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		<title>UMTS Reference Architecture</title>
		<link>https://engineersphere.com/umts-reference-architecture</link>
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		<dc:creator><![CDATA[Jeff Schuler]]></dc:creator>
		<pubDate>Sat, 16 Apr 2011 04:48:58 +0000</pubDate>
				<category><![CDATA[Cellular]]></category>
		<category><![CDATA[Wireless Hacking]]></category>
		<category><![CDATA[CDMA]]></category>
		<category><![CDATA[FDD]]></category>
		<category><![CDATA[GSM]]></category>
		<category><![CDATA[PSTN]]></category>
		<category><![CDATA[RNC]]></category>
		<category><![CDATA[RNS]]></category>
		<category><![CDATA[TDD]]></category>
		<category><![CDATA[UE]]></category>
		<category><![CDATA[UMTS architecture]]></category>
		<category><![CDATA[UTRAN]]></category>
		<guid isPermaLink="false">https://engineersphere.com/?p=3089</guid>

					<description><![CDATA[<p>Continuing on with wireless communications related subjects.  The architecture of a UMTS systems is the first thing to begin understanding if you want to undertake an education such as this.  It consists of three distinct components: The User Equipment (UE) The UMTS Terrestrial Radio Access Network (UTRAN) The Core Network (CN) As with any wireless [&#8230;]</p>
<p>The post <a href="https://engineersphere.com/umts-reference-architecture">UMTS Reference Architecture</a> first appeared on <a href="https://engineersphere.com">Engineersphere.com</a>.</p>]]></description>
										<content:encoded><![CDATA[<p>Continuing on with wireless communications related subjects.  The architecture of a UMTS systems is the first thing to begin understanding if you want to undertake an education such as this.  It consists of three distinct components:</p>
<ol>
<li><strong>The User Equipment (UE)</strong></li>
<li><strong>The UMTS Terrestrial Radio Access Network (UTRAN)</strong></li>
<li><strong>The Core Network (CN)</strong></li>
</ol>
<p>As with any wireless network out there, the main purpose is the provide access to services (data, voice, etc.)  The services network is divided into <strong>Public Switched Telephone Network (PSTN)</strong> that provides voice and special telephone related services (look it up on wiki), and the internet, which provides a wide range of packet data services such as email or access to the world wide web.  These things are probably all sounding very familiar to you, and they should, because they are critical in maintaining today&#8217;s society, and almost everyone living in the modern world uses at least one of these services sometimes in their daily lives.</p>
<p>The UMTS mobile, also known as the User Equipment (UE), interfaces with the UTRAN via the UMTS physical layer radio interface.  In addition to radio access, the UE provides the subscriber with access to services and profile information.  For example, the cell phone you carry in your pocket is the UE (user equipment) that interfaces with the cell phone towers that companies like Verizon, AT&amp;T, and Sprint provide for you.</p>
<p>In UMTS, there are two Core Network (CN) configurations, the Circuit Switched CN (CS-CN) and Packet Switched CN (PS-CN).  The CS-CN is based on the GSM Public Land Mobile Network (PLMN) and provides functions such as connectivity to the PSTN, circuit telephony services such as voice, and supplementary services such as call forwarding, call waiting, etc.  The PS-CN is based on the GSM General Packet Radio System (GPRS) PLMN, which provides access to the Internet and other packet data services.</p>
<p>Both core networks connect to the UMTS Terrestrial Radio Access Network.  The UTRAN has two options for its air interface operations.  One option is Time Division Duplex (TDD), which makes use of a single 5 MHz carrier for communication between the UE and the UTRAN.  The other option is the Frequency Division Duplex (FDD), which provides full duplex operation using 5 MHz of spectrum in each direction to and from the UTRAN.</p>
<p>How do all of these components fit together?  Check out the image below.</p>
<p><a href="https://engineersphere.com/stuff/uploads/2011/04/umts-reference-architecture.png"><img class="alignnone size-full wp-image-3090" title="umts-reference-architecture" src="https://engineersphere.com/stuff/uploads/2011/04/umts-reference-architecture.png" alt="umts-reference-architecture" width="500" height="500" srcset="https://engineersphere.com/stuff/uploads/2011/04/umts-reference-architecture.png 500w, https://engineersphere.com/stuff/uploads/2011/04/umts-reference-architecture-150x150.png 150w, https://engineersphere.com/stuff/uploads/2011/04/umts-reference-architecture-300x300.png 300w" sizes="(max-width: 500px) 100vw, 500px" /></a></p>
<h3>UTRAN Components</h3>
<p>The UTRAN consists of one or more <strong>Radio Network Subsystems (RNS)</strong>.  An RNS consists of one <strong>Radio Network Controller (RNC)</strong> and several Nodes.  The radio network controller and the nodes are two essential components of UTRAN.  Apart from these two component types, UTRAN requires Operation Maintenance Centers (OMC) to perform Operation Administration and Maintenance (OA&amp;M) functionality on the nodes and RNCs.  Yes, I know the acronyms are getting a little out of hand, but it is essential to learn them if you want to speak the language!  Engineers only speak to each other with acronyms and it is very very annoying indeed.</p>
<ul>
<li><strong>Radio Network Controller</strong>: The Radio Network Controller is the master of UTRAN.  It handles all aspects of radio resource management within the radio network subsystem.  The UMTS chose to use it instead of the base station controller in order to stress the independence of UTRAN from the Core Networks (CN).  It interfaces with the core network components such as the <strong>Mobile Switching Center (MSC)</strong> and <strong>Service GPRS Support Node (SGSN)</strong> to route signaling and traffic from the User Equipment (UE).  The RNC also interfaces with other RNC&#8217;s within UTRAN to provide it with wide mobility (very important!)</li>
</ul>
<ul>
<li><strong>Nodes</strong>: Within this network, a node is the radio transmission and reception unit within UTRAN (remember above, I explained that the UE is your cell phone you carry in your pocket).  It handles radio transmission and reception for multiple cells within a coverage area.  So if you think about the amount of area that a certain cell tower (cell) can transmit its signal with the proper quality of service (QoS), you can get a mental grasp on multiple cells being in one coverage area, say if the towers are close together.  The node implements CDMA-Specific functionality such as <a title="encoding techniques" href="https://engineersphere.com/digital-coding-techniques">encoding</a>, interleaving, spreading, scrambling &amp; modulation.  The nodes are also what used to be known as the <strong>Base Transceiver Subsystems (BTS)</strong> in second generation systems.</li>
</ul><p>The post <a href="https://engineersphere.com/umts-reference-architecture">UMTS Reference Architecture</a> first appeared on <a href="https://engineersphere.com">Engineersphere.com</a>.</p>]]></content:encoded>
					
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		<title>Digital Coding Techniques</title>
		<link>https://engineersphere.com/digital-coding-techniques</link>
					<comments>https://engineersphere.com/digital-coding-techniques#respond</comments>
		
		<dc:creator><![CDATA[Jeff Schuler]]></dc:creator>
		<pubDate>Thu, 07 Apr 2011 04:30:20 +0000</pubDate>
				<category><![CDATA[Cellular]]></category>
		<category><![CDATA[Wireless Hacking]]></category>
		<category><![CDATA[block interleaving]]></category>
		<category><![CDATA[convolutional coding]]></category>
		<category><![CDATA[digital coding techniques]]></category>
		<category><![CDATA[turbo encoding]]></category>
		<guid isPermaLink="false">https://engineersphere.com/?p=3015</guid>

					<description><![CDATA[<p>Hey!  This article has a simple purpose, to teach you basic functionality and terminology related to modern UMTS-CDMA digital coding techniques.  One swift read of this information-packed kick in the face will leave you trembling at the knees with curiosity for more.  Well, maybe not.  But the first step in the UMTS-CDMA system applies error [&#8230;]</p>
<p>The post <a href="https://engineersphere.com/digital-coding-techniques">Digital Coding Techniques</a> first appeared on <a href="https://engineersphere.com">Engineersphere.com</a>.</p>]]></description>
										<content:encoded><![CDATA[<p>Hey!  This article has a simple purpose, to teach you basic functionality and terminology related to modern UMTS-CDMA digital coding techniques.  One swift read of this information-packed kick in the face will leave you trembling at the knees with curiosity for more.  Well, maybe not.  But the first step in the UMTS-CDMA system applies error correction techniques so that any errors at the receiver can be corrected.  Various techniques can be employed to prevent errors at the transmitter.  You have probably studied one or more of these in your days if your an electrical or computer engineering student, professor, or professional.  One of these techniques is the use of <strong>Forward Error Codes</strong>, which are applied to the data before it is transmitted via the physical layer.</p>
<p>It is known that wireless is an inherently error-prone medium in which to operate our delicate signals.  Therefore, many error correction techniques are employed.  In the UMTS-CDMA systems, due to the large bandwidth available, a variety of coding techniques are employed.  The following three error correcting methods come to mind:</p>
<h3>Convolutional Encoding</h3>
<p>Convolutional encoding provides the ability to correct errors at the receiver.  So, the errors are removed from the signal by the receiver via <strong>convolutional encoding</strong>.  As a result, lower transmission power is required, which can result in more errors.  Some amount of errors can be tolerated since they can be recovered through convolutional encoding.  The convolutional encoder encodes input data bits int output symbols.  The data bits are entered into the first register at each clock cycle and the data bit in the last register is dumped out.  Data bits are tapped at various positions and XORed to provide encoded bits.  It is typically used for voice and low data rate applications.  Here are the main points to keep in mind:</p>
<ul>
<li>Provides the ability to detect and correct errors at the receiver.</li>
<li>10^(-3)  BER, typically used for voice and low data rates.</li>
<li>Uses history of bits to recover from errors.</li>
</ul>
<h3>Turbo Encoding</h3>
<p>Turbo codes are a new class of error correction codes used in digital comm systems.  Turbo codes have been shown to perform better for high-rate data services (which is what we crave for) with stringent error rate requirements on the order of 10-6 <strong>Bit Error Rate (BER)</strong>.  The turbo encoder consists of two constituent convolutional encoders.  Both constituent encoders use and code the same data.  The first one is fed data in the same order as the input data.  The second encoder uses a permuted form of the input data and the permuting is accomplished by the use of an <strong>interleaver</strong>, which will be discussed in detail in the following post(s).  Again, main points:</p>
<ul>
<li>10^(-6)  BER, suitable for high data rates.</li>
<li>Uses convolutional encoders in parallel to increase reliability.</li>
<li>Increased delays but better error correction capabilities.</li>
</ul>
<h3>Block Interleaving</h3>
<p>Block interleaving protects data against fading and prevents bursty errors (so imagine a sudden burst in the amplitude of a received signal, think that might saturate you and ruin your signal? You bet.)  This is accomplished by providing time diversity, where the bits are separated in time before transmission over the air.  This is typically used with FEC codes, since FEC codes are not well suited to handle these bursty errors.</p>
<ul>
<li>Method to shuffle bits to prevent errors during deep fade.</li>
<li>Provides time diversity.</li>
</ul>
<p>All of these techniques will be described in detail, separately, in the following three articles.</p><p>The post <a href="https://engineersphere.com/digital-coding-techniques">Digital Coding Techniques</a> first appeared on <a href="https://engineersphere.com">Engineersphere.com</a>.</p>]]></content:encoded>
					
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		<title>The Evolution of 3G Wireless Technologies</title>
		<link>https://engineersphere.com/the-evolution-of-3g-wireless-technologies</link>
					<comments>https://engineersphere.com/the-evolution-of-3g-wireless-technologies#respond</comments>
		
		<dc:creator><![CDATA[Jeff Schuler]]></dc:creator>
		<pubDate>Wed, 06 Apr 2011 02:29:41 +0000</pubDate>
				<category><![CDATA[Cellular]]></category>
		<category><![CDATA[Wireless Hacking]]></category>
		<category><![CDATA[3G]]></category>
		<category><![CDATA[CDMA2000]]></category>
		<category><![CDATA[EDGE]]></category>
		<category><![CDATA[GPRS]]></category>
		<category><![CDATA[GSM]]></category>
		<category><![CDATA[IS-136]]></category>
		<category><![CDATA[ITU]]></category>
		<category><![CDATA[UMTS]]></category>
		<guid isPermaLink="false">https://engineersphere.com/?p=3004</guid>

					<description><![CDATA[<p>There are several different types of 3G wireless technologies that are defined and planned to be up and working today.  There are also several that are on their way.  These are successors of course, to the previous 2G technologies that dominated the airwaves. CDMA2000 CDMA2000 is the successor to IS-95 systems.  CDMA2000 provides a definition [&#8230;]</p>
<p>The post <a href="https://engineersphere.com/the-evolution-of-3g-wireless-technologies">The Evolution of 3G Wireless Technologies</a> first appeared on <a href="https://engineersphere.com">Engineersphere.com</a>.</p>]]></description>
										<content:encoded><![CDATA[<p>There are several different types of<strong> 3G wireless technologies</strong> that are defined and planned to be up and working today.  There are also several that are on their way.  These are successors of course, to the previous 2G technologies that dominated the airwaves.</p>
<h3>CDMA2000</h3>
<p><strong>CDMA2000 </strong>is the successor to <strong>IS-95</strong> systems.  CDMA2000 provides a definition for two different options for <a href="https://engineersphere.com/category/wireless-hacking/cellular">3G technologies</a>.  IT Differs in the amount of the frequency spectrum that is used.  The <strong>Spreading Rate</strong> (SR1) operates in the 1.25 MHz band and is known as a 1x system.  Another proposal exists also which is referred to as 1xEV-DO.  The 1xEV-DO (1x Evolution for Data Optimized) solution is a data-only solution that enables a bandwidth of 2Mbps without any mechanism for voice.  This is the type of data rate that we are all familiar with, the 3G 2Mbps speed of data connection.</p>
<h3>The Universal Mobile Telecommunications System (UMTS)</h3>
<p>The Universal Mobile Telecommunications System (<a href="http://en.wikipedia.org/wiki/UMTS">UMTS</a>) is a successor to GSM/GPRS systems.  There are also two options for the UMTS networks.  The Frequency Division Duplex (FDD) option uses spectrum bands which are paired together.  For example, two different 5 MHz bands are used for uplink and downlink.  The Time Division Duplex (TDD) option uses an unpaired band.  In other words, the same 5 MHz band is shared between uplink and downlink for TDD.</p>
<h3>Universal Wireless Consortium for IS-136 systems</h3>
<p>The UWC-136 (Universal Wireless Consortium for IS-136 systems) was originally considered to be the evolution for IS-136 systems.  However, the IS-136 system operators eventually decided to follow the path of CDMA2000 or UMTS.</p>
<h3>Why did we need 3G Technology?</h3>
<p>Back in the late 1990&#8217;s, when most of the readers out there were still playing in the sandbox, the International Telecommunication Union (ITU) set the requirements for the next generation of wireless networks (that is why they are called Third Generation (3G)).  One of the many many requirements is to reach peak data rates of at least 2 Mbps.  This is more relvant to the Downlink since the majority of traffic comes from the server to the client in the<a href="http://google.com"> Internet World</a>.</p>
<p>To meet this new high speed requirement, the 2nd generation wireless networks came up with several different evolutions before eventually being replaced.  The <a href="http://en.wikipedia.org/wiki/GSM">GSM </a>evolution includes <a href="http://en.wikipedia.org/wiki/GPRS">GPRS</a> and <a href="http://en.wikipedia.org/wiki/EDGE">EDGE</a>, which provide packet data services and represent intermediate solutions until a UMTS Release 99 System is deployed.  The 1xEV-DO is one possible evolution path from 1xRTT, and HSDPA is a Release 5 feature of UMTS.</p>
<h3>So how did UMTS Evolve?</h3>
<p>UMTS is the network of choice these days.  Yes, UMTS is 3G&#8230;If you haven&#8217;t caught that yet.  For those nerds out there that are curious, the evolution of UMTS has progressed over the years in the following fashion:</p>
<p><strong>UMTS Release 99</strong></p>
<ul>
<li>2 Mbps theoretical peak packet data rates</li>
<li>384 kbps (practical)</li>
</ul>
<p><strong>UMTS Release 5</strong></p>
<ul>
<li>HSDPA (14 Mbps downlink theoretical)</li>
<li>IMS (IP Multimedia Subsystem for multimedia)</li>
<li>UP UTRAN (for scalability and lower cost)</li>
</ul>
<p><strong>UMTS Release 6</strong></p>
<ul>
<li>HSUPA (up to 5.76 Mbps uplink)</li>
<li>MBMS (Multimedia Broadcast Multicast Service)</li>
</ul>
<p><strong>UMTS Release 7</strong></p>
<ul>
<li>Multiple Input Multiple Output (MIMO) Antenna Systems</li>
</ul><p>The post <a href="https://engineersphere.com/the-evolution-of-3g-wireless-technologies">The Evolution of 3G Wireless Technologies</a> first appeared on <a href="https://engineersphere.com">Engineersphere.com</a>.</p>]]></content:encoded>
					
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		<title>The Fourier Integral / Transform Explained</title>
		<link>https://engineersphere.com/the-fourier-integral-transform-explained</link>
					<comments>https://engineersphere.com/the-fourier-integral-transform-explained#respond</comments>
		
		<dc:creator><![CDATA[Jeff Schuler]]></dc:creator>
		<pubDate>Sat, 02 Apr 2011 03:51:33 +0000</pubDate>
				<category><![CDATA[Electrical Engineering Concepts]]></category>
		<category><![CDATA[fourier]]></category>
		<category><![CDATA[fourier integral]]></category>
		<category><![CDATA[fourier transform]]></category>
		<category><![CDATA[fourier transform examples]]></category>
		<guid isPermaLink="false">https://engineersphere.com/?p=2824</guid>

					<description><![CDATA[<p>Magic is real, and it all comes from the Fourier Integral.&#160; But one doesn&#8217;t become a wizard without a little reading first &#8211; so, the purpose of this article is to explain the Fourier Integral theoretically and mathematically. Before reading any further, it is important to first understand this: in mathematics, there is a rule [&#8230;]</p>
<p>The post <a href="https://engineersphere.com/the-fourier-integral-transform-explained">The Fourier Integral / Transform Explained</a> first appeared on <a href="https://engineersphere.com">Engineersphere.com</a>.</p>]]></description>
										<content:encoded><![CDATA[<p>Magic is real, and it all comes from the <strong>Fourier Integral</strong>.&nbsp; But one doesn&#8217;t become a wizard without a little reading first &#8211; so, the purpose of this article is to explain the Fourier Integral theoretically and mathematically.</p>
<p>Before reading any further, it is important to first understand this:<em> in mathematics, there is a rule that states that any <a href="http://en.wikipedia.org/wiki/Periodic_function">periodic </a>function of time may be &#8220;reconstructed&#8221; exactly from the summation of an infinite series of harmonic sine-waves</em>.&nbsp; The generalized theory itself is referred to as a &#8220;<strong>Fourier Series</strong>.&#8221;&nbsp; For use with arbitrary electronic time-domain signals of period <img src='https://s0.wp.com/latex.php?latex=T_o&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='T_o' title='T_o' class='latex' />, it may be expressed as:</p>
<img src='https://s0.wp.com/latex.php?latex=f%28t%29+%3D+a_0+%2B+%5Cdisplaystyle%5Csum_%7Bn%3D1%7D%5E%7B%5Cinfty%7Da_ncos%28nw_ot%29+%2B+b_nsin%28nw_ot%29+&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='f(t) = a_0 + \displaystyle\sum_{n=1}^{\infty}a_ncos(nw_ot) + b_nsin(nw_ot) ' title='f(t) = a_0 + \displaystyle\sum_{n=1}^{\infty}a_ncos(nw_ot) + b_nsin(nw_ot) ' class='latex' />
<p>over the range:</p>
<img src='https://s0.wp.com/latex.php?latex=t_0+%5Cle+t+%5Cle+t_0+%2B+T_0&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='t_0 \le t \le t_0 + T_0' title='t_0 \le t \le t_0 + T_0' class='latex' />
<p>where:</p>
<p style="padding-left: 30px;"><img src='https://s0.wp.com/latex.php?latex=a_0&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='a_0' title='a_0' class='latex' /> is the magnitude of the 0th harmonic</p>
<p style="padding-left: 30px;"><img src='https://s0.wp.com/latex.php?latex=a_n&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='a_n' title='a_n' class='latex' /> represents the magnitude of the nth harmonic of cosine wave components</p>
<p style="padding-left: 30px;"><img src='https://s0.wp.com/latex.php?latex=b_n&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='b_n' title='b_n' class='latex' /> represents the magnitude of the nth harmonic of sine wave components</p>
<p style="padding-left: 30px;"><img src='https://s0.wp.com/latex.php?latex=w_o&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='w_o' title='w_o' class='latex' /> is the <strong>fundamental frequency</strong></p>
<p style="padding-left: 30px;"><img src='https://s0.wp.com/latex.php?latex=t&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='t' title='t' class='latex' /> is the variable that represents instances in time</p>
<p style="padding-left: 30px;"><img src='https://s0.wp.com/latex.php?latex=n&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='n' title='n' class='latex' /> is the variable that represents the specific harmonic, and is always an integer</p>
<p>This monumental discovery was first announced on December 21, 1807 by historic gentleman Baron Jean-Baptiste-Joseph Fourier.</p>
<figure style="width: 250px" class="wp-caption aligncenter"><img title="fourier" src="http://upload.wikimedia.org/wikipedia/commons/thumb/a/aa/Fourier2.jpg/250px-Fourier2.jpg" alt="" width="250" height="286"/><figcaption class="wp-caption-text">Joseph Fourier</figcaption></figure>
<p>In order to go from the Fourier Integral to the Fourier Transform, it is necessary to express the previous Fourier Series as a series of ever-lasting exponential functions.&nbsp; Using an <a href="http://en.wikipedia.org/wiki/Orthogonality">orthogonal basis set</a> of signals described by <img src='https://s0.wp.com/latex.php?latex=e%5E%7Bjnw_ot%7D&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='e^{jnw_ot}' title='e^{jnw_ot}' class='latex' /> of magnitude <img src='https://s0.wp.com/latex.php?latex=D_n&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='D_n' title='D_n' class='latex' />, we now write the Fourier Series as:</p>
<img src='https://s0.wp.com/latex.php?latex=f%28t%29+%3D+%5Cdisplaystyle%5Csum_%7Bn%3D-+%5Cinfty%7D%5E%7B%5Cinfty%7DD_ne%5E%7Bjnw_ot%7D&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='f(t) = \displaystyle\sum_{n=- \infty}^{\infty}D_ne^{jnw_ot}' title='f(t) = \displaystyle\sum_{n=- \infty}^{\infty}D_ne^{jnw_ot}' class='latex' />
<p>where <img src='https://s0.wp.com/latex.php?latex=j&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='j' title='j' class='latex' /> is <img src='https://s0.wp.com/latex.php?latex=%5Csqrt%7B-1%7D&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='\sqrt{-1}' title='\sqrt{-1}' class='latex' />.</p>
<h3>What is the Fourier Integral?<span style="text-decoration: underline;"><br />
</span></h3>
<p>The Fourier Integral, also referred to as Fourier Transform for electronic signals, is a mathematical method of turning any arbitrary function of time into a corresponding <em>function of frequency</em>.&nbsp; A signal, when transformed into a &#8220;function of frequency&#8221;, essentially becomes a function that expresses the <em>relative magnitudes of each harmonic</em><em> of a Fourier Series that would be summed to recreate the original time-domain signal. </em>To see this, observe the following figures:</p>
<figure style="width: 521px" class="wp-caption aligncenter"><img class=" " title="square-pulse-wave" src="http://i.imgur.com/2dsft.jpg" alt="square-pulse-wave" width="521" height="450"/><figcaption class="wp-caption-text">Figure 1. A Square Wave Pulse, in time</figcaption></figure>
<p>In order to rebuild a square wave with sines and cosines only, it is necessary to determine the magnitudes of each harmonic used in the Fourier Series, or rather, the Fourier Integral (for <strong>continuous </strong>time-domain signals).&nbsp; The relative magnitudes of these needed harmonics can be displayed graphically as a function of frequency (widely known as a signal&#8217;s <strong>frequency spectrum</strong>):</p>
<figure style="width: 565px" class="wp-caption aligncenter"><img class=" " title="sinc-wave" src="http://i.imgur.com/YR774.jpg" alt="sinc-wave" width="565" height="447"/><figcaption class="wp-caption-text">Figure 2. The Fourier Integral, aka Fourier Transform, of a square pulse is a Sinc function.&nbsp; The Sinc function is also known as the Frequency Spectrum of a Square Pulse.</figcaption></figure>
<p>Though the recreation of a signal using an infinite series of sines and cosines is impossible to achieve in the lab, one may get very close.&nbsp; Close enough that the most advanced lab equipment wouldn&#8217;t be able to calculate the error due to tolerance specifications.&nbsp; This allows engineers to use Fourier Analysis to work with time-domain signals, such as radio signals, television signals, satellite signals and just about any signal you can think of.&nbsp; By viewing a signal according to what frequency components are contained within it, electrical engineers may concern themselves with magnitude changes in frequency only, and may no longer worry about the signal&#8217;s magnitude-changes through time.&nbsp; Not only is this a very practical concept when working in the lab, it also greatly simplifies the mathematics behind signal conditioning in general.&nbsp; In fact, the entirety of the Communications industry owes its success to the Fourier Transform for not only antenna design, but a plethora of other applications.</p>
<h3>The math behind the Fourier Transform<span style="text-decoration: underline;"><br />
</span></h3>
<p>The derivations that follow have been summarized from Chapter 4 of the textbook &#8220;Signal Processing and Linear Systems&#8221; by B.P. Lathi, a fine book for students of Communication Systems.</p>
<p>We begin by considering some arbitrary, aperiodic time-domain signal.&nbsp; An example of this kind of wave would be the output of a microphone after a man speaks a few words into it.&nbsp; For the actual signal generated by the changes in voltage as the man spoke, we can use Fourier Analysis to describe it as a summation of exponential functions <em>if </em>we instead desire to reconstruct a <em>periodic </em>signal composed of the same voice signal <em>repeating </em>every <img src='https://s0.wp.com/latex.php?latex=T_0&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='T_0' title='T_0' class='latex' /> seconds.&nbsp; For an accurate description, it is important that <img src='https://s0.wp.com/latex.php?latex=T_0&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='T_0' title='T_0' class='latex' /> is long enough such that the repeating arbitrary signals do not overlap.&nbsp; However, if we let <img src='https://s0.wp.com/latex.php?latex=T_0&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='T_0' title='T_0' class='latex' /> approach <img src='https://s0.wp.com/latex.php?latex=%5Cinfty&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='\infty' title='\infty' class='latex' />, then this &#8220;periodic&#8221; signal is simply just the voice signal (or, any general arbitrary function) in time we wanted to describe initially.&nbsp; Mathematically, we express:</p>
<img src='https://s0.wp.com/latex.php?latex=%5Cdisplaystyle%5Clim_%7BT_0%5Cto%5Cinfty%7Df_%7BT_0%7D%28t%29+%3D+f%28t%29&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='\displaystyle\lim_{T_0\to\infty}f_{T_0}(t) = f(t)' title='\displaystyle\lim_{T_0\to\infty}f_{T_0}(t) = f(t)' class='latex' />
<p>where <img src='https://s0.wp.com/latex.php?latex=f%28t%29&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='f(t)' title='f(t)' class='latex' /> is the time-domain function we wish to apply the Fourier Transform on (here, the arbitrary &#8220;voice&#8221; signal).&nbsp; For the above equation to be true, <img src='https://s0.wp.com/latex.php?latex=f_%7BT_0%7D&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='f_{T_0}' title='f_{T_0}' class='latex' /> is equal to:</p>
<img src='https://s0.wp.com/latex.php?latex=f_%7BT_0%7D%28t%29+%3D+%5Cdisplaystyle%5Csum_%7Bn%3D-+%5Cinfty%7D%5E%7B%5Cinfty%7DD_ne%5E%7Bjnw_ot%7D&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='f_{T_0}(t) = \displaystyle\sum_{n=- \infty}^{\infty}D_ne^{jnw_ot}' title='f_{T_0}(t) = \displaystyle\sum_{n=- \infty}^{\infty}D_ne^{jnw_ot}' class='latex' />
<p style="padding-left: 30px;"><strong>where:</strong></p>
<p style="padding-left: 30px;"><img src='https://s0.wp.com/latex.php?latex=D_n+%3D+%5Cfrac%7B1%7D%7BT_0%7D+%5Cint%5E%5Cfrac%7BT_0%7D%7B2%7D_%5Cfrac%7B-T_0%7D%7B2%7D+f_%7BT_0%7D%28t%29e%5E%7B-jnw_ot%7Ddt&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='D_n = \frac{1}{T_0} \int^\frac{T_0}{2}_\frac{-T_0}{2} f_{T_0}(t)e^{-jnw_ot}dt' title='D_n = \frac{1}{T_0} \int^\frac{T_0}{2}_\frac{-T_0}{2} f_{T_0}(t)e^{-jnw_ot}dt' class='latex' /></p>
<p style="padding-left: 30px;"><img src='https://s0.wp.com/latex.php?latex=w_o+%3D+%5Cfrac%7B2%5Cpi%7D%7BT_o%7D&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='w_o = \frac{2\pi}{T_o}' title='w_o = \frac{2\pi}{T_o}' class='latex' /></p>
<p>It is important to note here that in practice, the <em>shape </em>(aka &#8220;envelope&#8221;) of a signal&#8217;s frequency spectrum is what is of main interest, and the magnitude of the components within the spectrum comes secondary.&nbsp; This is because amplifiers and other signal-conditioning circuits may be built to alter the magnitude in any way one wishes, and will not affect signal frequencies (so long as the circuits are LTI systems).&nbsp; Analyzing the envelope of a signal&#8217;s Fourier Transform allows one to use intuitive and mathematically-simplified approaches to signal-processing in general, which we shall see later.&nbsp; For this reason (and also as <img src='https://s0.wp.com/latex.php?latex=T_0&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='T_0' title='T_0' class='latex' /> approaches <img src='https://s0.wp.com/latex.php?latex=%5Cinfty&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='\infty' title='\infty' class='latex' />) let:</p>
<img src='https://s0.wp.com/latex.php?latex=F%28w%29+%3D+%5Cint%5E%7B%5Cinfty%7D_%7B-%5Cinfty%7D+f%28t%29e%5E%7B-jw_ot%7Ddt&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='F(w) = \int^{\infty}_{-\infty} f(t)e^{-jw_ot}dt' title='F(w) = \int^{\infty}_{-\infty} f(t)e^{-jw_ot}dt' class='latex' />
<p>Notice that <img src='https://s0.wp.com/latex.php?latex=F%28w%29&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='F(w)' title='F(w)' class='latex' /> is simply <img src='https://s0.wp.com/latex.php?latex=D_n&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='D_n' title='D_n' class='latex' /> without the constant multiplier <img src='https://s0.wp.com/latex.php?latex=%5Cfrac%7B1%7D%7BT_0%7D&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='\frac{1}{T_0}' title='\frac{1}{T_0}' class='latex' />, such that:</p>
<img src='https://s0.wp.com/latex.php?latex=D_n+%3D+%5Cfrac%7B1%7D%7BT_0%7DF%28nw_o%29&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='D_n = \frac{1}{T_0}F(nw_o)' title='D_n = \frac{1}{T_0}F(nw_o)' class='latex' />
<p>which implies that <img src='https://s0.wp.com/latex.php?latex=f_%7BT_0%7D%28t%29&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='f_{T_0}(t)' title='f_{T_0}(t)' class='latex' /> may be written:</p>
<img src='https://s0.wp.com/latex.php?latex=f_%7BT_0%7D%28t%29+%3D+%5Cdisplaystyle%5Csum_%7Bn%3D-+%5Cinfty%7D%5E%7B%5Cinfty%7D%5Cfrac%7BF%28nw_o%29%7D%7BT_0%7De%5E%7Bjnw_ot%7D&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='f_{T_0}(t) = \displaystyle\sum_{n=- \infty}^{\infty}\frac{F(nw_o)}{T_0}e^{jnw_ot}' title='f_{T_0}(t) = \displaystyle\sum_{n=- \infty}^{\infty}\frac{F(nw_o)}{T_0}e^{jnw_ot}' class='latex' />
<p>Observation of this fact reveals insight: The shorter the period, <img src='https://s0.wp.com/latex.php?latex=T_0&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='T_0' title='T_0' class='latex' />, the larger the magnitude of the coefficients.&nbsp; But, on the other hand, as <img src='https://s0.wp.com/latex.php?latex=T_0+%5Crightarrow+%5Cinfty&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='T_0 \rightarrow \infty' title='T_0 \rightarrow \infty' class='latex' />, the magnitudes of every frequency component approaches <img src='https://s0.wp.com/latex.php?latex=0&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='0' title='0' class='latex' /> &#8211; which is why engineers choose to analyze spectrum envelopes.&nbsp; So, instead of visualizing absolute frequency magnitudes, instead consider that the frequency spectrum simply expresses the <em>magnitude-density per unit of bandwidth, aka Hz. </em>And since:</p>
<img src='https://s0.wp.com/latex.php?latex=T_0+%3D+%5Cfrac%7B2%5Cpi%7D%7Bw_o%7D&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='T_0 = \frac{2\pi}{w_o}' title='T_0 = \frac{2\pi}{w_o}' class='latex' />
<p>then:</p>
<img src='https://s0.wp.com/latex.php?latex=w_o+%3D+%5Cfrac%7B2%5Cpi%7D%7BT_0%7D&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='w_o = \frac{2\pi}{T_0}' title='w_o = \frac{2\pi}{T_0}' class='latex' />
<p>and:</p>
<img src='https://s0.wp.com/latex.php?latex=%5CDelta+w_o+%3D+%5Cfrac%7B2%5Cpi%7D%7B%5CDelta+T_0%7D&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='\Delta w_o = \frac{2\pi}{\Delta T_0}' title='\Delta w_o = \frac{2\pi}{\Delta T_0}' class='latex' />
<p>so:</p>
<img src='https://s0.wp.com/latex.php?latex=f_%7BT_0%7D%28t%29+%3D+%5Cdisplaystyle%5Csum_%7Bn%3D-+%5Cinfty%7D%5E%7B%5Cinfty%7D%5Cfrac%7BF%28n%5CDelta+w_o%29%5CDelta+w_o%7D%7B2%5Cpi%7De%5E%7Bjn%5CDelta+w_ot%7D&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='f_{T_0}(t) = \displaystyle\sum_{n=- \infty}^{\infty}\frac{F(n\Delta w_o)\Delta w_o}{2\pi}e^{jn\Delta w_ot}' title='f_{T_0}(t) = \displaystyle\sum_{n=- \infty}^{\infty}\frac{F(n\Delta w_o)\Delta w_o}{2\pi}e^{jn\Delta w_ot}' class='latex' />
<p>In the limit as <img src='https://s0.wp.com/latex.php?latex=T_0+%5Crightarrow+%5Cinfty&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='T_0 \rightarrow \infty' title='T_0 \rightarrow \infty' class='latex' /> we see:</p>
<img src='https://s0.wp.com/latex.php?latex=f%28t%29+%3D+%5Cdisplaystyle%5Clim_%7BT_0%5Cto%5Cinfty%7Df_%7BT_0%7D%28t%29+%3D+%5Cfrac%7B1%7D%7B2%5Cpi%7D+%5Cint%5E%7B%5Cinfty%7D_%7B-%5Cinfty%7DF%28w%29e%5E%7Bjwt%7Ddw&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='f(t) = \displaystyle\lim_{T_0\to\infty}f_{T_0}(t) = \frac{1}{2\pi} \int^{\infty}_{-\infty}F(w)e^{jwt}dw' title='f(t) = \displaystyle\lim_{T_0\to\infty}f_{T_0}(t) = \frac{1}{2\pi} \int^{\infty}_{-\infty}F(w)e^{jwt}dw' class='latex' />
<p>which is referred to as the <strong>Fourier Integral</strong>.&nbsp; <img src='https://s0.wp.com/latex.php?latex=F%28w%29&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='F(w)' title='F(w)' class='latex' /> is referred to as the <strong>Fourier Transform</strong> of the original aperiodic function <img src='https://s0.wp.com/latex.php?latex=f%28t%29&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='f(t)' title='f(t)' class='latex' />, and we express this concept as:</p>
<p style="text-align: center;"><img src='https://s0.wp.com/latex.php?latex=f%28t%29+%5CLeftrightarrow+F%28w%29&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='f(t) \Leftrightarrow F(w)' title='f(t) \Leftrightarrow F(w)' class='latex' /></p>
<h3>A fourier transform example<span style="text-decoration: underline;"><br />
</span></h3>
<p>This example is from the same textbook as the previous derivation, and can be found on page 239.</p>
<p style="padding-left: 30px;">Find the Fourier Transform of: <img src='https://s0.wp.com/latex.php?latex=e%5E%7B-at%7Du%28t%29&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='e^{-at}u(t)' title='e^{-at}u(t)' class='latex' /> where <img src='https://s0.wp.com/latex.php?latex=a&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='a' title='a' class='latex' /> is an arbitrary constant.</p>
<p style="padding-left: 30px;">To do this, we apply the Fourier Integral to the function <img src='https://s0.wp.com/latex.php?latex=e%5E%7B-at%7Du%28t%29&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='e^{-at}u(t)' title='e^{-at}u(t)' class='latex' /> as follows:</p>
<p style="padding-left: 30px;"><img src='https://s0.wp.com/latex.php?latex=F%28w%29+%3D+%5Cint%5E%7B%5Cinfty%7D_%7B-%5Cinfty%7De%5E%7B-at%7Du%28t%29e%5E%7B-jwt%7Ddt&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='F(w) = \int^{\infty}_{-\infty}e^{-at}u(t)e^{-jwt}dt' title='F(w) = \int^{\infty}_{-\infty}e^{-at}u(t)e^{-jwt}dt' class='latex' /></p>
<p style="padding-left: 30px;">Because of the <img src='https://s0.wp.com/latex.php?latex=u%28t%29&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='u(t)' title='u(t)' class='latex' /> factor, we only integrate from <img src='https://s0.wp.com/latex.php?latex=0+%5Crightarrow+%5Cinfty&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='0 \rightarrow \infty' title='0 \rightarrow \infty' class='latex' />.&nbsp; We simplify for:</p>
<p style="padding-left: 30px;"><img src='https://s0.wp.com/latex.php?latex=F%28w%29+%3D+%5Cint%5E%7B%5Cinfty%7D_%7B0%7D+e%5E%7B-%28a%2Bjw%29t%7Ddt+%3D+%5Cfrac%7B-1%7D%7Ba%2Bjw%7De%5E%7B-%28a%2Bjw%29t%7D%5Cmid%5E%7B%5Cinfty%7D_%7B0%7D&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='F(w) = \int^{\infty}_{0} e^{-(a+jw)t}dt = \frac{-1}{a+jw}e^{-(a+jw)t}\mid^{\infty}_{0}' title='F(w) = \int^{\infty}_{0} e^{-(a+jw)t}dt = \frac{-1}{a+jw}e^{-(a+jw)t}\mid^{\infty}_{0}' class='latex' /></p>
<p style="padding-left: 30px;">Also, we know that <img src='https://s0.wp.com/latex.php?latex=%7Ce%5E%7B-jwt%7D%7C+%3DRe%5Be%5E%7B-jwt%7D%5D+%3D+1&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='|e^{-jwt}| =Re[e^{-jwt}] = 1' title='|e^{-jwt}| =Re[e^{-jwt}] = 1' class='latex' />.&nbsp; So, for <img src='https://s0.wp.com/latex.php?latex=a+%3E+0&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='a &gt; 0' title='a &gt; 0' class='latex' />, as <img src='https://s0.wp.com/latex.php?latex=t+%5Crightarrow+%5Cinfty&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='t \rightarrow \infty' title='t \rightarrow \infty' class='latex' />:</p>
<p style="padding-left: 30px;"><img src='https://s0.wp.com/latex.php?latex=e%5E%7B-%28a%2Bjw%29t%7D+%3D+e%5E%7B-at%7De%5E%7B-jwt%7D+%3D+0&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='e^{-(a+jw)t} = e^{-at}e^{-jwt} = 0' title='e^{-(a+jw)t} = e^{-at}e^{-jwt} = 0' class='latex' /></p>
<p style="padding-left: 30px;">So:</p>
<p style="padding-left: 30px;"><img src='https://s0.wp.com/latex.php?latex=F%28w%29+%3D+%5Cfrac%7B-1%7D%7Ba%2Bjw%7De%5E%7B-%28a%2Bjw%29t%7D%5Cmid%5E%7B%5Cinfty%7D_%7B0%7D+%3D+0+-+%5Cfrac%7B-1%7D%7Ba%2Bjw%7De%5E%7B-%28a%2Bjw%290%7D+%3D+%5Cfrac%7B1%7D%7Ba%2Bjw%7D&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='F(w) = \frac{-1}{a+jw}e^{-(a+jw)t}\mid^{\infty}_{0} = 0 - \frac{-1}{a+jw}e^{-(a+jw)0} = \frac{1}{a+jw}' title='F(w) = \frac{-1}{a+jw}e^{-(a+jw)t}\mid^{\infty}_{0} = 0 - \frac{-1}{a+jw}e^{-(a+jw)0} = \frac{1}{a+jw}' class='latex' /></p>
<p style="padding-left: 60px;">for: <img src='https://s0.wp.com/latex.php?latex=a%3E+0&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='a&gt; 0' title='a&gt; 0' class='latex' /></p>
<h3>Useful Fourier Transform Properties<span style="text-decoration: underline;"><br />
</span></h3>
<p>The relationship between <img src='https://s0.wp.com/latex.php?latex=f%28t%29&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='f(t)' title='f(t)' class='latex' /> and <img src='https://s0.wp.com/latex.php?latex=F%28w%29&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='F(w)' title='F(w)' class='latex' /> exhibit beautiful symmetry that help one to develop an intuitive approach to signal analysis.&nbsp; Among all the concepts within electrical engineering, the properties between a time-domain function and its Fourier transform are among the most important to understand.&nbsp; Observe these following properties <strong>that apply for all </strong><img src='https://s0.wp.com/latex.php?latex=f%28t%29+%5CLeftrightarrow+F%28w%29&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='f(t) \Leftrightarrow F(w)' title='f(t) \Leftrightarrow F(w)' class='latex' />:</p>
<p>1.) <strong>Fourier Transform: </strong>Gives an equation to solve for the time-domain function <img src='https://s0.wp.com/latex.php?latex=f%28t%29&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='f(t)' title='f(t)' class='latex' /> from <img src='https://s0.wp.com/latex.php?latex=F%28w%29&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='F(w)' title='F(w)' class='latex' />.</p>
<p style="padding-left: 30px; text-align: center;"><img src='https://s0.wp.com/latex.php?latex=f%28t%29+%3D+%5Cfrac%7B1%7D%7B2%5Cpi%7D+%5Cint%5E%7B%5Cinfty%7D_%7B-%5Cinfty%7DF%28w%29e%5E%7Bjwt%7Ddw&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='f(t) = \frac{1}{2\pi} \int^{\infty}_{-\infty}F(w)e^{jwt}dw' title='f(t) = \frac{1}{2\pi} \int^{\infty}_{-\infty}F(w)e^{jwt}dw' class='latex' /></p>
<p>2.) <strong>Inverse Fourier Transform: </strong>Gives an equation to solve for the frequency-domain function <img src='https://s0.wp.com/latex.php?latex=F%28w%29&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='F(w)' title='F(w)' class='latex' /> from <img src='https://s0.wp.com/latex.php?latex=f%28t%29&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='f(t)' title='f(t)' class='latex' />.</p>
<p style="padding-left: 30px; text-align: center;"><img src='https://s0.wp.com/latex.php?latex=F%28w%29+%3D+%5Cfrac%7B1%7D%7B2%5Cpi%7D+%5Cint%5E%7B%5Cinfty%7D_%7B-%5Cinfty%7Df%28t%29e%5E%7B-jwt%7Ddw&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='F(w) = \frac{1}{2\pi} \int^{\infty}_{-\infty}f(t)e^{-jwt}dw' title='F(w) = \frac{1}{2\pi} \int^{\infty}_{-\infty}f(t)e^{-jwt}dw' class='latex' /></p>
<p>3.) <strong>Symmetry Property:</strong> For a given pair of a time-domain signal and its Fourier transform, we note that the time-domain envelope is different in shape when compared to the frequency-domain envelope.&nbsp; However, switching the shape of the two functions with respect to domain (time or frequency), will result in the same envelopes except with different scaling coefficients.&nbsp; For example, a square pulse through time has a frequency spectrum described by a sinc function, and a sinc function through time results in a frequency spectrum described by a square pulse.</p>
<p style="text-align: center;"><img src='https://s0.wp.com/latex.php?latex=F%28t%29+%5CLeftrightarrow+2%5Cpi+f%28-w%29&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='F(t) \Leftrightarrow 2\pi f(-w)' title='F(t) \Leftrightarrow 2\pi f(-w)' class='latex' /></p>
<p>4.) <strong>Scaling Property:</strong> <a href="https://engineersphere.com/time-shifting-and-scaling-of-functions">Time-scaling</a> a time-domain signal (by a constant <img src='https://s0.wp.com/latex.php?latex=a&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='a' title='a' class='latex' />) will result in a magnitude-and-frequency-scaling of the signal&#8217;s corresponding frequency spectrum.&nbsp; Also signifies that the longer a signal exists through time, the narrower the bandwidth (collection of frequency components needed to rebuild the signal) of its frequency spectrum.</p>
<p style="text-align: center;"><img src='https://s0.wp.com/latex.php?latex=f%28at%29+%5CLeftrightarrow+%5Cfrac%7B1%7D%7B%7Ca%7C%7DF%28%5Cfrac%7Bw%7D%7Ba%7D%29&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='f(at) \Leftrightarrow \frac{1}{|a|}F(\frac{w}{a})' title='f(at) \Leftrightarrow \frac{1}{|a|}F(\frac{w}{a})' class='latex' /></p>
<p style="text-align: left;">5.) <strong>Time-Shifting Property: </strong>By time-shifting, or delaying/advancing, a time-domain signal results in a <em>phase delay</em> in each of the ever-lasting frequency-components needed to rebuild it.&nbsp; The frequency spectrum is otherwise unchanged &#8211; only the phase of each component is shifted.</p>
<p style="text-align: center;"><img src='https://s0.wp.com/latex.php?latex=f%28t-t_0%29+%5CLeftrightarrow+F%28w%29e%5E%7B-jwt_0%7D&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='f(t-t_0) \Leftrightarrow F(w)e^{-jwt_0}' title='f(t-t_0) \Leftrightarrow F(w)e^{-jwt_0}' class='latex' /></p>
<p style="text-align: left;">6.) <strong>Frequency-Shifting Property: </strong>Multiplying a time-domain signal by a sinusoidal signal of some frequency <img src='https://s0.wp.com/latex.php?latex=w_0&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='w_0' title='w_0' class='latex' />, a method which begets amplitude and frequency modulation (AM/FM), results in the frequency spectrum remains unchanged <strong>except for a shift in frequency for each individual frequency component by </strong><img src='https://s0.wp.com/latex.php?latex=w_0&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='w_0' title='w_0' class='latex' /><strong>.</strong></p>
<p style="text-align: center;"><img src='https://s0.wp.com/latex.php?latex=f%28t%29e%5E%7Bjw_0t%7D+%5CLeftrightarrow+F%28w-w_0%29&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='f(t)e^{jw_0t} \Leftrightarrow F(w-w_0)' title='f(t)e^{jw_0t} \Leftrightarrow F(w-w_0)' class='latex' /></p>
<p>Lastly, these tables (<a href="http://i.imgur.com/QDsT5.jpg">table 1</a>, <a href="http://i.imgur.com/TiRNu.jpg">table 2</a>) can greatly simplify Fourier analysis when used in signal processing.</p>
<p style="padding-left: 30px;">
<p style="padding-left: 30px;">
<p><span style="text-decoration: underline;">&nbsp;</span></p><p>The post <a href="https://engineersphere.com/the-fourier-integral-transform-explained">The Fourier Integral / Transform Explained</a> first appeared on <a href="https://engineersphere.com">Engineersphere.com</a>.</p>]]></content:encoded>
					
