ECE

EXAMPLE ON SUPER HETERODYNE RECEIVER

noreply@blogger.com (Anonymous) — Fri, 21 Jun 2013 09:01:00 +0530

In a broad cast super hetero dyne receiver having no RF amplifier, the load Q of the antenna coupling circuit is 100. If the IF frequency is 455 KHz, determine

A) The image frequency and its rejection ratio for tuning at 1.1 KHz station.

B) The image frequency and its rejection ratio for tuning at 25 MHz station.

Solution

Given IF frequency is 455 KHz

Quality factor Q = 100

a) f_S = 1.1 KHz

Image frequency f_Si = f_S + 2 f_i

= 1100 +( 2 * 455 )

= 2010 KHz

ρ = (f_Si/f_S) – (f_S/f_Si)

= (2010/1100) – (1100/2010)

=1.28

Image rejection ratio α = √( 1+ Q² ρ² )

= √( 1+ 100² 1.28² )

= 128

b) f_S = 25 MHz

Image frequency f_Si = f_S + 2 f_i

= 25 MHz +( 2 * 455 )

= 25.91 MHz

ρ = (f_Si/f_S) – (f_S/f_Si)

= (25.91 /25) – (25/25.91)

= 0.0715

Image rejection ratio α = √( 1+ Q² ρ² )

= √( 1+ 100² 0.0715² )

= 5.22

Thus rejection is poor for practical receiver in HF band

So, we can say that it is not a major problem for standard AM receiver without an RF stage, but it causes problem at high frequencies using same intermediate frequency.

We can eliminate this problem by introducing RF stage or by increasing intermediate frequency

SUPER HETERODYNE RECEIVER

noreply@blogger.com (Anonymous) — Thu, 20 Jun 2013 09:29:00 +0530

RF SECTION

The signal from the antenna is tuned and may be amplified in a so-called radio frequency (RF) amplifier, although this stage is often omitted. One or more tuned circuits at this stage block frequencies which are far removed from the intended reception frequency. In order to tune the receiver to a particular station, the frequency of the local oscillator is controlled by the tuning knob (for instance). Tuning of the local oscillator and the RF stage may use a variable capacitor, or varicap diode. The tuning of one (or more) tuned circuits in the RF stage must track the tuning of the local oscillator.

MIXER/CONVERTER SECTION

The signal is then fed into a circuit where it is mixed with a sine wave from a variable frequency oscillator known as the local oscillator (LO). The mixer uses a non-linear component to produce both sum and difference beat frequencies signals, each one containing the modulation contained in the desired signal. The output of the mixer may include the original RF signal at f_d, the local oscillator signal at f_LO, and the two new frequencies f_d+f_LO and f_d-f_LO. The mixer may inadvertently produce additional frequencies such as 3rd- and higher-order inter-modulation products. The undesired signals are removed by the IF band pass filter, leaving only the desired offset IF signal at f_IF which contains the original modulation (transmitted information) as the received radio signal had at f_d.

INTERMEDIATES FREQUENCY AMPLIFIER

Like the RF amplifier, the IF amplifier must effectively amplify the incoming signal. In the case of Intermediate Frequencies, amplification and filtering are often utilized simultaneously. It is also not uncommon for several stages of IF amplifiers to used in succession. The amplified IF output then proceeds to Limiter, Automatic Gain Control, and Discriminator circuitry.

WHY IF IS USED

Intermediate frequencies are used for three general reasons. At very high (gigahertz) frequencies, signal processing circuitry performs poorly. Active devices such as transistors cannot deliver much amplification (gain) Ordinary circuits using capacitors and inductors must be replaced with cumbersome high frequency techniques such as strip lines and waveguides. So a high frequency signal is converted to a lower IF for more convenient processing. For example, in satellite television receivers, converting the microwave down-link signal to a much lower intermediate frequency at the dish allows a relatively inexpensive coaxial cable to carry the signal to the rest of the receiver inside the building.

A second reason, in receivers that can be tuned to different frequencies, is to convert the various different frequencies of the stations to a common frequency for processing. It is difficult to build amplifiers, filters, and detectors that can be tuned to different frequencies, but easy to build tunable oscillators. Super hetero dyne receivers tune in different frequencies by adjusting the frequency of the local oscillator on the input stage, and all processing after that is done at the same fixed frequency, the IF. Without using an IF, all the complicated filters and detectors in a radio or television would have to be tuned in unison each time the frequency was changed, as was necessary in the early tuned radio frequency receivers.

The main reason for using an intermediate frequency is to improve frequency selectivity. In communication circuits, a very common task is to separate out or extract signals or components of a signal that are close together in frequency. This is called filtering. Some examples are, picking up a radio station among several that are close in frequency, or extracting the chrominance sub carrier from a TV signal. With all known filtering techniques the filter's bandwidth increases proportionately with the frequency. So a narrower bandwidth and more selectivity can be achieved by converting the signal to a lower IF and performing the filtering at that frequency.

