Test 3 - Communication Systems | Electronics and Communication (ECE)

Description: A test for Communication Systems of Electronics and Communication (ECE)
Number of Questions: 25
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Tags: Communication system
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Two independent random variables X and Y are uniformly distributed in the interval [-1, 1]. The probability that max [X, Y] is less than 1/2 is

  1. ¾

  2. 9/16

  3. ¼

  4. 2/3


Correct Option: B

c(t) and m(t) are used to generate an FM signal. If the peak frequency deviation of the generated FM signal is three times the transmission bandwidth of the AM signal, then the coefficient of the term cos[2$\pi$(1008 $\times$103t)] in the FM signal (in terms of the Bessel coefficients) is

  1. 5j4(3)

  2. $\dfrac{5}{2} j_\theta (3)$

  3. $\dfrac{5}{2} j_\theta (3)$

  4. 5j4(6)


Correct Option: D
Explanation:

Consider the pulse shape s(t) as shown. The impulse response h(t) of the filter matched to this pulse is


Correct Option: C
Explanation:

Let m (t) = cos [ (4$\pi$ $\times$ 103) t] be the message signal and c(t) = 5 cos[(2$\pi$ $\times$ 106)t] be the carrier. c(t) and m(t) are used to generate an AM signal. The modulation index of the generated AM signal is 0.5. Then the quantity $\dfrac{Total sideband power}{Carrier power}$ is

  1. $\dfrac{1}{2}$

  2. $\dfrac{1}{4}$

  3. $\dfrac{1}{3}$

  4. $\dfrac{1}{8}$


Correct Option: D
Explanation:

Match the following:||||| |---|---|---|---| | | Column-I| | Column-II| | P| Power efficient transmission of signals| 1| Conventional AM| | Q| Most bandwidth efficient transmission of voice signals| 2| FM| | R| Simplest receiver structure| 3| VSB| | S| Bandwidth efficient transmission of signals with significant dc component| 4| SSB-SC|

  1. P - 4, Q - 2, R - 1, S - 3

  2. P - 2, Q - 4, R - 1, S - 3

  3. P - 3, Q - 2, R - 1, S - 4

  4. P - 2, Q - 4, R - 3, S - 1


Correct Option: B
Explanation:

Four messages band limited to W, W, 2W and 3 W respectively are to be multiplexed using Time Division Multiplexing (TDM). The minimum bandwidth required for transmission of this TDM signal is

  1. 2 W

  2. 3 W

  3. 6 W

  4. W


Correct Option: D
Explanation:

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


Correct Option: B
Explanation:

Which of the following analog modulation schemes requires the minimum transmitted power and minimum channel bandwidth?

  1. VSB

  2. DSB-SC

  3. SSB

  4. AM


Correct Option: C
Explanation:

A binary symmetric channel (BSC) has a transition probability of 1/8. If the binary transmit symbol X is such that P(X = 0) = 9/10, then the probability of error for an optimum receiver will be

  1. 7/80

  2. 63/80

  3. 9/10

  4. 1/10


Correct Option: A

If E denotes expectation, then the variance of a random variable X is given by

  1. E [x2] – E2 [x]

  2. E [x2] + E2 [x]

  3. E [x2]

  4. E2 [x]


Correct Option: A
Explanation:

The variance of a random variable $x$ is given by $E[X^2] - D^Q[X]$

At a given probability of error, the binary coherent FSK is inferior to binary coherent PSK by

  1. 6 dB

  2. 3 dB

  3. 2 dB

  4. 0 dB


Correct Option: B
Explanation:

We have $P_c = \dfrac{1}{2} erfc \left( \sqrt {\dfrac{E_d}{2\eta}} \right)$

Since $P_\epsilon$ of Binary FSK is 3 dB inferior to binary PSK

An output of a communication channel is a random variable $\nu$ with the probability density function as shown in figure. The mean square value of $\nu$ is

  1. 4

  2. 6

  3. 8

  4. 9


Correct Option: C
Explanation:

A four-phase and an eight-phase signal constellation are shown in the figure below

Assuming high SNR and that all signals are equally probable, the additional average transmitted signal energy required by the 8-PSK signal to achieve the same error probability as the 4-PSK signal is

  1. 11.90 dB

  2. 8.73 dB

  3. 6.79 dB

  4. 5.33 dB


Correct Option: D
Explanation:

A DSB-SC signal is to be generated with a carrier frequency fc = 1 MHz using a nonlinear device with the input-output characteristic V0 = a0vi + a1vi3 where a0 and a1 are constants. The output of the nonlinear device can be filtered by an appropriate band-pass filter. Let vi = A'c cos (2$\pi$f'c t) + m(t) where m(t) is the message signal. The value of f'c (in MHz) is

  1. 1.0

  2. 0.333

  3. 0.5

  4. 3.0


Correct Option: C
Explanation:

A four-phase and an eight-phase signal constellation are shown in the figure below.