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		<title>The Evolution of Wireless Technologies</title>
		<link>https://engineersphere.com/the-evolution-of-wireless-technologies</link>
					<comments>https://engineersphere.com/the-evolution-of-wireless-technologies#respond</comments>
		
		<dc:creator><![CDATA[Jeff Schuler]]></dc:creator>
		<pubDate>Wed, 30 Mar 2011 04:28:19 +0000</pubDate>
				<category><![CDATA[Wireless Hacking]]></category>
		<category><![CDATA[UMTS]]></category>
		<category><![CDATA[W-CDMA]]></category>
		<category><![CDATA[wireless technology evolution]]></category>
		<guid isPermaLink="false">https://engineersphere.com/?p=2882</guid>

					<description><![CDATA[<p>Cellular systems have come a long way since their introduction in the 1980s.&#160; The evolution progressed from First Generation (1G) systems to Second Generation (2G) systems.&#160; Now, Third Generation (3G) systems are being deployed. 1G systems introduced the cellular concept, in which multiple antenna sites are used to serve an area.&#160; The coverage of a [&#8230;]</p>
<p>The post <a href="https://engineersphere.com/the-evolution-of-wireless-technologies">The Evolution of Wireless Technologies</a> first appeared on <a href="https://engineersphere.com">Engineersphere.com</a>.</p>]]></description>
										<content:encoded><![CDATA[<p>Cellular systems have come a long way since their introduction in the 1980s.&nbsp; The evolution progressed from <strong>First Generation (1G) systems</strong> to <strong>Second Generation (2G) systems</strong>.&nbsp; Now, <strong>Third Generation (3G) systems</strong> are being deployed.</p>
<p>1G systems introduced the cellular concept, in which multiple antenna sites are used to serve an area.&nbsp; The coverage of a single antenna site is called a cell.&nbsp; A cell can serve a certain number of users, and higher-system capacity can be achieved by creating more cells with smaller coverage areas.&nbsp; One distinguishing factor of 1G systems is that they make use of analog radio transmissions, so user information, such as voice, is never digitized.&nbsp; As such, they are best suited for voice communications, since data communications can be cumbersome.</p>
<p>The migration of 1G analog technologies toward 2G technologies began in the late 1980s and early 1990s.&nbsp; The primary motivation was increased system capacity.&nbsp; This was achieved by using more efficient digital radio techniques that enabled the transmission of digitized compressed speech signals.&nbsp; These digital radio techniques also supported data services with data rates as high as 14,400 bits per second (14.4 kbps) in some systems.&nbsp; 2G data communication is typically done using circuit-switched techniques, which are not very efficient for sending packet data such as that sent on the Internet.&nbsp; This inefficiency makes the use of wireless data more expensive f or the end user.</p>
<p>The next step in the evolution is from 2G to 3G, which started in the year 2000.&nbsp; The new key feature of 3G systems is the support of high-speed data services with data rates as high as 2 million bits per second (2 Mbps).&nbsp; Data can be transferred using packet-switching techniques rather than the circuit-switching approach.&nbsp; Therefore, it is more efficient and less expensive.&nbsp; This opens up the possibility of cost-effective Internet access, access to corporate intranets, and a host of multimedia services.</p>
<p>If you want to read more about the evolution of wireless networks and <a href="https://engineersphere.com/category/wireless-hacking">WCDMA</a> radio networks in general, please stay tuned for the next several editions where I will go into details.</p>
<p><strong>Upcoming including but not limited to:</strong></p>
<ul>
<li>Physical layer functions</li>
<li>W-CDMA Channels</li>
<li>Basic call setups</li>
<li>Data session setups</li>
<li>Service reconfigurations</li>
<li>UTRAN mobility management</li>
<li>Inter-system procedures</li>
<li>RF design &amp; analysis of UMTS radio networks</li>
<li>The evolution of UMTS</li>
<li>Architectures</li>
</ul>
<p>&nbsp;</p>
<p>&nbsp;</p><p>The post <a href="https://engineersphere.com/the-evolution-of-wireless-technologies">The Evolution of Wireless Technologies</a> first appeared on <a href="https://engineersphere.com">Engineersphere.com</a>.</p>]]></content:encoded>
					
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		<title>How to Determine if a Vector Set is Linearly Independent</title>
		<link>https://engineersphere.com/how-to-determine-if-a-vector-set-is-linearly-independent</link>
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		<dc:creator><![CDATA[Jeff Schuler]]></dc:creator>
		<pubDate>Fri, 25 Mar 2011 00:55:34 +0000</pubDate>
				<category><![CDATA[Electrical Engineering Concepts]]></category>
		<category><![CDATA[linear independence]]></category>
		<category><![CDATA[linear independent vectors]]></category>
		<category><![CDATA[linear vectors]]></category>
		<category><![CDATA[vectors]]></category>
		<guid isPermaLink="false">https://engineersphere.com/?p=2700</guid>

					<description><![CDATA[<p>Pre-(r)amble The odds favor that by the time someone has reached this article, myself included, they have spent at least the briefest of moments (frustratedly?) questioning the practical applications for  linear combination, linear independence and linear math. In a sentence, these concepts allow us to mathematically understand and represent multidimensional coordinate systems. If you&#8217;re looking [&#8230;]</p>
<p>The post <a href="https://engineersphere.com/how-to-determine-if-a-vector-set-is-linearly-independent">How to Determine if a Vector Set is Linearly Independent</a> first appeared on <a href="https://engineersphere.com">Engineersphere.com</a>.</p>]]></description>
										<content:encoded><![CDATA[<h3>Pre-(r)amble</h3>
<p>The odds favor that by the time someone has reached this article, myself included, they have spent at least the briefest of moments (frustratedly?) questioning the practical applications for  linear combination, linear independence and linear math. In a sentence, these concepts allow us to mathematically understand and represent multidimensional coordinate systems. If you&#8217;re looking for a quick explanation for a homework problem feel free to skim through the bolded topics for help in specific areas of concern. Otherwise, here&#8217;s something to think about.  Imagine maneuvering in three dimensional space. An instantaneous position can be described using a three dimensional coordinate system. When following a consistent pattern of movement, an instantaneous position can be described with a fourth dimension, time. Suppose you have just landed the snowball throw of a lifetime and hit a target moving across your view plane, increasing the distance between you, and uphill. You have properly estimated the intersection of two moving objects in four dimensions. This is not always an easy task to execute. Now make this throw using a fifth dimension. Most people can&#8217;t comprehend the existence of a fifth dimension without having to understand how to maneuver in it. With linear math we can attempt to understand and represent the relationships between these dimensions.</p>
<h3>Important Definitions</h3>
<p><strong>Linear Independence</strong><br />
A set of linearly independent vectors {<img src='https://s0.wp.com/latex.php?latex=V_1%2C+V_2%2C+.+.+.+%2C+V_k+&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='V_1, V_2, . . . , V_k ' title='V_1, V_2, . . . , V_k ' class='latex' />} has ONLY the zero (trivial) solution &lt;<img src='https://s0.wp.com/latex.php?latex=x_1%2C+x_2%2C+.+.+.+%2C+x_k+&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='x_1, x_2, . . . , x_k ' title='x_1, x_2, . . . , x_k ' class='latex' />&gt; <img src='https://s0.wp.com/latex.php?latex=%3D+&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='= ' title='= ' class='latex' /> &lt;<img src='https://s0.wp.com/latex.php?latex=0%2C+0%2C+.+.+.+%2C+0+&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='0, 0, . . . , 0 ' title='0, 0, . . . , 0 ' class='latex' />&gt;  for the equation <img src='https://s0.wp.com/latex.php?latex=x_1+%2A+V_1+%2B+x_2+%2A+V+_2+%2B+.+.+.+%2B+x_k+%2A+V_k+%3D+0+&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='x_1 * V_1 + x_2 * V _2 + . . . + x_k * V_k = 0 ' title='x_1 * V_1 + x_2 * V _2 + . . . + x_k * V_k = 0 ' class='latex' /></p>
<p><strong>Linear Dependence</strong><br />
Alternatively, if <img src='https://s0.wp.com/latex.php?latex=x_1%2C+x_2%2C+.+.+.+%2C+&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='x_1, x_2, . . . , ' title='x_1, x_2, . . . , ' class='latex' /> or <img src='https://s0.wp.com/latex.php?latex=x_k+%5Cneq+0+&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='x_k \neq 0 ' title='x_k \neq 0 ' class='latex' />, the set of vectors is said to be linearly dependent.</p>
<h3>Determining Linear Independence</h3>
<p>By row reducing a coefficient matrix created from our vectors {<img src='https://s0.wp.com/latex.php?latex=V_1%2C+V_2%2C+.+.+.+%2C+V_k+&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='V_1, V_2, . . . , V_k ' title='V_1, V_2, . . . , V_k ' class='latex' />}, we can determine our &lt;<img src='https://s0.wp.com/latex.php?latex=x_1%2C+x_2%2C+.+.+.+%2C+x_k+&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='x_1, x_2, . . . , x_k ' title='x_1, x_2, . . . , x_k ' class='latex' />&gt;. Then to classify a set of vectors as linearly independent or dependent, we compare to the definitions above.</p>
<p><strong>Example</strong><br />
Determine if the following set of vectors are linearly independent:</p>
<p><img src='https://s0.wp.com/latex.php?latex=%5Cbegin%7Bbmatrix%7D+1%5C%5C1%5C%5C1%5Cend%7Bbmatrix%7D&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='\begin{bmatrix} 1\\1\\1\end{bmatrix}' title='\begin{bmatrix} 1\\1\\1\end{bmatrix}' class='latex' /> , <img src='https://s0.wp.com/latex.php?latex=%5Cbegin%7Bbmatrix%7D+2%5C%5C2%5C%5C2%5Cend%7Bbmatrix%7D&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='\begin{bmatrix} 2\\2\\2\end{bmatrix}' title='\begin{bmatrix} 2\\2\\2\end{bmatrix}' class='latex' /> , <img src='https://s0.wp.com/latex.php?latex=%5Cbegin%7Bbmatrix%7D+0%5C%5C0%5C%5C5%5Cend%7Bbmatrix%7D&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='\begin{bmatrix} 0\\0\\5\end{bmatrix}' title='\begin{bmatrix} 0\\0\\5\end{bmatrix}' class='latex' /> , <img src='https://s0.wp.com/latex.php?latex=%5Cbegin%7Bbmatrix%7D+1%5C%5C2%5C%5C3%5Cend%7Bbmatrix%7D&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='\begin{bmatrix} 1\\2\\3\end{bmatrix}' title='\begin{bmatrix} 1\\2\\3\end{bmatrix}' class='latex' /></p>
<h3></h3>
<h3><strong>Setting up a Corresponding System of Equations and Finding it&#8217;s RREF Matrix</strong></h3>
<p>We need to understand that our vectors can be represented with a system of equations all equaling zero to satisfy the equation <img src='https://s0.wp.com/latex.php?latex=x_1+%2A+V_1+%2B+x_2+%2A+V+_2+%2B+.+.+.+%2B+x_k+%2A+V_k+%3D+0+&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='x_1 * V_1 + x_2 * V _2 + . . . + x_k * V_k = 0 ' title='x_1 * V_1 + x_2 * V _2 + . . . + x_k * V_k = 0 ' class='latex' /> from our definition of linear independence. These equations will look something like this:</p>
<p><img src='https://s0.wp.com/latex.php?latex=1%2Ax_1+%2B+2%2Ax_2+%2B+0%2Ax_3+%2B+1%2Ax_4+%3D+0+&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='1*x_1 + 2*x_2 + 0*x_3 + 1*x_4 = 0 ' title='1*x_1 + 2*x_2 + 0*x_3 + 1*x_4 = 0 ' class='latex' /><br />
<img src='https://s0.wp.com/latex.php?latex=1%2Ax_1+%2B+2%2Ax_2+%2B+0%2Ax_3+%2B+2%2Ax_4+%3D+0+&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='1*x_1 + 2*x_2 + 0*x_3 + 2*x_4 = 0 ' title='1*x_1 + 2*x_2 + 0*x_3 + 2*x_4 = 0 ' class='latex' /><br />
<img src='https://s0.wp.com/latex.php?latex=1%2Ax_1+%2B+2%2Ax_2+%2B+5%2Ax_3+%2B+3%2Ax_4+%3D+0+&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='1*x_1 + 2*x_2 + 5*x_3 + 3*x_4 = 0 ' title='1*x_1 + 2*x_2 + 5*x_3 + 3*x_4 = 0 ' class='latex' /></p>
<p>Notice that I have simply taken the coefficients from the given vectors and multiplied them by four variables (the number of variables will equal the number of vectors in the given set). They have been set equal to zero to allow us to test for linear independence. From here, create a coefficient matrix and perform row operations to reduce the matrix to reduced row echelon form (rref) .</p>
<p>rref <img src='https://s0.wp.com/latex.php?latex=%5Cleft%5B+%5Cbegin%7Barray%7D%7Bcccc%7Cc%7D+1%262%260%261%260%5C%5C1%262%260%262%260%5C%5C1%262%265%263%260%5Cend%7Barray%7D%5Cright%5D+&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='\left[ \begin{array}{cccc|c} 1&amp;2&amp;0&amp;1&amp;0\\1&amp;2&amp;0&amp;2&amp;0\\1&amp;2&amp;5&amp;3&amp;0\end{array}\right] ' title='\left[ \begin{array}{cccc|c} 1&amp;2&amp;0&amp;1&amp;0\\1&amp;2&amp;0&amp;2&amp;0\\1&amp;2&amp;5&amp;3&amp;0\end{array}\right] ' class='latex' /> = <img src='https://s0.wp.com/latex.php?latex=%5Cleft%5B+%5Cbegin%7Barray%7D%7Bcccc%7Cc%7D+1%262%260%260%260%5C%5C0%260%261%260%260%5C%5C0%260%260%261%260%5Cend%7Barray%7D%5Cright%5D+&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='\left[ \begin{array}{cccc|c} 1&amp;2&amp;0&amp;0&amp;0\\0&amp;0&amp;1&amp;0&amp;0\\0&amp;0&amp;0&amp;1&amp;0\end{array}\right] ' title='\left[ \begin{array}{cccc|c} 1&amp;2&amp;0&amp;0&amp;0\\0&amp;0&amp;1&amp;0&amp;0\\0&amp;0&amp;0&amp;1&amp;0\end{array}\right] ' class='latex' /></p>
<h3>Finding the Solution of the RREF Matrix</h3>
<p>Finding the solution of the rref matrix may be the more difficult step in this process. However, it may become trivial following <strong>a few simple steps</strong>.</p>
<p><strong>1) Identify the free variables in the matrix.</strong> Free variables are non-zero and located to the right of pivot variables. Pivot variables are the first non-zero entry in each row and since we have taken the rref of our matrix, all of the pivot variable coefficients are 1. By locating all free variables (or by eliminating all pivot variables) we find that <img src='https://s0.wp.com/latex.php?latex=x_2+&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='x_2 ' title='x_2 ' class='latex' /> is our only free variable.</p>
<p><strong>2) Write free variables into your solution. </strong>The variable <img src='https://s0.wp.com/latex.php?latex=x_2+&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='x_2 ' title='x_2 ' class='latex' /> can be written into our solution vector as itself but we will represent it with another variable name (i.e. <img src='https://s0.wp.com/latex.php?latex=t+&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='t ' title='t ' class='latex' />) so that our solution is in parametric form. Multiple free variables are represented with multiple variables names (i.e. <img src='https://s0.wp.com/latex.php?latex=s%2C+t+&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='s, t ' title='s, t ' class='latex' />). After this step your solution vector should look like this: &lt;<img src='https://s0.wp.com/latex.php?latex=x_1%2C+x_2%2C+x_3+%2C+x_4+&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='x_1, x_2, x_3 , x_4 ' title='x_1, x_2, x_3 , x_4 ' class='latex' />&gt; <img src='https://s0.wp.com/latex.php?latex=%3D+&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='= ' title='= ' class='latex' /> &lt;<img src='https://s0.wp.com/latex.php?latex=%3F+%2C+t%2C+%3F+%2C+%3F+&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='? , t, ? , ? ' title='? , t, ? , ? ' class='latex' />&gt;.</p>
<p><strong>3) Solve for pivot variables. </strong>The pivot variables should either be constant (i.e. 0, 6) or a function of your free variables (i.e. <img src='https://s0.wp.com/latex.php?latex=3+t-4+&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='3 t-4 ' title='3 t-4 ' class='latex' /> ). From the rref matrix we can see that <img src='https://s0.wp.com/latex.php?latex=x_1+%3D+-2+t+&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='x_1 = -2 t ' title='x_1 = -2 t ' class='latex' />, <img src='https://s0.wp.com/latex.php?latex=x_3+%3D+0+&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='x_3 = 0 ' title='x_3 = 0 ' class='latex' />, and <img src='https://s0.wp.com/latex.php?latex=x_4+%3D+0+&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='x_4 = 0 ' title='x_4 = 0 ' class='latex' />.</p>
<p><strong>4) Complete the solution vector. </strong>Placing the values we just calculated into our solution vector: &lt;<img src='https://s0.wp.com/latex.php?latex=x_1%2C+x_2%2C+x_3+%2C+x_4+&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='x_1, x_2, x_3 , x_4 ' title='x_1, x_2, x_3 , x_4 ' class='latex' />&gt; <img src='https://s0.wp.com/latex.php?latex=%3D+&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='= ' title='= ' class='latex' /> &lt;<img src='https://s0.wp.com/latex.php?latex=-2+t+%2C+t%2C+0+%2C+0+&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='-2 t , t, 0 , 0 ' title='-2 t , t, 0 , 0 ' class='latex' />&gt;</p>
<p><strong>Finally,</strong></p>
<p>Since not all of our <img src='https://s0.wp.com/latex.php?latex=X_i+%3D+0+&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='X_i = 0 ' title='X_i = 0 ' class='latex' />, the given set of vectors is said to be <span style="text-decoration: underline;">linearly dependent</span>. The linear dependence relation is written using our solution vector multiplied by the respective vector from the given set: <img src='https://s0.wp.com/latex.php?latex=-2+t+%2A+V_1+%2B+t+%2A+V_2+%2B+0+%2A+V_3+%2B+0+%2A+V_4+%3D+0+&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='-2 t * V_1 + t * V_2 + 0 * V_3 + 0 * V_4 = 0 ' title='-2 t * V_1 + t * V_2 + 0 * V_3 + 0 * V_4 = 0 ' class='latex' />. We can also conclude that any vectors with non-zero coefficients are linear combinations of each other. Therefore, <img src='https://s0.wp.com/latex.php?latex=V_1+&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='V_1 ' title='V_1 ' class='latex' /> and <img src='https://s0.wp.com/latex.php?latex=V_2+&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='V_2 ' title='V_2 ' class='latex' /> are a linear combination.</p>
<p>&nbsp;</p>
<p>&nbsp;</p><p>The post <a href="https://engineersphere.com/how-to-determine-if-a-vector-set-is-linearly-independent">How to Determine if a Vector Set is Linearly Independent</a> first appeared on <a href="https://engineersphere.com">Engineersphere.com</a>.</p>]]></content:encoded>
					