DETECTOR/DEMODULATOR SECTION

The received signal is now processed by the demodulator stage where the audio signal (or other baseband signal) is recovered and then further amplified. AM demodulation requires the simple rectification of the RF signal (so-called envelope detection), and a simple RC low pass filter to remove remnants of the intermediate frequency.FM signals may be detected using a discriminator, ratio detector, or phase-locked loop. CW (Morse code) and single sideband signals require a product detector using a so-called beat frequency oscillator, and there are other techniques used for different types of modulation. The resulting audio signal (for instance) is then amplified and drives a loudspeaker.

When so-called high-side injection has been used, where the local oscillator is at a higher frequency than the received signal (as is common), then the frequency spectrum of the original signal will be reversed. This must be taken into account by the demodulator (and in the IF filtering) in the case of certain types of modulation such as single sideband.

COMMONLY USED INTERMEDIATE FREQUENCIES

1)110 kHz was used in Long wave broadcast receivers

2) Analogue television receivers using system M: 41.25 MHz (audio) and 45.75 MHz (video). Note, the channel is flipped over in the conversion process in an inter carrier system, so the audio IF frequency is lower than the video IF frequency. Also, there is no audio local oscillator; the injected video carrier serves that purpose.

3) Analogue television receivers using system B and similar systems: 33.4 MHz. for aural and 38.9 MHz. for visual signal.(The discussion about the frequency conversion is the same as in system M)

4) FM radio receivers: 262 kHz, 455 kHz, 1.6 MHz, 5.5 MHz, 10.7 MHz, 10.8 MHz, 11.2 MHz, 11.7 MHz, 11.8 MHz, 21.4 MHz, 75 MHz and 98 MHz. In double-conversion super hetero dyne receivers, a first intermediate frequency of 10.7 MHz is often used, followed by a second intermediate frequency of 470 kHz. There are triple conversion designs used in police scanner receivers, high-end communications receivers, and many point-to-point microwave systems.

5)AM radio receivers: 450 kHz, 455 kHz, 460 kHz, 465 kHz, 470 kHz, 475 kHz, 480 kHz

6) Satellite unlink-downlink equipment: 70 MHz, 950-1450 Downlink first IF

7) Terrestrial microwave equipment: 250 MHz, 70 MHz or 75 MHz

8) Radar: 30 MHz

WAVE POLARIZATION

noreply@blogger.com (Anonymous) — Sat, 25 May 2013 08:31:00 +0530

Consider a wave propagating in Z direction. So the components present are E_X,E_Y,H_X,H_Y and the components E_Z = H_Z = 0

Polarization is not present in the direction of wave propagation

Consider E = E₁ cos (ωt-βz )a_X + E₂ cos (ωt-βz+φ )a_Y

The wave is propagating is in positive Z direction

Where φ indicates phase difference between x and y components of the field.

If φ = 0 then the wave is linearly polarized

If φ = 90, E₁ = E₂, then the wave is circularly polarized

If φ = 90, E₁ ≠ E₂, then the wave is elliptically polarized

If φ = some phase difference, E₁ ≠ E₂, then the wave is elliptically polarized with an inclination equal to phase difference

Example 1

E = sin (ωt-βz )a_X + sin (ωt-βz+90⁰ )a_Y

Solution

As there is a phase difference of 90 and E₁ = E₂, we can say it is circularly polarized

How to identify whether it is left or right handed.

First fix the orientation point by keeping propagation direction = 0 ie., Z= 0

Now substitute ωt =0,90,180,270,360

When ωt = 0

E = a_Y

When ωt = 90

E = a_X

When ωt = 180

E = - a_Y

When ωt = 270

E = - a_X

Now point your thumb finger in the direction of propagation and close your fingers according to the components appear on the graph, which hand suits the closing fingers with the direction of propagation gives left hand or right hand.

Positive Z direction upwards, negative Z direction downwards.

For the obtained graph left hand suits. So it is left circularly polarized.

Example 2

E = ( a_X + 2 e^j(∏/2) a_Y) e^{j(ωt+βz )}

Solution

Expand e^jθ = cos θ + jsinθ

E =( a_X + 2 ( cos 90 + jsin 90 ) a_Y) *( cos (ωt+βz ) + j sin ( ωt+βz )

E =( a_X + 2 j a_Y) *( cos (ωt+βz ) + j sin ( ωt+βz )

Know take real part of E

E = ( cos (ωt+βz ) a_X – 2 sin ( ωt+βz ) a_Y

Comparing with general expression of E

We can say that wave is going in negative Z direction

First fix the orientation point by keeping propagation direction = 0 ie., Z= 0

Now substitute ωt =0,90,180,270,360

When ωt = 0

E = a_Y

When ωt = 90

E = -2 a_X

When ωt = 180

E = - a_Y

When ωt = 270

E = + 2 a_X

Draw the graph as above example

For the obtained graph left hand suits. So it is left elliptically polarized.

Example 3

E = sin (ωt+βx)( aY + jaZ)

Solution

E = sin (ωt+βx) a_Y + sin (ωt+βx + 90)a_Z

We can say that wave is going in negative X direction

When ωt = 0

E = a_Z

When ωt = 90

E = a_Y

When ωt = 180

E = - a_Z

When ωt = 270

E = - a_Y

Draw the graph as above example

For the obtained graph left hand suits. So it is right circularly polarized.