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Exam Questions Papers - Exam Paper 1 (5)

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Exam Questions Papers - Exam Paper 1

33. 

What are the values of Emax and Emin displayed on the oscilloscope, when a 1 kV P-P carries is modulated to 50%?

A. 2 kV, 0.5 kV
B. 1 kV, 0.5 kV
C. 0.75 kV, 0.25 kV
D. 0.5 kV, 1.5 kV

Answer: Option C

Explanation:

3-10-45-1

Emax - Emin = 0.5 x 1 kV = 0.5 kV

2Emax = 1.5 kVi, Emax = 0.75 kV

Emin = 0.25 kV.

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34. 

Suppose that the modulating signal is m(t) = 2cos (2pfmt) and the carrier signal is xC(t) = AC cos(2pfct), which one of the following is a conventional AM signal without over modulation?

A. x(t) = Acm(t)cos(2pfct)
B. x(t) = Ac [1 + m(t)]cos (2pfct)
C. 7-03-20-3
D. x(t) = Accos(2pfmt)cos(2pfct) + Acsin(2pfmt)sin(2pfct)

Answer: Option C

Explanation:

7-08-20

7-08-20-1

(C) is without over modulation.

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35. 

Find 'X' in the circuit below :

14-02-20

f1(A, B, C, D) = Σ(6, 7, 13, 14);

f2(A, B, C, D) = Σ(3, 6, 7);

f3(A, B, C, D) = Σ(5, 6, 7, 14, 15)

A. p(0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 15)
B. 0
C. Σ(14)
D. 1

Answer: Option C

Explanation:

f1(A, B, C, D) = ∑(6, 7, 13, 14)

f2(A, B, C, D) = ∑(3, 6, 7)

f1 ⊕ f2 = ∑(3, 13, 14)

f3 x (f1 ⊕ f2) = ∑(14) = y

X = f1 x y ⇒ ∑14.

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36. 

In a 4 bit weighted resistor D/A converter, the resistor value corresponding to LSB is 32 kΩ. The resistor value corresponding to MSB will be

A. 32 KΩ
B. 16 KΩ
C. 8 KΩ
D. 4 KΩ

Answer: Option D

Explanation:

2n - 1 R = 32 K&Omeg;

n = 4

⇒ 8R = 32

R = 4 KΩ.

 
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37. 

Consider the amplitude modulated (AM) signal Ac cos ωct + 2 cos ωmt cos ωct For demodulating the signal using envelope detector, the minimum value of Ac should be

A. 2
B. 1
C. 0.5
D. 0

Answer: Option A

Explanation:

AC cos ωct + 2 cos ωmt cos ωct

AC cosωct 5-11-20

for envelope detection μ < 1 ⇒ 5-11-20-1< 1 ⇒ Ac should be at least-2.

 

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38. 

The inverse of given Laplace transform is

3-03-32

3-03-32-1

A. sin t
B. cos t
C. et
D. e2t

Answer: Option B

Explanation:

3-09-32

s = x2 - ex

3-09-32-1

x(t) = cos t.

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39. 

If the power spectral density of stationary random process is a sine-squared function of frequency, the shape of its autocorrelation is

A. 6-02-12-1
B. 6-02-12-2
C. 6-02-12-3
D. 6-02-12-4

Answer: Option B

Explanation:

Since autocorrelation function and power spectral density bears a Fourier transform relation, then since required in frequency domain will five rectangular convolutions in time domain thus it is triangular function.

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40. 

Radiation resistance of an antenna is 54 Ω and loss resistance is 6 Ω. If antenna has power gain of 10, then directivity is:

A. 9
B. 11.11
C. data not sufficient
D. 10

Answer: Option B

Explanation:

Efficiency of antenna 1-750-18-1

Power gain = 10

Thus directivity 1-750-18-2

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