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		<title>DC Biasing &#038; AC Performance Analysis of BJT &#038; FET Differential Amplifiers</title>
		<link>https://engineersphere.com/dc-biasing-ac-performance-analysis-of-bjt-fet-amplifiers</link>
					<comments>https://engineersphere.com/dc-biasing-ac-performance-analysis-of-bjt-fet-amplifiers#respond</comments>
		
		<dc:creator><![CDATA[Jeff Schuler]]></dc:creator>
		<pubDate>Sun, 20 Mar 2011 19:55:23 +0000</pubDate>
				<category><![CDATA[Electrical Engineering Concepts]]></category>
		<category><![CDATA[AC performance analysis]]></category>
		<category><![CDATA[CMRR]]></category>
		<category><![CDATA[common mode gain]]></category>
		<category><![CDATA[common mode input impedance]]></category>
		<category><![CDATA[common mode rejection ratio]]></category>
		<category><![CDATA[DC Biasing]]></category>
		<category><![CDATA[differential amplifier schematic]]></category>
		<category><![CDATA[differential amplifiers]]></category>
		<category><![CDATA[differential input stage]]></category>
		<category><![CDATA[differential mode gain]]></category>
		<category><![CDATA[input impedance]]></category>
		<guid isPermaLink="false">https://engineersphere.com/?p=2340</guid>

