Communication Engineering - MCQs from AMIE exams (Winter 2020)

Choose the correct option (10 x 2)

1. The laplace equation has

(a) Two solutions

(b) Infinite solutions

(d) One solution

2. A material has conductivity of 10⁻² mho/in and a relative permittivity of 4. The frequency at which conduction current is equal to displacement current is

(a) 45 MHz

(b) 90 MHz

(d) 900 MHz

3. The magnetic field in an ideal conductor is

(a) Zero

(b) Infinite

(d) Same as outside field

4. Statement 1 - Inside a conductor electric field is always zero.

Statement 2 - A conductor is always an equipotential surface.

(a) True, True

(b) False, False

(d) True, False

5. An integration of any vector around a closed path is always equal to the Curl of surface integral of that vector, this statement is known as :

(a) Ampere law

(b) Stoke's theorem

(d) Curl

6. The probability density function of the envelope of narrowband Gaussian noise is

(a) Poisson

(b) Gaussian

(d) Rician

7. Quantization noise is produced in

(a) All pulse modulation system

(b) PCM

(d) AM

8. Granular noise is associated with

(a) PCM

(b) DPCM

(d) QAM

9. A TDM link has 20 signal channels and each channel is sampled 8000 times/sec. Each sample is represented by 7 binary bits and contains an additional bit for synchronization. The total bit rate for the TDM link is

(a) 180 kbps

(b) 1280 kbps

(d) 1880 Mbps

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) 64 bps

(b) 16 kbps

(d) 256 kbps

Answers

1. (d)

2. (a)

$\omega = 2\pi f = \frac{\sigma }{{{\varepsilon _0}{\varepsilon _r}}}$

$\Rightarrow f = \frac{\sigma }{{2\pi {\varepsilon _0}{\varepsilon _r}}} = \frac{{{{10}^{ - 2}}}}{{2\pi (8.85x{{10}^{ - 12}})(4)}} = 45MHz$

3. (a) We know that magnetic field intensity B is directly proportional to the electric current in that wire, as we go inside the wire the current density J(=I/A) decrease so at the center it becomes zero.

4. (a)

On the surface of a perfect conductor, the potential distributed evenly over the surface. I.e. an equipotential surface. So Statement A follows.
The electric field inside a charged spherical conductor is always zero. The charges in a perfect conductor reside only on the surface of the conductor. So, the charge inside the conductor is zero.

5. (b)

This theorem states that the line integral of a vector around a closed path is equal to the surface integral of the normal component of its curl over the surface bounded by the path.

Mathematically,

where S is the surface enclosed by the path L.

6. (b)

7. (b) In pulse modulation, pulses result from sampling the modulating signal wave. In other words, the modulating wave is sliced into small units, the process is called quantizing or quantization. These quantum points are then converted into digital binary codes, which represents amplitude of the wave at that point.

8. (c) If the step size is made arbitrarily large to avoid slope-overload distortion, it may lead to ‘granular noise’. Imagine that the input speech signal is fluctuating but very close to zero over a limited time duration. This may happen due to pauses between sentences or else. During such moments, our delta modulator is likely to produce a fairly long sequence of 101010…., reflecting that the accumulator output is close but alternating around the input signal. This phenomenon is manifested at the output of the delta demodulator as a small but perceptible noisy background. This is known as ‘granular noise’. An expert listener can recognize the crackling sound.

9. (b) Time-division multiplexing is a type of digital multiplexing in which two or more signals are

transferred apparently simultaneously as subchannels in one communication channel, but are physically taking turns on the channel.

Given sampling rate of channel each = 8000 Hz = f_ch

Since 24 such signals are Multiplexed, we get a total pulse per second (frequency) of the multiplied signal as:

f_s = nf_ch

Putting on the respective values:

f_s = 20 × 8000 = 160,000 = 160 KHz

Now each sample is represented by 7 bits and contains an additional bit for synchronization.

The total number of bits per sample will be:

n = 7 + 1 = 8 bits

∴ The total bit rate of the TDM link will be:

R_b = nf_s = 8 × 160 Kbps = 1280 Kilo-bits/sec

10. (d)

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 kHz

With L = 256, the number of bits will be:

n = log₂ 256 = log₂ 2⁸

n = 8 bits

Now for 4 Voice signals

R_b = 4 × n × f_s = 4 × 8 × 8000 = 256 kbps

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Communication Engineering - MCQs from AMIE exams (Winter 2020)

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