)</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td></tr><tr><td>0</td><td>0</td><td>1</td><td>1</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td></tr></tbody></table></figure>



<ul class="wp-block-list"><li>(<em> p </em>&#8211;<em> </em>Re(<em> p</em><sub>1</sub><em> </em>)-<em> j </em>Im(<em> p</em><sub>1</sub><em> </em>))(<em> p </em>&#8211;<em> </em>Re(<em> p</em><sub>1</sub><em> </em>)+<em> j </em>Im(<em> p</em><sub>1</sub><em> </em>))</li></ul>



<ul class="wp-block-list"><li>(<em> p </em>&#8211;<em> </em>Re(<em> p</em><sub>1</sub><em> </em>))<sup>2</sup><em> </em>&#8211;<em> </em>(<em> j </em>Im(<em> p</em><sub>1</sub><em> </em>))<sup>2</sup></li></ul>



<ul class="wp-block-list"><li>(<em> p </em>&#8211;<em> </em>Re(<em> p</em><sub>1</sub><em> </em>))<sup>2</sup><em> </em>+<em> </em>(Im(<em> p</em><sub>1</sub><em> </em>))<sup>2</sup></li></ul>



<ul class="wp-block-list"><li>(<em> p </em>+<em> </em><em>xw</em><sub>0</sub><em> </em>)<sup>2</sup><em>&nbsp; </em>+<em> </em>(<em>w</em><em> </em><sub>0 </sub><img loading="lazy" decoding="async" src="" width="7" height="10"><img loading="lazy" decoding="async" src="" width="5" height="22">1 &#8211;<em>x</em> <sup>2</sup> )<sup>2</sup></li></ul>



<p class="wp-block-paragraph">On présente le dénominateur sous la forme ci-dessus car on connaît les transformées de</p>



<figure class="wp-block-table"><table><tbody><tr><td><sup>ì</sup><em> L </em>é<em>e</em><sup>&#8211;</sup><em><sup>at</sup></em><em> </em>sin</td><td>(</td><td><em>w</em><em>t u </em>(<em>t </em>)ù =</td><td>&nbsp;</td><td><em>w</em></td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td></tr><tr><td>2</td><td>&nbsp;</td><td>2</td><td>&nbsp;</td></tr><tr><td>ï</td><td>ë</td><td>)</td><td>û</td><td>&nbsp;</td><td>+ <em>w</em></td><td>&nbsp;</td></tr><tr><td>ï</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>(<em> p </em>+<em> a </em>)</td><td>&nbsp;</td><td>&nbsp;</td></tr><tr><td>Laplace suivantes : <sub>í</sub></td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td><em>p </em>+<em> a</em></td><td>&nbsp;</td><td>&nbsp;</td></tr><tr><td><sup>ï</sup><em>L </em>é<em>e</em><sup>&#8211;</sup><em><sup>at</sup></em><em> </em>cos(<em>w</em><em>t </em>)<em>u </em>(<em>t </em>)ù</td><td>=</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td></tr><tr><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td></tr><tr><td>&nbsp;</td><td>2</td><td>&nbsp;</td><td>2</td><td>&nbsp;</td></tr><tr><td>ï</td><td>ë</td><td>&nbsp;</td><td>&nbsp;</td><td>û</td><td>&nbsp;</td><td>&nbsp;</td><td>(<em> p </em>+<em> a</em>)</td><td>+ <em>w</em></td><td>&nbsp;</td></tr><tr><td>î</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td></tr></tbody></table></figure>



<p class="wp-block-paragraph">On peut réécrire la décomposition en éléments simples de la façon suivante :</p>



<p class="wp-block-paragraph">Page 8&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; <strong>Emmanuel FARGES</strong>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; Ó EduKlub S.A.</p>



<p class="wp-block-paragraph">Tous droits de l’auteur des œuvres réservés. Sauf autorisation, la reproduction ainsi que toute utilisation des œuvres autre que la consultation individuelle et privée sont interdites.</p>



<p class="wp-block-paragraph"><strong>Sciences Indusrielles</strong></p>



<p class="wp-block-paragraph"><strong>Systèmes linéaires continus invariants</strong></p>



<p class="wp-block-paragraph">Analyse temporelle</p>



<figure class="wp-block-table"><table><tbody><tr><td>é 1</td><td>&nbsp;</td><td>&nbsp;</td><td><em>p </em>+2<em>xw</em><sub>0</sub></td><td>ù</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td></tr><tr><td><em>S</em>(<em> p</em>)=<em> KA </em><sub>ê</sub></td><td>&nbsp;</td><td>&#8211;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>ú</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td></tr><tr><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td></tr><tr><td><sub>ë</sub><em> p&nbsp;&nbsp; p </em><sup>2</sup><em> </em>+2<em>xw</em><em> </em><sub>0</sub><em> p </em>+<em>w</em><sub>0</sub><sup>2</sup></td><td>û</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td></tr><tr><td>é</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>ù</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td></tr><tr><td>ê 1</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td><em>p </em>+<em>xw</em><sub>0</sub></td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td><em>xw</em><sub>0</sub></td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>ú</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td></tr><tr><td>= <em>KA</em> <sub>ê</sub></td><td>&nbsp;</td><td>&#8211;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&#8211;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>ú</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td></tr><tr><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>2</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>2</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td></tr><tr><td>&nbsp;</td><td>&nbsp;</td><td><em>p </em>+<em>xw</em></td><td>&nbsp;</td><td>&nbsp;</td><td>+ <em>w</em></td><td>1- <em>x</em> <sup>2</sup></td><td>&nbsp;</td><td><em>p </em>+<em>xw</em></td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>+ <em>w</em></td><td>&nbsp;</td><td>1 &#8211;<em>x</em> <sup>2</sup></td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td></tr><tr><td>ê <em><sup>p</sup></em></td><td>&nbsp;</td><td>(</td><td>0 )</td><td>2</td><td>)</td><td>&nbsp;</td><td>(</td><td>0 )</td><td>2</td><td>&nbsp;</td><td>)</td><td>&nbsp;</td><td>ú</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td></tr><tr><td>ë</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>( 0</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>( 0</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>û</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td></tr><tr><td>é</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>ù</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td></tr><tr><td>ê 1</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td><em>p </em>+<em>xw</em><sub>0</sub></td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td><em>x</em></td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td><em>w</em> <sub>0</sub>&nbsp;&nbsp; 1-<em>x</em> <sup>2</sup></td><td>&nbsp;</td><td>&nbsp;</td><td>ú</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td></tr><tr><td>= <em>KA</em> <sub>ê</sub></td><td>&nbsp;</td><td>&#8211;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&#8211;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>ú</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td></tr><tr><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>2</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>2</td><td>&nbsp;</td></tr><tr><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>1-<em>x</em> <sup>2</sup></td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td><sup>2</sup> + <em>w</em></td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td></tr><tr><td>ê <em><sup>p</sup></em></td><td>&nbsp;</td><td>(</td><td><em>p </em>+<em>xw</em></td><td>0 )</td><td>2</td><td>+ <em>w</em></td><td>1- <em>x</em> <sup>2</sup></td><td>)</td><td>&nbsp;</td><td>&nbsp;</td><td>(</td><td><em>p </em>+<em>xw</em></td><td>0 )</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>1-<em>x</em> <sup>2</sup></td><td>)</td><td>ú</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td></tr><tr><td>ë</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>( 0</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>(</td><td>&nbsp;</td><td>0</td><td>&nbsp;</td><td>&nbsp;</td><td>û</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td></tr><tr><td>Avec le rappel sur les transformées de Laplace ci-dessus ( <em>a</em> = <em>xw</em><sub>0</sub></td><td>et</td><td><em>w </em>=<em> w </em><sub>0</sub></td><td>1- <em>x</em> <sup>2</sup>&nbsp; ) , on</td><td>&nbsp;</td></tr></tbody></table></figure>



<p class="wp-block-paragraph">obtient la réponse d’un deuxième ordre faiblement amorti à un échelon de taille A :</p>



<figure class="wp-block-table"><table><tbody><tr><td>é</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>( 0</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>)</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td><em>x</em></td><td>&nbsp;</td><td>&nbsp;</td><td>2</td><td>&nbsp;</td><td>( 0</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>)</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>ù</td><td>&nbsp;</td><td>&nbsp;</td></tr><tr><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td></tr><tr><td>ê</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>1 &#8211;<em>x</em></td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>ú</td><td>&nbsp;</td><td>&nbsp;</td></tr><tr><td><em>s</em>(<em>t</em>)=<em> KA </em>ê1-cos</td><td><em>w</em></td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>1- <em>x</em> <sup>2</sup> <em>t</em></td><td>&nbsp;</td><td><em><sub>e</sub></em><sup>&#8211;<em>xw</em></sup>0 <em><sup>t</sup></em></td><td>&#8211;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>sin</td><td><em>w</em></td><td>1-<em>x</em></td><td>2 <em><sub>t</sub></em></td><td>&nbsp;</td><td><em><sub>e</sub></em><sup>&#8211;<em>xw</em></sup>0 <em><sup>t</sup></em> <sub>ú</sub></td><td>&nbsp;</td><td>&nbsp;</td></tr><tr><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td></tr><tr><td>ë</td><td><em>e</em></td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>û</td><td>&nbsp;</td><td>&nbsp;</td></tr><tr><td>é</td><td>&#8211;<em>xw</em><sub>0</sub><em>t</em></td><td>&nbsp;</td><td>é</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>ù</td><td>ù</td><td>&nbsp;</td><td>&nbsp;</td></tr><tr><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>2</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>2</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>2</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td></tr><tr><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td></tr><tr><td><em>s</em>(<em>t</em>)=<em> KA </em>ê1-</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>ê</td><td>1 &#8211;<em>x</em></td><td>&nbsp;</td><td>cos (<em>w</em><sub>0</sub>&nbsp;&nbsp; 1 &#8211;<em>x</em></td><td>&nbsp;</td><td><em>t </em>)+<em>x</em><em> </em>sin(<em>w</em><sub>0</sub></td><td>1 &#8211;<em>x</em></td><td>&nbsp;</td><td><em><sup>t </sup></em>)<sub>ú</sub></td><td>ú</td><td>&nbsp;</td><td>&nbsp;</td></tr><tr><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td></tr><tr><td>&nbsp;</td><td>1- <em>x</em></td><td>2</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td></tr><tr><td>ê</td><td>&nbsp;</td><td>&nbsp;</td><td>ë</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>û</td><td>ú</td><td>&nbsp;</td><td>&nbsp;</td></tr><tr><td>ë</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>û</td><td>&nbsp;</td><td>&nbsp;</td></tr><tr><td>Or, comme <em>x</em> &lt; 1 , on peut poser</td><td><em>x</em> = cos <em>j</em></td><td>.</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td></tr><tr><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td></tr><tr><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td></tr><tr><td>On remarquera alors que :</td><td>sin<em>j</em> =</td><td>1- cos<sup>2</sup> <em>j</em> =</td><td>1- <em>x</em> <sup>2</sup></td><td>d’où :</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td></tr><tr><td>é</td><td><em>e</em></td><td>&#8211;<em>xw</em><sub>0</sub><em>t</em></td><td>&nbsp;</td><td>é</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>ù</td><td>ù</td><td>&nbsp;</td></tr><tr><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>2</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>2</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td></tr><tr><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td></tr><tr><td><em>s</em>(<em>t</em>)=<em> KA </em>ê1-</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td><sub>ê</sub> sin<em>j</em> cos (<em>w</em><sub>0</sub></td><td>1- <em>x</em></td><td>&nbsp;</td><td><em>t </em>)+cos<em>j</em><em> </em>sin(<em>w</em><sub>0</sub></td><td>&nbsp;</td><td>&nbsp;</td><td>1- <em>x</em></td><td>&nbsp;</td><td>&nbsp;</td><td><em><sup>t </sup></em>)<sub>ú</sub></td><td>ú</td><td>&nbsp;</td></tr><tr><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td></tr><tr><td>&nbsp;</td><td>1- <em>x</em></td><td>2</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td></tr><tr><td>ê</td><td>&nbsp;</td><td>&nbsp;</td><td>ë</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>û</td><td>ú</td><td>&nbsp;</td></tr><tr><td>ë</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>û</td><td>&nbsp;</td></tr><tr><td>é</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&#8211;<em>xw</em><sub>0</sub><em> t</em></td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>ù</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td></tr><tr><td><em>s</em>(<em>t</em>)=<em> KA </em>ê1-</td><td><em>e</em></td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>sin <sub>ê</sub><sup>é</sup>(<em>w</em><sub>0</sub></td><td>1 &#8211;<em>x</em> <sup>2</sup> <em>t</em> )+<em>j</em> <sub>ú</sub><sup>ù</sup></td><td>ú</td><td>avec</td><td><em>x</em> = cos <em>j</em></td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td></tr><tr><td>&nbsp;</td><td>1- <em>x</em></td><td>2</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td></tr><tr><td>ê</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>ë</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>û</td><td>ú</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td></tr><tr><td>ë</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>û</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td></tr></tbody></table></figure>



<p class="wp-block-paragraph">C’est un système oscillant (fonction sinusoïdale) amorti (puisque le sinus est multiplié par une exponentielle décroissante)</p>



<p class="wp-block-paragraph">On parle alors de <strong><em><u>système oscillant amorti</u></em></strong>.</p>



<p class="wp-block-paragraph">Page 9&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; <strong>Emmanuel FARGES</strong>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; Ó EduKlub S.A.</p>



<p class="wp-block-paragraph">Tous droits de l’auteur des œuvres réservés. Sauf autorisation, la reproduction ainsi que toute utilisation des œuvres autre que la consultation individuelle et privée sont interdites.</p>



<p class="wp-block-paragraph"><strong>Systèmes linéaires continus invariants</strong></p>



<p class="wp-block-paragraph">Analyse temporelle</p>



<p class="wp-block-paragraph"><strong><em><u>Courbe réponse :</u></em></strong></p>



<figure class="wp-block-table"><table><tbody><tr><td><strong><em>Dépassement D</em></strong><strong><em><sub>1</sub></em></strong></td><td><strong><em>Réponse du système :</em></strong></td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td></tr><tr><td><em>H </em>(<em> p</em>)=</td><td>&nbsp;</td><td>10</td><td>&nbsp;</td><td>(<em>K </em>=10;<em>x</em><em> </em>=0,1;<em>w</em><sub>0</sub><em>&nbsp; </em>=0,1<em> rads</em><sup>&#8211;</sup><sup>1</sup><em> </em>)</td><td>&nbsp;</td></tr><tr><td>&nbsp;</td><td>1+ 2 <em>p</em> +100 <em>p</em></td><td>2</td><td>&nbsp;</td></tr><tr><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td></tr><tr><td></td><td></td><td></td><td></td><td></td><td></td><td></td></tr></tbody></table></figure>



<ul class="wp-block-list"><li><strong><em>un échelon de taille A=1. Dépassement D</em></strong><strong><em><sub>3</sub></em></strong><strong><em></em></strong></li></ul>



<figure class="wp-block-table"><table><tbody><tr><td><strong><em>Oscillations amorties</em></strong></td><td><strong>Asymptote KA</strong></td><td>&nbsp;</td></tr><tr><td>&nbsp;</td><td>&nbsp;</td></tr></tbody></table></figure>



<p class="wp-block-paragraph"><strong><em>Pseudopériode T</em></strong></p>



<p class="wp-block-paragraph"><strong><em>Pente nulle à l’origine</em></strong></p>



<p class="wp-block-paragraph"><strong><em><u>Caractéristiques essentielles :</u></em></strong></p>



<p class="wp-block-paragraph"><strong><em><u>Pseudopériode :</u></em></strong>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; Notée T, c’est la période des oscillations : <em>T</em> =</p>



<p class="wp-block-paragraph">2<em>p</em></p>



<p class="wp-block-paragraph"><em>w</em> <sub>0 </sub><img loading="lazy" decoding="async" src="" width="6" height="9"><img loading="lazy" decoding="async" src="" width="5" height="19">1 &#8211;<em>x</em> <sup>2</sup></p>



<p class="wp-block-paragraph">Elle dépend de la pulsation propre du système <em>w</em><sub>0</sub> mais aussi de son coefficient d’amortissement <em>x</em> .</p>



<p class="wp-block-paragraph"><strong><em><u>Valeur asymptotique :</u></em></strong><strong><em> </em></strong>Le système s’amortit pour atteindre la<strong><em> </em></strong><strong>valeur asymptotique KA</strong></p>



<p class="wp-block-paragraph"><strong><em><u>Dépassement :</u></em></strong>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; Les dépassements correspondent aux écarts maximum entre la courbe réponse (maximum et minimum d’oscillation) et la valeur</p>