					<description><![CDATA[<p>DC Biasing &#38; AC Performance Analysis of BJT and FET Differential Amplifier Sub-circuits with Active Loads Any op-amp worth its salt has a differential amplifier at its front end, and you&#8217;re nobody if you can&#8217;t design one yourself.&#160; So, this article presents a general method for biasing and analyzing the performance characteristics of single-stage BJT [&#8230;]</p>
<p>The post <a href="https://engineersphere.com/dc-biasing-ac-performance-analysis-of-bjt-fet-amplifiers">DC Biasing & AC Performance Analysis of BJT & FET Differential Amplifiers</a> first appeared on <a href="https://engineersphere.com">Engineersphere.com</a>.</p>]]></description>
										<content:encoded><![CDATA[<h2>DC Biasing &amp; AC Performance Analysis of BJT and FET Differential Amplifier Sub-circuits with Active Loads</h2>
<p>Any op-amp worth its salt has a differential amplifier at its front end, and you&#8217;re nobody if you can&#8217;t design one yourself.&nbsp; So, this article presents a general method for biasing and analyzing the performance characteristics of single-stage BJT and MOSFET differential amplifier circuits.&nbsp; The following images show the <em>general </em>schematic for both kinds of differential amplifiers, often referred to as a <strong>differential input stage</strong> when used in designing op-amps.&nbsp; Notice that these types of differential amplifiers use <strong><a href="http://en.wikipedia.org/wiki/Active_load">active loads</a> </strong>to achieve <em>wide swing</em> and <em>high gain</em>.</p>
<figure style="width: 598px" class="wp-caption alignnone"><img class=" " title="Differential Amplifiers with Active Loads" src="http://imgur.com/sruaL.jpg" alt="differential amplifiers with active loads" width="598" height="359"/><figcaption class="wp-caption-text">Figure 1. BJT and MOSFET differential amplifiers with active loads</figcaption></figure>
<p>Due to design processes and the nature of the devices involved, BJT circuits are &#8220;simpler&#8221; to analyze than their FET counterparts, whose circuits require a few extra steps when calculating performance parameters.&nbsp; For this reason, this tutorial will begin by biasing and analyzing a BJT differential amplifier circuit, and then will move on to do the same for a FET differential amplifier.&nbsp; But it should be noted that <strong>the procedures to analyze these types of differential amplifiers are virtually the same.</strong></p>
<h3>BJT Differential Amplifier<span style="text-decoration: underline;"><br />
</span></h3>
<p>The first thing needed is to configure the DC biasing.&nbsp; To accomplish this, a practical implementation of <img src='https://s0.wp.com/latex.php?latex=I_%7BBIAS%7D&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='I_{BIAS}' title='I_{BIAS}' class='latex' /> must be developed.&nbsp; A very popular method is to use a <strong><a href="http://users.ece.gatech.edu/mleach/ece3050/notes/bjt/bjtmirr.pdf">current mirror</a></strong>.&nbsp;&nbsp; A simple current mirror is shown below:</p>
<h3>BJT Current Mirror</h3>
<figure style="width: 307px" class="wp-caption alignnone"><img class=" " title="BJT Current Mirror" src="http://imgur.com/tjmAT.jpg" alt="bjt current mirror" width="307" height="283"/><figcaption class="wp-caption-text">Figure 2. BJT Current Mirror</figcaption></figure>
<p>It is easy to understand how a current mirror works.&nbsp; Observe the equation governing the amount of collector current in a BJT, denoted <img src='https://s0.wp.com/latex.php?latex=I_C&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='I_C' title='I_C' class='latex' />:</p>
<img src='https://s0.wp.com/latex.php?latex=I_C+%3D+I_S%28e%5E%7B%5Cfrac%7BV%7BBE%7D%7D%7BnV_T%7D%7D-1%29%281%2B%5Cfrac%7BV_%7BCB%7D%7D%7BV_A%7D%29&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='I_C = I_S(e^{\frac{V{BE}}{nV_T}}-1)(1+\frac{V_{CB}}{V_A})' title='I_C = I_S(e^{\frac{V{BE}}{nV_T}}-1)(1+\frac{V_{CB}}{V_A})' class='latex' />
<p><strong>where:</strong></p>
<ul>
<li><img src='https://s0.wp.com/latex.php?latex=I_C&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='I_C' title='I_C' class='latex' /> is the collector current</li>
<li><img src='https://s0.wp.com/latex.php?latex=I_S&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='I_S' title='I_S' class='latex' /> is the <a href="http://en.wikipedia.org/wiki/Saturation_current">scale current</a></li>
<li><img src='https://s0.wp.com/latex.php?latex=V_%7BBE%7D&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='V_{BE}' title='V_{BE}' class='latex' /> is the DC voltage across the base-emitter junction</li>
<li><img src='https://s0.wp.com/latex.php?latex=V_T&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='V_T' title='V_T' class='latex' /> is the thermal voltage, typically 25 mV</li>
<li><img src='https://s0.wp.com/latex.php?latex=n&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='n' title='n' class='latex' /> is the quality factor, typically between&nbsp; 1- 2 and is frequently assumed to be 1</li>
<li><img src='https://s0.wp.com/latex.php?latex=V_%7BCB%7D&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='V_{CB}' title='V_{CB}' class='latex' /> is the voltage across the collector-base junction</li>
<li><img src='https://s0.wp.com/latex.php?latex=V_A&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='V_A' title='V_A' class='latex' /> is the <a href="http://en.wikipedia.org/wiki/Early_effect">early voltage</a></li>
</ul>
<p><strong>Note: [</strong>This equation may look intimidating at first, but what is important to understand is that the point of designing &#8220;by hand&#8221; is to <em>get close.</em> One should aim simply to get a good <em>estimation </em>of such parameters as necessary bias current, gain, input impedance, etc.&nbsp; In this way, computer simulations can analyze the hand-designed circuit in much closer detail, which greatly aids in the process of designing a real-life differential amplifier.&nbsp; Knowing this, the equations to be used in this tutorial will be rough estimates, but are still invaluable when it comes to designing these types of circuits.<strong>]</strong></p>
<p>By assuming a very large equivalent resistance, one can estimate that the collector current through any BJT can be described by:</p>
<img src='https://s0.wp.com/latex.php?latex=I_C+%5Capprox+I_S+e%5E%7BV_%7BBE%7D%2FV_T%7D+&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='I_C \approx I_S e^{V_{BE}/V_T} ' title='I_C \approx I_S e^{V_{BE}/V_T} ' class='latex' />
<p>What can be noticed here is that the only controllable variable in that equation is <img src='https://s0.wp.com/latex.php?latex=V_%7BBE%7D&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='V_{BE}' title='V_{BE}' class='latex' />.&nbsp; All the other terms in the equation are constants that depend on either the environment or the actual physical size of the device.&nbsp; This means that for any two same-sized transistors, the currents through their collectors <em>will be the same as long as the voltage across their base-emitter junctions is the same. </em>By tying their bases and emitters together, we can mirror the currents between them!&nbsp; In order to implement a successful current mirror, one transistor (here, <img src='https://s0.wp.com/latex.php?latex=Q_5&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='Q_5' title='Q_5' class='latex' />) must have a current induced in it to mirror it to the differential amplifier&#8217;s current source (here, <img src='https://s0.wp.com/latex.php?latex=Q_6&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='Q_6' title='Q_6' class='latex' />).&nbsp; After adding this current mirror to our BJT differential amplifier, the resulting schematic is:</p>
<figure style="width: 406px" class="wp-caption alignnone"><img class=" " title="bjt diff amp with current mirror" src="http://i.imgur.com/HitQn.jpg" alt="bjt diff amp with current mirror" width="406" height="456"/><figcaption class="wp-caption-text">Figure 3. BJT differential amp with current mirror biasing</figcaption></figure>
<p>In order to properly bias this circuit, it is necessary to include <img src='https://s0.wp.com/latex.php?latex=R_%7BBIAS%7D&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='R_{BIAS}' title='R_{BIAS}' class='latex' />.&nbsp; Two things are accomplished by including <img src='https://s0.wp.com/latex.php?latex=R_%7BBIAS%7D&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='R_{BIAS}' title='R_{BIAS}' class='latex' /> in our circuit.&nbsp; One of them is that we can induce the current in <img src='https://s0.wp.com/latex.php?latex=Q_5&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='Q_5' title='Q_5' class='latex' />, and thus, the current in <img src='https://s0.wp.com/latex.php?latex=Q_6&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='Q_6' title='Q_6' class='latex' />.&nbsp; The other important thing this resistor does is drop a majority of the available voltage across itself, so that <img src='https://s0.wp.com/latex.php?latex=Q_5&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='Q_5' title='Q_5' class='latex' /> doesn&#8217;t have the entire voltage difference between the supplies across it!&nbsp; To bias this circuit, the first thing one must do is determine what the desired magnitude of the current source will be.&nbsp; This parameter depends on how you want the circuit to operate, and is usually a known value.&nbsp; In this tutorial, we will assume we want an <img src='https://s0.wp.com/latex.php?latex=I_%7BBIAS%7D&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='I_{BIAS}' title='I_{BIAS}' class='latex' /> of 1mA.&nbsp; In order to determine the necessary size of <img src='https://s0.wp.com/latex.php?latex=R_%7BBIAS%7D&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='R_{BIAS}' title='R_{BIAS}' class='latex' />, we analyze the loop that consists of:</p>
<p>&nbsp;</p>
<img src='https://s0.wp.com/latex.php?latex=VCC+%5Crightarrow+I_%7BBIAS%7D+%5Ccdot+R_%7BBIAS%7D+%5Crightarrow+V_%7BBE5%7D+%5Crightarrow+VEE&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='VCC \rightarrow I_{BIAS} \cdot R_{BIAS} \rightarrow V_{BE5} \rightarrow VEE' title='VCC \rightarrow I_{BIAS} \cdot R_{BIAS} \rightarrow V_{BE5} \rightarrow VEE' class='latex' />
<p>&nbsp;</p>
<p><a href="https://engineersphere.com/kirchoff-voltage-law">Kirchoff&#8217;s Voltage Law</a> (KVL) around this loop reveals:</p>
<p>&nbsp;</p>
<img src='https://s0.wp.com/latex.php?latex=0+%3D+-VCC+%2B+I_%7BBIAS%7D+%5Ccdot+R_%7BBIAS%7D+%2B+V_%7BBE5%7D+%2B+VEE&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='0 = -VCC + I_{BIAS} \cdot R_{BIAS} + V_{BE5} + VEE' title='0 = -VCC + I_{BIAS} \cdot R_{BIAS} + V_{BE5} + VEE' class='latex' />
<p>&nbsp;</p>
<p>These kinds of circuits are typically supplied rails of <img src='https://s0.wp.com/latex.php?latex=%5Cpm+10&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='\pm 10' title='\pm 10' class='latex' /> to <img src='https://s0.wp.com/latex.php?latex=15+V&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='15 V' title='15 V' class='latex' />.&nbsp; So, <strong>this tutorial will assume:</strong></p>
<p><strong><br />
</strong></p>
<p><img src='https://s0.wp.com/latex.php?latex=VCC+%3D+-+VEE+%3D+10+V&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='VCC = - VEE = 10 V' title='VCC = - VEE = 10 V' class='latex' />.</p>
<p>&nbsp;</p>
<p>For a given technology, all of the BJT transistors <strong>are designed to have the same turn-on voltage.</strong> This tutorial will assume .7 V for each BJT.&nbsp; That being the case, and rearranging the above equation, results in:</p>
<p>&nbsp;</p>
<img src='https://s0.wp.com/latex.php?latex=R_%7BBIAS%7D+%3D+%5Cfrac%7BVCC+-+VEE+-+V_%7BBE5%7D%7D%7BI_%7BBIAS%7D%7D+%3D+%5Cfrac%7B10V+-+%28-10V%29+-+.7V%7D%7B1+mA%7D+%3D+19.3+k%5COmega+&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='R_{BIAS} = \frac{VCC - VEE - V_{BE5}}{I_{BIAS}} = \frac{10V - (-10V) - .7V}{1 mA} = 19.3 k\Omega ' title='R_{BIAS} = \frac{VCC - VEE - V_{BE5}}{I_{BIAS}} = \frac{10V - (-10V) - .7V}{1 mA} = 19.3 k\Omega ' class='latex' />
<p>&nbsp;</p>
<p>By introducing a resistor <img src='https://s0.wp.com/latex.php?latex=R_%7BBIAS%7D&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='R_{BIAS}' title='R_{BIAS}' class='latex' /> of <img src='https://s0.wp.com/latex.php?latex=19.3k%5COmega&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='19.3k\Omega' title='19.3k\Omega' class='latex' /> to the above schematic, the bias current is now established at 1 mA.&nbsp; Due to symmetry, the currents through transistors <img src='https://s0.wp.com/latex.php?latex=Q_1&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='Q_1' title='Q_1' class='latex' /> and <img src='https://s0.wp.com/latex.php?latex=Q_2&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='Q_2' title='Q_2' class='latex' /> are each half of the bias current, described by:</p>
<p>&nbsp;</p>
<img src='https://s0.wp.com/latex.php?latex=I_%7BC1%7D+%3D+I_%7BC2%7D+%3D+%5Cfrac%7BI_%7BBIAS%7D%7D%7B2%7D+%3D%5Cfrac%7B1+mA%7D%7B2%7D+%3D+.5+mA&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='I_{C1} = I_{C2} = \frac{I_{BIAS}}{2} =\frac{1 mA}{2} = .5 mA' title='I_{C1} = I_{C2} = \frac{I_{BIAS}}{2} =\frac{1 mA}{2} = .5 mA' class='latex' />
<p>&nbsp;</p>
<p>Now that we know the collector currents through <img src='https://s0.wp.com/latex.php?latex=Q_1&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='Q_1' title='Q_1' class='latex' /> and <img src='https://s0.wp.com/latex.php?latex=Q_2&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='Q_2' title='Q_2' class='latex' />, characterizing the performance of this differential amplifier is a breeze.&nbsp; Since the parameters we are interested in (gain, <a href="https://engineersphere.com/amplifiers-part-ii">CMRR</a>, etc) are <em>small-signal</em> parameters, the <a href="https://engineersphere.com/frequency-response-for-mosfetbjt"><em>small-signal</em> model</a> of this circuit is needed.&nbsp; To obtain this, a nice trick is to &#8220;cut the amplifier in half&#8221; (lengthwise, such that you only analyze the output side of the amplifier) to obtain:</p>
<figure style="width: 738px" class="wp-caption alignnone"><img class=" " title="bjt small signal model" src="http://i.imgur.com/WSpJ1.jpg" alt="bjt small signal model" width="738" height="203"/><figcaption class="wp-caption-text">Figure 4. Small-signal model for above differential amplifer</figcaption></figure>
<p><strong>Note: [</strong>even though the output signal is single-ended here, the output is still a result of the entire input signal, and not just half of it.&nbsp; This is because the small-signal changes in the currents flowing through <img src='https://s0.wp.com/latex.php?latex=Q_%7B2%2C4%7D&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='Q_{2,4}' title='Q_{2,4}' class='latex' /> are impeded from traveling down the branches controlled by current sources <img src='https://s0.wp.com/latex.php?latex=Q_%7B2%2C6%7D&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='Q_{2,6}' title='Q_{2,6}' class='latex' />.&nbsp; Also note that the connections between <img src='https://s0.wp.com/latex.php?latex=R_%7B%5Cpi%7D&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='R_{\pi}' title='R_{\pi}' class='latex' /> and the voltage-controlled current source (VCCS) indicate that the voltage that controls the VCCS is the voltage across <img src='https://s0.wp.com/latex.php?latex=R_%7B%5Cpi%7D&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='R_{\pi}' title='R_{\pi}' class='latex' />.&nbsp; This is because the resistance in the emitter of these transistors has been omitted, due to its typically small value (10 to 25 <img src='https://s0.wp.com/latex.php?latex=%5COmega&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='\Omega' title='\Omega' class='latex' />).&nbsp; In addition to this, <img src='https://s0.wp.com/latex.php?latex=Q_6&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='Q_6' title='Q_6' class='latex' /> is assumed to be a small signal (AC) open-circuit.&nbsp; The frequency response has also been omitted, and the amplifier is assumed to be <a href="http://en.wikipedia.org/wiki/Electronic_amplifier#Unilateral_or_bilateral">unilateral</a>.<strong>]</strong></p>
<h3>Differential Mode Gain</h3>
<p>It is simple to see that <img src='https://s0.wp.com/latex.php?latex=v_%7Bout%7D&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='v_{out}' title='v_{out}' class='latex' /> (the small-signal output voltage) is equal to the current across the parallel combination of the resistors <img src='https://s0.wp.com/latex.php?latex=r_%7Bo2%7D&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='r_{o2}' title='r_{o2}' class='latex' /> and <img src='https://s0.wp.com/latex.php?latex=r_%7Bo1%7D&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='r_{o1}' title='r_{o1}' class='latex' /> multiplied by the size of the same parallel combination.&nbsp; Since we know the value of the current through this combination is equal to the input voltage multiplied by <img src='https://s0.wp.com/latex.php?latex=g_m&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='g_m' title='g_m' class='latex' /> (the <em>transconductance </em>parameter):</p>
<img src='https://s0.wp.com/latex.php?latex=v_%7Bout%7D+%3D-+g_mv_%7Bin%7D+%5Ccdot+%28r_%7Bo2%7D+%5C%7C+r_%7Bo4%7D%29&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='v_{out} =- g_mv_{in} \cdot (r_{o2} \| r_{o4})' title='v_{out} =- g_mv_{in} \cdot (r_{o2} \| r_{o4})' class='latex' />
<p>The transconductance parameter is a ratio of <em>output current </em>to <em>input voltage.</em> It is described mathematically as:</p>
<img src='https://s0.wp.com/latex.php?latex=g_m+%3D+%5Cfrac%7B%5Cpartial+i_c%7D%7B%5Cpartial+v_%7Bbe%7D%7D+%3D+%5Cfrac%7B%5Cpartial+i_%7Bout%7D%7D%7B%5Cpartial+v_%7Bin%7D%7D+&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='g_m = \frac{\partial i_c}{\partial v_{be}} = \frac{\partial i_{out}}{\partial v_{in}} ' title='g_m = \frac{\partial i_c}{\partial v_{be}} = \frac{\partial i_{out}}{\partial v_{in}} ' class='latex' />
<p>and can be solved for thusly:</p>
<img src='https://s0.wp.com/latex.php?latex=%5Cfrac%7B%5Cpartial+I_C%7D%7B%5Cpartial+V_%7BBE%7D%7D+%3D+%5Cfrac%7B%5Cpartial+%28I_Se%5E%7B%5Cfrac%7BV_%7BBE%7D%7D%7BV_T%7D%7D%29%7D%7B%5Cpartial+V_%7BBE%7D%7D+%3D+%5Cfrac%7BI_Se%5E%7B%5Cfrac%7BV_%7BBE%7D%7D%7BV_T%7D%7D%7D%7BV_T%7D+%3D+%5Cfrac+%7BI_C%7D%7BV_T%7D&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='\frac{\partial I_C}{\partial V_{BE}} = \frac{\partial (I_Se^{\frac{V_{BE}}{V_T}})}{\partial V_{BE}} = \frac{I_Se^{\frac{V_{BE}}{V_T}}}{V_T} = \frac {I_C}{V_T}' title='\frac{\partial I_C}{\partial V_{BE}} = \frac{\partial (I_Se^{\frac{V_{BE}}{V_T}})}{\partial V_{BE}} = \frac{I_Se^{\frac{V_{BE}}{V_T}}}{V_T} = \frac {I_C}{V_T}' class='latex' />
<p>In this example, <img src='https://s0.wp.com/latex.php?latex=I_C&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='I_C' title='I_C' class='latex' /> is .5 mA and <img src='https://s0.wp.com/latex.php?latex=V_T&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='V_T' title='V_T' class='latex' /> is 25 mV.&nbsp; With these values, we compute:</p>
<img src='https://s0.wp.com/latex.php?latex=g_m+%3D+%5Cfrac%7BI_C%7D%7BV_T%7D+%3D+%5Cfrac%7B.5+mA%7D%7B25+mV%7D+%3D+20+%5Cfrac%7BmA%7D%7BV%7D&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='g_m = \frac{I_C}{V_T} = \frac{.5 mA}{25 mV} = 20 \frac{mA}{V}' title='g_m = \frac{I_C}{V_T} = \frac{.5 mA}{25 mV} = 20 \frac{mA}{V}' class='latex' />
<p>Now that the transconductance parameter is known, the only other values needed to compute the differential mode gain are <img src='https://s0.wp.com/latex.php?latex=r_%7Bo2%7D&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='r_{o2}' title='r_{o2}' class='latex' /> and <img src='https://s0.wp.com/latex.php?latex=r_%7Bo4%7D&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='r_{o4}' title='r_{o4}' class='latex' />.&nbsp; <a href="https://engineersphere.com/bjt-circuit-and-symbol-conventions"><img src='https://s0.wp.com/latex.php?latex=Q_2&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='Q_2' title='Q_2' class='latex' /> is an npn transistor, while <img src='https://s0.wp.com/latex.php?latex=Q_4&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='Q_4' title='Q_4' class='latex' /> is a pnp transistor</a>, so they will not have the same small-signal resistance, but the procedure to find these two values are nearly identical.&nbsp; The following equation describes the small-signal output resistance of any BJT:</p>
<img src='https://s0.wp.com/latex.php?latex=r_%7Bo_%7Bn%2Cp%7D%7D+%3D+%5Cfrac%7B%7CV_%7BA_%7Bn%2Cp%7D%7D%7C+%2B+V_%7BCE%7D%7D%7BI_C%7D+%5Capprox+%5Cfrac%7B%7CV_%7BA_%7Bn%2Cp%7D%7D%7C%7D%7BI_C%7D&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='r_{o_{n,p}} = \frac{|V_{A_{n,p}}| + V_{CE}}{I_C} \approx \frac{|V_{A_{n,p}}|}{I_C}' title='r_{o_{n,p}} = \frac{|V_{A_{n,p}}| + V_{CE}}{I_C} \approx \frac{|V_{A_{n,p}}|}{I_C}' class='latex' />
<p>The parameter <img src='https://s0.wp.com/latex.php?latex=V_%7BA_%7Bn%2Cp%7D%7D&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='V_{A_{n,p}}' title='V_{A_{n,p}}' class='latex' /> is typically given, and in this tutorial:</p>
<p style="padding-left: 30px;"><img src='https://s0.wp.com/latex.php?latex=V_%7BA_n%7D+%3D+130+V&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='V_{A_n} = 130 V' title='V_{A_n} = 130 V' class='latex' /></p>
<p style="padding-left: 30px;"><img src='https://s0.wp.com/latex.php?latex=V_%7BA_p%7D+%3D+50+V&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='V_{A_p} = 50 V' title='V_{A_p} = 50 V' class='latex' /></p>
<p>Which would result in:</p>
<p><img src='https://s0.wp.com/latex.php?latex=r_%7Bo2%7D+%3D+%5Cfrac%7BV_%7BA_n%7D%7D%7BI_C%7D+%3D+%5Cfrac%7B130+V%7D%7B.5+mA%7D+%3D+260+k%5COmega&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='r_{o2} = \frac{V_{A_n}}{I_C} = \frac{130 V}{.5 mA} = 260 k\Omega' title='r_{o2} = \frac{V_{A_n}}{I_C} = \frac{130 V}{.5 mA} = 260 k\Omega' class='latex' /> and</p>
<img src='https://s0.wp.com/latex.php?latex=r_%7Bo4%7D+%3D+%5Cfrac%7BV_%7BA_p%7D%7D%7BI_C%7D+%3D+%5Cfrac%7B50+V%7D%7B.5+mA%7D+%3D+100+k%5COmega&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='r_{o4} = \frac{V_{A_p}}{I_C} = \frac{50 V}{.5 mA} = 100 k\Omega' title='r_{o4} = \frac{V_{A_p}}{I_C} = \frac{50 V}{.5 mA} = 100 k\Omega' class='latex' />
<p>Now that the small-signal resistances are known, along with the transconductance parameter, the differential mode gain (<img src='https://s0.wp.com/latex.php?latex=A_%7Bv%2CDM%7D&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='A_{v,DM}' title='A_{v,DM}' class='latex' />) may be calculated:</p>
<img src='https://s0.wp.com/latex.php?latex=A_%7Bv%2CDM%7D+%3D-+g_m+%5Ccdot+%28r_%7Bo2%7D+%5C%7C+r_%7Bo4%7D%29+%3D-+20+%5Cfrac%7BmA%7D%7BV%7D+%5Ccdot+%5Cfrac%7B260+k%5COmega+%5Ccdot+100+k%5COmega%7D%7B%28260%2B100%29+k%5COmega%7D+%3D+-1444.4+%5Cfrac%7Bv%7D%7Bv%7D+&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='A_{v,DM} =- g_m \cdot (r_{o2} \| r_{o4}) =- 20 \frac{mA}{V} \cdot \frac{260 k\Omega \cdot 100 k\Omega}{(260+100) k\Omega} = -1444.4 \frac{v}{v} ' title='A_{v,DM} =- g_m \cdot (r_{o2} \| r_{o4}) =- 20 \frac{mA}{V} \cdot \frac{260 k\Omega \cdot 100 k\Omega}{(260+100) k\Omega} = -1444.4 \frac{v}{v} ' class='latex' />
<p>or, in decibels (dB):</p>
<img src='https://s0.wp.com/latex.php?latex=A_%7Bv%2CDM%28dB%29%7D+%3D+20log%28%7CA_%7Bv%2CDM%7D%7C%29+%3D+63.2+dB&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='A_{v,DM(dB)} = 20log(|A_{v,DM}|) = 63.2 dB' title='A_{v,DM(dB)} = 20log(|A_{v,DM}|) = 63.2 dB' class='latex' />
<h3>Differential Input Impedance</h3>
<p>The differential input impedance of a differential amplifier <strong>is the impedance a &#8220;seen&#8221; by any &#8220;differential&#8221; signal. </strong>A &#8220;differential signal&#8221; is any and all signals that <em>aren&#8217;t shared by </em><img src='https://s0.wp.com/latex.php?latex=V_%7Bin-%7D&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='V_{in-}' title='V_{in-}' class='latex' /> and <img src='https://s0.wp.com/latex.php?latex=V_%7Bin%2B%7D&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='V_{in+}' title='V_{in+}' class='latex' />.&nbsp; For instance, if:</p>
