# Topic: Communications (Test 7)

Topic: Communications
Q.1

For a message signal and carrier of frequency fc which of thefollowing represents a single side-band (SSB) signal ?

A. B. C. D. Explaination / Solution: Workspace
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Q.2
The Nyquist sampling rate for the signal is given by
A. 400 Hz
B. 600 Hz
C. 1200 Hz
D. 1400 Hz
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

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Q.3
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
A. Sy (f ) > 0 for all f
B. Sy (f ) = 0 for |f| > 1 kHz
C. Sy (f ) = 0 for f = nf0, f0 = 2 kHz kHz, n any integer
D. Sy (f ) = 0 for f = (2n + 1) f= 1 kHz, n any integer
Explaination / Solution: Workspace
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Q.4
In the circuit shown, all the transmission line sections are lossless. The Voltage Standing Wave Ration(VSWR) on the 60 Ω line is A. 1.00
B. 1.64
C. 2.50
D. 3.00
Explaination / Solution:  Workspace
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Q.5
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).
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. The value of the parameter α (in V-1) is
A. 1010
B. 107
C. 1. 414 × 10-10
D. × 10-20
Explaination / Solution:

Let response of LPF filters Noise variance (power) is given as Workspace
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Q.6
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).
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. The probability of bit error is
A. 0.5 × e-3.5
B. 0.5 × e-5
C. 0.5 × e-7
D. 0.5 × e-10
Explaination / Solution:

Probability of error is given by Workspace
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Q.7
The block diagram of a frequency synthesizer consisting of a Phase Locked Loop (PLL) and a divide-by-𝑁 counter (comprising ÷2, ÷4, ÷8, ÷16 outputs) is sketched below. The synthesizer is excited with a 5 kHz signal (Input 1). The free-running frequency of the PLL is set to20 kHz. Assume that the commutator switch makes contacts repeatedly in the order 1-2-3-4. The corresponding frequencies synthesized are:
A. 10 kHz, 20 kHz, 40 kHz, 80 kHz
B. 20 kHz, 40 kHz, 80 kHz, 160 kHz
C. 80 kHz, 40 kHz, 20 kHz, 10 kHz
D. 160 kHz, 80 kHz, 40 kHz, 20 kHz
Explaination / Solution:
No Explaination.

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Q.8
The propagation constant of a lossy transmission line is (2 + 𝑗5) m-1 and its characteristic impedance is (50 + j0)𝜔 at 𝜔 = 106 rad s-1. The values of the line constants L, C, R, G are, respectively,
A. L = 200 μH/m, C = 0.1 μH/m, R = 50 Ω/m, G = 0.02 S/m
B. L = 250 μH/m, C = 0.1 μH/m, R = 100 Ω/m, G = 0.024S/m
C. L = 200 μH/m, C = 0.2 μH/m, R = 100 Ω/m, G = 0.02 S/m
D. L = 250 μH/m, C = 0.2 μH/m, R = 50 Ω/m, G = 0.04 S/m
Explaination / Solution: Workspace
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Q.9
An antenna pointing in a certain direction has a noise temperature of 50K. The ambient temperature is 290 K. The antenna is connected to a pre-amplifier that has a noise figure of 2 dB and an available gain of 40 dB over an effective bandwidth of 12 MHz. The effective input noise temperature Te for the amplifier and the noise power Pao at the output of the preamplifier, respectively, are
A. Te = 169.36K and Pao = 3.73 × 10-10W
B. Te = 1.36K and Pao = 3.73 × 10-10W
C. Te = 182.5K and Pao = 3.83 × 10-10W
D. Te = 160.62K and Pao = 4.6 × 10-10W
Explaination / Solution:
No Explaination.

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Q.10
The electric field of a uniform plane wave travelling along the negative z direction is given by the following equation: This wave is incident upon a receiving antenna placed at the origin and whose radiated electric field towards the incident wave is given by the following equation: The polarization of the incident wave, the polarization of the antenna and losses due to the polarization mismatch are, respectively,
A. Linear, Circular (clockwise), −5dB
B. Circular (clockwise), Linear, −5dB
C. Circular (clockwise), Linear, −3dB
D. Circular (anti clockwise), Linear, −3dB