<figure class="wp-block-table"><table><tbody><tr><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>asymptotique KA. Ils se situent donc aux instants <em>t</em><em><sub>k</sub></em> tels que</td><td>&nbsp;</td></tr><tr><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td><em>s</em>¢(<em>t</em><em><sub>k</sub></em></td><td>)= 0.</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td></tr><tr><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td><strong>Le dépassement se situant à l’instant </strong><em>t</em><em><sub>k</sub></em><strong>&nbsp; est noté </strong><em>D</em><em><sub>k</sub></em></td><td>&nbsp;</td></tr><tr><td>Déterminons les instants</td><td><em>t</em><em><sub>k</sub></em><em>&nbsp;&nbsp; </em>:</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td></tr><tr><td>&nbsp;</td><td>é</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&#8211;<em>xw</em><sub>0</sub><em> t</em></td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>ù</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td></tr><tr><td><em>s</em>(<em>t</em>)=<em> KA </em>ê1-</td><td>&nbsp;</td><td>&nbsp;</td><td><em>e</em></td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>sin</td><td><sub>ê</sub><sup>é</sup>(<em><sup>w</sup></em>0</td><td>1 &#8211;<em>x</em></td><td><sup>2</sup><em> t </em>)</td><td>+<em>j</em> <sub>ú</sub><sup>ù</sup></td><td>ú</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td></tr><tr><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td></tr><tr><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>1- <em>x</em></td><td>2</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td></tr><tr><td>&nbsp;</td><td>ê</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>ë</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>û</td><td>ú</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td></tr><tr><td>&nbsp;</td><td>ë</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>û</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td></tr><tr><td>&nbsp;</td><td>&nbsp;</td><td><em>KA</em></td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>é</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>ù</td><td>&nbsp;</td></tr><tr><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td><em><sub>e</sub></em><sup>&#8211;<em>xw</em></sup>0 <em><sup>t</sup></em></td><td>ê</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>ú</td><td>&nbsp;</td></tr><tr><td><em>s</em>¢(<em>t </em>)=</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td><em>x</em></td><td><em>w</em></td><td>&nbsp;</td><td>sin</td><td>&nbsp;</td><td><em>w</em></td><td>&nbsp;</td><td>&nbsp;</td><td>1 &#8211;<em>x</em> <sup>2</sup> <em>t</em> +<em>j</em></td><td>&nbsp;</td><td>&#8211; <em>w</em></td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>1- <em>x</em> <sup>2</sup></td><td>cos <em>w</em></td><td>&nbsp;</td><td>&nbsp;</td><td>1- <em>x</em> <sup>2</sup> <em>t</em> +<em>j</em></td><td>&nbsp;</td></tr><tr><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td></tr><tr><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td></tr><tr><td>&nbsp;</td><td>1-<em>x</em> <sup>2</sup></td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>ê</td><td>{</td><td>&nbsp;</td><td>0</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>(</td><td>&nbsp;</td><td>&nbsp;</td><td>0</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>)</td><td>&nbsp;</td><td>0</td><td><s>14</s>243</td><td>(</td><td>&nbsp;</td><td>0</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>)ú</td><td>&nbsp;</td></tr><tr><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>ê</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>sin <em>j</em></td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>ú</td><td>&nbsp;</td></tr><tr><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td><sub>ë</sub> cos<em>j</em></td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>û</td><td>&nbsp;</td></tr><tr><td>&nbsp;</td><td><em>KA</em><em>w</em><sub>0</sub></td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&#8211;<em>xw</em><sub>0</sub><em> t</em></td><td>é</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>ù</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td></tr><tr><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>2</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>2</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td></tr><tr><td><em>s</em>¢(<em>t </em>)=</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td><em>e</em></td><td>&nbsp;</td><td><sub>ê</sub> cos<em>j</em> sin (<em>w</em><sub>0</sub></td><td>1-<em>x</em></td><td>&nbsp;</td><td><em>t </em>+<em>j</em>)-sin<em>j</em><em> </em>cos(<em>w</em><sub>0</sub></td><td>1 &#8211; <em>x</em></td><td>&nbsp;</td><td><em>t</em></td><td><sup>+</sup><em><sup>j</sup></em><sup> </sup>)<sub>ú</sub></td><td>&nbsp;</td><td>&nbsp;</td></tr><tr><td>&nbsp;</td><td>&nbsp;</td><td>2</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td></tr><tr><td>&nbsp;</td><td>1-<em>x</em></td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>ë</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>û</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td></tr><tr><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td></tr><tr><td>Page 10</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td><strong>Emmanuel FARGES</strong></td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>Ó EduKlub S.A.</td><td>&nbsp;</td></tr></tbody></table></figure>



<p class="wp-block-paragraph">Tous droits de l’auteur des œuvres réservés. Sauf autorisation, la reproduction ainsi que toute utilisation des œuvres autre que la consultation individuelle et privée sont interdites.</p>



<p class="wp-block-paragraph"><strong>Sciences Indusrielles</strong></p>



<p class="wp-block-paragraph"><strong>Systèmes linéaires continus invariants</strong></p>



<p class="wp-block-paragraph">Analyse temporelle</p>



<figure class="wp-block-table"><table><tbody><tr><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td><em>k</em><em>p</em></td><td>&nbsp;</td><td>&nbsp;</td></tr><tr><td><em>s</em>¢(<em>t</em><em><sub>k</sub></em><em>&nbsp; </em>)=0Û<em>w</em><sub>0</sub><em>&nbsp;&nbsp; </em>1-<em>x</em><em> </em><sup>2</sup><em> t</em><em><sub>k</sub></em><em>&nbsp; </em>=<em> k </em><em>p</em><em>&nbsp; </em>avec<em> k </em>Î<strong>Z</strong><sup>+</sup><em>&nbsp; </em>soit</td><td><em>t</em><em><sub>k</sub></em></td><td>=</td><td>&nbsp;</td></tr><tr><td><em>w</em> <sub>0</sub>&nbsp;&nbsp; 1- <em>x</em> <sup>2</sup></td><td>&nbsp;</td></tr><tr><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td></tr><tr><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td></tr><tr><td></td><td></td><td></td><td></td><td></td><td></td><td></td><td></td><td></td><td></td></tr></tbody></table></figure>



<p class="wp-block-paragraph"><strong>On remarque que à t = 0 on a une dérivée nulle, c’est à dire que la courbe réponse présente une pente nulle à l’origine.</strong></p>



<p class="wp-block-paragraph">Calcul des dépasseme nts <em>D</em><em><sub>k</sub></em></p>



<figure class="wp-block-table"><table><tbody><tr><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td><em>KA</em></td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td></tr><tr><td><em>D</em><em><sub>k</sub></em><em>&nbsp; </em>=</td><td><em>s</em>(<em>t</em><em><sub>k</sub></em><em> </em>)-<em> KA</em></td><td>=</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td><em><sub>e</sub></em> <sup>&#8211;<em>xw</em></sup>0 <em><sup>t</sup></em><em>k</em></td><td>sin (<em>w</em> <sub>0</sub>&nbsp;&nbsp; 1- <em>x</em> <sup>2</sup> <em>t</em><em><sub>k</sub></em>&nbsp; + <em>j</em>)</td><td>&nbsp;</td></tr><tr><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td></tr><tr><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td></tr><tr><td>1 &#8211;<em>x</em></td><td>2</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td></tr><tr><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td></tr><tr><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>æ</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>ö</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td></tr><tr><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td><em>x</em><em> k</em><em>p</em></td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td></tr><tr><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&#8211;<sup>ç</sup></td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>÷</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td></tr><tr><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td></tr><tr><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>ç</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>2</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>÷</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td></tr><tr><td>&nbsp;</td><td>&nbsp;</td><td><em>KA</em></td><td>ç</td><td>1</td><td>&#8211; <em>x</em></td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>÷</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td></tr><tr><td>=</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td><em>e </em><sup>è</sup></td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>ø</td><td>&nbsp;</td><td>&nbsp;</td><td>± sin <em>j</em></td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td></tr><tr><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td></tr><tr><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td></tr><tr><td>&nbsp;</td><td>&nbsp;</td><td>1- <em>x</em></td><td>2</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>14243</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td></tr><tr><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td></tr><tr><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td></tr><tr><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>1-<em>x</em> <sup>2</sup></td><td>&nbsp;</td></tr><tr><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td></tr><tr><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td></tr><tr><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>æ</td><td>&nbsp;</td><td>&nbsp;</td><td><em>x</em><em> k</em><em>p</em></td><td>&nbsp;</td><td>&nbsp;</td><td>ö</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td></tr><tr><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&#8211; <sup>ç</sup></td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>÷</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td></tr><tr><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>ç</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>2</td><td>&nbsp;</td><td>÷</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td></tr><tr><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>ç</td><td>&nbsp;</td><td>1- <em>x</em></td><td>÷</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td></tr><tr><td>D’où :</td><td><em>D</em><em><sub>k</sub></em><em>&nbsp; </em>=<em> KA </em><em>e</em><em> </em><sup>è</sup></td><td>&nbsp;</td><td>ø</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td></tr></tbody></table></figure>



<p class="wp-block-paragraph">On remarque qu’à la valeur asymptotique KA près, les dépassements ne dépendent que de l’amortissement <em>x</em> . La mesure de D<sub>1</sub> sur la réponse du deuxième ordre faiblement amorti</p>



<p class="wp-block-paragraph">permet ainsi d’identifier (déterminer) l’amortissement de celui-ci. On mesure en général D<sub>1</sub> car c’est le plus marqué, c’est à dire celui sur lequel on fait le moins d’erreur de mesure. En effet l’amortissement est très difficilement prévisible à priori. Pour le quantifier avec précision il faut donc réaliser cet essai (réponse indiciel).</p>



<figure class="wp-block-table"><table><tbody><tr><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>æ</td><td>&nbsp;</td><td><em>xp</em></td><td>&nbsp;</td><td>ö</td><td>&nbsp;</td></tr><tr><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&#8211; ç</td><td>&nbsp;</td><td>&nbsp;</td><td>÷</td><td>&nbsp;</td></tr><tr><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>ç</td><td>&nbsp;</td><td>&nbsp;</td><td>2</td><td>&nbsp;</td><td>÷</td><td>&nbsp;</td></tr><tr><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>ç</td><td>1 &#8211;<em>x</em></td><td>÷</td><td>&nbsp;</td></tr><tr><td><strong><em>Premier dépassement D</em></strong></td><td><strong><em>1</em></strong></td><td><strong><em>:</em></strong></td><td><em>D</em><sub>1</sub><em> </em>=<em> KA </em><em>e</em><em> </em><sup>è</sup></td><td>&nbsp;</td><td>&nbsp;</td><td><sup>ø</sup>&nbsp; avecAtailledel&rsquo;échelon</td><td>&nbsp;</td></tr><tr><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td></tr></tbody></table></figure>



<p class="wp-block-paragraph"><strong><em><u>Temps à 5% :</u></em></strong></p>



<p class="wp-block-paragraph">Il quantifie encore une fois la rapidité du système. Il existe des courbes (fournies avec les sujets de concours si il y a lieu) qui donne la valeur de t<sub>5%</sub> en fonction du coefficient d’amortissement. En effet pour un système oscillant le temps à 5% prend toute sa signification : C’est <strong><em>le temps à partir duquel les oscillations restent comprises dans une</em></strong> <strong><em>bande à + ou – 5% par rapport à la valeur asymptotique KA :</em></strong></p>



<p class="wp-block-paragraph">Page 11&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; <strong>Emmanuel FARGES</strong>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; Ó EduKlub S.A.</p>



<p class="wp-block-paragraph">Tous droits de l’auteur des œuvres réservés. Sauf autorisation, la reproduction ainsi que toute utilisation des œuvres autre que la consultation individuelle et privée sont interdites.</p>



<p class="wp-block-paragraph"><strong>Sciences Indusrielles</strong></p>



<p class="wp-block-paragraph"><strong>Systèmes linéaires continus invariants</strong></p>



<p class="wp-block-paragraph">Analyse temporelle</p>



<p class="wp-block-paragraph"><strong><em>Asymptote KA</em></strong></p>



<p class="wp-block-paragraph"><strong><em>KA+5%=1,05KA</em></strong></p>



<p class="wp-block-paragraph"><strong>KA-5%=0,95KA</strong></p>



<p class="wp-block-paragraph"><strong><em>Temps à 5% : </em></strong><strong><em>t</em></strong><strong><em><sub>5%</sub></em></strong></p>



<p class="wp-block-paragraph"><strong><em><u>Influence de l’amortissement d’un deuxième ordre :</u></em></strong></p>



<p class="wp-block-paragraph">Regardons ci-dessous les réponses de systèmes de gain (K=10) et de pulsation propre</p>



<p class="wp-block-paragraph">(<em>w</em><sub>0</sub> = 0,1 <em>rads</em> <sup>&#8211;</sup><sup>1</sup> ) identiques à une entrée identique (échelon de taille A=1) mais de coefficient d’amortissement différent</p>



<ul class="wp-block-list"><li>= 0,1 <em>x</em> = 0 , 5<em></em></li></ul>



<p class="wp-block-paragraph"><strong><em>Asymptote commune KA</em></strong></p>



<p class="wp-block-paragraph"><em>x</em> = 2</p>



<ul class="wp-block-list"><li>= 1<em></em></li></ul>



<h3 class="wp-block-heading"><strong>2.4 Réponse à une rampe</strong></h3>



<p class="wp-block-paragraph">Le but de cette étude est d’évaluer un système du deuxième ordre au niveau dynamique. Fondamentalement, les différents types de réponses ne différant pas énormément, nous ne développerons pas autant les calculs que pour les réponses à un échelon.</p>



<p class="wp-block-paragraph">Page 12&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; <strong>Emmanuel FARGES</strong>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; Ó EduKlub S.A.</p>



<p class="wp-block-paragraph">Tous droits de l’auteur des œuvres réservés. Sauf autorisation, la reproduction ainsi que toute utilisation des œuvres autre que la consultation individuelle et privée sont interdites.</p>



<p class="wp-block-paragraph"><strong>Sciences Indusrielles</strong></p>



<p class="wp-block-paragraph"><strong>Systèmes linéaires continus invariants</strong></p>



<p class="wp-block-paragraph">Analyse temporelle</p>



<h4 class="wp-block-heading"><strong>2.4.1 Réponse dans le domaine fréquentielle</strong></h4>



<p class="wp-block-paragraph">On soumet le système à une entrée rampe de pente V. L’entrée est donc la fonction du temps :</p>



<p class="wp-block-paragraph"><strong>e(t)= V t u(t) </strong>avec u(t) l’échelon unitaire.</p>



<p class="wp-block-paragraph">e (t)</p>



<figure class="wp-block-table"><table><tbody><tr><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>Vt</td><td>&nbsp;</td></tr><tr><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td></tr><tr><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>t=0</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td></tr><tr><td>e(t)=V t u(t) donc&nbsp; <em>E</em> ( <em>p</em>) =</td><td><em>V</em></td><td>&nbsp;</td><td>&nbsp;</td><td>or</td><td>&nbsp;</td></tr><tr><td><em>p</em><sup>2</sup></td><td>&nbsp;</td><td>&nbsp;</td></tr><tr><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td></tr><tr><td><em>H </em>(<em> p</em>)=</td><td><em>S</em>(<em> p</em>)</td><td>=</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td><em>K</em></td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>Þ <em>S</em> ( <em>p</em>) =</td><td>&nbsp;</td></tr><tr><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>2<em>x</em></td><td>&nbsp;</td><td>&nbsp;</td><td><em>p</em><sup>2</sup></td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td></tr><tr><td>&nbsp;</td><td>&nbsp;</td><td><em>E </em>(<em> p</em>)</td><td>1+</td><td><em>p </em>+</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td></tr><tr><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td><em>w</em><sub>0</sub></td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td></tr><tr><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td><em>w</em><sub>0</sub></td><td>2</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td></tr></tbody></table></figure>



<p class="wp-block-paragraph">Temps&nbsp;&nbsp;&nbsp; t</p>



<p class="wp-block-paragraph"><em>K V</em></p>



<figure class="wp-block-table"><table><tbody><tr><td><em>p </em><sup>2</sup></td><td>æ</td><td>+</td><td>2<em>x</em></td><td><em>p </em>+</td><td><em>p</em><sup>2</sup></td><td>ö</td><td>&nbsp;</td></tr><tr><td>ç <sup>1</sup></td><td>&nbsp;</td><td>&nbsp;</td><td>÷</td><td>&nbsp;</td></tr><tr><td><em>w</em><sub>0</sub></td><td>2</td><td>&nbsp;</td></tr><tr><td>&nbsp;</td><td>è</td><td>&nbsp;</td><td>&nbsp;</td><td><em>w</em><sub>0</sub></td><td>ø</td><td>&nbsp;</td></tr></tbody></table></figure>



<p class="wp-block-paragraph">Il reste à trouver la décomposition en éléments simples de ce rapport polynomial.</p>



<h4 class="wp-block-heading"><strong>2.4.2 Première décomposition en éléments simples</strong></h4>



<figure class="wp-block-table"><table><tbody><tr><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>é</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>ù</td><td>&nbsp;</td></tr><tr><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td><em>K V</em></td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>ê</td><td><em>a</em></td><td>&nbsp;</td><td><em>b</em></td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td><em>cp </em>+<em> d</em></td><td>&nbsp;</td><td>&nbsp;</td><td>ú</td><td>&nbsp;</td></tr><tr><td><em>S</em>(<em> p</em>)=</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>= <em>KV</em> <sup>ê</sup></td><td>+</td><td>+</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>ú</td><td>&nbsp;</td></tr><tr><td>&nbsp;</td><td>æ</td><td>&nbsp;</td><td>2<em>x</em></td><td>&nbsp;</td><td><em>p</em><sup>2</sup><em>&nbsp; </em>ö</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>2<em>x</em></td><td>&nbsp;</td><td><em>p</em><sup>2</sup></td><td>&nbsp;</td><td>&nbsp;</td></tr><tr><td>&nbsp;</td><td><em>p </em><sup>2</sup></td><td>+</td><td><em>p </em>+</td><td>ê <em>p</em> <sup>2</sup></td><td>&nbsp;</td><td><em>p</em></td><td>1+</td><td><em>p </em>+</td><td>ú</td><td>&nbsp;</td></tr><tr><td>&nbsp;</td><td>ç <sup>1</sup></td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>÷</td><td>ê</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>ú</td><td>&nbsp;</td></tr><tr><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td><em>w</em><sub>0</sub></td><td><em>w</em></td><td>2</td><td>&nbsp;</td></tr><tr><td><em>w</em></td><td><em>w</em></td><td>2</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td></tr><tr><td>&nbsp;</td><td>&nbsp;</td><td>è</td><td>&nbsp;</td><td>&nbsp;</td><td>ø</td><td>ë</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>0</td><td>û</td><td>&nbsp;</td></tr><tr><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>0</td><td>&nbsp;</td><td>0</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td></tr></tbody></table></figure>



<p class="wp-block-paragraph"><em>KVa </em>=lim<em> p </em><sup>2</sup><em> S</em>(<em> p</em>)=<em> KV </em>Þ<em>a</em><em> </em>=1</p>



<ul class="wp-block-list"><li>®0</li></ul>



<figure class="wp-block-table"><table><tbody><tr><td>é</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>ù</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>2<em>x</em></td><td>&nbsp;</td><td><em>p </em><sup>2</sup></td><td>&nbsp;</td><td>æ</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>2<em>x</em></td><td>&nbsp;</td><td>&nbsp;</td><td><em>p</em><sup>2</sup></td><td>ö</td><td>&nbsp;</td><td>3</td><td>&nbsp;</td><td>2</td><td>&nbsp;</td></tr><tr><td>ê</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>ú</td><td>&nbsp;</td><td>1 +</td><td>&nbsp;</td><td><em>p </em>+</td><td>&nbsp;</td><td>+ <em>bp</em> <sub>ç</sub> 1</td><td>+</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td><em>p </em>+</td><td>&nbsp;</td><td>&nbsp;</td><td>÷</td><td>+ <em>cp</em></td><td>&nbsp;</td><td>+ <em>dp</em></td><td>&nbsp;</td><td>&nbsp;</td></tr><tr><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td><em>w </em><sub>0</sub></td><td>2</td><td>&nbsp;</td><td><em>w</em><sub>0</sub></td><td>&nbsp;</td><td>2</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td></tr><tr><td><em>S</em>(<em> p</em>)=<em> KV </em><sup>ê</sup></td><td>1</td><td>+</td><td><em>b</em></td><td>+</td><td>&nbsp;</td><td><em>cp </em>+<em> d</em></td><td>&nbsp;</td><td>&nbsp;</td><td>ú</td><td>= <em>KV</em></td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td><em>w </em><sub>0</sub></td><td>&nbsp;</td><td>è</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td><em>w</em><sub>0</sub></td><td>ø</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td></tr><tr><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>2<em>x</em></td><td>&nbsp;</td><td><em>p</em><sup>2</sup></td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>æ</td><td>&nbsp;</td><td>2<em>x</em></td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td><em>p</em><sup>2</sup></td><td>ö</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td></tr><tr><td>ê <em>p</em> <sup>2</sup></td><td>&nbsp;</td><td><em>p</em></td><td>1+</td><td><em>p </em>+</td><td>ú</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td><em>p </em><sup>2</sup></td><td>+</td><td>&nbsp;</td><td><em>p </em>+</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td></tr><tr><td>ê</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>ú</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>ç <sup>1</sup></td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>÷</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td></tr><tr><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td><em>w</em><sub>0</sub></td><td>2</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td></tr><tr><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>2</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td></tr><tr><td>ë</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td><em>w</em><sub>0</sub></td><td>û</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>è</td><td>&nbsp;</td><td><em>w</em><sub>0</sub></td><td>&nbsp;</td><td><em>w</em><sub>0</sub></td><td>ø</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td></tr></tbody></table></figure>



<p class="wp-block-paragraph">D’où le système à trois équations, trois inconnues :</p>



<figure class="wp-block-table"><table><tbody><tr><td>ì</td><td><em>c </em>+</td><td><em>b</em></td><td>= 0</td><td>ì</td><td><em>c</em></td><td>=</td><td>2<em>x</em></td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td></tr><tr><td>ï</td><td>&nbsp;</td><td>&nbsp;</td><td>ï</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td></tr><tr><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td><em>w</em><sub>0</sub><sup>3</sup></td><td>&nbsp;</td></tr><tr><td><em>w</em><sup>2</sup></td><td>&nbsp;</td></tr><tr><td>ï</td><td>&nbsp;</td><td>2<em>x</em> <em>b</em></td><td>0</td><td>1</td><td>&nbsp;</td><td>ï</td><td>&nbsp;</td><td>&nbsp;</td><td>4<em>x</em> <sup>2</sup> -1</td><td>&nbsp;</td></tr><tr><td><strong>ï</strong></td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>ï</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td></tr><tr><td>í <em>d</em> +</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>+</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>= 0 Þ í<em>d</em> =</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td></tr><tr><td>&nbsp;</td><td><em>w</em><sub>0</sub></td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td><em>w</em><sub>0</sub><sup>2</sup></td><td>&nbsp;</td><td>&nbsp;</td><td><em>w</em><sub>0</sub></td><td>2</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td></tr><tr><td>ï</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>ï</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td></tr><tr><td>ï</td><td><em>b </em>+</td><td>2<em>x</em></td><td>= 0</td><td>ï</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>2<em>x</em></td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td></tr><tr><td>ï</td><td>ï <em>b</em> = &#8211;</td><td>&nbsp;</td><td>&nbsp;</td></tr><tr><td><em>w</em><sub>0</sub></td><td>&nbsp;</td><td>&nbsp;</td></tr><tr><td>ï</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>ï</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td><em>w</em></td><td>0</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td></tr><tr><td>î</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>î</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td></tr><tr><td>D’où la première décomposition en éléments</td><td>&nbsp;</td></tr></tbody></table></figure>



<figure class="wp-block-table"><table><tbody><tr><td>é</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>2<em>x</em></td><td><em>p </em>+4<em>x</em></td><td>2</td><td>-1</td><td>&nbsp;</td><td>ù</td><td>&nbsp;</td><td>&nbsp;</td></tr><tr><td>ê <sub>1</sub></td><td>&nbsp;</td><td>2<em>x</em></td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>ú</td><td>&nbsp;</td><td>&nbsp;</td></tr><tr><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td><em>w</em></td><td>0</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td></tr><tr><td>simples <em>S</em>( <em>p</em>) = <em>KV</em> ê</td><td>&nbsp;</td><td>&nbsp;</td><td>&#8211;</td><td>&nbsp;</td><td>+</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>ú</td><td>&nbsp;</td><td>&nbsp;</td></tr><tr><td>&nbsp;</td><td>2</td><td><em>w</em><sub>0</sub><em> p</em></td><td>2</td><td>+ 2<em>xw</em><sub>0</sub> <em>p</em> + <em>p</em></td><td>2</td><td>&nbsp;</td><td>&nbsp;</td></tr><tr><td>ê <em><sup>p</sup></em></td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td><em>w</em><sub>0</sub></td><td>&nbsp;</td><td>ú</td><td>&nbsp;</td><td>&nbsp;</td></tr><tr><td>ê</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>ú</td><td>&nbsp;</td><td>&nbsp;</td></tr><tr><td>ë</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>û</td><td>&nbsp;</td><td>&nbsp;</td></tr><tr><td>Page 13</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td><strong>Emmanuel FARGES</strong></td><td>Ó EduKlub S.A.</td><td>&nbsp;</td></tr></tbody></table></figure>



<p class="wp-block-paragraph">Tous droits de l’auteur des œuvres réservés. Sauf autorisation, la reproduction ainsi que toute utilisation des œuvres autre que la consultation individuelle et privée sont interdites.</p>



<p class="wp-block-paragraph"><strong>Sciences Indusrielles</strong></p>



<p class="wp-block-paragraph"><strong>Systèmes linéaires continus invariants</strong></p>



<p class="wp-block-paragraph">Analyse temporelle</p>



<p class="wp-block-paragraph"><strong><em>On peut d’ores et déjà déterminer pour tous les cas de figures (amortissement faible, limite et fort) l’asymptote vers laquelle tend la réponse du système et en déduire les deux critères de performances dynamique que sont le temps de retard </em></strong><em>t</em><em><sub>d</sub></em><strong><em> et l’erreur dynamique (ou de</em></strong></p>



<p class="wp-block-paragraph"><strong><em>traînage) </em></strong><em>e</em><em><sub>t</sub></em></p>



<p class="wp-block-paragraph">D’après le théorème de la valeur finale , ce qui se passe à l’infini dans le domaine temporel est équivalent à ce qui ce passe en 0 dans le domaine de Laplace. Donc si on veut un équivalent pour s(t) quand t tend vers l’infini, il nous suffit de trouver un équivalent pour <em>S </em>(<em> p</em>)quand p tend vers 0.</p>



<figure class="wp-block-table"><table><tbody><tr><td>é</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>2<em>x</em></td><td><em>p </em>+(4<em>x</em></td><td>2</td><td>-1)</td><td>ù</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td></tr><tr><td>ê</td><td>1</td><td>&nbsp;</td><td>2<em>x</em></td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>ú</td><td>&nbsp;</td><td>&nbsp;</td><td>é</td><td>1</td><td>&nbsp;</td><td>2<em>x</em> ù</td><td>&nbsp;</td></tr><tr><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td><em>w</em></td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td></tr><tr><td><em>S</em>(<em> p</em>)=<em> KV </em>ê</td><td>&nbsp;</td><td>&nbsp;</td><td>&#8211;</td><td>&nbsp;</td><td>+</td><td>0</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>ú</td><td>»</td><td><em>KV</em></td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&#8211;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td></tr><tr><td>&nbsp;</td><td>2</td><td>&nbsp;</td><td>2</td><td>&nbsp;</td><td>&nbsp;</td><td>2</td><td>ê</td><td>&nbsp;</td><td>2</td><td>ú</td><td>&nbsp;</td></tr><tr><td>ê <em><sup>p</sup></em></td><td>&nbsp;</td><td><em>p</em><em>w</em><sub>0</sub></td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>ú</td><td>{</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td></tr><tr><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td><em>w</em><sub>0</sub></td><td>+ 2<em>xw</em><sub>0</sub> <em>p</em> + <em>p</em></td><td><em>p </em>®0</td><td>&nbsp;</td><td>ë <em><sup>p</sup></em></td><td>&nbsp;</td><td>&nbsp;</td><td><em>p</em><em>w</em><sub>0</sub><em> </em><sub>û</sub></td><td>&nbsp;</td></tr><tr><td>ê</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>ú</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td></tr><tr><td>ë</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>û</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td></tr></tbody></table></figure>



<figure class="wp-block-table"><table><tbody><tr><td>Or d’après les transformées inverses de Laplace : <em>L</em><sup>&#8211;</sup><sup>1</sup></td><td>é</td><td>æ</td><td>1</td><td>&nbsp;</td><td>2<em>x</em> öù</td><td>æ</td><td>&nbsp;</td><td>2<em>x</em> ö</td><td>&nbsp;</td></tr><tr><td>ê <em>KV</em></td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&#8211;</td><td>&nbsp;</td><td>ú = <em>KV</em></td><td>&nbsp;</td><td><em>t </em>&#8211;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td></tr><tr><td>ç</td><td>&nbsp;</td><td>2</td><td>&nbsp;</td><td>ç</td><td>÷</td><td>&nbsp;</td></tr><tr><td>&nbsp;</td><td>ë</td><td><em>p</em></td><td>&nbsp;</td><td>÷</td><td>&nbsp;</td><td>&nbsp;</td></tr><tr><td>&nbsp;</td><td>è</td><td>&nbsp;</td><td>&nbsp;</td><td><em><sup>p</sup></em><em><sup>w</sup></em> 0&nbsp; øû</td><td>è</td><td>&nbsp;</td><td><em><sup>w</sup></em>0&nbsp; ø</td><td>&nbsp;</td></tr><tr><td></td><td></td><td></td><td></td><td></td><td></td><td></td><td></td><td></td><td></td><td></td><td></td><td></td></tr></tbody></table></figure>



<figure class="wp-block-table"><table><tbody><tr><td>Donc</td><td><em>s</em>(<em> t</em>)<em>&nbsp;&nbsp; </em>»</td><td><em>KV </em><sup>æ</sup><em> t </em>&#8211;</td><td>2<em>x</em> ö</td><td>&nbsp;</td><td>c’est à dire l’équation de l’asymptote de pente KV, coupant</td><td>&nbsp;</td><td>&nbsp;</td></tr><tr><td>&nbsp;</td><td>&nbsp;</td><td>÷</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td></tr><tr><td>&nbsp;</td><td>{</td><td>ç</td><td><em>w</em><sub>0</sub></td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td></tr><tr><td>&nbsp;</td><td><em>t </em>® ¥</td><td>è</td><td>ø</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td></tr><tr><td>l’axe des abcisses à la date</td><td><em>t</em></td><td>=</td><td>2<em>x</em></td><td>&nbsp;</td><td><strong><em>temps de retard du système</em></strong></td><td>.</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td></tr><tr><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td></tr><tr><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td><em>d</em></td><td>&nbsp;</td><td><em>w</em><sub>0</sub></td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td></tr><tr><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td></tr><tr><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td></tr><tr><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>é</td><td>æ</td><td>2<em>x</em> öù</td><td>ì ¥&nbsp; si <em>K</em> ¹ 1</td><td>&nbsp;</td></tr><tr><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>ï</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td></tr><tr><td><strong><em>L’erreur dynamique vaut :</em></strong></td><td>&nbsp;</td><td><em>e</em><em><sub>t</sub></em></td><td>&nbsp;</td><td>= lim [<em>e</em>(<em>t</em>) &#8211; <em>s</em>(<em>t</em> )] = lim <sub>ê</sub><em>Vt</em> &#8211; <em>K V</em> <sub>ç</sub> <em>t</em> &#8211;</td><td>&nbsp;</td><td>÷ú <sup>=</sup> í</td><td>2<em>x</em><em>V</em></td><td>si <em>K</em></td><td>= 1</td><td>&nbsp;</td></tr><tr><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td></tr><tr><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td><em>t </em>®¥</td><td><em>t</em>®¥<em> </em><sub>ë</sub></td><td>è</td><td><em><sup>w</sup></em>0&nbsp; øûï</td><td><em>w</em><sub>0</sub></td><td>&nbsp;</td></tr><tr><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>î</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td></tr></tbody></table></figure>



<h4 class="wp-block-heading"><strong>2.4.3 Seconde décomposition en éléments simples</strong></h4>



<h5 class="wp-block-heading"><strong>2.4.3.1 Deuxième ordre fortement amorti</strong></h5>



<p class="wp-block-paragraph"><strong><em>On tend vers l’asymptote sans oscillations, voir figure ci-dessous :</em></strong></p>



<p class="wp-block-paragraph"><strong><em>Entrée rampe avec V=1</em></strong></p>



<p class="wp-block-paragraph"><strong><em><u>Réponse du système</u></em></strong></p>



<p class="wp-block-paragraph">1</p>



<p class="wp-block-paragraph"><em>H </em>(<em> p</em>)=</p>



<p class="wp-block-paragraph">1+ 100 <em>p</em> +100 <em>p</em><sup>2</sup></p>



<p class="wp-block-paragraph"><strong><em>(<u>K= 1</u>, amortissement 5, pulsation propre 0,1 rads</em></strong><strong><em><sup>-1</sup></em></strong><strong><em>)</em></strong></p>



<p class="wp-block-paragraph"><strong><em>Temps de retard du système </em></strong><em>t</em><em><sub>d</sub></em></p>



<p class="wp-block-paragraph"><em>e</em><em><sub>t</sub></em>&nbsp; = 100 ¹ ¥ <strong><em>car le</em></strong></p>



<p class="wp-block-paragraph"><strong><em>gain du système</em></strong></p>



<p class="wp-block-paragraph"><strong><em>est unitaire Þ</em></strong></p>



<p class="wp-block-paragraph"><strong><em>l’asymptote et</em></strong></p>



<p class="wp-block-paragraph"><strong><em>l’entrée sont</em></strong></p>



<p class="wp-block-paragraph"><strong><em>parallèles</em></strong></p>



<p class="wp-block-paragraph"><strong><em>Asymptote parallèle</em></strong></p>



<ul class="wp-block-list"><li><strong><em>l’entrée car le gain est égal à 1</em></strong></li></ul>



<p class="wp-block-paragraph">Page 14&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; <strong>Emmanuel FARGES</strong>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; Ó EduKlub S.A.</p>



<p class="wp-block-paragraph">Tous droits de l’auteur des œuvres réservés. Sauf autorisation, la reproduction ainsi que toute utilisation des œuvres autre que la consultation individuelle et privée sont interdites.</p>



<p class="wp-block-paragraph"><strong>Sciences Indusrielles</strong></p>



<p class="wp-block-paragraph"><strong>Systèmes linéaires continus invariants</strong></p>



<p class="wp-block-paragraph">Analyse temporelle</p>



<h5 class="wp-block-heading"><strong>2.4.3.2 Deuxième ordre limite : amortissement unitaire</strong></h5>



<p class="wp-block-paragraph"><strong><em>La courbe est similaire à celle du cas précédent, voir ci-dessous :</em></strong></p>



<p class="wp-block-paragraph">1</p>



<p class="wp-block-paragraph"><strong><em><u>Réponse du système</u></em></strong><strong><em> </em></strong><em>H</em>(<em>p</em>)=</p>



<p class="wp-block-paragraph">1+ 20 <em>p</em> +100 <em>p</em><sup>2</sup></p>



<p class="wp-block-paragraph"><strong><em>K=1 donc asymptote de même pente que</em></strong></p>



<p class="wp-block-paragraph"><strong><em>l’entrée, </em></strong><em>x</em><strong><em> =1 et </em></strong><em>w</em><sub>0</sub><strong><em> =0,1 rad s</em></strong><strong><em><sup>-1</sup></em></strong></p>