<p><img src='https://s0.wp.com/latex.php?latex=V_%7Bin-%7D+%3D+%282+%2B+sin%282+%5Cpi+ft%29%29+V&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='V_{in-} = (2 + sin(2 \pi ft)) V' title='V_{in-} = (2 + sin(2 \pi ft)) V' class='latex' /> and</p>
<img src='https://s0.wp.com/latex.php?latex=V_%7Bin%2B%7D+%3D+%282+%2B+cos%282+%5Cpi+ft%29%29+V&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='V_{in+} = (2 + cos(2 \pi ft)) V' title='V_{in+} = (2 + cos(2 \pi ft)) V' class='latex' />
<p>then the common mode signal and differential mode signals are:</p>
<p style="padding-left: 30px;"><img src='https://s0.wp.com/latex.php?latex=V_%7Bin%2CCM%7D+%3D+2V&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='V_{in,CM} = 2V' title='V_{in,CM} = 2V' class='latex' /> and</p>
<p style="padding-left: 30px;"><img src='https://s0.wp.com/latex.php?latex=V_%7Bin%2CDM%7D+%3D+cos%282+%5Cpi+ft%29+-+sin%282+%5Cpi+ft%29+&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='V_{in,DM} = cos(2 \pi ft) - sin(2 \pi ft) ' title='V_{in,DM} = cos(2 \pi ft) - sin(2 \pi ft) ' class='latex' /></p>
<p>To find the differential input impedance, begin by following the loop consisting of:</p>
<p><img src='https://s0.wp.com/latex.php?latex=V_%7Bin-%7D+%5Crightarrow+V_%7BBE1%7D+%5Crightarrow+-V_%7BBE2%7D+%5Crightarrow+V_%7Bin%2B%7D&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='V_{in-} \rightarrow V_{BE1} \rightarrow -V_{BE2} \rightarrow V_{in+}' title='V_{in-} \rightarrow V_{BE1} \rightarrow -V_{BE2} \rightarrow V_{in+}' class='latex' />, as illustrated below:</p>
<figure style="width: 538px" class="wp-caption alignnone"><img class=" " title="bjt diff amp Rin" src="http://i.imgur.com/ZUnFT.jpg" alt="bjt diff amp Rin" width="538" height="519"/><figcaption class="wp-caption-text">Figure 5. Loop analyzed in order to determine Rin(DM)</figcaption></figure>
<p>We see that, in the differential signal mode, the path to ground only consists of <img src='https://s0.wp.com/latex.php?latex=r_%7B%5Cpi%7D&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='r_{\pi}' title='r_{\pi}' class='latex' /> <em>of each input transistor.</em> Since this is the case, the differential mode input impedance of any BJT diff-amp may be expressed as (<strong>omitting emitter resistance and assuming</strong> <img src='https://s0.wp.com/latex.php?latex=Q_%7B1%2C2%7D&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='Q_{1,2}' title='Q_{1,2}' class='latex' /><strong> matched</strong>):</p>
<img src='https://s0.wp.com/latex.php?latex=R_%7Bin%2CDM%7D+%3D+r_%7B%5Cpi+1%7D%2Br_%7B%5Cpi+2%7D+%3D+2r_%7B%5Cpi%7D&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='R_{in,DM} = r_{\pi 1}+r_{\pi 2} = 2r_{\pi}' title='R_{in,DM} = r_{\pi 1}+r_{\pi 2} = 2r_{\pi}' class='latex' />
<p style="padding-left: 30px;">where: <img src='https://s0.wp.com/latex.php?latex=r_%7B%5Cpi%7D+%3D+%5Cfrac%7B%5Cbeta%7D%7Bg_m%7D+&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='r_{\pi} = \frac{\beta}{g_m} ' title='r_{\pi} = \frac{\beta}{g_m} ' class='latex' /></p>
<p style="padding-left: 30px;"><img src='https://s0.wp.com/latex.php?latex=%5Cbeta+%3D+%5Cfrac%7Bi_c%7D%7Bi_b%7D&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='\beta = \frac{i_c}{i_b}' title='\beta = \frac{i_c}{i_b}' class='latex' /> (current gain factor)</p>
<p>A typical value for <img src='https://s0.wp.com/latex.php?latex=%5Cbeta&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='\beta' title='\beta' class='latex' /> is 100, and knowing <img src='https://s0.wp.com/latex.php?latex=g_m&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='g_m' title='g_m' class='latex' /> allows one to compute:</p>
<img src='https://s0.wp.com/latex.php?latex=R_%7Bin%2CDM%7D+%3D+2+%5Ccdot+%5Cfrac%7B%5Cbeta%7D%7Bg_m%7D+%3D+2+%5Ccdot+%5Cfrac%7B100%7D%7B20+%5Cfrac%7BmA%7D%7BV%7D%7D+%3D+10+k+%5COmega&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='R_{in,DM} = 2 \cdot \frac{\beta}{g_m} = 2 \cdot \frac{100}{20 \frac{mA}{V}} = 10 k \Omega' title='R_{in,DM} = 2 \cdot \frac{\beta}{g_m} = 2 \cdot \frac{100}{20 \frac{mA}{V}} = 10 k \Omega' class='latex' />
<p>So, for the BJT differential amplifier in this tutorial, the <strong>differential mode input impedance</strong> is:</p>
<p><img src='https://s0.wp.com/latex.php?latex=R_%7Bin%2CDM%7D+%3D+10+k+%5COmega&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='R_{in,DM} = 10 k \Omega' title='R_{in,DM} = 10 k \Omega' class='latex' /> (<a href="https://engineersphere.com/internal-resistance-and-the-effects-of-loading">what impact will this have?</a>)</p>
<h3>Common Mode Gain</h3>
<p>The CM gain (<img src='https://s0.wp.com/latex.php?latex=A_%7Bv%2CCM%7D&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='A_{v,CM}' title='A_{v,CM}' class='latex' />) is the &#8220;gain&#8221; that common mode signals &#8220;see,&#8221; or rather, is the <em><a href="http://en.wikipedia.org/wiki/Attenuation">attenuation </a>applied to signals present on both differential inputs.</em> A good op amp attempts to eliminate all common mode signals, but this is obviously not possible in the real world.&nbsp; However, one may compute the common mode gain by &#8220;cutting the amplifier in half&#8221; by observing one of the loops in the following diagram.&nbsp; The path differs from that of differential signals because common mode signals make it so that the two signal sources don&#8217;t &#8220;see&#8221; each other.&nbsp; Notice:<span style="text-decoration: underline;"><br />
</span></p>
<p>&nbsp;</p>
<figure style="width: 538px" class="wp-caption alignnone"><img class=" " title="common mode voltage gain" src="http://i.imgur.com/g7Pmp.jpg" alt="common mode voltage gain" width="538" height="519"/><figcaption class="wp-caption-text">Figure 6. Loop(s) analyzed to determine common mode voltage gain and input impedance</figcaption></figure>
<p>We choose a loop and draw the small-signal model to obtain:</p>
<figure style="width: 593px" class="wp-caption alignnone"><img class=" " title="common mode gain BJT diff amp" src="http://i.imgur.com/nvmrN.jpg" alt="common mode gain BJT diff amp" width="593" height="441"/><figcaption class="wp-caption-text">Figure 7. Small-signal model for common mode input signals</figcaption></figure>
<p>Similar to the output voltage of the differential mode small signal model, we can see that <img src='https://s0.wp.com/latex.php?latex=V_o&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='V_o' title='V_o' class='latex' /> is the voltage across <img src='https://s0.wp.com/latex.php?latex=r_%7Bo4%7D&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='r_{o4}' title='r_{o4}' class='latex' />.&nbsp; We also know the current running through this resistance, and may equate the output voltage to:</p>
<img src='https://s0.wp.com/latex.php?latex=V_o+%3D+-+g_mv_%7B%5Cpi%7D+%5Ccdot+r_%7Bo4%7D&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='V_o = - g_mv_{\pi} \cdot r_{o4}' title='V_o = - g_mv_{\pi} \cdot r_{o4}' class='latex' />
<p>This time, though, <img src='https://s0.wp.com/latex.php?latex=v_%7Bin%2CCM%7D&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='v_{in,CM}' title='v_{in,CM}' class='latex' /> isn&#8217;t distributed entirely over the resistances at the base.&nbsp; Instead, a fraction of the input common mode input signal is across the base-emitter junction.&nbsp; Referring back to the small signal model, we see that the loop composed of:</p>
<img src='https://s0.wp.com/latex.php?latex=V_%7Bin%7D+%5Crightarrow+v_%7B%5Cpi2%7D+%5Crightarrow+%28i_b+%2B+g_mv_%7B%5Cpi+2%7D%29+%5Ccdot+2+%5Ccdot+r_%7Bo6%7D&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='V_{in} \rightarrow v_{\pi2} \rightarrow (i_b + g_mv_{\pi 2}) \cdot 2 \cdot r_{o6}' title='V_{in} \rightarrow v_{\pi2} \rightarrow (i_b + g_mv_{\pi 2}) \cdot 2 \cdot r_{o6}' class='latex' />
<p>reveals that:</p>
<img src='https://s0.wp.com/latex.php?latex=V_%7Bin%7D+%3D+v_%7B%5Cpi2%7D+%2B+2+%5Ccdot+r_%7Bo6%7D+%5Ccdot+%28i_b+%2B+g_mv_%7B%5Cpi2%7D%29+&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='V_{in} = v_{\pi2} + 2 \cdot r_{o6} \cdot (i_b + g_mv_{\pi2}) ' title='V_{in} = v_{\pi2} + 2 \cdot r_{o6} \cdot (i_b + g_mv_{\pi2}) ' class='latex' />
<p>but <img src='https://s0.wp.com/latex.php?latex=i_b&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='i_b' title='i_b' class='latex' /> is negligible compared to the current supplied by the collector, so we say:</p>
<img src='https://s0.wp.com/latex.php?latex=V_%7Bin%7D+%3D+v_%7B%5Cpi2%7D+%2B+2+%5Ccdot+r_%7Bo6%7D+%5Ccdot+g_mv_%7B%5Cpi2%7D+%3D+v_%7B%5Cpi2%7D+%5Ccdot+%281+%2B+2r_%7Bo6%7Dg_m%29&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='V_{in} = v_{\pi2} + 2 \cdot r_{o6} \cdot g_mv_{\pi2} = v_{\pi2} \cdot (1 + 2r_{o6}g_m)' title='V_{in} = v_{\pi2} + 2 \cdot r_{o6} \cdot g_mv_{\pi2} = v_{\pi2} \cdot (1 + 2r_{o6}g_m)' class='latex' />
<p>which we use to solve for <img src='https://s0.wp.com/latex.php?latex=v_%7B%5Cpi2%7D&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='v_{\pi2}' title='v_{\pi2}' class='latex' />:</p>
<img src='https://s0.wp.com/latex.php?latex=v_%7B%5Cpi2%7D+%3D%5Cfrac%7B+v_%7Bin%7D%7D%7B1+%2B+2+%5Ccdot+r_%7Bo6%7D+%5Ccdot+g_m%7D&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='v_{\pi2} =\frac{ v_{in}}{1 + 2 \cdot r_{o6} \cdot g_m}' title='v_{\pi2} =\frac{ v_{in}}{1 + 2 \cdot r_{o6} \cdot g_m}' class='latex' />
<p>Which we then plug back into the equation for <img src='https://s0.wp.com/latex.php?latex=V_o&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='V_o' title='V_o' class='latex' />:</p>
<img src='https://s0.wp.com/latex.php?latex=V_o+%3D+-+g_mv_%7B%5Cpi%7D+%5Ccdot+r_%7Bo4%7D+%3D+-+%5Cfrac%7Br_%7Bo4%7Dg_m%7D%7B1%2B2+%5Ccdot+r_%7Bo6%7D+%5Ccdot+g_m%7D+%5Ccdot+V_%7Bin%7D&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='V_o = - g_mv_{\pi} \cdot r_{o4} = - \frac{r_{o4}g_m}{1+2 \cdot r_{o6} \cdot g_m} \cdot V_{in}' title='V_o = - g_mv_{\pi} \cdot r_{o4} = - \frac{r_{o4}g_m}{1+2 \cdot r_{o6} \cdot g_m} \cdot V_{in}' class='latex' />
<p>From this we can solve directly for the common mode gain:</p>
<img src='https://s0.wp.com/latex.php?latex=A_%7Bv%2CCM%7D+%3D+%5Cfrac%7BV_o%7D%7BV_%7Bin%7D%7D+%3D+-%5Cfrac%7Br_%7Bo4%7Dg_m%7D%7B1+%2B+2r_%7Bo6%7Dg_m%7D&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='A_{v,CM} = \frac{V_o}{V_{in}} = -\frac{r_{o4}g_m}{1 + 2r_{o6}g_m}' title='A_{v,CM} = \frac{V_o}{V_{in}} = -\frac{r_{o4}g_m}{1 + 2r_{o6}g_m}' class='latex' />
<p>Here, the <strong>common mode gain</strong> is:</p>
<img src='https://s0.wp.com/latex.php?latex=A_%7Bv%2CCM%7D+%3D+-.3845+%3D+-8.3+dB&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='A_{v,CM} = -.3845 = -8.3 dB' title='A_{v,CM} = -.3845 = -8.3 dB' class='latex' />
<h3>Common Mode Input Impedance</h3>
<p>The common-mode input impedance <em>is the impedance that common-mode input signals &#8220;see.&#8221;</em> One can analyze the common mode input impedance (<img src='https://s0.wp.com/latex.php?latex=R_%7Bin_%7BCM%7D%7D&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='R_{in_{CM}}' title='R_{in_{CM}}' class='latex' />) by, again, &#8220;cutting the differential amplifier in half&#8221; and analyzing one side the resulting schematic, assuming a common mode signal.&nbsp; This can be found by observing the figure 6, above.</p>
<p>Choosing one of these paths, we construct the corresponding small-signal model for common mode signals (<strong>assuming <img src='https://s0.wp.com/latex.php?latex=r_%7Bo2%7D+%3D+%5Cinfty&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='r_{o2} = \infty' title='r_{o2} = \infty' class='latex' /></strong>), which is shown in figure 7.&nbsp; From this figure, deriving <img src='https://s0.wp.com/latex.php?latex=R_%7Bin%2CCM%7D&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='R_{in,CM}' title='R_{in,CM}' class='latex' /> is simple.&nbsp; Notice the currents flowing in the loop that consists of:</p>
<img src='https://s0.wp.com/latex.php?latex=V_%7Bin%7D+%5Crightarrow+i_b+%5Ccdot+r_%7B%5Cpi2%7D+%5Crightarrow+%28i_b+%2B+g_mv_%7B%5Cpi2%7D%29+%5Ccdot+2+%5Ccdot+r_%7Bo6%7D+&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='V_{in} \rightarrow i_b \cdot r_{\pi2} \rightarrow (i_b + g_mv_{\pi2}) \cdot 2 \cdot r_{o6} ' title='V_{in} \rightarrow i_b \cdot r_{\pi2} \rightarrow (i_b + g_mv_{\pi2}) \cdot 2 \cdot r_{o6} ' class='latex' />
<p>from this loop, one may compute:</p>
<img src='https://s0.wp.com/latex.php?latex=0+%3D+-V_%7Bin%7D+%2B+i_%7Bb%7D+%5Ccdot+r_%7B%5Cpi+2%7D+%2B+%28i_%7Bb%7D+%2B+g_mv_%7B%5Cpi+2%7D%29+%5Ccdot+2%5Ccdot+r_%7Bo6%7D+&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='0 = -V_{in} + i_{b} \cdot r_{\pi 2} + (i_{b} + g_mv_{\pi 2}) \cdot 2\cdot r_{o6} ' title='0 = -V_{in} + i_{b} \cdot r_{\pi 2} + (i_{b} + g_mv_{\pi 2}) \cdot 2\cdot r_{o6} ' class='latex' />
<p>which is used to find an equation for <img src='https://s0.wp.com/latex.php?latex=V_%7Bin%7D&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='V_{in}' title='V_{in}' class='latex' /></p>
<img src='https://s0.wp.com/latex.php?latex=V_%7Bin%7D+%3D+i_%7Bb%7D+%5Ccdot+r_%7B%5Cpi+2%7D+%2B+%28i_%7Bb%7D+%2B+g_mv_%7B%5Cpi+2%7D%29+%5Ccdot+2%5Ccdot+r_%7Bo6%7D&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='V_{in} = i_{b} \cdot r_{\pi 2} + (i_{b} + g_mv_{\pi 2}) \cdot 2\cdot r_{o6}' title='V_{in} = i_{b} \cdot r_{\pi 2} + (i_{b} + g_mv_{\pi 2}) \cdot 2\cdot r_{o6}' class='latex' />
<p>and since:</p>
<p style="padding-left: 30px;"><img src='https://s0.wp.com/latex.php?latex=g_mv_%7B%5Cpi+2%7D+%3D+%5Cbeta+%5Ccdot+i_%7Bb%7D+&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='g_mv_{\pi 2} = \beta \cdot i_{b} ' title='g_mv_{\pi 2} = \beta \cdot i_{b} ' class='latex' /></p>
<p style="padding-left: 30px;">and <img src='https://s0.wp.com/latex.php?latex=i_b+%3D+i_%7Bin%7D&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='i_b = i_{in}' title='i_b = i_{in}' class='latex' /></p>
<p>So:</p>
<img src='https://s0.wp.com/latex.php?latex=V_%7Bin%7D+%3D+i_b+%5Ccdot+%28r_%7B%5Cpi+2%7D+%2B+2+%5Ccdot+r_%7Bo6%7D+%5Ccdot+%28+%5Cbeta+%2B+1%29%29+&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='V_{in} = i_b \cdot (r_{\pi 2} + 2 \cdot r_{o6} \cdot ( \beta + 1)) ' title='V_{in} = i_b \cdot (r_{\pi 2} + 2 \cdot r_{o6} \cdot ( \beta + 1)) ' class='latex' />
<p>which is the same as:</p>
<img src='https://s0.wp.com/latex.php?latex=V_%7Bin%7D+%3D+i_%7Bin%7D+%5Ccdot+%28r_%7B%5Cpi+2%7D+%2B+2+%5Ccdot+r_%7Bo6%7D+%5Ccdot+%28+%5Cbeta+%2B+1%29%29+&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='V_{in} = i_{in} \cdot (r_{\pi 2} + 2 \cdot r_{o6} \cdot ( \beta + 1)) ' title='V_{in} = i_{in} \cdot (r_{\pi 2} + 2 \cdot r_{o6} \cdot ( \beta + 1)) ' class='latex' />
<p>which can be rearranged for:</p>
<img src='https://s0.wp.com/latex.php?latex=R_%7Bin%2CCM%7D+%3D+%5Cfrac%7BV_%7Bin%7D%7D%7Bi_%7Bin%7D%7D+%3D+r_%7B%5Cpi+2%7D+%2B+2+%5Ccdot+r_%7Bo6%7D+%5Ccdot+%28%5Cbeta+%2B+1%29+&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='R_{in,CM} = \frac{V_{in}}{i_{in}} = r_{\pi 2} + 2 \cdot r_{o6} \cdot (\beta + 1) ' title='R_{in,CM} = \frac{V_{in}}{i_{in}} = r_{\pi 2} + 2 \cdot r_{o6} \cdot (\beta + 1) ' class='latex' />
<p style="padding-left: 30px;">where: <img src='https://s0.wp.com/latex.php?latex=r_%7B%5Cpi%7D+%3D+%5Cfrac%7B%5Cbeta%7D%7Bg_m%7D+&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='r_{\pi} = \frac{\beta}{g_m} ' title='r_{\pi} = \frac{\beta}{g_m} ' class='latex' /></p>
<p>Which, in this tutorial, results in:</p>
<img src='https://s0.wp.com/latex.php?latex=R_%7Bin%2CCM%7D+%3D+r_%7B%5Cpi+2%7D+%2B+2+%5Ccdot+r_%7Bo6%7D+%5Ccdot+%28%5Cbeta+%2B+1%29+%3D+%5Cfrac%7B1+mA%7D%7B25+mV%7D+%2B+2+%5Ccdot+%5Cfrac%7B130+V%7D%7B1+mA%7D+%5Ccdot+%28100+%2B+1%29+%3D+26.26+M%5COmega&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='R_{in,CM} = r_{\pi 2} + 2 \cdot r_{o6} \cdot (\beta + 1) = \frac{1 mA}{25 mV} + 2 \cdot \frac{130 V}{1 mA} \cdot (100 + 1) = 26.26 M\Omega' title='R_{in,CM} = r_{\pi 2} + 2 \cdot r_{o6} \cdot (\beta + 1) = \frac{1 mA}{25 mV} + 2 \cdot \frac{130 V}{1 mA} \cdot (100 + 1) = 26.26 M\Omega' class='latex' />
<h3>Common Mode Rejection Ratio (CMRR)</h3>
<p>The common mode rejection ratio (CMRR) is simply a ratio of the differential mode gain to the common mode gain, and is defined as:</p>
<img src='https://s0.wp.com/latex.php?latex=CMRR+%3D+%5Cfrac%7BA_%7Bv%2CDM%7D%7D%7BA_%7Bv%2CCM%7D%7D&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='CMRR = \frac{A_{v,DM}}{A_{v,CM}}' title='CMRR = \frac{A_{v,DM}}{A_{v,CM}}' class='latex' />
<p>Here, the CMRR is:</p>
<img src='https://s0.wp.com/latex.php?latex=CMRR+%3D+%5Cfrac%7B-1444.4+v%2Fv%7D%7B-.384+v%2Fv%7D+%3D+3761.46+%3D+71.5+dB&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='CMRR = \frac{-1444.4 v/v}{-.384 v/v} = 3761.46 = 71.5 dB' title='CMRR = \frac{-1444.4 v/v}{-.384 v/v} = 3761.46 = 71.5 dB' class='latex' />
<h3>Analysis of FET Differential Amplifiers</h3>
<p>As stated before, the analysis of these performance parameters are done virtually the same for FET diff amps as they are for BJT diff amps.&nbsp; There are, however, a few key differences.&nbsp; For one, all BJT transistors are typically built to be the same size on a given IC device.&nbsp; But for an IC device that uses FETs, this is not the case.&nbsp; Each FET has an adjustable length and width that affects how much current it will pass for a given voltage-drop across the device.&nbsp; In fact, observe the equation for the <em>drain current </em>in a FET:</p>
<img src='https://s0.wp.com/latex.php?latex=I_D+%3D+%5Cfrac%7Bk_%7Bn%2Cp%7D%7D%7B2%7D+%5Cfrac%7BW%7D%7BL%7D+%28%7CV_%7BGS%7D%7C+-+%7CV_%7Bth_%7Bn%2Cp%7D%7D%7C%29%5E2&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='I_D = \frac{k_{n,p}}{2} \frac{W}{L} (|V_{GS}| - |V_{th_{n,p}}|)^2' title='I_D = \frac{k_{n,p}}{2} \frac{W}{L} (|V_{GS}| - |V_{th_{n,p}}|)^2' class='latex' />
<p>From this, the gate-source voltage is:</p>
<img src='https://s0.wp.com/latex.php?latex=V_%7BGS%7D+%3D+%5Csqrt%7B%5Cfrac%7B2I_DL%7D%7BkW%7D%7D+-+V_%7Bth_%7Bn%2Cp%7D%7D&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='V_{GS} = \sqrt{\frac{2I_DL}{kW}} - V_{th_{n,p}}' title='V_{GS} = \sqrt{\frac{2I_DL}{kW}} - V_{th_{n,p}}' class='latex' />
<p>where:</p>
<ul>
<li><img src='https://s0.wp.com/latex.php?latex=k&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='k' title='k' class='latex' /> is the process conductivity parameter, and is equal to:<br />
<blockquote><p><img src='https://s0.wp.com/latex.php?latex=k+%3D+%5Cmu_%7Bn%2Cp%7D+C_%7Box%7D&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='k = \mu_{n,p} C_{ox}' title='k = \mu_{n,p} C_{ox}' class='latex' /> , which is the <a href="http://en.wikipedia.org/wiki/Electron_mobility">electron mobility</a> multiplied by the oxide capacitance</p></blockquote>
</li>
<li><img src='https://s0.wp.com/latex.php?latex=W%2C+L&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='W, L' title='W, L' class='latex' /> are the width and length of the device, respectively</li>
<li><img src='https://s0.wp.com/latex.php?latex=V_%7BGS%7D&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='V_{GS}' title='V_{GS}' class='latex' /> is the gate-to-source voltage</li>
<li><img src='https://s0.wp.com/latex.php?latex=V_%7Bth%7D&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='V_{th}' title='V_{th}' class='latex' /> is the <a href="http://en.wikipedia.org/wiki/Threshold_voltage">threshold voltage</a> of the FET</li>
</ul>
<p>Analyzing BJTs in a circuit is more simple because all base-emitter voltages are assumed to be equal.&nbsp; But this is not the case for mosfets, and one must analyze the above equation (or others) to find device voltages.&nbsp; But there is the threshold voltage &#8211; the minimum gate-to-source voltage that will allow for any conduction whatsoever.&nbsp; The threshold voltage is a result of the FET fabrication process, and is typically provided on datasheets for each FET gender.</p>
<p>For a differential amplifier composed of FETs to work, it is imperative that all the FETs be in <strong>saturation mode</strong>.&nbsp; For a FET to be in saturation implies:</p>
<img src='https://s0.wp.com/latex.php?latex=%7CV_%7BDS%7D%7C+%5Cge+%7CV_%7BGS%7D%7C+-+%7CV_%7Bth%7D%7C&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='|V_{DS}| \ge |V_{GS}| - |V_{th}|' title='|V_{DS}| \ge |V_{GS}| - |V_{th}|' class='latex' />
<p>So this must be checked when analyzing these types of circuits.</p>
<p>Another important difference is the derivation of the transconductance parameter, <img src='https://s0.wp.com/latex.php?latex=g_m&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='g_m' title='g_m' class='latex' />.&nbsp; When analyzed for a BJT, it was defined as the ratio of the change in collector current to the change in the base-emitter voltage.&nbsp; For a FET there is a similar procedure, as the transconductance is defined as the ratio of the change in drain current to the change in gate-source voltage.&nbsp; Mathematically, the transconductance parameter is:</p>
<img src='https://s0.wp.com/latex.php?latex=g_m+%3D+%5Cfrac%7B%5Cpartial%7Bi_D%7D%7D%7B%5Cpartial%7Bv_%7BGS%7D%7D%7D+%3D%5Cfrac%7B%5Cpartial%28+%5Cfrac%7Bk_%7Bn%2Cp%7D%7D%7B2%7D+%5Cfrac%7BW%7D%7BL%7D+%28%7CV_%7BGS%7D%7C+-+%7CV_%7Bth_%7Bn%2Cp%7D%7D%7C%29%5E2%29%7D%7B%5Cpartial%7Bv_%7BGS%7D%7D%7D+%3D+%5Csqrt%7B2I_Dk%5Cfrac%7BW%7D%7BL%7D%7D&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='g_m = \frac{\partial{i_D}}{\partial{v_{GS}}} =\frac{\partial( \frac{k_{n,p}}{2} \frac{W}{L} (|V_{GS}| - |V_{th_{n,p}}|)^2)}{\partial{v_{GS}}} = \sqrt{2I_Dk\frac{W}{L}}' title='g_m = \frac{\partial{i_D}}{\partial{v_{GS}}} =\frac{\partial( \frac{k_{n,p}}{2} \frac{W}{L} (|V_{GS}| - |V_{th_{n,p}}|)^2)}{\partial{v_{GS}}} = \sqrt{2I_Dk\frac{W}{L}}' class='latex' />
<p>The last notable difference is the computation for a FET&#8217;s small-signal resistance.&nbsp; The equation describing <img src='https://s0.wp.com/latex.php?latex=r_%7Bo%7D&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='r_{o}' title='r_{o}' class='latex' /> is:</p>
<img src='https://s0.wp.com/latex.php?latex=r_o+%3D+%5Cfrac%7B1%7D%7B%5Clambda+I_D%7D&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='r_o = \frac{1}{\lambda I_D}' title='r_o = \frac{1}{\lambda I_D}' class='latex' />
<p>where <img src='https://s0.wp.com/latex.php?latex=%5Clambda&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='\lambda' title='\lambda' class='latex' /> is the <a href="http://en.wikipedia.org/wiki/Channel_length_modulation">channel-length modulation parameter</a>.</p>
<p>From this little discussion, you should be able to apply the principles used to analyze the BJT differential amplifier to the analysis of a FET-based differential amplifier.&nbsp; But, of course, if you would like to see a FET differential amplifier explained in more detail, do not hesitate to <a href="https://engineersphere.com/ask-a-question">ask a question</a>!</p>
<h3>Credit &amp; Acknowledgment</h3>
<p>This post was created in March 2011 by Kansas State University Electrical Engineering student Safa Khamis.&nbsp; A million thank yous extended to Safa for taking the time to document this important process for everyone else to learn from.&nbsp; Please leave questions, comments, or ask a question in the questions section of the website.</p>
<p><span style="text-decoration: underline;"><br />
</span></p>
<div id="_mcePaste" class="mcePaste" style="position: absolute; left: -10000px; top: 2962px; width: 1px; height: 1px; overflow: hidden;">
<img src='https://s0.wp.com/latex.php?latex=V_A_n+%3D+%5Cfrac%7BV_A_n%7D%7BI_C%7D+%3D+%5Cfrac%7B130+V%7D%7B.5+mA%7D+%3D+260+k%5COmega&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='V_A_n = \frac{V_A_n}{I_C} = \frac{130 V}{.5 mA} = 260 k\Omega' title='V_A_n = \frac{V_A_n}{I_C} = \frac{130 V}{.5 mA} = 260 k\Omega' class='latex' />
<img src='https://s0.wp.com/latex.php?latex=V_A_n+%3D+%5Cfrac%7BV_A_n%7D%7BI_C%7D+%3D+%5Cfrac%7B130+V%7D%7B.5+mA%7D+%3D+260+k%5COmega&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='V_A_n = \frac{V_A_n}{I_C} = \frac{130 V}{.5 mA} = 260 k\Omega' title='V_A_n = \frac{V_A_n}{I_C} = \frac{130 V}{.5 mA} = 260 k\Omega' class='latex' />
</div>
<p><script src="http://connect.facebook.net/en_US/all.js#xfbml=1"></script></p><p>The post <a href="https://engineersphere.com/dc-biasing-ac-performance-analysis-of-bjt-fet-amplifiers">DC Biasing & AC Performance Analysis of BJT & FET Differential Amplifiers</a> first appeared on <a href="https://engineersphere.com">Engineersphere.com</a>.</p>]]></content:encoded>
					
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		<title>The Convolution Integral Explained</title>
		<link>https://engineersphere.com/the-convolution-integral-explained</link>
					<comments>https://engineersphere.com/the-convolution-integral-explained#respond</comments>
		
		<dc:creator><![CDATA[Jeff Schuler]]></dc:creator>
		<pubDate>Thu, 10 Mar 2011 06:57:38 +0000</pubDate>
				<category><![CDATA[Electrical Engineering Concepts]]></category>
		<category><![CDATA[convolution integral example]]></category>
		<category><![CDATA[convolution integrals]]></category>
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					<description><![CDATA[<p>Introduction to the convolution Amongst the concepts that cause the most confusion to electrical engineering students, the Convolution Integral stands as a repeat offender.&#160; As such, the point of this article is to explain what a convolution integral is, why engineers need it, and the math behind it. In essence, the &#8220;convolution&#8221; of two functions [&#8230;]</p>
<p>The post <a href="https://engineersphere.com/the-convolution-integral-explained">The Convolution Integral Explained</a> first appeared on <a href="https://engineersphere.com">Engineersphere.com</a>.</p>]]></description>
										<content:encoded><![CDATA[<h3>Introduction to the convolution<span style="text-decoration: underline;"><br />
</span></h3>
<p>Amongst the concepts that cause the most confusion to electrical engineering students, the <a href="https://engineersphere.com/the-convolution-integral-explained">Convolution Integral</a> stands as a repeat offender.&nbsp; As such, the point of this article is to explain what a convolution integral is, why engineers need it, and the math behind it.</p>
<p>In essence, the &#8220;convolution&#8221; of two functions (over the same variable, e.g. <img src='https://s0.wp.com/latex.php?latex=f_1%28t%29&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='f_1(t)' title='f_1(t)' class='latex' /> and <img src='https://s0.wp.com/latex.php?latex=f_2%28t%29&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='f_2(t)' title='f_2(t)' class='latex' />) is an operation that produces a separate third function that describes how the first function &#8220;modifies&#8221; the second one.&nbsp; Conversely, the resulting function can be seen as how the second function &#8220;modifies&#8221; the first function.&nbsp; Sometimes the result is used to describe how much the first two functions &#8220;have in common.&#8221;&nbsp; In all honesty, the concept of the convolution of two functions is quite abstract, but the frequency at which it appears in nature grants its importance to scientists and engineers.&nbsp; Ultimately the aim here is to identify its use to electrical engineers &#8211; so for now do not dwell solely on its mathematical significance.</p>
<p>A convolution of two functions is denoted with the operator &#8220;<img src='https://s0.wp.com/latex.php?latex=%2A+&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='* ' title='* ' class='latex' />&#8220;, and is written as:</p>
<figure style="width: 281px" class="wp-caption alignnone"><img class=" " title="Convolution Integral" src="http://mathurl.com/4r2zkod.png" alt="convolution integral" width="281" height="40"/><figcaption class="wp-caption-text">Convolution of f1(t) and f2(t)</figcaption></figure>
<p>Where <img src='https://s0.wp.com/latex.php?latex=%5Ctau&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='\tau' title='\tau' class='latex' /> is used as a &#8220;dummy variable.&#8221;&nbsp; To aid in understanding this equation, observe the following graphic:</p>
<figure style="width: 468px" class="wp-caption alignnone"><img class=" " title="Convolution of 2 square pulses" src="http://upload.wikimedia.org/wikipedia/commons/6/6a/Convolution_of_box_signal_with_itself2.gif" alt="Convolution of 2 square pulses" width="468" height="147"/><figcaption class="wp-caption-text">Convolution of two square pulses, resulting in a triangular pulse</figcaption></figure>
<p>Before diving any further into the math, let us first discuss the relevance of this equation to the realm of electrical engineering.</p>
<h3>Why is the convolution integral relevant?</h3>
<p>Most electrical circuits are designed to be <em>linear, time-invariant </em>(<a href="http://en.wikipedia.org/wiki/LTI_system_theory">LTI</a>) systems.&nbsp; Being &#8220;linear&#8221; implies that the magnitude of a circuit&#8217;s output signal is a <strong>scaled </strong>version of the input signal&#8217;s magnitude.&nbsp; Further, an LTI system that is excited by two independent signal sources will output the <strong>sum </strong>of the <strong>scaled </strong>versions of each signal.&nbsp; This is extended for an infinite number of independent signal sources, and gives rise to the concept of <em>superposition</em>.&nbsp; Put in another way, if a function <img src='https://s0.wp.com/latex.php?latex=x_1%28t%29&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='x_1(t)' title='x_1(t)' class='latex' /> causes an LTI system to output <img src='https://s0.wp.com/latex.php?latex=y_1%28t%29&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='y_1(t)' title='y_1(t)' class='latex' />, then:</p>
<p style="text-align: center;"><img src='https://s0.wp.com/latex.php?latex=a_1+%5Ccdot+x_1%28t%29+%5Cto+a_1+%5Ccdot+y_1%28t%29&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='a_1 \cdot x_1(t) \to a_1 \cdot y_1(t)' title='a_1 \cdot x_1(t) \to a_1 \cdot y_1(t)' class='latex' /></p>
<p>Where <img src='https://s0.wp.com/latex.php?latex=a_1&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='a_1' title='a_1' class='latex' /> is a multiplicative constant.&nbsp; In addition to this, superposition allows us to say:</p>
<p style="text-align: center;"><img src='https://s0.wp.com/latex.php?latex=a_1+%5Ccdot+x_1%28t%29+%2B+a_2+%5Ccdot+x_2%28t%29+%2B+%5Cldots+%5Cto+a_1+%5Ccdot+y_1%28t%29+%2B+a_2+%5Ccdot+y_2%28t%29+%2B+%5Cldots+&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='a_1 \cdot x_1(t) + a_2 \cdot x_2(t) + \ldots \to a_1 \cdot y_1(t) + a_2 \cdot y_2(t) + \ldots ' title='a_1 \cdot x_1(t) + a_2 \cdot x_2(t) + \ldots \to a_1 \cdot y_1(t) + a_2 \cdot y_2(t) + \ldots ' class='latex' /></p>
<p>Being a &#8220;time-invariant&#8221; system means <em>it does not matter when the input signal is applied</em> &#8211; a <em>specific </em>input signal will always result in <em>the same </em>output signal for a given LTI system.&nbsp; Put mathematically, time-invariance can be expressed as:</p>
<p style="text-align: center;"><img src='https://s0.wp.com/latex.php?latex=x_1%28t%29+%5Cto+y_1%28t%29+%5CLeftrightarrow+x_1%28t%2B%5Ctau%29+%5Cto+y_1%28t%2B%5Ctau%29+&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='x_1(t) \to y_1(t) \Leftrightarrow x_1(t+\tau) \to y_1(t+\tau) ' title='x_1(t) \to y_1(t) \Leftrightarrow x_1(t+\tau) \to y_1(t+\tau) ' class='latex' /></p>
<p>where <img src='https://s0.wp.com/latex.php?latex=%5Ctau&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='\tau' title='\tau' class='latex' /> can be viewed as a time delay when dealing with signals through time (i.e. &#8220;time-domain signals&#8221;).&nbsp; Though not directly, this concept also signifies that <em>an output signal cannot contain frequency components not inherent in the input signal (</em>causality).</p>
<p>The vast majority of circuits are <a href="https://engineersphere.com/the-convolution-integral-explained">LTI systems</a>, each with a specific <em>impulse response. </em>The &#8220;impulse response&#8221; of a system is a system&#8217;s output when its input is fed with an <em>impulse signal</em> &#8211; a signal of infinitesimally short duration.&nbsp; A real-world &#8220;impulse signal&#8221; would be something like a lightning bolt &#8211; or any form of ESD (electro-static dischage).&nbsp;&nbsp; Basically, any voltage or current that spikes in magnitude for a <em>relatively</em> short period of time may be viewed as an impulse signal.&nbsp; The impulse response of a circuit will always be a time-domain signal, and exists because no signal can propagate through a circuit in zero time; each individual electron involved can only move so quickly through each component.&nbsp; Typically, real-world electronic LTI systems exhibit an impulse response that consists of an initial spike in magnitude, followed by an everlasting and ever-decreasing exponential relationship in signal magnitude.&nbsp; The following image describes this graphically.</p>
<figure style="width: 560px" class="wp-caption alignnone"><img title="Typical Unit Impulse Response" src="http://www.me.cmu.edu/ctms/modeling/tutorial/transferfunction/tutorial_tf_impulse.gif" alt="" width="560" height="420"/><figcaption class="wp-caption-text">Typical Unit Impulse Response</figcaption></figure>
<p>So, here&#8217;s the big deal: the fact that each LTI circuit has a specific impulse response function (here, referred to as <img src='https://s0.wp.com/latex.php?latex=h%28t%29&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='h(t)' title='h(t)' class='latex' />) is very useful in predicting its behavior given a particular input signal (here, referred to as <img src='https://s0.wp.com/latex.php?latex=x%28t%29&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='x(t)' title='x(t)' class='latex' />).&nbsp; This is because the input signal itself may be viewed as an <em>impulse train &#8211; </em>a stream of continuous impulse functions, with infinitesimally short durations of time between each impulse.&nbsp; This fact, along with superposition, allows one to find the output of an LTI system given an arbitrary input signal <em>by summing the LTI system&#8217;s impulse response to each impulse function that make up the input signal.</em> By allowing the time between each &#8220;impulse&#8221; of the input signal to go to zero, this approach can be used to determine the output time-domain signal of an LTI system for any time-domain input signal.&nbsp; For example, the following graphic shows the output of an RC circuit when fed with a square pulse:</p>
<figure style="width: 468px" class="wp-caption alignnone"><img title="RC square wave convolution" src="http://upload.wikimedia.org/wikipedia/commons/b/b9/Convolution_of_spiky_function_with_box2.gif" alt="" width="468" height="135"/><figcaption class="wp-caption-text">Convolution of RC network impulse response and square wave input to find the output signal.</figcaption></figure>
<p>What is seen here is the integral of the impulse response and the input square wave <em>as the square wave is stepped through time.</em> In the above convolution equation, it is seen that the operation is done with respect to <img src='https://s0.wp.com/latex.php?latex=%5Ctau&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='\tau' title='\tau' class='latex' />, a dummy variable.&nbsp; In reality, we are taking an input signal, flipping it vertically through the origin (not evident with a square wave), and determining what the integral is at each value of <img src='https://s0.wp.com/latex.php?latex=%5Ctau&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='\tau' title='\tau' class='latex' />, which here is <em>delay through time.</em> Since the output of any LTI system is non-causal (meaning it cannot exist until the signal that excites the output has been applied), we must mathematically step through time to see how each impulse signal of the input affects the LTI system&#8217;s impulse response &#8211; again, achieved by stepping through <img src='https://s0.wp.com/latex.php?latex=%5Ctau&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='\tau' title='\tau' class='latex' /> &#8211; the &#8220;time-delay&#8221; dummy variable.</p>
<h3>A Convolution Example</h3>
<p>To see how the convolution integral can be used to predict the output of an LTI circuit, observe the following example:</p>
<p style="padding-left: 30px;">For an LTI system with an impulse response of <img src='https://s0.wp.com/latex.php?latex=h%28t%29+%3D+e%5E%7B-2t%7Du%28t%29+&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='h(t) = e^{-2t}u(t) ' title='h(t) = e^{-2t}u(t) ' class='latex' />, calculate the output, <img src='https://s0.wp.com/latex.php?latex=y%28t%29&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='y(t)' title='y(t)' class='latex' />, given the input of:</p>
<p style="padding-left: 30px;"><img src='https://s0.wp.com/latex.php?latex=f%28t%29+%3D+e%5E%7B-t%7Du%28t%29&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='f(t) = e^{-t}u(t)' title='f(t) = e^{-t}u(t)' class='latex' /></p>
<p style="padding-left: 30px;">The output of this system is found by solving:</p>
<p style="padding-left: 30px;"><img src='https://s0.wp.com/latex.php?latex=y%28t%29+%3D+h%28t%29%2Af%28t%29+%3D+%5Cint_%7B0%7D%5E%7B%5Cinfty%7D+h%28%5Ctau%29+%5Ccdot+f%28t-%5Ctau%29+d%5Ctau&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='y(t) = h(t)*f(t) = \int_{0}^{\infty} h(\tau) \cdot f(t-\tau) d\tau' title='y(t) = h(t)*f(t) = \int_{0}^{\infty} h(\tau) \cdot f(t-\tau) d\tau' class='latex' /></p>
<p style="padding-left: 30px;">We only integrate between 0 and +<img src='https://s0.wp.com/latex.php?latex=%5Cinfty&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='\infty' title='\infty' class='latex' /> because, if we define <img src='https://s0.wp.com/latex.php?latex=t+%3D+0&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='t = 0' title='t = 0' class='latex' /> as the time that the input signal <img src='https://s0.wp.com/latex.php?latex=f%28t%29&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='f(t)' title='f(t)' class='latex' /> is applied, then both <img src='https://s0.wp.com/latex.php?latex=h%28t%29&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='h(t)' title='h(t)' class='latex' /> and <img src='https://s0.wp.com/latex.php?latex=f%28t%29&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='f(t)' title='f(t)' class='latex' /> have zero magnitude at any time <img src='https://s0.wp.com/latex.php?latex=t%3C0&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='t&lt;0' title='t&lt;0' class='latex' />.</p>
<p style="padding-left: 30px;">From there, we calculate:</p>
<p style="padding-left: 30px;"><img src='https://s0.wp.com/latex.php?latex=y%28t%29+%3D+%5Cint_%7B0%7D%5E%7B%5Cinfty%7D+e%5E%7B-2%5Ctau%7Du%28%5Ctau%29+%5Ccdot+e%5E%7B-%5Ctau%7D+d%5Ctau%3D+%5Cint_%7B0%7D%5E%7Bt%7De%5E%7B-%5Ctau%7D+%5Ccdot+e%5E%7B-2%28t-%5Ctau%29%7Dd%5Ctau+&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='y(t) = \int_{0}^{\infty} e^{-2\tau}u(\tau) \cdot e^{-\tau} d\tau= \int_{0}^{t}e^{-\tau} \cdot e^{-2(t-\tau)}d\tau ' title='y(t) = \int_{0}^{\infty} e^{-2\tau}u(\tau) \cdot e^{-\tau} d\tau= \int_{0}^{t}e^{-\tau} \cdot e^{-2(t-\tau)}d\tau ' class='latex' /></p>
<p style="padding-left: 30px;">Next, we can simplify and compute the integral:</p>
<p style="padding-left: 30px;"><img src='https://s0.wp.com/latex.php?latex=y%28t%29+%3D+%5Cint_%7B0%7D%5E%7Bt%7De%5E%7B-%5Ctau%7D+%5Ccdot+e%5E%7B-2%28t-%5Ctau%29%7Dd%5Ctau+%3D+e%5E%7B-2t%7D+%5Cint_%7B0%7D%5E%7Bt%7De%5E%7B%5Ctau%7Dd%5Ctau+%3D+e%5E%7B-2t%7D%28e%5Et-1%29+%3D+e%5E%7B-t%7D+-+e%5E%7B-2t%7D&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='y(t) = \int_{0}^{t}e^{-\tau} \cdot e^{-2(t-\tau)}d\tau = e^{-2t} \int_{0}^{t}e^{\tau}d\tau = e^{-2t}(e^t-1) = e^{-t} - e^{-2t}' title='y(t) = \int_{0}^{t}e^{-\tau} \cdot e^{-2(t-\tau)}d\tau = e^{-2t} \int_{0}^{t}e^{\tau}d\tau = e^{-2t}(e^t-1) = e^{-t} - e^{-2t}' class='latex' /></p>
<p style="padding-left: 30px;">Since <img src='https://s0.wp.com/latex.php?latex=y%28t%29+%3D+0&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='y(t) = 0' title='y(t) = 0' class='latex' /> for all <img src='https://s0.wp.com/latex.php?latex=t+%3C+0&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='t &lt; 0' title='t &lt; 0' class='latex' />, we can write the output <img src='https://s0.wp.com/latex.php?latex=y%28t%29+&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='y(t) ' title='y(t) ' class='latex' /> as:</p>
<p style="padding-left: 30px;"><img src='https://s0.wp.com/latex.php?latex=y%28t%29+%3D+%28e%5E%7B-t%7D-e%5E%7B-2t%7D%29u%28t%29+&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='y(t) = (e^{-t}-e^{-2t})u(t) ' title='y(t) = (e^{-t}-e^{-2t})u(t) ' class='latex' /></p>
<p style="padding-left: 30px;">This result <img src='https://s0.wp.com/latex.php?latex=y%28t%29&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='y(t)' title='y(t)' class='latex' /> <em>describes the output function for an LTI system with an impulse response </em><img src='https://s0.wp.com/latex.php?latex=h%28t%29&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='h(t)' title='h(t)' class='latex' /> <em>when fed the input signal </em><img src='https://s0.wp.com/latex.php?latex=f%28t%29&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='f(t)' title='f(t)' class='latex' />.</p>
<h3>5 Steps to perform mathematical convolution</h3>
<p>Often, one may wish to compute the convolution of two signals that can&#8217;t be described with one function of time alone.&nbsp; For arbitrary signals, such as pulse trains or PCM signals, the convolution <em>at any time t</em> can be computed graphically.&nbsp; For signals <em>whose individual &#8220;sections&#8221; can be described mathematically</em>, follow these steps to perform a convolution:</p>
<p style="padding-left: 30px;">1.) Choose one of the two funtions (<img src='https://s0.wp.com/latex.php?latex=h%28%5Ctau%29&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='h(\tau)' title='h(\tau)' class='latex' /> or <img src='https://s0.wp.com/latex.php?latex=f%28%5Ctau%29&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='f(\tau)' title='f(\tau)' class='latex' />), and leave it fixed in <img src='https://s0.wp.com/latex.php?latex=%5Ctau&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='\tau' title='\tau' class='latex' />-space.</p>
<p style="padding-left: 30px;">2.) Flip the <em>other </em>function vertically across the origin, so that it is <em>time-inverted</em>.</p>
<p style="padding-left: 30px;">3.) Shift the inverted signal through the <img src='https://s0.wp.com/latex.php?latex=%5Ctau+&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='\tau ' title='\tau ' class='latex' /> axis by <img src='https://s0.wp.com/latex.php?latex=t_0+&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='t_0 ' title='t_0 ' class='latex' /> seconds.&nbsp; Choose to shift the signal to the first &#8220;section&#8221; of the fixed function that is described by the same equation.&nbsp; The inverted signal (say, <img src='https://s0.wp.com/latex.php?latex=f%28-+%5Ctau%29+&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='f(- \tau) ' title='f(- \tau) ' class='latex' />), now shifted, represents <img src='https://s0.wp.com/latex.php?latex=f%28t_0+-+%5Ctau%29+&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='f(t_0 - \tau) ' title='f(t_0 - \tau) ' class='latex' />, which is basically a &#8220;freeze frame&#8221; of the output after the input signal has been fed to the LTI system for <img src='https://s0.wp.com/latex.php?latex=t_0&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='t_0' title='t_0' class='latex' /> seconds.</p>
<p style="padding-left: 30px;">4.) The integral of the two functions, after shifting the inverted function by <img src='https://s0.wp.com/latex.php?latex=t_0&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='t_0' title='t_0' class='latex' /> seconds, is the value of the convolution integral (i.e. output signal) at <img src='https://s0.wp.com/latex.php?latex=t+%3D+t_0&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='t = t_0' title='t = t_0' class='latex' />.</p>
<p style="padding-left: 30px;">5.) Repeat this procedure through all &#8220;sections&#8221; of the function fixed in <img src='https://s0.wp.com/latex.php?latex=%5Ctau&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='\tau' title='\tau' class='latex' />-space.&nbsp; By doing this, you can compute the value of the output at any time <img src='https://s0.wp.com/latex.php?latex=t&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='t' title='t' class='latex' />!</p>
<h3>Useful Properties</h3>
<p>&nbsp;</p>
<p>The following is a list of useful properties of the convolution integral that can help in developing an intuitive approach to solving problems:<span style="text-decoration: underline;"><br />
</span></p>
<p>1.) Commutative Property:</p>
<p style="padding-left: 30px;"><img src='https://s0.wp.com/latex.php?latex=f_1%28t%29%2Af_2%28t%29+%3D+f_2%28t%29%2Af_1%28t%29+&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='f_1(t)*f_2(t) = f_2(t)*f_1(t) ' title='f_1(t)*f_2(t) = f_2(t)*f_1(t) ' class='latex' /></p>
<p>2.) Distributive Property:</p>
<p style="padding-left: 30px;"><img src='https://s0.wp.com/latex.php?latex=f_1%28t%29%2A%5Bf_2%28t%29%2Bf_3%28t%29%5D+%3D+f_1%28t%29%2Af_2%28t%29+%2B+f_1%28t%29%2Af_3%28t%29&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='f_1(t)*[f_2(t)+f_3(t)] = f_1(t)*f_2(t) + f_1(t)*f_3(t)' title='f_1(t)*[f_2(t)+f_3(t)] = f_1(t)*f_2(t) + f_1(t)*f_3(t)' class='latex' /></p>
<p>3.) Associative Property:</p>
<p style="padding-left: 30px;"><img src='https://s0.wp.com/latex.php?latex=f_1%28t%29%2A%5Bf_2%28t%29%2Af_3%28t%29%5D+%3D+%5Bf_1%28t%29%2Af_2%28t%29%5D%2Af_3%28t%29&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='f_1(t)*[f_2(t)*f_3(t)] = [f_1(t)*f_2(t)]*f_3(t)' title='f_1(t)*[f_2(t)*f_3(t)] = [f_1(t)*f_2(t)]*f_3(t)' class='latex' /></p>
<p>4.) Shift Property:</p>
<p style="padding-left: 30px;">if <img src='https://s0.wp.com/latex.php?latex=f_1%28t%29%2Af_2%28t%29+%3D+c%28t%29&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='f_1(t)*f_2(t) = c(t)' title='f_1(t)*f_2(t) = c(t)' class='latex' /></p>
<p style="padding-left: 30px;">then <img src='https://s0.wp.com/latex.php?latex=f_1%28t-T_1%29%2Af_2%28t-T_2%29+%3D+c%28t-T_1-T_2%29&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='f_1(t-T_1)*f_2(t-T_2) = c(t-T_1-T_2)' title='f_1(t-T_1)*f_2(t-T_2) = c(t-T_1-T_2)' class='latex' /></p>
<p>5.) Convolution with an Impulse results in the original function:</p>
<p style="padding-left: 30px;"><img src='https://s0.wp.com/latex.php?latex=f%28t%29%2A+%5Cdelta+%28t%29+%3D+f%28t%29&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='f(t)* \delta (t) = f(t)' title='f(t)* \delta (t) = f(t)' class='latex' /> where <img src='https://s0.wp.com/latex.php?latex=%5Cdelta+%28t%29&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='\delta (t)' title='\delta (t)' class='latex' /> is the unit impulse function</p>
<p>6.) Width Property:</p>
<p style="padding-left: 30px;"><em>The convolution of a signal of duration </em><img src='https://s0.wp.com/latex.php?latex=T_1&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='T_1' title='T_1' class='latex' /><em> and a signal of duration </em><img src='https://s0.wp.com/latex.php?latex=T_2&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='T_2' title='T_2' class='latex' /> <em>will result in a signal of duration</em> <img src='https://s0.wp.com/latex.php?latex=T_3+%3D+T_1+%2B+T_2&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='T_3 = T_1 + T_2' title='T_3 = T_1 + T_2' class='latex' /></p>
<h3>Convolution Table</h3>
<p>Finally, here is a<a href="http://i.imgur.com/nTgs9.jpg"> Convolution Table</a> that can <em>greatly </em>reduce the difficulty in solving convolution integrals.</p>
<p>Thank you so much to <a href="https://engineersphere.com">Safa Khamis</a> @ Kansas State University for taking the time to write this tutorial for Engineersphere and the <a href="http://www.ieee.org/index.html">electrical engineering community</a>.</p>
<p>&nbsp;</p>
<p>&nbsp;</p><p>The post <a href="https://engineersphere.com/the-convolution-integral-explained">The Convolution Integral Explained</a> first appeared on <a href="https://engineersphere.com">Engineersphere.com</a>.</p>]]></content:encoded>
					
