# Communication System - GATE practice papers - Electronics & Communication

Here, you can read Communication System multiple choice questions and answers with explanation.

1)   How is the process of non-uniform quantization equivalently & combinationally  performed?

A. Passage of baseband signal through the compressor
B. Application of compressed signal to a uniform quantizer
C. Passage of passband signal through the compressor
D. Application of compressed signal to a non-uniform quantizer
(Marks : 02)

- Published on 19 Oct 15

a. A & B
b. C & D
c. A & D
d. C & B
 Answer  Explanation ANSWER: A & B Explanation: Non-uniform quantizer has has non-linear characteristics while the step-size consistently keeps on varying & depends on the amplitude of input signal. This process of non-uniform quantization resemble to the mechanisms of passing the baseband signal through compressor and then eventually applying the compressed signal to a uniform quantizer.

2)   Which conditions justify the mutual orthogonality of two random signals X(t) & Y(t)?
(Marks : 02)

- Published on 19 Oct 15

a. RXY , (t1, t2 ) = 0 for every t1 and t2
b. RXY , (t1, t2 ) = 1 for every t1 and t2
c. RXY = 0, (t1, t2 ) = 1 for t1 and t2 instants respectively
d. RXY = 1 , (t1, t2 ) = 0 for t1 and t2 instants respectively
 Answer  Explanation ANSWER: RXY , (t1, t2 ) = 0 for every t1 and t2 Explanation: Two random X(t) & Y(t) processes are usually said to be mutually orthogonal only and only if RXY, (t1,t2) becomes exactly equal to zero; wheret1 & t2 are the two time instants at which the processes are observed.

3)   A signal m(t) = 20cos (2π200t) along with carrier generates an FM signal x(t) = 40cos {2π x 106t  + 10 sin (2π 200t)}. What will be the value of approximate bandwidth of the frequency modulation? (Marks : 02)
- Published on 19 Oct 15

a. 200 kHz
b. 2 kHz
c. 4.4 kHz
d. None of these
 Answer  Explanation ANSWER: 4.4 kHz Explanation: Given data : Frequency Deviation = Δf = 10 x ( 2π 200) radDeviation ratio = D = [10 x ( 2π 200)] / [( 2π 200)]Therefore, D = 10 Bandwidth of Frequency Modulated Signal is given by, B = 2 (D + 1) fm = 2 (10 +1 ) x 200 = 2 (11) x 200 = 22 x 200 = 4400 = 4.4 kHz

4)   Assume that super heterodyne receiver consumes an IF frequency of 350 kHz by high side tuning to a transmitter with a carrier frequency of 1200 kHz . What would be the value of an image frequency ? (Marks : 01)
- Published on 19 Oct 15

a. 1550kHz
b. 8400 kHz
c. 1900 kHz
d. 1000kHz
 Answer  Explanation ANSWER: 1900 kHz Explanation: Given data :IF = 350 kHzCarrier Frequency = fc = 1200 kHzWe know that,Image Frequency = 2 IF + fc = 2x 350 + 1200 = 1900 kHz

5)   Consider the statements given below:

A. All SSS (Stationary in Strict Sense) processes are also WSS (Stationary in Wide Sense)

B. All the processes that are WSS  (Stationary in Wide Sense) are also absolutely  SSS (Stationary in Strict Sense)

Which of them are correct? (Marks : 01)

- Published on 19 Oct 15

a. A is true & B is false
b. A is false & B is true
c. Both A & B are true
d. Both A & B are false
 Answer  Explanation ANSWER: A is true & B is false Explanation: A random process is said to be Stationary in Strict Sense (SSS) if the joint probability distribution factor remains invariant to the translation of time origin. On the contrary, it is said to be Stationary in Wide Sense if mean value mx (t) is independent of time and the autocorrelation function Rx (tk, ti) depends only on the time difference (tk - ti). Hence, it is obvious that all SSS are definitely WSS but all WSS are not necessarily SSS.

6)   When can a random process is said to be an ergodic process? (Marks : 01)
- Published on 19 Oct 15

a. Only when time averages are less than the ensemble averages
b. Only when time averages are equal to the ensemble averages
c. Only when time averages are greater than the ensemble averages
d. none of the above
 Answer  Explanation ANSWER: Only when time averages are equal to the ensemble averages Explanation: Random Processes can be generally and completely specified on the basis of ensemble and the time statistics since only PDFs are insufficient to describe them completely. So, any random process possessing the time averages exactly equal to averages [Probability Distribution Factor PDF (fx(t) (x)] which are derived from ensemble, then it is said to be an ergodic process.

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