<figure class="wp-block-table"><table><tbody><tr><td><strong><em>Entrée rampe</em></strong></td><td><strong><em>Temps de retard </em></strong><em>t</em><em><sub>d</sub></em></td></tr><tr><td><strong><em>avec V=1</em></strong></td><td>&nbsp;</td></tr></tbody></table></figure>



<h5 class="wp-block-heading"><strong>2.4.3.3 Deuxième ordre faiblement amorti</strong></h5>



<p class="wp-block-paragraph"><strong><em>On tend vers l’asymptote avec des oscillations autour d’elle, voir ci-dessous </em></strong>:</p>



<p class="wp-block-paragraph"><strong><em>Erreur de traînage </em></strong><em>e</em><em><sub>t</sub></em></p>



<p class="wp-block-paragraph">1</p>



<p class="wp-block-paragraph"><strong><em>Réponse du système </em></strong><em>H</em>(<em>p</em>)=</p>



<p class="wp-block-paragraph">1+ 2 <em>p</em> +100 <em>p</em><sup>2</sup></p>



<p class="wp-block-paragraph"><strong><em>K=1 donc asymptote de même pente que l’entrée </em></strong>(<em>e</em><em><sub>t</sub></em><strong><em> </em></strong>¹ ¥)<strong><em> , </em></strong><em>x</em><strong><em> </em></strong>=<strong><em> </em></strong>0,1<strong><em> et </em></strong><em>w</em><sub>0</sub><strong><em> =0,1 rad s</em></strong><strong><em><sup>-1</sup></em></strong></p>



<p class="wp-block-paragraph"><strong><em>Entrée rampe</em></strong></p>



<p class="wp-block-paragraph"><strong><em>avec V=1</em></strong></p>



<p class="wp-block-paragraph"><strong><em>Erreur dynamique&nbsp; </em></strong><em>e</em><em>t</em><strong><em>&nbsp; </em></strong>=<strong><em> </em></strong><sup>2</sup><em><sup>x</sup></em><em><sup>V</sup></em><strong><em> </em></strong><em>w</em><sub>0</sub></p>



<figure class="wp-block-table"><table><tbody><tr><td>æ</td><td>2<em>x</em> ö</td><td>&nbsp;</td></tr><tr><td><strong><em>Temps de retard </em></strong><sub>ç</sub></td><td>&nbsp;</td><td>÷</td><td>&nbsp;</td></tr><tr><td>&nbsp;</td><td>&nbsp;</td></tr><tr><td>è <em><sup>w</sup></em>0</td><td>ø</td><td>&nbsp;</td></tr></tbody></table></figure>



<h5 class="wp-block-heading"><strong>2.4.3.3 Influence du gain</strong></h5>



<p class="wp-block-paragraph">1</p>



<p class="wp-block-paragraph"><strong><em><u>Réponse du système</u></em></strong><strong><em> </em></strong><em>H</em>(<em>p</em>)=</p>



<p class="wp-block-paragraph">1+ 2 <em>p</em> +100 <em>p</em><sup>2</sup></p>



<ul class="wp-block-list"><li>= 1<strong><em>donc asymptote de même pente que</em></strong> <strong><em>l’entrée ( </em></strong><em>e</em><em><sub>t</sub></em><strong><em> </em></strong>¹ ¥<strong><em> ), </em></strong><em>x</em><strong><em> </em></strong>=<strong><em> </em></strong>0,1et<strong><em> </em></strong><em>w</em><sub>0</sub><strong><em> </em></strong>=<strong><em> </em></strong>0,1rads<sup>-1</sup><em></em></li></ul>



<p class="wp-block-paragraph">0,5</p>



<p class="wp-block-paragraph"><strong><em><u>Réponse du système</u></em></strong><strong><em> </em></strong><em>H</em>(<em>p</em>)=</p>



<p class="wp-block-paragraph">1+ 2 <em>p</em> +100 <em>p</em><sup>2</sup></p>



<ul class="wp-block-list"><li>= 0 , 5 <strong><em>donc asymptote de pente différente</em></strong></li></ul>



<p class="wp-block-paragraph"><strong><em>que l’entrée ( </em></strong><em>e</em><em><sub>t</sub></em><strong><em> </em></strong>= ¥<strong><em> ), </em></strong><em>x</em> = 0,1et <em>w</em><sub>0</sub> = 0,1rads<sup>-1</sup><em></em></p>



<p class="wp-block-paragraph">Page 15&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; <strong>Emmanuel FARGES</strong></p>



<p class="wp-block-paragraph"><strong><em>Entrée identique pour les deux systèmes : rampe avec V=1</em></strong></p>



<ul class="wp-block-list"><li><em><sub>t</sub></em>&nbsp; ¥ <strong><em>car le</em></strong><em></em></li></ul>



<p class="wp-block-paragraph"><strong><em>gain K de ce</em></strong></p>



<p class="wp-block-paragraph"><strong><em>système est</em></strong></p>



<p class="wp-block-paragraph"><strong><em>différent de 1</em></strong></p>



<ul class="wp-block-list"><li>EduKlub S.A.</li></ul>



<p class="wp-block-paragraph">Tous droits de l’auteur des œuvres réservés. Sauf autorisation, la reproduction ainsi que toute utilisation des œuvres autre que la consultation individuelle et privée sont interdites.</p>



<p class="has-text-align-center wp-block-paragraph"><strong><a href="https://www.mediafire.com/file/mw651ugx5n0venx/Telechargercours.com+Systèmes+linéaires+continus+invariants+ANALYSE+TEMPORELLE+Partie+2+SI-CO06-AAS-EF.pdf/file" target="_blank" rel="noreferrer noopener">Telecharger</a></strong></p>
]]></content:encoded>
					
		
		
			</item>
		<item>
		<title>SYSTEMES LINEAIRES CONTINUS INVARIANTS ANALYSE TEMPORELLE (Partie1)</title>
		<link>https://telechargercours.com/systemes-lineaires-continus-invariants-analyse-temporelle-partie1/</link>
		
		<dc:creator><![CDATA[Cours Zilla]]></dc:creator>
		<pubDate>Thu, 01 Jul 2021 15:35:29 +0000</pubDate>
				<category><![CDATA[Science & Technologie]]></category>
		<guid isPermaLink="false">https://telechargercours.com/?p=18088</guid>

					<description><![CDATA[Sciences Indusrielles Systèmes linéaires continus invariants Analyse temporelle SYSTEMES LINEAIRES CONTINUS INVARIANTS ANALYSE TEMPORELLE (Partie 1) L’étude se limite aux systèmes de bases, c’est à dire aux systèmes du premier ordre et du second ordre. En effet le comportement d’une grande partie des systèmes peut être raisonnablement assimilé au comportement d’un système du premier ou [&#8230;]]]></description>
										<content:encoded><![CDATA[
<p class="has-text-align-center wp-block-paragraph"><strong>Sciences Indusrielles</strong></p>



<p class="has-text-align-center wp-block-paragraph"><strong>Systèmes linéaires continus invariants</strong></p>



<p class="has-text-align-center wp-block-paragraph">Analyse temporelle</p>



<p class="has-text-align-center wp-block-paragraph"><strong>SYSTEMES LINEAIRES CONTINUS INVARIANTS</strong></p>



<p class="has-text-align-center wp-block-paragraph"><em>ANALYSE TEMPORELLE (Partie 1)</em></p>



<p class="wp-block-paragraph">L’étude se limite aux systèmes de bases, c’est à dire aux systèmes du premier ordre et du second ordre. En effet le comportement d’une grande partie des systèmes peut être raisonnablement assimilé au comportement d’un système du premier ou du second ordre.</p>



<p class="wp-block-paragraph">On va regarder la réponse de ces systèmes à des entrées typiques (échelon et rampe). Ces réponses vont nous permettre d’évaluer un certain nombre de performances associées à ces systèmes.</p>



<p class="wp-block-paragraph">La connaissance des résultats présentés dans ce chapitre est très utile pour répondre aux problèmes d’asservissement.</p>



<h2 class="wp-block-heading"><strong>1</strong>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; <strong>Système du premier ordre.</strong></h2>



<h3 class="wp-block-heading"><strong>1.1</strong>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; <strong>Définition d’un système du premier ordre</strong></h3>



<p class="wp-block-paragraph"><strong>Un système du premier ordre a son comportement régi par une équation différentielle du premier ordre de la forme :</strong></p>



<figure class="wp-block-table"><table><tbody><tr><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>ì</td><td><em>e </em>=<em> e</em>(<em>t </em>)&nbsp;&nbsp; entréedusystème</td><td>&nbsp;</td></tr><tr><td>&nbsp;</td><td><em>ds</em></td><td>&nbsp;</td><td>ï</td><td><em>s </em>=<em> s </em>(<em>t </em>)&nbsp;&nbsp; réponsedusystèmeàl&rsquo;entréee(t)</td><td>&nbsp;</td></tr><tr><td><em>s </em>+<em>t</em></td><td>= <em>Ke</em></td><td>ï</td><td>&nbsp;</td></tr><tr><td>&nbsp;</td><td>í</td><td>K:gainstatique(&gt;0)</td><td>&nbsp;</td></tr><tr><td><em>dt</em></td><td>&nbsp;</td></tr><tr><td>&nbsp;</td><td>&nbsp;</td><td>ï</td><td>&nbsp;</td></tr></tbody></table></figure>



<p class="wp-block-paragraph"><sup>ï</sup><sub>î</sub><em>t</em> :constantedetemps(&gt;0,homogèneà untemps(seconde (s))</p>



<h3 class="wp-block-heading"><strong>1.2</strong>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; <strong>Fonction de transfert globale d’un premier ordre</strong></h3>



<p class="wp-block-paragraph">On applique la transformée de Laplace à l’ensemble de l’équation différentielle ci-dessus, avec des conditions initiales nulles :</p>



<p class="wp-block-paragraph"><strong><em><u>Rappel :</u></em></strong>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; La transformée de Laplace d’une fonction f(t) est une fonction de p, notée par</p>



<p class="wp-block-paragraph">convention avec la lettre majuscule de la fonction du temps transformée : F(p)</p>



<p class="wp-block-paragraph">Donc : <em>s</em> +<em>t</em> <em><sup>ds</sup></em> = <em>Ke</em>&nbsp;&nbsp;&nbsp;&nbsp; transforméedeLaplace&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; <em>S </em>+<em> </em><em>t</em><em> pS </em>=<em> KE</em></p>



<p class="wp-block-paragraph"><em><sub>dt</sub></em>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; uuuuuuuuuuuuuuuuuuuuuuuuuuru</p>



<p class="wp-block-paragraph">On peut alors présenter le rapport de la sortie S(p) sur l’entrée E(p), c’est à dire la fonction de</p>



<figure class="wp-block-table"><table><tbody><tr><td>transfert globale du système :</td><td><em>H </em>(<em> p</em>)=</td><td><em>S</em>(<em> p </em>)</td><td>=</td><td><em>K</em></td><td>&nbsp;</td><td>&nbsp;</td></tr><tr><td>&nbsp;</td><td>1+<em>t</em> <em>p</em></td><td>&nbsp;</td></tr><tr><td>&nbsp;</td><td>&nbsp;</td><td><em>E</em>(<em> p</em>)</td><td>&nbsp;</td></tr></tbody></table></figure>



<p class="wp-block-paragraph"><strong><em><u>On notera :</u></em></strong></p>



<ul class="wp-block-list"><li>Le gain statique vaut : K=H(0)</li></ul>



<ul class="wp-block-list"><li>Pour identifier les caractéristiques d’un système du premier ordre (c’est à dire <em>K </em>et<em>t</em><em> </em>), on veillera bien à présenter la fonction de transfert globale H(p) avec le<em> </em>coefficient en p<sup>0</sup> du polynôme au dénominateur égal à 1. Ainsi le numérateur peut être</li></ul>



<p class="wp-block-paragraph">Page 1&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; <strong>Emmanuel FARGES</strong>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; Ó EduKlub S.A.</p>



<p class="wp-block-paragraph">Tous droits de l’auteur des œuvres réservés. Sauf autorisation, la reproduction ainsi que toute utilisation des œuvres autre que la consultation individuelle et privée sont interdites.</p>



<p class="wp-block-paragraph"><strong>Sciences Indusrielles</strong></p>



<p class="wp-block-paragraph"><strong>Systèmes linéaires continus invariants</strong></p>



<p class="wp-block-paragraph">Analyse temporelle</p>



<p class="wp-block-paragraph">identifié au gain statique K et le coefficient en p<sup>1</sup> du polynôme au dénominateur peut être identifié à la constante de temps <em>t</em></p>



<h3 class="wp-block-heading"><strong>1.3 Réponse à un échelon (réponse dite indicielle)</strong></h3>



<h4 class="wp-block-heading"><strong>1.3.1 Réponse dans le domaine fréquentielle (en p)</strong></h4>



<p class="wp-block-paragraph">On soumet le système à une entrée échelon de taille A. L’entrée est donc la fonction du temps : <strong>e(t)= A u(t)</strong> avec u(t) l’échelon unitaire tel que u(t&lt;0)=0 et u(t³0)=1.</p>



<p class="wp-block-paragraph">e (t)</p>



<figure class="wp-block-table"><table><tbody><tr><td>&nbsp;</td><td>A</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>Temps&nbsp;&nbsp; t</td><td>&nbsp;</td></tr><tr><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td></tr><tr><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td></tr><tr><td>&nbsp;</td><td>t=0</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td></tr><tr><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td></tr><tr><td>e(t)=A u(t) donc&nbsp; <em>E</em> ( <em>p</em>) =</td><td><em>A</em></td><td>or <em>H</em> ( <em>p</em>) =</td><td><em>S</em>(<em> p</em>)</td><td>=</td><td><em>K</em></td><td>Þ <em>S</em> ( <em>p</em>) =</td><td><em>K A</em></td><td>&nbsp;</td><td>&nbsp;</td></tr><tr><td><em>p</em></td><td>&nbsp;</td><td>&nbsp;</td><td><em>p </em>(1+<em>t</em><em> p</em>)</td><td>&nbsp;</td></tr><tr><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td><em>E</em>(<em> p</em>)&nbsp;&nbsp; 1+<em>t</em><em> p</em></td><td>&nbsp;</td></tr></tbody></table></figure>



<p class="wp-block-paragraph">On obtient donc très rapidement la solution (réponse à une entrée) dans le domaine fréquentielle (solution en p).</p>



<p class="wp-block-paragraph">Le problème est que l’on ne sait pas prendre la transformée de Laplace inverse d’une fonction quelconque. Il va donc falloir décomposer la solution trouvée en somme de termes dont on connaît les transformées inverses. Cela revient à effectuer la décomposition en éléments simples de ce rapport polynomial.</p>



<h4 class="wp-block-heading"><strong>1.3.2 Décomposition en éléments simples</strong></h4>



<p class="wp-block-paragraph">Tout rapport polynomial peut s’écrire comme somme de rapport du type :</p>



<figure class="wp-block-table"><table><tbody><tr><td><em>a</em></td><td>ì</td><td>auncoefficientàdéterminer</td><td>&nbsp;</td></tr><tr><td>&nbsp;</td><td><em>avec </em>í</td><td>&nbsp;</td><td>&nbsp;</td></tr><tr><td><em>p </em>&#8211;<em> p</em><em><sub>i</sub></em></td><td>&nbsp;</td><td>&nbsp;</td></tr><tr><td>î<sup>p</sup>i</td><td>leszérosdupolynomeaudénominateur</td><td>&nbsp;</td></tr></tbody></table></figure>



<p class="wp-block-paragraph">La décomposition en éléments simples revient à chercher dans notre cas particulier à mettre</p>



<figure class="wp-block-table"><table><tbody><tr><td>S(p) sous la forme : <em>S</em> ( <em>p</em>) =</td><td><em>K A</em></td><td>=</td><td><em>a</em></td><td>+</td><td><em>b</em></td><td>avec a et b deux coefficients à déterminer.</td><td>&nbsp;</td></tr><tr><td><em>p</em>(1+<em>t</em><em> p</em>)</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td></tr><tr><td>&nbsp;</td><td>&nbsp;</td><td><em>p&nbsp;&nbsp; </em>1+<em>t</em><em> p</em></td><td>&nbsp;</td><td>&nbsp;</td></tr></tbody></table></figure>



<p class="wp-block-paragraph">A partir de la on peut employer deux méthodes pour déterminer a et b :</p>



<p class="wp-block-paragraph"><strong><em><u>1</u></em></strong><strong><em><u><sup>ère</sup></u></em></strong><strong><em><u> méthode : (vue en mathématique) :</u></em></strong></p>



<p class="wp-block-paragraph"><em>a </em>=<em>&nbsp;&nbsp; pS</em>(<em> p</em>)=123</p>



<p class="wp-block-paragraph">pour p=0</p>



<figure class="wp-block-table"><table><tbody><tr><td>&nbsp;</td><td><em>KA</em></td><td>&nbsp;</td><td>= <em>KA</em></td><td>et<em>b </em>=(1+<em>t</em><em> p </em>)<em> S </em>(<em> p </em>)=</td><td>&nbsp;</td></tr><tr><td>&nbsp;</td><td>1+<em>t</em> <em>p</em></td><td>&nbsp;</td></tr><tr><td>&nbsp;</td><td>&nbsp;</td><td>1<strong>4</strong>2443</td><td>&nbsp;</td></tr><tr><td>123</td><td>&nbsp;</td><td>&nbsp;</td><td>pour p=-</td><td>1</td><td>&nbsp;</td><td>&nbsp;</td></tr><tr><td>pourp=0</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td></tr><tr><td>&nbsp;</td><td>t</td><td>&nbsp;</td></tr><tr><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td></tr></tbody></table></figure>



<p class="wp-block-paragraph"><em>KA</em></p>



<p class="wp-block-paragraph">= &#8211;<em>KA</em><em>t</em></p>



<p class="wp-block-paragraph"><em>p</em></p>



<p class="wp-block-paragraph">{</p>



<p class="wp-block-paragraph"><sup>1</sup></p>



<p class="wp-block-paragraph">pourp=-</p>



<figure class="wp-block-table"><table><tbody><tr><td><strong><em>2</em></strong><strong><em><sup>ème</sup></em></strong><strong><em> méthode : identification polynomiale :</em></strong></td><td>&nbsp;</td><td><em>a </em>+<em> p</em>(<em>b </em>+<em> a</em><em>t</em><em> </em>)</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td></tr><tr><td><em>S </em>(<em> p</em>)=</td><td><em>K A</em></td><td>=</td><td><em>a</em></td><td>+</td><td><em>b</em></td><td>=</td><td><em>a</em>(1+<em>t</em><em> p</em>)+</td><td><em>bp</em></td><td>=</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td></tr><tr><td><em>p</em>(1+<em>t</em><em> p</em>)</td><td>&nbsp;</td><td>&nbsp;</td><td><em>p</em>(1+<em>t</em><em> p</em>)</td><td><em>p</em>(1+<em>t</em><em> p</em>)</td><td>&nbsp;</td><td>&nbsp;</td></tr><tr><td>&nbsp;</td><td>&nbsp;</td><td><em>p&nbsp;&nbsp; </em>1+<em>t</em><em> p</em></td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td></tr><tr><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>ì <em>a</em> = <em>KA</em></td><td>ì <em>a</em> = <em>KA</em></td><td>&nbsp;</td></tr><tr><td>On en déduit donc en identifiant les numérateurs que : <sub>í</sub></td><td><em>soit </em><sub>í</sub></td><td>&nbsp;</td></tr><tr><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td><sub>î</sub><em> b </em>+<em> a</em><em>t</em><em> </em>=0</td><td><sub>î</sub><em>b </em>= &#8211;<em>KA</em><em>t</em></td><td>&nbsp;</td></tr></tbody></table></figure>



<p class="wp-block-paragraph">Quelle que soit la méthode employée (il est bon de savoir vérifier son calcul par identification), on aboutit à la décomposition de S(p) en éléments simples sous la forme :</p>



<p class="wp-block-paragraph">Page 2&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; <strong>Emmanuel FARGES</strong>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; Ó EduKlub S.A.</p>



<p class="wp-block-paragraph">Tous droits de l’auteur des œuvres réservés. Sauf autorisation, la reproduction ainsi que toute utilisation des œuvres autre que la consultation individuelle et privée sont interdites.</p>



<p class="wp-block-paragraph"><strong>Sciences Indusrielles</strong></p>



<p class="wp-block-paragraph"><strong>Systèmes linéaires continus invariants</strong></p>



<p class="wp-block-paragraph">Analyse temporelle</p>



<figure class="wp-block-table"><table><tbody><tr><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>é</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>ù</td><td>&nbsp;</td></tr><tr><td>&nbsp;</td><td><em>KA</em></td><td>&nbsp;</td><td><em>KA</em><em>t</em></td><td>é 1</td><td>&nbsp;</td><td>&nbsp;</td><td><em>t</em>&nbsp; ù</td><td>ê 1</td><td>&nbsp;</td><td>1&nbsp;&nbsp; ú</td><td>&nbsp;</td></tr><tr><td><em>S</em>(<em> p</em>)=</td><td>&nbsp;</td><td>&#8211;</td><td>&nbsp;</td><td>&nbsp;</td><td>= <em>KA</em><sub>ê</sub></td><td>&nbsp;</td><td>&#8211;</td><td>&nbsp;</td><td>&nbsp;</td><td>ú</td><td>= <em>KA</em> <sub>ê</sub></td><td>&nbsp;</td><td>&#8211;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>ú</td><td>&nbsp;</td></tr><tr><td>&nbsp;</td><td>&nbsp;</td><td>+ <em>t</em> <em>p</em></td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>1</td><td>&nbsp;</td></tr><tr><td>&nbsp;</td><td><em>p&nbsp;&nbsp; </em>1</td><td><sub>ë</sub><em> p&nbsp;&nbsp; </em>1</td><td>+<em>t</em> <em>p</em> <sub>û</sub></td><td>ê <em><sup>p</sup></em></td><td>&nbsp;</td><td><em>p </em>+</td><td>ú</td><td>&nbsp;</td></tr><tr><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td></tr><tr><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>ë</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td><em>t</em> û</td><td>&nbsp;</td></tr></tbody></table></figure>



<h4 class="wp-block-heading"><strong>1.3.3 Réponse dans le domaine temporel</strong></h4>



<figure class="wp-block-table"><table><tbody><tr><td>é</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>ù</td><td>&nbsp;</td></tr><tr><td>ê 1</td><td>&nbsp;</td><td>1</td><td>&nbsp;</td><td>ú</td><td>&nbsp;</td></tr><tr><td>On vient donc de montrer que&nbsp; <em>S</em>( <em>p</em>) = <em>KA</em> ê</td><td>&nbsp;</td><td>&#8211;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>ú or on a déjà vu lors du cours sur les</td><td>&nbsp;</td></tr><tr><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td></tr><tr><td>ê <em><sup>p</sup></em></td><td>&nbsp;</td><td>1 +</td><td>1</td><td>ú</td><td>&nbsp;</td></tr><tr><td>&nbsp;</td><td><em>t</em></td><td>&nbsp;</td></tr><tr><td>ë</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>û</td><td>&nbsp;</td></tr></tbody></table></figure>



<figure class="wp-block-table"><table><tbody><tr><td>ì</td><td>&nbsp;</td><td>1</td><td>&nbsp;</td><td>&nbsp;</td></tr><tr><td><sub>ï</sub><em>U </em>(<em> p</em>)=<em> L </em>[<em> u</em>(<em> t</em>)]=</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td></tr><tr><td><em><sup>p</sup></em>&nbsp; (par abus de notation on omet souvent de</td><td>&nbsp;</td></tr><tr><td>transformées de Laplace que <sub>í</sub><sup>ï</sup></td><td>1</td><td>&nbsp;</td></tr><tr><td><sup>ï</sup><em> L</em>[<em>e</em><sup>&#8211;</sup><em><sup>at</sup></em><em>u </em>(<em>t </em>)]=</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td></tr><tr><td><em>p </em>+<em> a</em></td><td>&nbsp;</td></tr><tr><td>ï</td><td>&nbsp;</td></tr><tr><td>î</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td></tr></tbody></table></figure>



<p class="wp-block-paragraph">multiplier toute les fonctions du temps par u(t) l’échelon unitaire ; ce qui revient à écrire :</p>



<p class="wp-block-paragraph"><em>L</em>[<em> e</em><sup>&#8211;</sup><em><sup>at</sup></em><em> </em>]=&nbsp; <sup>1</sup>&nbsp;&nbsp;&nbsp; )</p>



<p class="wp-block-paragraph"><em>p </em>+<em> a</em></p>



<p class="wp-block-paragraph">En lisant ces transformées dans l’autre sens, c’est à dire en prenant les transformées inverses de Laplace, on obtient :</p>



<figure class="wp-block-table"><table><tbody><tr><td>&nbsp;</td><td>&nbsp;</td><td>é</td><td>é</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>ùù</td><td>é</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>æ</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>öù</td><td>&nbsp;</td></tr><tr><td>-1</td><td>-1</td><td>ê</td><td>ê 1</td><td>&nbsp;</td><td>1&nbsp; úú</td><td>ê</td><td>-1</td><td>æ 1 ö</td><td>-1</td><td>ç</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>1&nbsp; ÷ú</td><td>&nbsp;</td></tr><tr><td><em>s</em>(<em>t</em>)=<em> L</em></td><td>[ <em>S</em> ( <em>p</em>)] = <em>L</em></td><td>ê<em>KA</em>ê</td><td>&nbsp;</td><td>&#8211;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>úú = <em>KA</em>ê <em>L</em></td><td>ç</td><td>&nbsp;</td><td><sub>÷</sub>&#8211; <em>L</em></td><td>&nbsp;</td><td>ç</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>÷ú</td><td>&nbsp;</td></tr><tr><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>1</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>1</td><td>&nbsp;</td><td>&nbsp;</td></tr><tr><td>&nbsp;</td><td>&nbsp;</td><td>ê</td><td>ê <em><sup>p</sup></em></td><td>&nbsp;</td><td>1 +</td><td>úú</td><td>ê</td><td>&nbsp;</td><td>è <em><sup>p</sup></em> ø</td><td>&nbsp;</td><td>&nbsp;</td><td>ç</td><td>1+</td><td>&nbsp;</td><td>÷</td><td>ú</td><td>&nbsp;</td></tr><tr><td>&nbsp;</td><td>&nbsp;</td><td>ê</td><td>ë</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>ú</td><td>ê</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>ú</td><td>&nbsp;</td></tr><tr><td>&nbsp;</td><td>&nbsp;</td><td>ë</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td><em>t</em> ûû</td><td>ë</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>è</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td><em>t</em> ø</td><td>û</td><td>&nbsp;</td></tr><tr><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>æ</td><td>&nbsp;</td><td>&#8211;</td><td><em>t</em></td><td>&nbsp;</td><td>ö</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td></tr><tr><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td><em>u </em>(<em>t </em>)</td><td>&nbsp;</td><td>&nbsp;</td></tr><tr><td>Soit la solution dans le domaine temporel : <em>s</em>(<em>t</em>) = <em>KA</em> 1</td><td>&#8211; <em>e</em></td><td>&nbsp;</td><td><em>t</em></td><td>÷</td><td>&nbsp;</td><td>&nbsp;</td></tr><tr><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>ç</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td></tr><tr><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>è</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>ø</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td></tr><tr><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>æ</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&#8211;</td><td><em>t</em></td><td>ö</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td></tr><tr><td>Ce qui revient à écrire par abus de notation :</td><td>&nbsp;</td><td>&nbsp;</td><td>&#8211; <em>e</em></td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td></tr><tr><td><em>s</em>(<em>t</em>)=<em> KA </em>ç1</td><td>&nbsp;</td><td><em>t</em></td><td>÷</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td></tr><tr><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>è</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>ø</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td></tr></tbody></table></figure>



<h4 class="wp-block-heading"><strong>1.3.4 Analyse temporelle</strong></h4>



<h5 class="wp-block-heading"><strong>1.3.4.1 Rapidité</strong></h5>



<p class="wp-block-paragraph">On cherche à évaluer le temps que met le système à répondre à l’entrée, c’est à dire le temps que met le système à passer du régime transitoire au régime permanent. Le régime permanent est la solution particulière de l’équation différentielle que l’on recherche de la même forme que l’entrée, c’est à dire ici une constante.</p>



<figure class="wp-block-table"><table><tbody><tr><td>æ</td><td>&nbsp;</td><td>&#8211;</td><td><em>t</em></td><td>ö</td><td>&nbsp;</td></tr><tr><td>1- <em>e</em></td><td>&nbsp;</td><td>&nbsp;</td></tr><tr><td>La réponse du système <em>s</em>(<em>t</em>) = <em>KA</em> ç</td><td><em><sup>t</sup></em> ÷ est une courbe qui tend vers la valeur constante</td><td>&nbsp;</td></tr></tbody></table></figure>



<ul class="wp-block-list"><li>ø</li></ul>



<p class="wp-block-paragraph">KA (asymptote horizontale) en restant toujours inférieure à cette valeur KA.</p>



<p class="wp-block-paragraph"><strong>La rapidité est évaluée dans la pratique, par le temps que met le système à atteindre des</strong></p>



<p class="wp-block-paragraph"><strong>valeurs ne s’écartant pas de </strong><strong>±</strong><strong> 5 % de sa valeur asymptotique.Ce temps se note </strong><strong>t</strong><strong><sub>5%</sub></strong><strong> </strong>Pratiquement, pour un premier ordre le temps à 5% est le temps à -5% puisque</p>



<figure class="wp-block-table"><table><tbody><tr><td>æ</td><td>&#8211;</td><td><em>t</em></td><td>ö</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td></tr><tr><td>&nbsp;</td><td>&lt; <em>KA</em></td><td>&nbsp;</td><td>&nbsp;</td></tr><tr><td>« <em>t</em> , <em>s</em> (<em>t</em> ) = <em>KA</em> 1</td><td>&#8211; <em>e</em> <em><sup>t</sup></em></td><td>÷</td><td>&nbsp;</td><td>&nbsp;</td></tr><tr><td>ç</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td></tr><tr><td>è</td><td>&nbsp;</td><td>&nbsp;</td><td>ø</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td></tr><tr><td>Page 3</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td><strong>Emmanuel FARGES</strong></td><td>Ó EduKlub S.A.</td><td>&nbsp;</td></tr></tbody></table></figure>



<p class="wp-block-paragraph">Tous droits de l’auteur des œuvres réservés. Sauf autorisation, la reproduction ainsi que toute utilisation des œuvres autre que la consultation individuelle et privée sont interdites.</p>



<p class="wp-block-paragraph"><strong>Sciences Indusrielles</strong></p>



<p class="wp-block-paragraph"><strong>Systèmes linéaires continus invariants</strong></p>



<p class="wp-block-paragraph">Analyse temporelle</p>



<figure class="wp-block-table"><table><tbody><tr><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>æ</td><td>&#8211;</td><td><em><sup>t</sup></em>5%</td><td>ö</td><td>&nbsp;</td></tr><tr><td>On cherche donc <em>t</em><sub>5%</sub>&nbsp; telque <em>s</em>(<em>t</em><sub>5%</sub> ) = <em>KA</em> ç 1 &#8211; <em>e</em></td><td>&nbsp;</td><td><em>t</em></td><td>÷ = 0,95 <em>KA</em></td><td>&nbsp;</td></tr><tr><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>è</td><td>&nbsp;</td><td>&nbsp;</td><td>ø</td><td>&nbsp;</td></tr><tr><td>æ</td><td>&#8211; <em>e</em></td><td>&#8211;</td><td><em><sup>t</sup></em>5%</td><td>ö</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td></tr><tr><td>0,95 <em>KA</em> = <em>KA</em> 1</td><td>&nbsp;</td><td><em>t</em></td><td>÷</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td></tr><tr><td>ç</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td></tr><tr><td>è</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>ø</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td></tr></tbody></table></figure>



<p class="wp-block-paragraph"><sub>&#8211;</sub><em><sup>t</sup></em>5 %</p>



<p class="wp-block-paragraph">0,05 = <em>e</em>&nbsp;&nbsp; <em><sup>t</sup></em></p>



<p class="wp-block-paragraph">ln(0,05) = &#8211; <em><sup>t</sup></em><sup>5%</sup>&nbsp;&nbsp;&nbsp; avec&nbsp;&nbsp;&nbsp; ln0,05 ; -3</p>



<p class="wp-block-paragraph"><em><sub>t</sub></em></p>



<h5 class="wp-block-heading"><strong>1.3.4.2 Précision statique</strong></h5>



<p class="wp-block-paragraph">La précision est l’écart entre ce que l’on voulait (la consigne : e(t)) et ce que l’on a (la réponse s(t)). La précision est ici qualifiée de statique puisque au bout d’un certain temps ( <em>t</em> ® ¥ ) l’entrée est constante (c’est un échelon) et la sortie aussi (on a vu que lim <em>s</em>(<em>t</em> ) = <em>KA</em> ).</p>



<p class="wp-block-paragraph"><em>t</em>®¥</p>



<p class="wp-block-paragraph"><strong><em><u>La précision statique vaut donc </u></em></strong><strong><u>:</u></strong></p>



<p class="wp-block-paragraph"><em>e</em><em><sub>S</sub></em>&nbsp; = lim [<em>e</em>(<em>t</em> ) &#8211; <em>s</em> (<em>t</em> )] avece(t)uneentréeéchelonets(t)laréponsedu systèmeà cetéchelon</p>



<p class="wp-block-paragraph"><em>t </em>®¥</p>



<p class="wp-block-paragraph">Dans le cas d’un système du premier ordre, on trouve donc : <em>e</em><em><sub>S</sub></em>&nbsp; = <em>A</em> (1 &#8211; <em>K</em>)</p>



<p class="wp-block-paragraph"><strong>C’est à dire que le système est précis ( </strong><em>e</em><em><sub>S</sub></em><strong> </strong>=<strong> </strong>0<strong> ) si le gain de la fonction de transfert globale du système est unitaire.</strong></p>



<p class="wp-block-paragraph"><strong><em><u>Remarque :</u></em></strong>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; On peut trouver ce résultat directement à partir de la fonction de transfert globale du système, en utilisant le théorème de la valeur finale :</p>