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			</item>
		<item>
		<title>Finding the Inverse of a Matrix</title>
		<link>https://engineersphere.com/finding-the-inverse-of-a-matrix</link>
					<comments>https://engineersphere.com/finding-the-inverse-of-a-matrix#respond</comments>
		
		<dc:creator><![CDATA[Jeff Schuler]]></dc:creator>
		<pubDate>Sun, 06 Mar 2011 03:26:46 +0000</pubDate>
				<category><![CDATA[Electrical Engineering Concepts]]></category>
		<category><![CDATA[matrix inversion]]></category>
		<category><![CDATA[matrix math]]></category>
		<category><![CDATA[solve matrices]]></category>
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					<description><![CDATA[<p>Matrix manipulations and properties Finding the inverse of a matrix is much more complex than finding the inverse of a number. All real numbers have an inverse (i.e. ). However, not all matrices have an inverse. There are several characteristics that allow us to visibly determine whether a matrix has an inverse but we will [&#8230;]</p>
<p>The post <a href="https://engineersphere.com/finding-the-inverse-of-a-matrix">Finding the Inverse of a Matrix</a> first appeared on <a href="https://engineersphere.com">Engineersphere.com</a>.</p>]]></description>
										<content:encoded><![CDATA[<h3>Matrix manipulations and properties</h3>
<p>Finding the inverse of a matrix is much more complex than finding the inverse of a number. All real numbers have an inverse (i.e. <img src='https://s0.wp.com/latex.php?latex=6%5E%7B-1%7D%3D+%5Cfrac%7B1%7D%7B6%7D+&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='6^{-1}= \frac{1}{6} ' title='6^{-1}= \frac{1}{6} ' class='latex' />). However, not all matrices have an inverse. There are several characteristics that allow us to visibly determine whether a matrix has an inverse but we will only focus on one. A matrix must be square (i.e. 2&#215;2, 3&#215;3, etc.) to have an inverse. Performing the following manipulations will be a waste of time if a matrix is not square. It is also important to know the inverse matrix property. Using my example above, <img src='https://s0.wp.com/latex.php?latex=6%5E%7B-1%7D+%2A+%5Cfrac%7B1%7D%7B6%7D+%3D+1+&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='6^{-1} * \frac{1}{6} = 1 ' title='6^{-1} * \frac{1}{6} = 1 ' class='latex' /> and similarly with matrices, <img src='https://s0.wp.com/latex.php?latex=A+%2A+A%5E%7B-1%7D+%3D+In+&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='A * A^{-1} = In ' title='A * A^{-1} = In ' class='latex' /> where In is the identity matrix (diagonal from top left to bottom right contains all 1&#8217;s, and everything else is 0) . We take advantage of this property when solving systems of matrices.</p>
<p>In words, the general algorithm for determining the existence of an inverse matrix is to manipulate the matrix into row reduced echelon form (rref). If the rref matrix is an identity matrix, then the inverse matrix exists. Hang on now, earlier I mentioned that there were other, visible characteristics that allow us to determine the existence of an inverse matrix, but now I&#8217;m asking you to perform a tedious process (without a calculator) with the same goal? Wouldn&#8217;t it be easier to first determine if finding the rref of the matrix is worthwhile? You&#8217;re right, except we are going to make a simple manipulation, and at the same time that we finish our rref process and determine that an inverse matrix exists, we will have found the inverse matrix! How do we do that? We will create an augmented matrix between our matrix in question, <img src='https://s0.wp.com/latex.php?latex=A&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='A' title='A' class='latex' />, and the appropriate identity matrix where the size of matrix <img src='https://s0.wp.com/latex.php?latex=A+&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='A ' title='A ' class='latex' /> is equal to the size of matrix <img src='https://s0.wp.com/latex.php?latex=In+&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='In ' title='In ' class='latex' />. We will perform the same rref process to the augmented matrix <img src='https://s0.wp.com/latex.php?latex=%7C+A+&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='| A ' title='| A ' class='latex' /> <img src='https://s0.wp.com/latex.php?latex=In+%7C+&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='In | ' title='In | ' class='latex' />. If the portion of our augmented matrix previously belonging to matrix <img src='https://s0.wp.com/latex.php?latex=A+&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='A ' title='A ' class='latex' /> reduces to an identity matrix (indicating the existence of <img src='https://s0.wp.com/latex.php?latex=A%5E%7B-1%7D+&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='A^{-1} ' title='A^{-1} ' class='latex' /> ), then the portion previously belonging to the identity matrix, will equal <img src='https://s0.wp.com/latex.php?latex=A%5E%7B-1%7D+&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='A^{-1} ' title='A^{-1} ' class='latex' />.</p>
<h3>Some matrix math</h3>
<p>Now, for the math&#8230;</p>
<p>Suppose we are asked to find the inverse of the following matrix:</p>
<img src='https://s0.wp.com/latex.php?latex=%5Cbegin%7Bbmatrix%7D1%263%263%5C%5C1%264%263%5C%5C1%263%264%5Cend%7Bbmatrix%7D+&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='\begin{bmatrix}1&amp;3&amp;3\\1&amp;4&amp;3\\1&amp;3&amp;4\end{bmatrix} ' title='\begin{bmatrix}1&amp;3&amp;3\\1&amp;4&amp;3\\1&amp;3&amp;4\end{bmatrix} ' class='latex' />
<p>First, we must set up the augmented matrix discussed above. Notice that I have simply placed the identity matrix (of the same size as <img src='https://s0.wp.com/latex.php?latex=A+&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='A ' title='A ' class='latex' /> ) on the right of matrix <img src='https://s0.wp.com/latex.php?latex=A+&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='A ' title='A ' class='latex' />.</p>
<img src='https://s0.wp.com/latex.php?latex=%5Cbegin%7Bbmatrix%7D1%263%263%261%260%260%5C%5C1%264%263%260%261%260%5C%5C1%263%264%260%260%261%5Cend%7Bbmatrix%7D+&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='\begin{bmatrix}1&amp;3&amp;3&amp;1&amp;0&amp;0\\1&amp;4&amp;3&amp;0&amp;1&amp;0\\1&amp;3&amp;4&amp;0&amp;0&amp;1\end{bmatrix} ' title='\begin{bmatrix}1&amp;3&amp;3&amp;1&amp;0&amp;0\\1&amp;4&amp;3&amp;0&amp;1&amp;0\\1&amp;3&amp;4&amp;0&amp;0&amp;1\end{bmatrix} ' class='latex' />
<h3>Finding the rref of an augmented matrix</h3>
<p>Next, we will attempt to find the rref of the augmented matrix. If the portion of the augmented matrix previously belonging to <img src='https://s0.wp.com/latex.php?latex=A+&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='A ' title='A ' class='latex' /> yields an identity matrix, <img src='https://s0.wp.com/latex.php?latex=A+&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='A ' title='A ' class='latex' /> is invertible.</p>
<p>rref <img src='https://s0.wp.com/latex.php?latex=%5Cbegin%7Bbmatrix%7D1%263%263%261%260%260%5C%5C1%264%263%260%261%260%5C%5C1%263%264%260%260%261%5Cend%7Bbmatrix%7D+&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='\begin{bmatrix}1&amp;3&amp;3&amp;1&amp;0&amp;0\\1&amp;4&amp;3&amp;0&amp;1&amp;0\\1&amp;3&amp;4&amp;0&amp;0&amp;1\end{bmatrix} ' title='\begin{bmatrix}1&amp;3&amp;3&amp;1&amp;0&amp;0\\1&amp;4&amp;3&amp;0&amp;1&amp;0\\1&amp;3&amp;4&amp;0&amp;0&amp;1\end{bmatrix} ' class='latex' /> = <img src='https://s0.wp.com/latex.php?latex=%5Cbegin%7Bbmatrix%7D1%260%260%267%26-3%26-3%5C%5C0%261%260%26-1%261%260%5C%5C0%260%261%26-1%260%261%5Cend%7Bbmatrix%7D+&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='\begin{bmatrix}1&amp;0&amp;0&amp;7&amp;-3&amp;-3\\0&amp;1&amp;0&amp;-1&amp;1&amp;0\\0&amp;0&amp;1&amp;-1&amp;0&amp;1\end{bmatrix} ' title='\begin{bmatrix}1&amp;0&amp;0&amp;7&amp;-3&amp;-3\\0&amp;1&amp;0&amp;-1&amp;1&amp;0\\0&amp;0&amp;1&amp;-1&amp;0&amp;1\end{bmatrix} ' class='latex' /></p>
<p>Ok great! The left half of our augmented matrix reduced to an identity matrix. That means two things to us: the matrix has an inverse <em>and</em> we&#8217;ve already found the inverse. If you recall from above, <img src='https://s0.wp.com/latex.php?latex=A%5E%7B-1%7D+&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='A^{-1} ' title='A^{-1} ' class='latex' /> is the right half of the augmented matrix (after finding it&#8217;s rref, of course). So we can conclude:</p>
<img src='https://s0.wp.com/latex.php?latex=%5Cbegin%7Bbmatrix%7D1%263%263%5C%5C1%264%263%5C%5C1%263%264%5Cend%7Bbmatrix%7D%5E%7B-1%7D+%3D+%5Cbegin%7Bbmatrix%7D7%26-3%26-3%5C%5C-1%261%260%5C%5C-1%260%261%5Cend%7Bbmatrix%7D+&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='\begin{bmatrix}1&amp;3&amp;3\\1&amp;4&amp;3\\1&amp;3&amp;4\end{bmatrix}^{-1} = \begin{bmatrix}7&amp;-3&amp;-3\\-1&amp;1&amp;0\\-1&amp;0&amp;1\end{bmatrix} ' title='\begin{bmatrix}1&amp;3&amp;3\\1&amp;4&amp;3\\1&amp;3&amp;4\end{bmatrix}^{-1} = \begin{bmatrix}7&amp;-3&amp;-3\\-1&amp;1&amp;0\\-1&amp;0&amp;1\end{bmatrix} ' class='latex' />
<p>If our rref of the augmented matrix had yielded anything other than an identity matrix, we would conclude that <img src='https://s0.wp.com/latex.php?latex=A%5E%7B-1%7D+&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='A^{-1} ' title='A^{-1} ' class='latex' /> does not exist. This method will simply allow us to determine the existence of and entries to <img src='https://s0.wp.com/latex.php?latex=A%5E%7B-1%7D+&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='A^{-1} ' title='A^{-1} ' class='latex' /> for any size matrix.</p><p>The post <a href="https://engineersphere.com/finding-the-inverse-of-a-matrix">Finding the Inverse of a Matrix</a> first appeared on <a href="https://engineersphere.com">Engineersphere.com</a>.</p>]]></content:encoded>
					
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		<title>Solving a Linear System Using the Inverse Matrix</title>
		<link>https://engineersphere.com/solving-a-linear-system-using-the-inverse-matrix</link>
					<comments>https://engineersphere.com/solving-a-linear-system-using-the-inverse-matrix#respond</comments>
		
		<dc:creator><![CDATA[Jeff Schuler]]></dc:creator>
		<pubDate>Thu, 03 Mar 2011 22:49:47 +0000</pubDate>
				<category><![CDATA[Electrical Engineering Concepts]]></category>
		<category><![CDATA[coefficient matrix]]></category>
		<category><![CDATA[commutative property]]></category>
		<category><![CDATA[inverse matrix math]]></category>
		<category><![CDATA[linear systems]]></category>
		<category><![CDATA[solving linear systems]]></category>
		<guid isPermaLink="false">https://engineersphere.com/?p=2009</guid>

					<description><![CDATA[<p>Describing the process of solving a linear system using the adjacent matrix is best done while performing an example. Suppose we have a system where is the coefficient matrix of our system, is the column vector containing our variables, and is the solution column vector. We are asked to solve for the column vector made [&#8230;]</p>
<p>The post <a href="https://engineersphere.com/solving-a-linear-system-using-the-inverse-matrix">Solving a Linear System Using the Inverse Matrix</a> first appeared on <a href="https://engineersphere.com">Engineersphere.com</a>.</p>]]></description>
										<content:encoded><![CDATA[<p>Describing the process of solving a linear system using the adjacent matrix is best done while performing an example. Suppose we have a system <img src='https://s0.wp.com/latex.php?latex=A%2Ax+%3D+B+&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='A*x = B ' title='A*x = B ' class='latex' /> where <img src='https://s0.wp.com/latex.php?latex=A+&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='A ' title='A ' class='latex' /> is the <a href="http://en.wikipedia.org/wiki/Coefficient_matrix">coefficient matrix</a> of our system, <img src='https://s0.wp.com/latex.php?latex=x+&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='x ' title='x ' class='latex' /> is the column vector containing our variables, and <img src='https://s0.wp.com/latex.php?latex=B+&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='B ' title='B ' class='latex' /> is the solution column vector. We are asked to solve for the column vector <img src='https://s0.wp.com/latex.php?latex=x+&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='x ' title='x ' class='latex' /> made up of variables <img src='https://s0.wp.com/latex.php?latex=x_1&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='x_1' title='x_1' class='latex' />, <img src='https://s0.wp.com/latex.php?latex=x_2&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='x_2' title='x_2' class='latex' />, and <img src='https://s0.wp.com/latex.php?latex=x_3&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='x_3' title='x_3' class='latex' />.</p>
<img src='https://s0.wp.com/latex.php?latex=%5Cbegin%7Bbmatrix%7D1%263%263%5C%5C1%264%263%5C%5C1%263%264%5Cend%7Bbmatrix%7D+%5Cbegin%7Bbmatrix%7D+x_1%5C%5Cx_2%5C%5Cx_3%5Cend%7Bbmatrix%7D+%3D+%5Cbegin%7Bbmatrix%7D+12%5C%5C-10%5C%5C16%5Cend%7Bbmatrix%7D&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='\begin{bmatrix}1&amp;3&amp;3\\1&amp;4&amp;3\\1&amp;3&amp;4\end{bmatrix} \begin{bmatrix} x_1\\x_2\\x_3\end{bmatrix} = \begin{bmatrix} 12\\-10\\16\end{bmatrix}' title='\begin{bmatrix}1&amp;3&amp;3\\1&amp;4&amp;3\\1&amp;3&amp;4\end{bmatrix} \begin{bmatrix} x_1\\x_2\\x_3\end{bmatrix} = \begin{bmatrix} 12\\-10\\16\end{bmatrix}' class='latex' />
<p>Typically, we would divide <img src='https://s0.wp.com/latex.php?latex=B+&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='B ' title='B ' class='latex' /> by <img src='https://s0.wp.com/latex.php?latex=A+&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='A ' title='A ' class='latex' /> to solve for <img src='https://s0.wp.com/latex.php?latex=x+&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='x ' title='x ' class='latex' />, however there is no method for performing division between matrices. By taking advantage of the inverse matrix property <img src='https://s0.wp.com/latex.php?latex=A%5E%7B-1%7D%2AA+%3D+1+&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='A^{-1}*A = 1 ' title='A^{-1}*A = 1 ' class='latex' />, we can simply the formula to solve for the column vector <img src='https://s0.wp.com/latex.php?latex=x+&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='x ' title='x ' class='latex' />. The <a href="http://en.wikipedia.org/wiki/Commutativity">commutative property</a> does not apply in matrix multiplication so <img src='https://s0.wp.com/latex.php?latex=A%5E%7B-1%7D%2AB+%5Cnot%3D+B%2AA%5E%7B-1%7D&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='A^{-1}*B \not= B*A^{-1}' title='A^{-1}*B \not= B*A^{-1}' class='latex' />.&nbsp; <em>Therefore we have have to be aware of the &#8216;order&#8217; in which we multiply</em>:</p>
<p><img src='https://s0.wp.com/latex.php?latex=%28A%5E%7B-1%7D%29+%2A+A+%2A+x+%3D+%28A%5E%7B-1%7D%29+%2A+B+&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='(A^{-1}) * A * x = (A^{-1}) * B ' title='(A^{-1}) * A * x = (A^{-1}) * B ' class='latex' />&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; simplifies to&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; <img src='https://s0.wp.com/latex.php?latex=x+%3D+A%5E%7B-1%7D%2AB+&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='x = A^{-1}*B ' title='x = A^{-1}*B ' class='latex' /></p>
<p>Notice that since we multiplied by <img src='https://s0.wp.com/latex.php?latex=A%5E%7B-1%7D+&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='A^{-1} ' title='A^{-1} ' class='latex' /> &#8216;first&#8217; on the left side of the equation, we also multiply &#8216;first&#8217; on the right side. Now, multiplying the inverse of matrix <img src='https://s0.wp.com/latex.php?latex=A+&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='A ' title='A ' class='latex' /> by matrix <img src='https://s0.wp.com/latex.php?latex=B+&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='B ' title='B ' class='latex' /> will yield a column vector matching our <img src='https://s0.wp.com/latex.php?latex=x_1&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='x_1' title='x_1' class='latex' />, <img src='https://s0.wp.com/latex.php?latex=x_2&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='x_2' title='x_2' class='latex' />, and <img src='https://s0.wp.com/latex.php?latex=x_3&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='x_3' title='x_3' class='latex' />. Below, I have used the equation <img src='https://s0.wp.com/latex.php?latex=x+%3D+A%5E%7B-1%7D%2AB+&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='x = A^{-1}*B ' title='x = A^{-1}*B ' class='latex' /> and plugged the values for <img src='https://s0.wp.com/latex.php?latex=A%5E%7B-1%7D+&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='A^{-1} ' title='A^{-1} ' class='latex' /> into the equation. The product between <img src='https://s0.wp.com/latex.php?latex=A%5E%7B-1%7D+&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='A^{-1} ' title='A^{-1} ' class='latex' /> and <img src='https://s0.wp.com/latex.php?latex=B+&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='B ' title='B ' class='latex' /> is shown on the far right. Note: This article assumes you know how to find the inverse of a matrix. This process is described in my article <a href="https://engineersphere.com/finding-the-inverse-of-a-matrix">Finding The Inverse of a Matrix</a>.</p>
<img src='https://s0.wp.com/latex.php?latex=%5Cbegin%7Bbmatrix%7D+x_1%5C%5Cx_2%5C%5Cx_3%5Cend%7Bbmatrix%7D+%3D+%5Cbegin%7Bbmatrix%7D7%26-3%26-3%5C%5C-1%261%260%5C%5C-1%260%261%5Cend%7Bbmatrix%7D+%5Cbegin%7Bbmatrix%7D+12%5C%5C-10%5C%5C16%5Cend%7Bbmatrix%7D+%3D+%5Cbegin%7Bbmatrix%7D+66%5C%5C-22%5C%5C4%5Cend%7Bbmatrix%7D&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='\begin{bmatrix} x_1\\x_2\\x_3\end{bmatrix} = \begin{bmatrix}7&amp;-3&amp;-3\\-1&amp;1&amp;0\\-1&amp;0&amp;1\end{bmatrix} \begin{bmatrix} 12\\-10\\16\end{bmatrix} = \begin{bmatrix} 66\\-22\\4\end{bmatrix}' title='\begin{bmatrix} x_1\\x_2\\x_3\end{bmatrix} = \begin{bmatrix}7&amp;-3&amp;-3\\-1&amp;1&amp;0\\-1&amp;0&amp;1\end{bmatrix} \begin{bmatrix} 12\\-10\\16\end{bmatrix} = \begin{bmatrix} 66\\-22\\4\end{bmatrix}' class='latex' />
<p>Therefore, <img src='https://s0.wp.com/latex.php?latex=x_1%3D66&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='x_1=66' title='x_1=66' class='latex' />, <img src='https://s0.wp.com/latex.php?latex=x_2%3D-22&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='x_2=-22' title='x_2=-22' class='latex' />, and <img src='https://s0.wp.com/latex.php?latex=x_3%3D4&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='x_3=4' title='x_3=4' class='latex' />. Simple systems (i.e. this 3&#215;3 system) are much easier to solve with algebra instead of finding the inverse of the coefficient matrix and performing matrix multiplication. This application is more practical for larger systems or while working on Matrix Theory homework.</p>
<p><strong>Please leave comments by <a href="https://engineersphere.com/wp-login.php">signing in</a> and then clicking on the &#8220;sticky note&#8221; located in the top right corner of this post to show your appreciation to the author!</strong></p><p>The post <a href="https://engineersphere.com/solving-a-linear-system-using-the-inverse-matrix">Solving a Linear System Using the Inverse Matrix</a> first appeared on <a href="https://engineersphere.com">Engineersphere.com</a>.</p>]]></content:encoded>
					
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