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 nonuniform 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 nonuniform 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: Nonuniform quantizer has has nonlinear characteristics while the stepsize consistently keeps on varying & depends on the amplitude of input signal. This process of nonuniform 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. R_{XY} , (t_{1}, t_{2} ) = 0 for every t_{1} and t_{2}
b. R_{XY} , (t_{1}, t_{2} ) = 1 for every t_{1} and t_{2}
c. R_{XY} = 0, (t_{1}, t_{2} ) = 1 for t_{1} and t_{2} instants respectively
d. R_{XY } = 1 , (t_{1}, t_{2} ) = 0 for t_{1} and t_{2} instants respectively
Answer
Explanation

ANSWER: R_{XY} , (t_{1}, t_{2} ) = 0 for every t_{1} and t_{2}
Explanation: Two random X(t) & Y(t) processes are usually said to be mutually orthogonal only and only if R_{XY}, (t_{1},t_{2}) becomes exactly equal to zero; where t_{1} & t_{2} 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 10^{6t} + 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) rad Deviation ratio = D = [10 x ( 2π 200)] / [( 2π 200)] Therefore, D = 10
Bandwidth of Frequency Modulated Signal is given by, B = 2 (D + 1) f_{m} = 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 : I_{F} = 350 kHz Carrier Frequency = f_{c} = 1200 kHz
We know that, Image Frequency = 2 I_{F} + f_{c} = 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 m_{x} (t) is independent of time and the autocorrelation function Rx (t_{k}, t_{i}) depends only on the time difference (t_{k}  t_{i}). 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 (f_{x(t)} (x)] which are derived from ensemble, then it is said to be an ergodic process.

