Analog and Digital Communication Viva Questions

Analog and Digital Communication Viva Questions

AM Generation:

1. Experimentally find out the side band power that would be delivered to an antenna of 600 Ω characteristic impedance when an AM wave is fed to it.

Use the expression PSB = m2Vc2 /4R where,
PSB = power of two side bands
Vc = carrier amplitude of AM
m = modulation index
R = impedance

2. Generate an AM signal with 30 % modulation. Modulating signal is given.

Select the amplitude of carrier signal and modulating signal in such a way that the amplitude of the modulating signal is 30% of that of carrier signal.

3. Draw the frequency spectrum of an AM signal obtained by modulating a sinusoidal signal by another sinusoidal signal.

4. Write the expression for power of an AM signal.

Pt = Pc (1 + m2/2) where Pt is the power of the AM signal, Pc is the power of the carrier signal and m is the modulation index.

5. Compare AM-DSB, AM-SC and AM-SSB.

AM-DSB is the system in which both the side-Bands are transmitted. SC stands for suppressed carrier. In AM-SC system both sidebands only are transmitted. In AM-SSB system, carrier and a single side-hand is transmitted. In this case, power can be saved but the receiver circuit will become complex because the carrier and two sidebands must be regenerated in the receiver. This can be achieved using a product detector or by sending a pilot carrier from transmitter to receiver along with the transmitted signal.

6. Write the expression for resultant modulation index in terms of individual modulation indices.

mt =   (m12 + m22 + ….)

7. What is VSB? Where is it used?

VSB stands for vestigial side band. One side band and a part of other side band is transmitted in this system. This transmission system is used in TV transmission.

8. Discuss band width requirements of AM and FM?

Band width of AM signal is twice the modulating signal frequency while that of FM system is twice the sum of frequency deviation and modulating signal frequency.
i.e., BWAM = 2fm and BWFM = 2(δ + fm)

9. Discuss high level and low level modulation systems.

In high level modulation system, modulation is done at the higher power level of the carrier signal while in the low level modulation system, it is done at low power level. Emitter and collector modulations are the examples of high level modulation and base modulation is an example of low level modulation. Low level modulation has an advantage of low power requirement for the modulating signal.

10. What is the AM broadcast spectrum?

540 kHz to 1650 kHz

11. What are the uses of an RF amplifier?

RFA provides the tuning to the desired signal frequency. It provides high gain and high signal to noise ratio. It avoids the reradiation of the local oscillator signal through antenna.

IF Tuned Amplifier:

12. Why BF194 transistor is used instead of BC107 in this circuit?

BF194 is meant for high frequency applications like AM and FM generation.

13. What are the criteria in choosing intermediate frequency?

Following facts are considered in choosing IF:

(a) Very high value for IF will cause poor selectivity and adjacent channel rejection.
(b) If IF is too low, image signal rejection will be poor.
(c) If IF is very low, frequency stability of local oscillator must be maintained very high.
(d) High value of IF will cause tracking difficulties.
(e) IF must not be in the broadcast spectrum to avoid direct pick up.

14. What is image signal frequency?

Image signal frequency is defined as fsi = fo+fi where fsi = image signal frequency, fo = local oscillator frequency and fi = intermediate frequency. It can also be defined as fsi = fs + 2fi: since fo = fs + fi where fs = signal/station frequency.

15. Why is image signal frequency called so?

Super heterodyne receiver receives the station frequency fs= (fofi) and generates intermediate frequency fi by mixing with local oscillator frequency fo. If another frequency fo + fi manages to enter the receiver, it will also generate a frequency fi. Though the frequencies are same in both the cases, the demodulated signals are different. The net effect is the reception of two transmitter station frequencies at a time.


16. How PWM is generated? How is it demodulated?

PWM is generated using monostable multivibrator, Trigger is applied to control the starting time of pulses and modulating signal is fed to control the duration of pulses. Using an integrator, original signal can be detected from PWM signal.

17. How is PPM generated? How is it demodulated?

PPM can be generated from PWM by differentiating it and rectifying the positive going pulses. To demodulate PPM, PPM must he converted first to PWM using a bistable multivibrator. Trigger pulses sent to the receiver as pilot carrier will initiate the pulses and width of the pulses will be controlled by PPM pulses. The resulting PWM is integrated to get original signal.

18. Prove sampling theorem experimentally.

Shannon-Nyquist sampling theorem demands that the sampling frequency should be greater than the double of the input signal frequency. Set up a PAM circuit and demodulator circuit, Use 1 kHz sine wave as input signal and > 2 kHz square wave as sampling signal. The demodulated wave will be original signal. If the square wave frequency is less than 2 KHz, the quality of the demodulated output will be poor.

Delta Modulation:

19. What is adaptive delta modulation?

In adaptive delta modulation, step size is automatically varied according to the slope of m(t) to avoid slope overload error.

20. How the DM can be demodulated?

Using an integrator

First Order low pass filter

21. Compare the response of Butterworth and Chebyshev filters.

Butterworth filter has fiat pass band but the roll off rate in the attenuation hand is not as steep as that of chehyshev filter. Chebyshev filter has rippled pass band and steeper roll off.

22. What dB/decade and dB/octave mean '?

dB/decade gives the change of gain in dB for a tenfold rise in frequency. dB/octave is the change in dB for a two fold rise in frequency. For instance, - 20dB/decade means that the gain falls by 20 dB when frequency increases from 10 kHz to 100 kHz. 20 dB/decade is equivalent to 6 dB/octave. -6dB/octave means a decrease in gain by 6 dB if frequency increases from 2 kHz to 4 kHz.

23. What are the advantages of active filters over passive filters '?

In active filters, inductances which are bulky can be eliminated. Also they provide gain.

24. What is the frequency scaling?

It is the process of converting the original cut off frequency to a new value by changing the values of resistances by multiplying or dividing by a factor instead of redesigning the circuit.

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