<figure class="wp-block-table"><table><tbody><tr><td><em>e</em></td><td>&nbsp;</td><td>= lim</td><td>(</td><td><em>e</em>(<em>t</em>)-<em> s</em>(<em>t</em>)</td><td>)</td><td>= lim <em>p</em></td><td>(</td><td><em>E</em>(<em> p</em>)-<em> S </em>(<em> p</em>)</td><td>)</td><td>= lim <em>pE</em>( <em>p</em>) <sup>é</sup>1-</td><td><em>S </em>(<em> p</em>)</td><td>ù</td><td>= lim <em>pE</em>( <em>p</em>) 1 &#8211; <em>H</em> ( <em>p</em>)</td><td>]</td><td>&nbsp;</td></tr><tr><td><em>s</em></td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td></tr><tr><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>ê</td><td>&nbsp;</td><td>&nbsp;</td><td>ú</td><td>&nbsp;</td><td>&nbsp;</td><td>[</td><td>&nbsp;</td><td>&nbsp;</td></tr><tr><td>&nbsp;</td><td>&nbsp;</td><td><em>t </em>®¥</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td><em>p </em>®0</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td><em>p </em>®<em> </em>0</td><td>ë</td><td>&nbsp;</td><td><em>E </em>(<em> p</em>)<sub>û</sub></td><td>&nbsp;</td><td><em>p</em>®0</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td></tr><tr><td>Remarque :</td><td>tant que l’on a pas expliciter E(p) et H(p), cette relation peut être appliquer à</td><td>&nbsp;</td></tr><tr><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>n’importe quel système pour n’importe quel type d’entrée.</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td></tr><tr><td>Dans le cas qui nous intéresse, pour un premier ordre&nbsp; <em>H</em> ( <em>p</em>) =</td><td>&nbsp;</td><td><em>K</em></td><td>&nbsp;</td><td>, et pour une entrée</td><td>&nbsp;</td><td>&nbsp;</td></tr><tr><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td></tr><tr><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>1+<em>t</em> <em>p</em></td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td></tr><tr><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td><em>A</em></td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>é</td><td><em>K&nbsp;&nbsp; </em>ù</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>é1- <em>K</em> +<em>t</em> <em>p</em> ù</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td></tr><tr><td>échelon,</td><td>&nbsp;</td><td><em>E </em>(<em> p</em>)=</td><td>&nbsp;</td><td>, on retrouve : <em>e</em><em><sub>s</sub></em>&nbsp; = lim <em>A</em><sub>ê</sub>1-</td><td>&nbsp;</td><td>ú</td><td>= lim <em>A</em> <sub>ê</sub></td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td><sub>ú</sub> = <em>A</em> (1</td><td>&#8211; <em>K</em>) , sans</td><td>&nbsp;</td></tr><tr><td>&nbsp;</td><td><em>p</em></td><td>&nbsp;</td><td>&nbsp;</td><td>1+<em>t</em> <em>p</em></td><td>&nbsp;</td></tr><tr><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td><em>p </em>®0</td><td>ë</td><td>1+ <em>t</em> <em>p</em> <sub>û</sub></td><td><em>p</em>®0</td><td>ë</td><td>&nbsp;</td><td>û</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td></tr><tr><td>avoir eu besoin de calculer la réponse en fonction du temps.</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td></tr><tr><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td><strong>1.3.4.3 Pente à l’origine.</strong></td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td></tr><tr><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>æ</td><td>&nbsp;</td><td>&#8211;</td><td><em>t</em></td><td>ö</td><td>, donc : <em>s</em>¢(<em>t</em> ) =</td><td><em>KA</em></td><td>&#8211;</td><td><em>t</em></td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td></tr><tr><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td></tr><tr><td><em>s</em>(<em>t</em>)=<em> KA </em>ç1</td><td>&#8211; <em>e</em> <em><sup>t</sup></em> ÷</td><td>&nbsp;</td><td><em>e </em><em><sup>t</sup></em></td><td>&nbsp;</td><td>, soit une pente à l’origine de :</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td></tr><tr><td><em>t</em></td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td></tr><tr><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>è</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>ø</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td></tr><tr><td><em>s</em>¢(0)=</td><td><em>KA</em></td><td>, c’est à dire une droite qui passe par l’origine et par le point d’abscisse <em>t</em> et</td><td>&nbsp;</td><td>&nbsp;</td></tr><tr><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td></tr><tr><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td><em>t</em></td><td>&nbsp;</td><td>&nbsp;</td><td><em>KA </em>.</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td></tr><tr><td>d’ordonnée</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td></tr></tbody></table></figure>



<p class="wp-block-paragraph">Page 4&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; <strong>Emmanuel FARGES</strong>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; Ó EduKlub S.A.</p>



<p class="wp-block-paragraph">Tous droits de l’auteur des œuvres réservés. Sauf autorisation, la reproduction ainsi que toute utilisation des œuvres autre que la consultation individuelle et privée sont interdites.</p>



<p class="wp-block-paragraph"><strong>Sciences Indusrielles</strong></p>



<p class="wp-block-paragraph"><strong>Systèmes linéaires continus invariants</strong></p>



<p class="wp-block-paragraph">Analyse temporelle</p>



<h5 class="wp-block-heading"><strong>1.3.4.4 Valeur à la constante de temps </strong><em>t</em><strong> .</strong></h5>



<figure class="wp-block-table"><table><tbody><tr><td>æ</td><td>&nbsp;</td><td>&#8211;</td><td><em>t</em></td><td>ö</td><td>-1</td><td>æ <em>e</em> -1ö</td><td>&nbsp;</td><td>&nbsp;</td></tr><tr><td>&nbsp;</td><td><em>t</em></td><td>&nbsp;</td><td>&nbsp;</td></tr><tr><td><em>s</em>(<em>t</em>)=<em> KA </em>ç1</td><td>&#8211; <em>e</em></td><td>&nbsp;</td><td>&nbsp;</td><td>÷ donc <em>s</em>(<em>t</em> ) = <em>KA</em> (1- <em>e</em></td><td>&nbsp;</td><td>) = <em>KA</em> ç</td><td>&nbsp;</td><td>÷</td><td>; 0,63<em>KA</em></td><td>&nbsp;</td></tr><tr><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td><em>e</em></td><td>&nbsp;</td></tr><tr><td>è</td><td>&nbsp;</td><td>ø</td><td>&nbsp;</td><td>è</td><td>ø</td><td>&nbsp;</td><td>&nbsp;</td></tr></tbody></table></figure>



<p class="wp-block-paragraph"><strong><em>La valeur de la réponse à la constante de temps vaut 63% de la valeur asymptotique.</em></strong></p>



<h5 class="wp-block-heading"><strong>1.3.4.5 Courbe réponse caractéristique</strong></h5>



<p class="wp-block-paragraph">KA</p>



<p class="wp-block-paragraph"><strong><em>Réponse d’un</em></strong></p>



<p class="wp-block-paragraph"><strong><em>premier ordre à</em></strong>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; <sup>0,95 KA</sup></p>



<p class="wp-block-paragraph"><strong><em>un échelon de</em></strong></p>



<p class="wp-block-paragraph"><strong><em>taille A</em></strong></p>



<p class="wp-block-paragraph">0,63 KA</p>



<p class="wp-block-paragraph">Pente à l’origine : <em><sup>KA</sup></em></p>



<p class="wp-block-paragraph"><em>t</em></p>



<p class="wp-block-paragraph"><em>t</em>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; <img loading="lazy" decoding="async" src="" width="12" height="24">&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; 3<em>t</em> : <em>rapidité</em></p>



<p class="wp-block-paragraph"><em>Régime transitoire</em>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; <em>Régime permanent</em></p>



<p class="wp-block-paragraph"><strong><em>Evolution de la réponse d’un premier ordre</em></strong></p>



<p class="wp-block-paragraph"><strong><em>avec la constante de temps </em></strong><em>t</em><strong><em> . Plus </em></strong><em>t</em><strong><em> est grand</em></strong></p>



<p class="wp-block-paragraph"><strong><em>plus les systèmes sont lents</em></strong></p>



<h3 class="wp-block-heading"><strong>1.4 Réponse à une rampe</strong></h3>



<h4 class="wp-block-heading"><strong>1.4.1 Réponse dans le domaine fréquentiel</strong></h4>



<p class="wp-block-paragraph">On soumet le système à une entrée rampe de pente V. L’entrée est donc la fonction du temps :</p>



<p class="wp-block-paragraph"><strong>e(t)= V t u(t) </strong>avec u(t) l’échelon unitaire.</p>



<p class="wp-block-paragraph">Page 5&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; <strong>Emmanuel FARGES</strong>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; Ó EduKlub S.A.</p>



<p class="wp-block-paragraph">Tous droits de l’auteur des œuvres réservés. Sauf autorisation, la reproduction ainsi que toute utilisation des œuvres autre que la consultation individuelle et privée sont interdites.</p>



<p class="wp-block-paragraph"><strong>Sciences Indusrielles</strong></p>



<p class="wp-block-paragraph"><strong>Systèmes linéaires continus invariants</strong></p>



<p class="wp-block-paragraph">Analyse temporelle</p>



<figure class="wp-block-table"><table><tbody><tr><td>&nbsp;</td><td>&nbsp;</td><td>e (t)</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td></tr><tr><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>Vt</td><td>&nbsp;</td><td>&nbsp;</td></tr><tr><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>Temps&nbsp;&nbsp; t</td><td>&nbsp;</td><td>&nbsp;</td></tr><tr><td>&nbsp;</td><td>t=0</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td></tr><tr><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td></tr><tr><td>e(t)=V t u(t) donc&nbsp; <em>E</em> ( <em>p</em>) =</td><td><em>V</em></td><td>or <em>H</em> ( <em>p</em>) =</td><td><em>S</em>(<em> p</em>)</td><td>=</td><td><em>K</em></td><td>Þ <em>S</em> ( <em>p</em>) =</td><td><em>K V</em></td><td>&nbsp;</td></tr><tr><td><em>p</em><sup>2</sup></td><td>&nbsp;</td><td>&nbsp;</td><td><em>p </em><sup>2</sup><em> </em>(1+<em>t</em><em> p</em>)</td><td>&nbsp;</td></tr><tr><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td><em>E</em>(<em> p</em>)&nbsp;&nbsp; 1+<em>t</em><em> p</em></td><td>&nbsp;</td></tr></tbody></table></figure>



<p class="wp-block-paragraph">Il reste à décomposer la solution trouvée en somme de termes dont on connaît les transformées inverses. Cela revient à effectuer la décomposition en éléments simples de ce rapport polynomial.</p>



<h4 class="wp-block-heading"><strong>1.4.2 Décomposition en éléments simples</strong></h4>



<p class="wp-block-paragraph">La décomposition en éléments simples revient à chercher dans notre cas particulier à mettre</p>



<figure class="wp-block-table"><table><tbody><tr><td>S(p) sous la forme : <em>S</em> ( <em>p</em>) =</td><td><em>K A</em></td><td>=</td><td><em>a</em></td><td>+</td><td><em>b</em></td><td>+</td><td><em>c</em></td><td>avec a, b et c trois coefficients à</td><td>&nbsp;</td></tr><tr><td><em>p</em><sup>2</sup><em> </em>(1+<em>t</em><em> p </em>)</td><td><em>p </em><sup>2</sup></td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td></tr><tr><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td><em>p&nbsp;&nbsp; </em>1+<em>t</em><em> p</em></td><td>&nbsp;</td><td>&nbsp;</td></tr></tbody></table></figure>



<p class="wp-block-paragraph">déterminer. Le terme en <em>p</em><sup>2</sup> du dénominateur signifie que 0 est une racine double du dénominateur. On a donc deux coefficients (a et b) associés à cette racine double dans la décomposition en éléments simples.</p>



<p class="wp-block-paragraph">A partir de la on peut employer les deux méthodes pour déterminer a, b et c :</p>



<p class="wp-block-paragraph"><strong><em><u>1</u></em></strong><strong><em><u><sup>ère</sup></u></em></strong><strong><em><u> méthode : (vue en mathématique) :</u></em></strong></p>



<p class="wp-block-paragraph"><em>a </em>=<em> p </em><sup>2</sup><em> S</em>(<em> p </em>)14243</p>



<p class="wp-block-paragraph">pourp=0</p>



<p class="wp-block-paragraph">et</p>



<figure class="wp-block-table"><table><tbody><tr><td>=</td><td>&nbsp;</td><td><em>KV</em></td><td>&nbsp;</td><td>= <em>KV</em></td><td>,</td><td><em>c </em>=(1+<em>t</em><em> p</em>)<em> S</em>(<em> p</em>)=</td><td>&nbsp;</td><td><em>KV</em></td><td>&nbsp;</td><td>= <em>KV</em><em>t</em> <sup>2</sup></td><td>&nbsp;</td></tr><tr><td>1+<em>t</em> <em>p</em></td><td>&nbsp;</td><td><em>p</em><sup>2</sup></td><td>&nbsp;</td></tr><tr><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>1<strong>4</strong>2443</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td></tr><tr><td>&nbsp;</td><td>123</td><td>&nbsp;</td><td>&nbsp;</td><td>1</td><td>&nbsp;</td><td>{</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td></tr><tr><td>&nbsp;</td><td>pour p=0</td><td>&nbsp;</td><td>&nbsp;</td><td>pourp=-</td><td>&nbsp;</td><td>&nbsp;</td><td>pourp=-</td><td>1</td><td>&nbsp;</td><td>&nbsp;</td></tr><tr><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>t</td><td>&nbsp;</td><td>&nbsp;</td></tr><tr><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>t</td><td>&nbsp;</td><td>&nbsp;</td></tr><tr><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td></tr></tbody></table></figure>



<figure class="wp-block-table"><table><tbody><tr><td>lim <em>ps</em>( <em>p</em>) = lim</td><td>é <em>a</em></td><td>+ <em>b</em> +</td><td>&nbsp;</td><td><em>cp&nbsp; </em>ù</td><td>= <em>b</em> +</td><td><em>c</em></td><td>= lim</td><td><em>KV</em></td><td>= 0.&nbsp;&nbsp; Donc <em>b</em> = &#8211;</td><td>&nbsp;</td><td><em>c</em></td><td>= &#8211;<em>KV</em><em>t</em></td><td>&nbsp;</td></tr><tr><td>ê</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>ú</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td></tr><tr><td>&nbsp;</td><td>&nbsp;</td><td><em>t</em></td><td>&nbsp;</td><td><em>t</em></td><td>&nbsp;</td></tr><tr><td><em>p</em>®¥</td><td><em>p </em>®¥</td><td>ë <em><sup>p</sup></em></td><td>&nbsp;</td><td>1 + <em>t</em> <em>p</em> <sub>û</sub></td><td>&nbsp;</td><td><em>p</em>®¥<em>&nbsp; </em><em>p</em><em> </em>(1+<em>t</em><em> </em><em>p</em>)</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td></tr><tr><td></td><td></td><td></td><td></td><td></td><td></td><td></td><td></td><td></td><td></td><td></td><td></td><td></td><td></td><td></td><td></td><td></td></tr></tbody></table></figure>



<figure class="wp-block-table"><table><tbody><tr><td>&nbsp;</td><td><strong><em>2</em></strong><strong><em><sup>ème</sup></em></strong><strong><em> méthode : identification polynomiale :</em></strong></td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td></tr><tr><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td></tr><tr><td><em>S </em>(<em> p</em>)=</td><td><em>K V</em></td><td>=</td><td><em>a</em></td><td>+</td><td><em>b</em></td><td>+</td><td><em>c</em></td><td>=</td><td><em>a</em>(1+<em>t</em><em> p</em>)+<em> bp</em>(1+<em>t</em><em> p</em>)+<em> cp</em><sup>2</sup></td><td>=</td><td><em>a </em>+<em> p</em>(<em>b</em>+<em> a</em><em>t</em><em> </em>)+<em> p </em><sup>2</sup></td><td>(<em>c</em> + <em>b</em><em>t</em> )</td><td>&nbsp;</td></tr><tr><td><em>p</em><sup>2</sup><em> </em>(1+<em>t</em><em> p </em>)</td><td><em>p </em><sup>2</sup></td><td>&nbsp;</td><td>&nbsp;</td><td><em>p</em><sup>2</sup><em> </em>(1+<em>t</em><em> p</em>)</td><td>&nbsp;</td><td><em>p </em><sup>2</sup><em> </em>(1+<em>t</em><em> p</em>)</td><td>&nbsp;</td><td>&nbsp;</td></tr><tr><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td><em>p&nbsp;&nbsp; </em>1+<em>t</em><em> p</em></td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td></tr><tr><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>ì <em>a</em> = <em>KV</em></td><td>&nbsp;</td><td>&nbsp;</td><td>ì <em>a</em> = <em>KV</em></td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td></tr><tr><td>On en déduit donc en identifiant les numérateurs que :</td><td>ï</td><td><em>soit</em></td><td>ï</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td></tr><tr><td>í <em>b</em> + <em>a</em><em>t</em> = 0</td><td>í<em>b</em> = &#8211;<em>KV</em><em>t</em></td><td>&nbsp;</td><td>&nbsp;</td></tr><tr><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>ï</td><td>&nbsp;</td><td>&nbsp;</td><td>ï</td><td>2</td><td>&nbsp;</td><td>&nbsp;</td></tr><tr><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td><sub>î</sub><em> c </em>+<em> b</em><em>t</em><em> </em>=0</td><td>&nbsp;</td><td>&nbsp;</td><td><sub>î</sub><em>c </em>=<em> KV</em><em>t</em></td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td></tr></tbody></table></figure>



<p class="wp-block-paragraph">Quelle que soit la méthode employée (il est bon de savoir vérifier son calcul par identification), on aboutit à la décomposition de S(p) en éléments simples sous la forme :</p>