For the constraint that the minimum distance between pairs of signal points be d for both constellations, the radii r1, and r2 of the circles are

  1. r1 = 0.707d, r2 = 2.782d

  2. r1 = 0.707d, r2 = 1.932d

  3. r1 = 0.707d, r2 = 1.545d

  4. r1 = 0.707d, r2 = 1.307d


Correct Option: D
Explanation:

During transmission over a certain binary communication channel, bit errors occur independently with probability p. The probability of AT MOST one bit in error in a block of n bits is given by

  1. pn

  2. 1 – pn

  3. np(1 – p)n-1 + (1 – p)n

  4. 1 – (1 – p)n


Correct Option: C
Explanation:

A uniformly distributed random variable X with probability density function fx (x) = $\dfrac{1}{10}$(u (x + 5) - u (x - 5)) Where u (.) is the unit step function is passed through a transformation given in the figure below. The probability density function of the transformed random variable Y would be

  1. fy =$\dfrac{1}{5}$(u (y + 2.5) - u (y - 2.5))

  2. fy = 0.5$\delta$ (y) + 0.5$\delta$ (y -1)

  3. fy = 0.25$\delta$ (y + 2.5) + 0.25$\delta$ (y -2.5) + 0.5$\delta$(y)

  4. fy = 0.25$\delta$ (y + 2.5) + 0.25$\delta$ (y -2.5) + $\dfrac{1}{10}$(u (y + 2.5) - u (y - 2.5))


Correct Option: B
Explanation:

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A zero-mean white Gaussian noise signal is passed through an ideal lowpass filter of bandwidth 10 kHz. The output is then uniformly sampled with sampling period 0.03 m. The samples so obtained would be

  1. correlated

  2. statistically independent

  3. uncorrelated

  4. orthogonal


Correct Option: A
Explanation:

The sampling frequency is

$$ f_n = \dfrac{1}{0.03m} = 33 KHz $$

Since $f_n \ge 2 f_m$ the signal can be recovered and are correlated.

Consider the frequency modulated signal 10 cos [2$\pi$ x 105 t + 5 sin (2$\pi$ x 1500 t) + 7.5 sin (2$\pi$ x 1000 t)] with carrier frequency of 105 Hz. The modulation index is

  1. 12.5

  2. 10

  3. 7.5

  4. 5


Correct Option: B
Explanation:

The distribution function Fx (x) of a random variable x is shown in the figure. The probability that x = 1 is

  1. zero

  2. 0.25

  3. 0.55

  4. 0.30


Correct Option: D
Explanation:

In the output of a DM speech encoder, the consecutive pulses are of opposite polarities during time interval t1 $\le$ t $\le$ t2. This indicates that during this interval,

  1. the input to the modulator is essentially constant

  2. the modulator is going through slope overload

  3. the accumulator is in saturation

  4. the speech signal is being sampled at the Nyquist rate


Correct Option: A
Explanation:

Consecutive pulses are of the same polarity when modulator is in slope overload. Consecutive pulses are of opposite polarities when the input is constant.

A 100 MHz carrier of 1 V amplitude and a 1 MHz modulating signal of 1 V amplitude are fed to a balanced modulator. The output of the modulator is passed through an ideal high-pass filter with cut-off frequency of 100 MHz. The output of the filter is added with 100 MHz signal of 1 V amplitude and 900 phase shift as shown in the figure. The envelope of the resultant signal is

  1. constant

  2. $\sqrt{1+sin(2\pi \times 10^6 t)}$

  3. $\sqrt{ \dfrac{5}{4} - sin(2\pi - 10^6 t)}$

  4. $\sqrt{ \dfrac{5}{4} + cos(2\pi - 10^6 t)}$


Correct Option: C
Explanation:

Directions :The amplitude of a random signal is uniformly distributed between - 5 V and 5 V.

If the signal to quantization noise ratio required in uniformly quantizing the signal is 43.5 dB, the step of the quantization is approximately

  1. 0.033 V

  2. 0.05 V

  3. 0.0667 V

  4. 0.10 V


Correct Option: C
Explanation:

Directions :The amplitude of a random signal is uniformly distributed between - 5 V and 5 V.

If the positive values of the signal are uniformly quantized with a step size of 0.05 V, and the negative values are uniformly quantized with a step size of 0.1 V, the resulting signal to quantization noise ration is approximately

  1. 46 dB

  2. 43.8 dB

  3. 42 dB

  4. 40 dB


Correct Option: B
Explanation:

A communication channel with AWGN operating at a signal to noise ration SNR >> 1 and bandwidth B has capacity C1. If the SNR is doubled keeping constant, the resulting capacity C2 is given by

  1. C2 $\approx$ 2C1

  2. C2 $\approx$ C1 + B

  3. C2 $\approx$ C1 + 2B

  4. C2 $\approx$ C1 + 0.3 B


Correct Option: B
Explanation:

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