# Communication Engineering (Test 6)

## Tancet Anna University : Ece Electronics And Communication Engineering

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Q.1
In the following scheme, if the spectrum M(f) of m(t) is as shown, then the spectrum Y(f) of y(t) will be A. B. C. D. Answer : Option B
Explaination / Solution:   Workspace
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Q.2
Two 4-array signal constellations are shown. It is given that φ1 and φ2 constitute an orthonormal basis for the two constellation. Assume that the four symbols in both the constellations are equiprobable. Let N0/2 denote the power spectral density of white Gaussian noise. The if ratio or the average energy of Constellation 1 to the average energy of Constellation 2 is
A. 4a2
B. 4
C. 2
D. 8
Answer : Option B
Explaination / Solution: Workspace
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Q.3
In the design of a single mode step index optical fibre close to upper cut-off, the single-mode operation is not preserved if
A. radius as well as operating wavelength are halved
B. radius as well as operating wavelength are doubled
C. radius is halved and operating wavelength is doubled
D. radius is doubled and operating wavelength is halved
Answer : Option C
Explaination / Solution:
No Explaination.

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Q.4
In the following network, the switch is closed at t = 0- and the sampling starts from t = 0. The sampling frequency is 10 Hz. The samples x(n), n = (0, 1, 2, ...) are given by
A.

5(1 - e-0.05n)

B. 5e-0.05n
C. 5(1 e-5n)
D. 5e-5n
Answer : Option B
Explaination / Solution:
No Explaination.

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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
Answer : Option B
Explaination / Solution:

Let response of LPF filters Noise variance (power) is given as Workspace
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Q.6
The time domain behavior of an RL circuit is represented by For an initial current of
For an initial current of the steady state value of the current is given by
A. B. C. D. Answer : Option A
Explaination / Solution:

Steady state all transient effect die out and inductor act as short circuits and forced response acts only. It doesn’t depend on initial current state. From the given time domain behavior we get that circuit has only R and L in series with vThus at steady state Workspace
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Q.7
A transmission line of characteristic impedance 50Ω is terminated by a 50 W load. When excited by a sinusoidal voltage source at 10 GHz, the phase difference between two points spaced 2 mm apart on the line is found to be π/4 radians. The phase velocity of the wave along the line is
A.

0.8 × 108 m/s

B.

1.2 × 108 m/s

C.

1.6 × 108 m/s

D.

3 × 108 m/s

Answer : Option C
Explaination / Solution:

We have d = 2 mm and f = 10 GHz Workspace
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Q.8
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
Answer : Option A
Explaination / Solution:
No Explaination.

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Q.9
Let the Laplace transform of a function F(t) which exists for t > 0 be F1(s) and the Laplace transform of its delayed version be the complex conjugate of F1(s) with the Laplace variable set as s=σ + jw. If G(s) = , then the inverse Laplace transform of G(s) is
A. B. C. An ideal step function u(t)
D. Answer : Option B
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Q.10
The impulse response of a system is h(t) = tu(t). For an input u(t − 1), the output is
A. B. C. D. Answer : Option C
Explaination / Solution:
No Explaination.

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