### Communication Engineering - MCQs from AMIE exams (Summer 2018)

*Choose the correct alternative for the following: 10 x 2*

1. The autocorrelation function R(τ) of the signal X(t) = Vsinωt is given by

(a) 0.5V

^{2}cosωt(b) V

^{2}cosωt(c) V

^{2}cos^{2}ωt(d) 2V

^{2}cos^{2}ωt2. Which of the following cannot be the Fourier series expansion of a periodic signal?

(a) x(t) = 2cos6t + 3cos3t

(b) x(t) = 2cosπt + 7 cost

(c) x(t) = cost + 0.5

(d) x(t) = 2 cos1.5πt + sin3.5πt

3. Consider a white noise of 2-sided spectral density 2 x 10⁻⁶ Hz applied to a simple RC low pass filter whose 3 dB cut-off frequency is 4 kHz. The mean squared value of the noise output will be

(a) 4π x 10⁻³ V²

(b) 2π x 10⁻³ V²

(c) 4π x 10⁻³ V²

(d) π x 10⁻³ V²

4. Find the magnetic field intensity due to an infinite sheet of current 5 A and charge density of 12j units in the positive y-direction and the 2 component is below the sheet.

(a) -6

(b) 12k

(c) 6

(d) 60

5. In a VSB system the power contained in the sidebands due to a modulating frequency of 4 MHz is 50 W, the power in the sidebands due to a 0.5 MHz frequency is

(a) 50W

(b) 100W

(c) 25 W

(d) Cannot be calculated from given data

6. What is the change in the value of transmitted power when the modulation index changes from 0 to 1?

(a) 100%

(b) Remains unchanged

(c) 50%

(d) 80%

7. DPCM encodes the PCM values based on

(a) Difference between the current and predicted value

(b) Quantization level

(c) Interval between levels

(d) None of the mentioned

8. Granular noise occurs when

(a) Step size is too small

(b) Step size is too large

(c) Three is interference from the adjacent channel

(d) Bandwidth is too large

9. The demodulator in delta modulation technique is

(a) Differentiator

(b) Integrator

(c) Quantizer

(d) None of the mentioned

10. Four voice signals, each limited to 4 kHz and sampled at Nyquist rate, are converted into binary PCM signals using 256 quantization levels. The bit transmission rate for the time-division multiplexed signal will be

(a) 8 kbps

(b) 64 kbps

(c) 256 kbps

(d) 512 kbps

### Answers

1. (a)

Rx(T) =E [x(t)x(t+T)]

= E[vsin(ωt)vsin(ωt+ωT)]

= E[v

^{2}sin(ωt)sin(ωt+ωT)]= v

^{2}E[sin(ωt)sin(ωt+ωT)]= v

^{2}E[(cos(ωT)-cos(2ωt+ωT))/2]now E[cos(2ωt+ωT)]=0 as average of cosine function over one cycle is zero.

2. (b)

For Fourier series expression, x(t)x(t), should be periodic for x(t) = 2cosπt + 7 cost

T

_{1}= 2π/π = 2T

_{2}= 2πT

_{1}/T_{2}= 1/π (not rational hence not periodic)3.

4. (c) The magnetic field intensity when the normal component is above the sheet is H(x) = 0.5 K, where K = 12. Thus we get H = 0.5 x 12 = 6 units.

5. (b)

6. (c)

We know

Hence

and

The % increase in the modulated signal power is given by:

7. (a)

- Differential pulse code modulation (DPCM) is a procedure of converting an analog into a digital signal in which an analog signal is sampled and then the difference between the actual sample value and its predicted value (predicted value is based on previous sample or samples) is quantized and then encoded forming a digital value.
- DPCM codewords represent differences between samples, unlike PCM where codewords represented a sample value.

8. (b)

9. (b)

10. (c)

Since the sampling frequency is not mentioned, we'll assume it to be sampled at the Nyquist rate, i.e.

f

_{s}= 2f_{m}f

_{m}= Maximum frequency present at the modulating signal.∴ For the given band-limited signal with a frequency of 4 kHz, the sampling frequency will be:

f

_{s}= 2 × 4 = 8 kHzWith L = 256, the number of bits will be:

n = log

_{2}256 = log_{2}2^{8}n = 8 bits

Now for 4 Voice signals

R

_{b}= 4 × n × f_{s}= 4 × 8 × 8000 = 256 kbps---

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