<figure class="wp-block-table"><table><tbody><tr><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>é</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>ù</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td></tr><tr><td>é 1</td><td>&nbsp;</td><td><em>t</em></td><td>&nbsp;</td><td>&nbsp;</td><td><em>t</em> <sup>2</sup>&nbsp;&nbsp;&nbsp; ù</td><td>ê 1</td><td>&nbsp;</td><td><em>t</em></td><td>&nbsp;</td><td><em>t</em>&nbsp; ú</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td></tr><tr><td><em>S</em>(<em> p</em>)=<em> KV </em><sub>ê</sub></td><td>&nbsp;</td><td>&nbsp;</td><td>&#8211;</td><td>&nbsp;</td><td>+</td><td>&nbsp;</td><td>&nbsp;</td><td>ú</td><td>= <em>KV</em> <sub>ê</sub></td><td>&nbsp;</td><td>&nbsp;</td><td>&#8211;</td><td>&nbsp;</td><td>+</td><td>&nbsp;</td><td>&nbsp;</td><td>ú</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td></tr><tr><td>&nbsp;</td><td>2</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>2</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td></tr><tr><td>ë <em><sup>p</sup></em></td><td>&nbsp;</td><td><em>p&nbsp;&nbsp; </em>1</td><td>+<em>t</em> <em>p</em> <sub>û</sub></td><td>ê <em><sup>p</sup></em></td><td>&nbsp;</td><td><em>p</em></td><td>&nbsp;</td><td><em>p </em>+</td><td>1</td><td>ú</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td></tr><tr><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td></tr><tr><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>ë</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td><em>t</em> û</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td></tr><tr><td>Page 6</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td><strong>Emmanuel FARGES</strong></td><td>Ó EduKlub S.A.</td><td>&nbsp;</td></tr><tr><td></td><td></td><td></td><td></td><td></td><td></td><td></td><td></td><td></td><td></td><td></td><td></td><td></td><td></td><td></td><td></td><td></td><td></td><td></td><td></td><td></td><td></td></tr></tbody></table></figure>



<p class="wp-block-paragraph">Tous droits de l’auteur des œuvres réservés. Sauf autorisation, la reproduction ainsi que toute utilisation des œuvres autre que la consultation individuelle et privée sont interdites.</p>



<p class="wp-block-paragraph"><strong>Sciences Indusrielles</strong></p>



<p class="wp-block-paragraph"><strong>Systèmes linéaires continus invariants</strong></p>



<p class="wp-block-paragraph">Analyse temporelle</p>



<h4 class="wp-block-heading"><strong>1.4.3 Réponse dans le domaine temporel</strong></h4>



<figure class="wp-block-table"><table><tbody><tr><td>é</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>ù</td><td>&nbsp;</td></tr><tr><td>ê 1</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td><em>t</em></td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td><em>t</em></td><td>&nbsp;</td><td>&nbsp;</td><td>ú</td><td>&nbsp;</td></tr><tr><td>On vient donc de montrer que&nbsp; <em>S</em>( <em>p</em>) = <em>KV</em> <sub>ê</sub></td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&#8211;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>+</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td><sub>ú</sub> or on a déjà vu lors du cours sur</td><td>&nbsp;</td></tr><tr><td>&nbsp;</td><td>2</td><td>&nbsp;</td><td>&nbsp;</td><td><em>p</em></td><td>&nbsp;</td><td>&nbsp;</td><td>1</td><td>&nbsp;</td><td>&nbsp;</td></tr><tr><td>ê <em><sup>p</sup></em></td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>1+</td><td>ú</td><td>&nbsp;</td></tr><tr><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td></tr><tr><td>ë</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td><em>t</em> û</td><td>&nbsp;</td></tr><tr><td>ì</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>1</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td></tr><tr><td><sub>ï</sub><em>U </em>(<em> p</em>)=<em> L </em>[<em>u </em>(<em>t</em>) ]=</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td></tr><tr><td>&nbsp;</td><td><em>p</em></td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td></tr><tr><td>ï</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td></tr><tr><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td></tr><tr><td>les transformées de Laplace que <sub>í</sub><sup>ï</sup> <em>L</em>[<em>e</em><sup>&#8211;</sup><em><sup>at</sup></em><em>u</em> (<em>t</em> )] =</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>1</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>En lisant ces transformées dans l’autre</td><td>&nbsp;</td></tr><tr><td>&nbsp;</td><td>&nbsp;</td><td><em>p </em>+<em> a</em></td><td>&nbsp;</td><td>&nbsp;</td></tr><tr><td>ï</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td></tr><tr><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td></tr><tr><td>ï</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>1</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td></tr><tr><td><sub>ï</sub><em>&nbsp; L</em>[<em> t u</em>(<em> t</em>)]=</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td></tr><tr><td>&nbsp;</td><td>&nbsp;</td><td>2</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td></tr><tr><td>î</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td><em>p</em></td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td></tr></tbody></table></figure>



<p class="wp-block-paragraph">sens, c’est à dire en prenant les transformées inverses de Laplace, on obtient :</p>



<figure class="wp-block-table"><table><tbody><tr><td>&nbsp;</td><td>&nbsp;</td><td>é</td><td>é</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>ùù</td><td>é</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>æ</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>öù</td><td>&nbsp;</td></tr><tr><td>-1</td><td>-1</td><td>ê</td><td>ê 1&nbsp;&nbsp;&nbsp; <em>t</em></td><td>&nbsp;</td><td><em>t</em>&nbsp; úú</td><td>ê</td><td>-1</td><td>æ 1 ö</td><td>-1</td><td>&nbsp;</td><td>æ <em>t</em> ö</td><td>-1</td><td>ç</td><td><em>t</em>&nbsp; ÷ú</td><td>&nbsp;</td></tr><tr><td><em>s</em>(<em>t</em>)=<em> L</em></td><td>[ <em>S</em> ( <em>p</em>)] = <em>L</em></td><td>ê<em>KV</em> ê</td><td>&nbsp;</td><td>&nbsp;</td><td>&#8211;</td><td>&nbsp;</td><td>+</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>úú = <em>KV</em> ê <em>L</em></td><td>ç</td><td>&nbsp;</td><td>&nbsp;</td><td><sub>÷</sub> &#8211; <em>L</em></td><td>ç</td><td>&nbsp;</td><td>&nbsp;</td><td><sub>÷</sub> &#8211; <em>L</em></td><td>ç</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>÷ú</td><td>&nbsp;</td></tr><tr><td>&nbsp;</td><td>2</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>1</td><td>&nbsp;</td><td><em>p</em></td><td>2</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>1</td><td>&nbsp;</td><td>&nbsp;</td></tr><tr><td>&nbsp;</td><td>&nbsp;</td><td>ê</td><td>ê <em><sup>p</sup></em></td><td>&nbsp;</td><td>&nbsp;</td><td><em>p</em></td><td>1 +</td><td>úú</td><td>ê</td><td>&nbsp;</td><td>è</td><td>&nbsp;</td><td>ø</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>è</td><td>&nbsp;</td><td><em>p </em><sub>ø</sub></td><td>&nbsp;</td><td>ç</td><td>1+</td><td>÷</td><td>ú</td><td>&nbsp;</td></tr><tr><td>&nbsp;</td><td>&nbsp;</td><td>ê</td><td>ë</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>ú</td><td>ê</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>ú</td><td>&nbsp;</td></tr><tr><td>&nbsp;</td><td>&nbsp;</td><td>ë</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td><em>t</em> ûû</td><td>ë</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>è</td><td>&nbsp;</td><td><em>t</em> ø</td><td>û</td><td>&nbsp;</td></tr><tr><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>æ</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&#8211;</td><td><em>t</em></td><td>&nbsp;</td><td>ö</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td></tr><tr><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td></tr><tr><td>Soit la solution dans le domaine temporel : <em>s</em>(<em>t</em>) = <em>KV</em> ç <em>t</em> &#8211; <em>t</em> +<em>t</em> <em>e</em> <em><sup>t</sup></em></td><td>÷<em>u</em> (<em>t</em> )</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td></tr><tr><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>è</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>ø</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td></tr><tr><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>æ</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&#8211;</td><td><em>t</em></td><td>&nbsp;</td><td>ö</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td></tr><tr><td>Ce qui revient à écrire par abus de notation :</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td></tr><tr><td><em>s</em>(<em>t</em>)=<em> KV </em>ç<em> t </em>&#8211;<em>t</em><em> </em>+<em>t</em><em> e</em></td><td>&nbsp;</td><td><em>t</em></td><td>÷</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td></tr><tr><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>è</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>ø</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td></tr></tbody></table></figure>



<h4 class="wp-block-heading"><strong>1.4.4 Analyse temporelle</strong></h4>



<h5 class="wp-block-heading"><strong>1.4.4.1 Précision dynamique</strong></h5>



<p class="wp-block-paragraph">La précision est l’écart entre ce que l’on voulait (la consigne : e(t)) et ce que l’on a (la réponse s(t)). La précision est ici qualifiée de dynamique puisque l’entrée et la sortie varient au cours du temps.</p>



<p class="wp-block-paragraph">On qualifie ici la capacité qu’a le système à suivre une entrée variable dans le temps.</p>



<p class="wp-block-paragraph">On parle aussi pour désigner la précision dynamique, d’erreur de suivi ou encore d’erreur de traînage. C’est pourquoi on la note : <em>e</em><em><sub>t</sub></em></p>



<p class="wp-block-paragraph"><strong><em><u>L’erreur dynamique, de suivi ou encore de traînage vaut donc </u></em></strong><strong><u>:</u></strong></p>



<p class="wp-block-paragraph"><em>e</em><em><sub>t</sub></em>&nbsp; = lim [<em>e</em>(<em>t</em> ) &#8211; <em>s</em>( <em>t</em>)] avece(t)uneentréerampeets(t)laréponsedusystèmeàcette rampe</p>



<p class="wp-block-paragraph"><em>t </em>®¥</p>



<p class="wp-block-paragraph">On remarquera que la définition est la même que pour l’erreur statique. Seule l’entrée à laquelle on soumet le système change (et donc la réponse du système aussi).</p>



<p class="wp-block-paragraph">Dans le cas d’un système du premier ordre, on trouve donc :</p>



<figure class="wp-block-table"><table><tbody><tr><td>&nbsp;</td><td>æ</td><td>æ</td><td>&#8211;</td><td><em>t</em></td><td>öö</td><td>ì ¥ si <em>K</em> ¹ 1</td><td>&nbsp;</td></tr><tr><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td></tr><tr><td><em>e</em><em><sub>t</sub></em>&nbsp; = lim ç<em>Vt</em> &#8211; <em>KV</em> ç <em>t</em> &#8211;<em>t</em> + <em>t</em> <em>e</em></td><td><em><sup>t</sup></em> ÷÷ = í</td><td>&nbsp;</td></tr><tr><td><em>t </em>®¥</td><td>ç</td><td>è</td><td>&nbsp;</td><td>÷</td><td><sub>î</sub><em>V</em><em>t</em><em> </em>si<em> K </em>=1</td><td>&nbsp;</td></tr><tr><td>&nbsp;</td><td>è</td><td>&nbsp;</td><td>øø</td><td>&nbsp;</td></tr><tr><td></td><td></td><td></td><td></td><td></td><td></td><td></td><td></td><td></td></tr></tbody></table></figure>



<figure class="wp-block-table"><table><tbody><tr><td><strong><em>Remarque :</em></strong></td><td>On peut trouver ce résultat directement à partir de la fonction de transfert</td><td>&nbsp;</td></tr><tr><td>&nbsp;</td><td>&nbsp;</td><td>globale du système, en utilisant le théorème de la valeur finale :</td><td>&nbsp;</td></tr><tr><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>é</td><td>&nbsp;</td><td><em>S </em>(<em> p</em>)ù</td><td>= lim <em>pE</em>( <em>p</em>)[1- <em>H</em> ( <em>p</em>)]</td><td>&nbsp;</td></tr><tr><td><em>e</em><em><sub>t</sub></em>&nbsp; = lim (<em>e</em>(<em>t</em> ) &#8211; <em>s</em>(<em>t</em>)) = lim <em>p</em> (<em>E</em> ( <em>p</em>) &#8211; <em>S</em>( <em>p</em>) ) = lim <em>pE</em>( <em>p</em>) <sub>ê</sub>1</td><td>&#8211;</td><td>&nbsp;</td><td>ú</td><td>&nbsp;</td></tr><tr><td>&nbsp;</td><td>&nbsp;</td></tr><tr><td><em>t </em>®¥</td><td><em>p</em>®0</td><td><em>p </em>®0</td><td>ë</td><td>&nbsp;</td><td><em>E </em>(<em> p</em>)<sub>û</sub></td><td><em>p</em>®0</td><td>&nbsp;</td></tr><tr><td>Page 7</td><td>&nbsp;</td><td><strong>Emmanuel FARGES</strong></td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>Ó EduKlub S.A.</td><td>&nbsp;</td></tr></tbody></table></figure>



<p class="wp-block-paragraph">Tous droits de l’auteur des œuvres réservés. Sauf autorisation, la reproduction ainsi que toute utilisation des œuvres autre que la consultation individuelle et privée sont interdites.</p>



<p class="wp-block-paragraph"><strong>Sciences Indusrielles</strong></p>



<p class="wp-block-paragraph"><strong>Systèmes linéaires continus invariants</strong></p>



<p class="wp-block-paragraph">Analyse temporelle</p>



<figure class="wp-block-table"><table><tbody><tr><td>Dans le cas qui nous intéresse, c’est à dire pour un premier ordre&nbsp; <em>H</em> ( <em>p</em>) =</td><td><em>K</em></td><td>, et pour une</td><td>&nbsp;</td></tr><tr><td>&nbsp;</td><td>&nbsp;</td></tr><tr><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>1+<em>t</em> <em>p</em></td><td>&nbsp;</td><td>&nbsp;</td></tr><tr><td>entrée rampe ,</td><td>&nbsp;</td><td><em>E </em>(<em> p</em>)=</td><td><em>V</em></td><td>, on obtient :</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td></tr><tr><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td></tr><tr><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td><em>p</em><sup>2</sup></td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td></tr><tr><td><em>e</em></td><td>&nbsp;</td><td>= lim</td><td><em>V</em></td><td><sup>é</sup>1-</td><td>&nbsp;</td><td><em>K&nbsp; </em>ù</td><td>= lim</td><td><em>V</em></td><td>é</td><td>1- <em>K</em> +<em>t</em> <em>p</em></td><td>ù</td><td>=</td><td>ì ¥ si <em>K</em> ¹ 1</td><td>&nbsp;</td><td>&nbsp;</td></tr><tr><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td><sup>í</sup><em>V </em><em>t</em><em> </em>si<em> K </em>=1</td><td>&nbsp;</td><td>&nbsp;</td></tr><tr><td>&nbsp;</td><td>&nbsp;</td><td>1+ <em>t</em> <em>p</em><sup>ú</sup></td><td>&nbsp;</td><td>1+<em>t</em> <em>p</em></td><td>&nbsp;</td><td>&nbsp;</td></tr><tr><td>&nbsp;</td><td><em>t</em></td><td><em>p </em>®<em> </em>0<em>&nbsp; </em><em><sub>p</sub></em><em> </em><sup>ê</sup></td><td><em>p</em>®0<em>&nbsp; </em><em><sub>p</sub></em><em> </em><sup>ê</sup></td><td>ú</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td></tr><tr><td>&nbsp;</td><td>&nbsp;</td><td>ë</td><td>û</td><td>ë</td><td>&nbsp;</td><td>û</td><td>&nbsp;</td><td>î</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td></tr></tbody></table></figure>



<h5 class="wp-block-heading"><strong>1.4.4.2 Courbe réponse caractéristique</strong></h5>



<figure class="wp-block-table"><table><tbody><tr><td><em>e</em></td><td><em>t</em></td><td>®<strong><em> </em></strong><em>t</em><strong><em> car K=1</em></strong></td><td>&nbsp;</td></tr><tr><td><strong><em>Entrée : rampe de pente V=1</em></strong></td><td><em>t</em>®¥</td><td>&nbsp;</td></tr></tbody></table></figure>



<p class="wp-block-paragraph"><strong><em>Asymptote de la réponse : de</em></strong></p>



<p class="wp-block-paragraph"><strong><em>même pente que l’entrée car K=1</em></strong></p>



<p class="wp-block-paragraph">1</p>



<p class="wp-block-paragraph"><strong><em>Réponse du système </em></strong><em>H</em>(<em>p</em>)=</p>



<p class="wp-block-paragraph">1 + 2 <em>p</em></p>



<p class="wp-block-paragraph"><strong><em>(gain unitaire) à une rampe de pente</em></strong></p>



<p class="wp-block-paragraph"><strong><em>Entrée rampe avec V=1</em></strong></p>



<p class="wp-block-paragraph"><strong><em>Evolution des réponses</em></strong></p>



<p class="wp-block-paragraph"><strong><em>de systèmes de gain</em></strong></p>



<p class="wp-block-paragraph"><strong><em>unitaire (K=1) avec </em></strong><em>t</em><strong><em> </em></strong>Z</p>



<p class="wp-block-paragraph">Page 8&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; <strong>Emmanuel FARGES</strong>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; Ó EduKlub S.A.</p>



<p class="wp-block-paragraph">Tous droits de l’auteur des œuvres réservés. Sauf autorisation, la reproduction ainsi que toute utilisation des œuvres autre que la consultation individuelle et privée sont interdites.</p>



<p class="wp-block-paragraph"><strong>Sciences Indusrielles</strong></p>



<p class="wp-block-paragraph"><strong>Systèmes linéaires continus invariants</strong></p>



<p class="wp-block-paragraph">Analyse temporelle</p>



<figure class="wp-block-table"><table><tbody><tr><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td><em>e</em><em><sub>t</sub></em><em> &nbsp;</em>= ¥</td><td>&nbsp;</td></tr><tr><td>&nbsp;</td><td><em>e</em><em><sub>t</sub></em></td><td><strong><em>constant</em></strong></td><td><strong><em>car asymptote</em></strong></td><td><strong><em>car asymptote de</em></strong></td><td>&nbsp;</td></tr><tr><td>&nbsp;</td><td><strong><em>de même pente que l’entrée</em></strong></td><td><strong><em>pente différente</em></strong></td><td>&nbsp;</td></tr><tr><td>&nbsp;</td><td><strong><em>que l’entrée</em></strong></td><td>&nbsp;</td></tr><tr><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td></tr><tr><td>&nbsp;</td><td><strong><em>Entrée rampe avec V=1</em></strong></td><td>&nbsp;</td><td>&nbsp;</td></tr><tr><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td></tr><tr><td><strong><em>Comparaison des</em></strong></td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td></tr><tr><td><strong><em>réponses en</em></strong></td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td></tr><tr><td><strong><em>fonction des</em></strong></td><td>&nbsp;</td><td>&nbsp;</td><td><strong><em>Réponse d’un système</em></strong></td><td>&nbsp;</td><td>&nbsp;</td></tr><tr><td><strong><em>différents gains</em></strong></td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td></tr><tr><td>&nbsp;</td><td>&nbsp;</td><td><strong><em>de gain </em></strong><em>K</em><strong><em> </em></strong>¹<strong><em> </em></strong>1</td><td>&nbsp;</td><td>&nbsp;</td></tr><tr><td><strong><em>des systèmes</em></strong></td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td></tr><tr><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td></tr><tr><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td></tr></tbody></table></figure>



<p class="wp-block-paragraph"><strong><em>Réponse d’un</em></strong></p>



<p class="wp-block-paragraph"><strong><em>système de gain K=1</em></strong></p>



<p class="wp-block-paragraph">Page 9&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; <strong>Emmanuel FARGES</strong>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; Ó EduKlub S.A.</p>



<p class="wp-block-paragraph">Tous droits de l’auteur des œuvres réservés. Sauf autorisation, la reproduction ainsi que toute utilisation des œuvres autre que la consultation individuelle et privée sont interdites.</p>



<p class="has-text-align-center wp-block-paragraph"><strong><a href="https://www.mediafire.com/file/0b3fz1lqffmj41a/Telechargercours.com_SYSTEMES_LINEAIRES_CONTINUS_INVARIANTS_SI-CO05-AAS-EF.pdf/file" target="_blank" rel="noreferrer noopener nofollow">Telecharger</a></strong></p>
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		<title>TD n°4 : CORRECTION VAN, TRI, Délai de récupération</title>
		<link>https://telechargercours.com/td-n4-correction-van-tri-delai-de-recuperation/</link>
		
		<dc:creator><![CDATA[Cours Zilla]]></dc:creator>
		<pubDate>Wed, 12 May 2021 16:01:27 +0000</pubDate>
				<category><![CDATA[Economie & Entreprise]]></category>
		<guid isPermaLink="false">https://telechargercours.com/?p=15032</guid>

					<description><![CDATA[TD n°4&#160;: CORRECTION VAN, TRI, Délai de récupération La valeur actuelle nette (VAN) permet de mesurer la profitabilité d’un investissement. La VAN correspond à l’investissement de départ avec les cash-flow (flux de trésorerie) générés par cet investissement de départ. Le taux de rentabilité interne (TRI) est le taux d’actualisation pour lequel la valeur actuelle nette [&#8230;]]]></description>
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<p class="wp-block-paragraph"><strong><u>TD n°4</u></strong>&nbsp;: <strong>CORRECTION</strong> <strong>VAN, TRI, Délai de récupération</strong></p>



<p class="wp-block-paragraph">La valeur actuelle nette (VAN) permet de mesurer la profitabilité d’un investissement. La VAN correspond à l’investissement de départ avec les cash-flow (flux de trésorerie) générés par cet investissement de départ.</p>



<p class="wp-block-paragraph">Le taux de rentabilité interne (TRI) est le taux d’actualisation pour lequel la valeur actuelle nette est égale à 0. C’est le taux qui équilibre le montant de l’investissement et les flux de trésorerie dégagés par cet investissement.</p>



<p class="wp-block-paragraph">Le délai de récupération du capital investi (DRCI) permet de mesurer le délai nécessaire pour que la somme actualisée des flux de trésorerie prévisionnels puisse permettre la récupération du coût subi par l’investissement. À partir de cette date, l’investissement commence à créer de la valeur.</p>



<ul class="wp-block-list"><li><strong><u>Exercice 1</u></strong>&nbsp;: <strong>Projet (VAN&nbsp;, TRI&nbsp;, DELAI DE RECUPERATION)</strong></li></ul>



<p class="wp-block-paragraph">De l’analyse d’un projet d’investissement P, on retient les informations suivantes&nbsp;:</p>



<ul class="wp-block-list"><li>Capital investi&nbsp;: 900 de matériels amortissables linéairement en 5 ans&nbsp;;</li><li>Durée de vie&nbsp;: 5 ans&nbsp;;</li><li>Valeur résiduelle, nette d’impôts, au terme des 5 ans&nbsp;: 10.</li></ul>



<p class="wp-block-paragraph">Les prévisions d’exploitation sont données par le tableau ci-dessous&nbsp;:</p>



<figure class="wp-block-table"><table><tbody><tr><td><strong>Années</strong></td><td><strong>2006</strong></td><td><strong>De 2007 à 2010</strong></td></tr><tr><td>Chiffre d’affaire HT</td><td>900</td><td>1200</td></tr><tr><td>Charges d’exploitation variables</td><td>360</td><td>480</td></tr><tr><td>Charges d’exploitation fixes (hors amortissements)</td><td>300</td><td>300</td></tr></tbody></table></figure>



<ol class="wp-block-list" type="1"><li>Calculer les flux nets de liquidités attendus du projet (taux de l’IS&nbsp;: 33 1/3%)</li><li>Calculer la VAN, le TRI et le délai de récupération, sachant que le taux de rentabilité minimum exigé est de 8%. Conclure.</li><li>Sachant que le besoin en fond de roulement représente un mois de CAHT, expliquez comment cette information sera prise en compte dans les calculs et calculez le VAN à 8%.</li></ol>



<p class="wp-block-paragraph"><strong><u>Solution</u></strong><strong>&nbsp;:</strong></p>



<p class="wp-block-paragraph">1°) Flux nets de liquidités prévisionnels.</p>



<figure class="wp-block-table"><table><tbody><tr><td><strong>Années</strong></td><td><strong>2006</strong></td><td><strong>De 2007 à 2010</strong></td></tr><tr><td>CAHT</td><td>900</td><td>1200</td></tr><tr><td>Charges variables</td><td>360</td><td>480</td></tr><tr><td>Charges fixes</td><td>300</td><td>300</td></tr><tr><td>Dot. Aux amortissements</td><td>= 180</td><td>= 180</td></tr><tr><td>Résultat avant impôts</td><td>900 – 360 – 300 – 180 = 60</td><td>1200-480-300-180 = 240</td></tr><tr><td>Résultat net</td><td>60´ = 40</td><td>240´ = 160</td></tr><tr><td>Flux nets de liquidités</td><td>40 + 180 = 220</td><td>160 + 180 = 340</td></tr></tbody></table></figure>



<p class="wp-block-paragraph">2°)</p>



<ul class="wp-block-list"><li><strong><u>VAN au taux de 8%</u></strong><strong><br></strong>VAN = 220´(1,08)<sup> – 1</sup> + 340´(1,08)<sup> – 2</sup> + 340´(1,08)<sup> – 3</sup> + 340´(1,08)<sup> – 4</sup> + (340 + 10)´(1,08)<sup> – 5</sup> – 900<br></li><li><strong><u>TRI</u></strong><br>900 = 220´(1+x)<sup> – 1</sup> + 340´(1+x)<sup> – 2</sup> + 340´(1+x)<sup> – 3</sup> + 340´(1+x)<sup> – 4</sup> + (340 + 10)´( 1+x)<sup> – 5</sup><br>On trouve</li><li>Délai de récupération&nbsp;:</li></ul>



<figure class="wp-block-table"><table><tbody><tr><td>Années</td><td>1</td><td>2</td><td>3</td><td>4</td><td>5</td></tr><tr><td>Cash-flows</td><td>220</td><td>340</td><td>340</td><td>340</td><td>350</td></tr><tr><td>Cash-flows actualisés</td><td>204</td><td>291</td><td>270</td><td>250</td><td>238</td></tr><tr><td>Cumul</td><td>204</td><td>495</td><td>765</td><td>1 015</td><td>1 253</td></tr><tr><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td><td>&nbsp;</td></tr><tr><td>délai de récupération</td><td>&nbsp;</td><td>&nbsp;</td><td>3.539786541</td><td>&nbsp;</td><td>&nbsp;</td></tr><tr><td>an</td><td>&nbsp;</td><td>&nbsp;</td><td>3</td><td>&nbsp;</td><td>&nbsp;</td></tr><tr><td>mois</td><td>&nbsp;</td><td>&nbsp;</td><td>6</td><td>&nbsp;</td><td>&nbsp;</td></tr><tr><td>jours</td><td>&nbsp;</td><td>&nbsp;</td><td>14</td><td>&nbsp;</td><td>&nbsp;</td></tr></tbody></table></figure>



<p class="wp-block-paragraph">La VAN est largement positive et le TRI est très au-dessus du taux de rentabilité exigé par l’entreprise&nbsp;; le projet est donc rentable.</p>



<p class="wp-block-paragraph">Le délai de récupération doit être comparé à une durée maximum fixée par l’entreprise.</p>



<p class="wp-block-paragraph">3°)</p>



<p class="wp-block-paragraph">Si l’exploitation de l’investissement entraîne une augmentation durable du BFR, celle-ci doit être prise en compte comme un investissement dans les calculs de rentabilité.</p>



<p class="wp-block-paragraph">La diminution du BFR libère des fonds jusque là investis et doit être assimilée à un complément de cash-flows.</p>



<p class="wp-block-paragraph">Les variations du BFR sont sensées se produire <strong><u>en début d’année</u></strong>.</p>



<figure class="wp-block-table"><table><tbody><tr><td><strong>Années</strong></td><td><strong>2006</strong></td><td><strong>2007</strong></td><td><strong>2008 à 2010</strong></td></tr><tr><td>BFR</td><td>= 75</td><td>= 100</td><td>= 100</td></tr><tr><td>Variation du BFR</td><td>75</td><td>100-75 = 25</td><td>0</td></tr></tbody></table></figure>



<p class="wp-block-paragraph">Donc on a le diagramme des flux (<strong>fin d’année</strong>)</p>



<figure class="wp-block-table"><table><tbody><tr><td>220 &#8211; 25 &nbsp;</td></tr></tbody></table></figure>



<figure class="wp-block-table"><table><tbody><tr><td>350 + 100 &nbsp;</td></tr></tbody></table></figure>



<figure class="wp-block-table"><table><tbody><tr><td>340 &nbsp;</td></tr></tbody></table></figure>



<figure class="wp-block-table"><table><tbody><tr><td>340 &nbsp;</td></tr></tbody></table></figure>



<figure class="wp-block-table"><table><tbody><tr><td>340 &nbsp;</td></tr></tbody></table></figure>



<p class="wp-block-paragraph"><strong></strong></p>



<p class="wp-block-paragraph"><strong><u>Calcul de la VAN à 8%</u></strong></p>



<p class="wp-block-paragraph">VAN = -975 + 195´(1,08)<sup> – 1</sup> + 340´(1,08)<sup> – 2</sup> + 340´(1,08)<sup> – 3</sup> + 340´(1,08)<sup> – 4</sup> + (340 + 10+100)´(1,08)<sup> – 5</sup></p>



<p class="wp-block-paragraph"><strong><u>Exercice 2</u></strong>&nbsp;: (d’après DECF)</p>



<p class="wp-block-paragraph">Afin d’accroître&nbsp; sa capacité de production, un société envisage de réaliser un programme d’investissement. Le projet comprend.</p>



<ul class="wp-block-list"><li>Un bâtiment pour 3&nbsp;000&nbsp;000 HT, amortissable en linéaire sur 15 ans</li><li>Un matériel et du mobilier, amortissable en linéaire sur 5 ans,<br>pour un montant de 1&nbsp;000&nbsp;000 HT</li></ul>



<p class="wp-block-paragraph">Pour l’étude de la rentabilité de ce projet, il a été décidé de ne considérer que les 5 premières années d’exploitation (1 à 5).</p>



<p class="wp-block-paragraph">Les quantités supplémentaires vendues seraient les suivantes&nbsp;:</p>



<figure class="wp-block-table"><table><tbody><tr><td>Années</td><td>1</td><td>2</td><td>3</td><td>4</td><td>5</td></tr><tr><td>Quantités</td><td>25 000</td><td>30 000</td><td>30 000</td><td>30 000</td><td>30 000</td></tr></tbody></table></figure>



<ul class="wp-block-list"><li>Le prix de vente unitaire est fixé à 400 euros HT.</li><li>Les charges variables unitaires s’élevaient à 300 euros HT.</li><li>Les charges fixes annuelles (hors amortissement) seraient de 600&nbsp;000 euros.</li><li>La réalisation de cet investissement nécessitant une durée de 16 mois, son financement serait fractionné de la façon suivante&nbsp;: ¼ fin d’année – 1&nbsp;; le reste fin d’année 0&nbsp;; l’investissement étant opérationnel dès le début de l’année 1.</li></ul>



<p class="wp-block-paragraph">Calculez le taux de rentabilité interne de ce projet (pour ce calcul, on négligera la valeur résiduelle des bâtiments ainsi que celle du matériel et du mobilier). Taux d’impôt sur les bénéfices&nbsp;: 33 1/3 %</p>



<p class="wp-block-paragraph">Correction</p>



<ul class="wp-block-list"><li>Calcul des amortissements par an:&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; + = <strong><u>400&nbsp;000 euros</u></strong></li></ul>



<figure class="wp-block-table"><table><tbody><tr><td><strong>Années</strong></td><td><strong>1</strong></td><td><strong>2</strong></td><td><strong>3</strong></td><td><strong>4</strong></td><td><strong>5</strong></td></tr><tr><td>Marge sur coûts variables</td><td>(400-300)´25&nbsp;000 = 2&nbsp;500 000</td><td>3&nbsp;000 000</td><td>3&nbsp;000 000</td><td>3&nbsp;000 000</td><td>3&nbsp;000 000</td></tr><tr><td>Charges fixes + amortissements</td><td>600&nbsp;000 + 400&nbsp;000 = 1&nbsp;000 000</td><td>1&nbsp;000 000</td><td>1&nbsp;000 000</td><td>1&nbsp;000 000</td><td>1&nbsp;000 000</td></tr><tr><td>Résultat avant impôts</td><td>2&nbsp;500&nbsp;000 – 1&nbsp;000&nbsp;000 = 1&nbsp;500 000</td><td>2&nbsp;000 000</td><td>2&nbsp;000 000</td><td>2&nbsp;000 000</td><td>2&nbsp;000 000</td></tr><tr><td>Résultat net</td><td>1&nbsp;500&nbsp;000 ´ = 1&nbsp;000 000</td><td>1&nbsp;333 333</td><td>1&nbsp;333 333</td><td>1&nbsp;333 333</td><td>1&nbsp;333 333</td></tr><tr><td>Flux nets de liquidités</td><td>1&nbsp;000&nbsp;000 + 400&nbsp;000 = 1&nbsp;400 000</td><td>1&nbsp;733 333</td><td>1&nbsp;733 333</td><td>1&nbsp;733 333</td><td>1&nbsp;733 333</td></tr></tbody></table></figure>



<p class="wp-block-paragraph">Il faut maintenant établir le diagramme des flux. Attention, la réalisation de cet investissement nécessitant une durée de 16 mois, son financement est fractionné de la façon suivante&nbsp;:</p>



<ul class="wp-block-list"><li>¼ fin d’année – 1, soit&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; = 1&nbsp;000 000</li><li>soit &nbsp;le reste fin d’année 0, soit &nbsp; 3&nbsp;000 000</li></ul>



<figure class="wp-block-table"><table><tbody><tr><td>1&nbsp;733 333 &nbsp;</td></tr></tbody></table></figure>



<figure class="wp-block-table"><table><tbody><tr><td>1&nbsp;733 333 &nbsp;</td></tr></tbody></table></figure>



<figure class="wp-block-table"><table><tbody><tr><td>1&nbsp;733 333 &nbsp;</td></tr></tbody></table></figure>



<figure class="wp-block-table"><table><tbody><tr><td>1&nbsp;733 333 &nbsp;</td></tr></tbody></table></figure>



<figure class="wp-block-table"><table><tbody><tr><td>1&nbsp;400 000 &nbsp;</td></tr></tbody></table></figure>



<figure class="wp-block-table"><table><tbody><tr><td>&#8211; 3&nbsp;000 000 &nbsp;</td></tr></tbody></table></figure>



<p class="wp-block-paragraph"><strong>Le TRI vérifie donc l’équation</strong></p>



<p class="wp-block-paragraph">1&nbsp;000&nbsp;000 = &#8211; 3&nbsp;000&nbsp;000´(1+TRI)<sup> – 1</sup> + 1&nbsp;400 000´(1+TRI)<sup> – 2</sup> + 1&nbsp;733 333´(1+TRI)<sup> – 3</sup> +…+1&nbsp;733 333´(1+TRI)<sup> – 6</sup></p>



<p class="wp-block-paragraph">Soit&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;</p>



<p class="wp-block-paragraph"><strong><u>Exercice 3</u></strong>&nbsp;: <strong>TRI</strong> (d’après ITB)</p>



<p class="wp-block-paragraph">1°) Dessiner le diagramme des flux.</p>



<p class="wp-block-paragraph">2°)</p>



<p class="wp-block-paragraph">Correction.</p>



<p class="has-text-align-center wp-block-paragraph"><strong><a href="https://www.mediafire.com/file/ebiavzmk5x59oiv/Telechargercours.com+fiche-td_4_van_tri_delai-corr.pdf" class="rank-math-link">Telecharger</a></strong></p>
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