Communications - Online Test
Q1.
For a message signal 
and carrier of frequency fc which of thefollowing represents a single side-band (SSB) signal ?
Answer : Option C
Explaination / Solution:
Q2. The Nyquist sampling rate for the signal 
is given by
is given by
Answer : Option C
Explaination / Solution:
S(t) = sinc(500t) sinc(700t)
S(f ) is convolution of two signals whose spectrum covers f1 = 250 Hz and f2 = 350 Hz. So convolution extends
f = 25 + 350 = 600 Hz
Nyquist sampling rate
N = 2f = 2#600 = 1200 Hz
Q3. X(t) is a stationary process with the power spectral density Sx(f)>0, for all f . The process is passed through a system shown below
Let Sy (f ) be the power spectral density of Y(t). Which one of the following statements is correct
Answer : Option D
Explaination / Solution:
Q4. 
In the circuit shown, all the transmission line sections are lossless. The Voltage Standing Wave Ration(VSWR) on the 60 Ω line is
Answer : Option B
Explaination / Solution:
Q5. 
Consider a baseband binary PAM receiver shown below. The additive channel noise n(t) is with power spectral density 
The lowpass filter is ideal with unity gain and cut-off frequency 1 MHz. Let Yk represent the random variable y(tk).
The lowpass filter is ideal with unity gain and cut-off frequency 1 MHz. Let Yk represent the random variable y(tk).Yk = Nk , if transmitted bit bk = 0
Yk = a + Nk if transmitted bit bk = 1
Where Nk represents the noise sample value. The noise sample has a probability density function, 
(This has mean zero and variance 2/α2). Assume transmitted bits to be equiprobable and threshold z is set to a/2
= 10-6 V.
(This has mean zero and variance 2/α2). Assume transmitted bits to be equiprobable and threshold z is set to a/2
= 10-6 V.The value of the parameter α
(in V-1) is
Answer : Option B
Explaination / Solution:
Let response of LPF filters
Noise variance (power) is given as
Q6.
Consider a baseband binary PAM receiver shown below. The additive channel noise n(t) is with power spectral density The lowpass filter is ideal with unity gain and cut-off frequency 1 MHz. Let Yk represent the random variable y(tk).
The lowpass filter is ideal with unity gain and cut-off frequency 1 MHz. Let Yk represent the random variable y(tk).Yk = Nk , if transmitted bit bk = 0
Yk = a + Nk if transmitted bit bk = 1
Where Nk represents the noise sample value. The noise sample has a probability density function, (This has mean zero and variance 2/α2). Assume transmitted bits to be equiprobable and threshold z is set to a/2 = 10-6 V.
(This has mean zero and variance 2/α2). Assume transmitted bits to be equiprobable and threshold z is set to a/2 = 10-6 V.The probability of bit error is
Answer : Option D
Explaination / Solution:
Probability of error is given by
Q7. In a digital communication system, the overall pulse shape p(t) at the receiver before the sampler has the Fourier transform P(f). If the symbols are transmitted at the rate of 2000 symbols per second, for which of the following cases is inter symbol interference zero?
Answer : Option B
Explaination / Solution:
For ISI free pulse,
If P(t) is having spectrum P(f)
Then 
= constant
= constantRs = 2KSpa
Thin condition is met by pulse given in option B.
Q8. Which one of the following statements about differential pulse code modulation (DPCM) is true?
Answer : Option D
Explaination / Solution:
DPCM Block diagram
eq[n] is quantized e[n]
e[n] is difference of message signal sample with its prediction.
Q9. Consider a wireless communication link between a transmitter and a receiver located in free space, with finite and strictly positive capacity. If the effective areas of the transmitter and the receiver antennas, and the distance between them are all doubled, and everything else remains unchanged, the maximum capacity of the wireless link
Answer : Option C
Explaination / Solution:
Channelcpacity remain same.
Q10. In binary frequency shift keying (FSK), the given signal waveform are
u0 (t) = 5cos (20000 πt); 0 ≤ t ≤ T, and
u1 (t) = 5cos (22000 πt); 0 ≤ t ≤ T,
Where T is the bit-duration interval and t is in seconds. Both u0 (t) and u1 (t) are zero outside the interval 0 ≤ t ≤ T. With a matched filter (correlator) based receiver, the smallest positive value of T (in milliseconds) required to have u0 (t) and u1 (t) uncorrelated is
Answer : Option B
Explaination / Solution:
