Test 4 - Communication Systems | Electronics and Communication (ECE4

Description: A test for Communication Systems of Electronics and Communication (ECE)
Number of Questions: 33
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Tags: Communications systems Signals and System
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A carrier is phase modulated (PM) with frequency deviation of 10 kHz by a single tone frequency of 1 kHz. If the single tone frequency is increased to 2 kHz, assuming that phase deviation remains unchanged, the bandwidth of the PM signal is

  1. 21 kHz

  2. 22 kHz

  3. 42 kHz

  4. 44 kHz


Correct Option: D
Explanation:

Two sinusoidal signals of same amplitude and frequencies 10 kHz and 10.1 kHz are added together. The combined signal is given to an ideal frequency detector. The output of the detector is

  1. 0.1 kHz sinusoid

  2. 20.1 kHz sinusoid

  3. a linear function of time

  4. a constant


Correct Option: A
Explanation:

X(t) is a stationary random process with autocorrelation function Rx($\tau$) = exp $(\pi r^2)$. This process is passed through the system shown below. The power spectral density of the output process Y(t) is

  1. $(4 \pi ^2 f^2 + 1) exp (-\pi t^2)$

  2. $(4 \pi ^2 f^2 - 1) exp (-\pi t^2)$

  3. $(4 \pi ^2 f^2 + 1) exp (-\pi f)$

  4. $(4 \pi ^2 f^2 - 1) exp (-\pi f)$


Correct Option: A
Explanation:

A sinusoidal signal with peak-to-peak amplitude of 1.536 V is quantised into 128 levels using a mid-rise uniform quantiser. The quantisation noise power is

  1. 0.768 V2

  2. 48 $\times$10-6 V2

  3. 12 $\times$10-6 V2

  4. 3.072 V2


Correct Option: C
Explanation:

The Nyquist sampling rate for the signal s(t) = $\dfrac{sin(500 \pi t)}{\pi t}$ $\times$ $\dfrac{sin(700 \pi t)}{\pi t}$ is given by

  1. 400 Hz

  2. 600 Hz

  3. 1200Hz

  4. 1400 Hz


Correct Option: C
Explanation:

A Hilbert transformer is a

  1. non-linear system

  2. non-causal system

  3. time-varying system

  4. low-pass system


Correct Option: A
Explanation:

A Hilbert transformer is a non-linear system.

The signal cos$\omega$c t + 0.5 cos$\omega_m$t sin$\omega_c$t + is

  1. FM only

  2. AM only

  3. AM and FM both

  4. None of these


Correct Option: C
Explanation:

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In the following figure, the minimum value of the constant “C”, which is to be added to y1 (t) such that y1 (t) and y2 (t) are different, is

  1. $\Delta$

  2. $\dfrac{\Delta}{2}$

  3. $\dfrac{\Delta ^ 2}{12}$

  4. $\dfrac{\Delta}{L}$


Correct Option: B
Explanation:

When $\Delta / 2$ is added to $y(t)$ then signal will move to next quantization level. Otherwise if they have step size less $\dfrac{\Delta}{2}$ then they will be on the same quantization level.

A signal m(t )with bandwidth 500 Hz is first multiplied by a signal g (t ) where g (t) = $\displaystyle \sum_{R = - \infty}^\infty (-1)^k \delta (t - 0.5 \times O^{-4}k )$ The resulting signal is then passed through an ideal low pass filter with bandwidth 1 kHz. The output of the low pass filter would be

  1. $\delta$ (t )

  2. m (t )

  3. 0

  4. m (t )$\delta$ (t )


Correct Option: B
Explanation:

Match the following:

Group 1 Group 2 1. FM P. Slope overload 2. DM Q. $\mu$-law 3. PSK R. Envelope detector 4. PCM S. Capture effect T. Hilbert transform U. Matched filter

  1. 1 - T, 2 - P, 3 - U, 4 - S

  2. 1 - S, 2 - U, 3 - P, 4 - T

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

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


Correct Option: C
Explanation:

A random variable X with uniform density in the interval 0 to 1 is quantized as follows : If 0$\le$x$\le$ 0.3, xq = 0 If 0.3 < X$\le$1, xq = 0.7 where xq is the quantized value of X. The root-mean square value of the quantization noise is

  1. 0.573

  2. 0.198

  3. 2.205

  4. 0.266


Correct Option: B
Explanation:

If Eb, the energy per bit of a binary digital signal, is 10-6 watt-sec and the onesided power spectral density of the white noise, N0 = 10-5 W/Hz, then the output SNR of the matched filter is

  1. 26 dB

  2. 10 dB

  3. 20 dB

  4. 13 dB


Correct Option: D
Explanation:

A message signal with bandwidth 10 kHz is Lower-Side Band SSB modulated with carrier frequency fc1 = 106 Hz. The resulting signal is then passed through a Narrow-Band Frequency Modulator with carrier frequency fc2 = 109 Hz. The bandwidth of the output would be

  1. 4 x 104 Hz

  2. 2 x 106 Hz

  3. 2 x 109 Hz

  4. 2 x 1010 Hz


Correct Option: C
Explanation:

The input to a linear delta modulator having a step-size $\Delta$ = 0.628 is a sine wave with frequency fm and peak amplitude Em. If the sampling frequency fs = 40 kHz, the combination of the sine wave frequency and the peak amplitude, where slope overload will take place, is

  1. Em = 0.3 V, fm = 8 kHz

  2. Em = 1.5 V, fm = 4 kHz

  3. Em = 1.5 V, fm = 2 kHz

  4. Em = 3.0 V, fm = 1 kHz


Correct Option: B
Explanation:

 

A speed signal, band limited to 4 kHz and peak voltage varying between + 5 V and - 5 V, is sampled at the Nyquist rate. Each sample is quantized and represented by 8 bits.

If the bits 0 and 1 are transmitted using bipolar pulses, the minimum bandwidth required for distortion free transmission is

  1. 64 kHz

  2. 32 kHz

  3. 8 kHz

  4. 4 kHz


Correct Option: B
Explanation:

A symmetric three-level midtread quantizer is to be designed assuming equiprobable occurrence of all quantization levels.

The quantization noise power for the quantization region between -a and +a in the figure is

  1. $\dfrac{4}{81}$

  2. $\dfrac{1}{9}$

  3. $\dfrac{5}{81}$

  4. $\dfrac{2}{81}$


Correct Option: A
Explanation:

A device with input x(t) and output y(t) is characterized by: y(t) = x2(t). An FM signal with frequency deviation of 90 kHz and modulating signal bandwidth of 5 kHz is applied to this device. The bandwidth of the output signal is

  1. 370 kHz

  2. 190 kHz

  3. 380 kHz

  4. 95 kHz


Correct Option: A
Explanation:

A super heterodyne receiver is to operate in the frequency range 550 kHz - 1650 kHz, with the intermediate frequency of 450 kHz. Let R = $\dfrac{C_{max}}{C_{min}}$ denote the required capacitance ratio of the local oscillator and I denote the image frequency (in kHz) of the incoming signal. If the receiver is tuned to 700 kHz, then

  1. R = 4.41, I = 1600

  2. R = 2.10, I = 1150

  3. R = 3.0, I = 1600

  4. R = 9.0, I = 1150


Correct Option: A
Explanation:

A speed signal, band limited to 4 kHz and peak voltage varying between + 5 V and - 5 V are sampled at the Nyquist rate. Each sample is quantized and represented by 8 bits. What is the number of quantization levels required to reduce the quantization noise by a factor of 4?

  1. 1024

  2. 512

  3. 256

  4. 64


Correct Option: B
Explanation:

A symmetric three-level midtread quantizer is to be designed assuming equiprobable occurrence of all quantization levels.

If the input probability density function is divided into three regions as shown in figure, the value of a in the figure is

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

  2. $\dfrac{2}{3}$

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

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


Correct Option: B
Explanation:

As the area under pdf curve must be unity, all three regions are equivaprobable. Thus area under each region must be $\dfrac{1}{3}$ $2a \times \dfrac{1}{4} = \dfrac{1}{3} \rightarrow a = \dfrac{2}{3}$

 

In a Direct Sequence CDMA System, the chip rate is 1.2288 $\times$ 106 chips per second. If the processing gain is desired to be at least 100, then the data rate

  1. must be less than or equal to 12.288 $\times$ 103 bits per sec

  2. must be greater than 12.288 $\times$ 103 bits per sec

  3. must be exactly equal to 12.288 $\times$ 103 bits per sec

  4. can take any value less than 122.88 $\times$ 103 bits per sec


Correct Option: A
Explanation:

The minimum sampling frequency (in samples/sec) required to reconstruct the following signal x (t) = 5 $\left( \dfrac{sin2\pi 1000 t}{\pi t} \right) ^ 3 $ + 7$\left( \dfrac{sin2\pi 1000 t}{\pi t} \right) ^ 2 $ from its samples without distortion would be

  1. 2 x 103

  2. 4 x 103

  3. 6 x 103

  4. 8 x 103


Correct Option: C
Explanation:

An input to a 6-level quantizer has the probability density function f(x) as shown in the figure. Decision boundaries of the quantizer are chosen so as t maximize the entropy of the quantizer output. It is given that 3 consecutive decision boundaries are -1, 0 and 1.

Assuming that the reconstruction levels of the quantizer are the mid-points of the decision boundaries, the ratio of signal power to quantization noise power is

  1. $\dfrac{152}{9}$

  2. $\dfrac{64}{3}$

  3. $\dfrac{76}{3}$

  4. 28


Correct Option: A
Explanation:

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An input to a 6-level quantizer has the probability density function f(x) as shown in the figure. Decision boundaries of the quantizer are chosen so as to maximize the entropy of the quantizer output. It is given that 3 consecutive decision boundaries are -1, 0 and 1.

The values of a and b are

  1. a = $\dfrac{1}{6}$and b = $\dfrac{1}{12}$

  2. a = $\dfrac{1}{5}$and b = $\dfrac{3}{40}$

  3. a = $\dfrac{1}{4}$and b = $\dfrac{1}{16}$

  4. a = $\dfrac{1}{3}$and b = $\dfrac{1}{24}$


Correct Option: A
Explanation:

Consider a system shown in the figure. Let X (f) and Y( f) denote the Fourier transforms of x (t) and y (t) respectively. The ideal HPF has the cutoff frequency 10 kHz. The positive frequencies, where Y (f) has spectral peaks are

  1. 1 kHz and 24 kHz

  2. 2 kHz and 244 kHz

  3. 1 kHz and 14 kHz

  4. 2 kHz and 14 kHz


Correct Option: B
Explanation:

If S represents the carrier synchronization at the receiver and $\rho$ represents the bandwidth efficiency, then the correct statement for the coherent binary PSK is

  1. $\rho$ = 0.5, S is required

  2. $\rho$ = 1.0, S is required

  3. $\rho$ = 0.5, S is not required

  4. $\rho$ = 1.0, S is not required


Correct Option: A
Explanation:

A speed signal, band limited to 4 kHz and peak voltage varying between + 5 V and - 5 V, is sampled at the Nyquist rate. Each sample is quantised and represented by 8 bits.

Assuming the signal to be uniformly distributed between its peak to peak value, the signal to noise ratio at the quantiser output is

  1. 16 dB

  2. 32 dB

  3. 48 dB

  4. 4 dB


Correct Option: C
Explanation:

Three analog signals, having bandwidths 1200 Hz, 600 Hz and 600 Hz, are sampled at their respective Nyquist rates, encoded with 12 bit words, and time division multiplexed. The bit rate for the multiplexed signal is

  1. 115.2 kbps

  2. 28.8 kbps

  3. 57.6 kbps

  4. 38.4 kbps


Correct Option: C
Explanation:

A discrete random variable X takes values from 1 to 5 with probabilities as shown in the table. A student calculates the mean X as 3.5 and her teacher calculates the variance of X as 1.5. Which of the following statements is true?

  1. Both the student and the teacher are right.

  2. Both the student and the teacher are wrong.

  3. The student is wrong but the teacher is right.

  4. The student is right but the teacher is wrong.


Correct Option: B
Explanation:

X(t) is a stationary process with the power spectral density Sx(f)>0 for all f. The process is passed through a system shown below. Let Sy(f) be the power spectral density of Y(t). Which one of the following statements is correct?

Let Sy(f) be the power spectral density of Y(t). Which one of the following statements is correct?

  1. Sy(f)>0 for all f

  2. Sy(f)=0 for |f|>1kHz

  3. Sy(f)=0 for f=nf0, f0=2kHz, n any integer

  4. Sy(f)=0 for f=(2n+1)f0=1kHz, n any integer


Correct Option: D
Explanation:

A signal is sampled at 8 kHz and is quantized using 8-bit uniform quantizer. Assuming SNRq for a sinusoidal signal, the correct statement for PCM signal with a bit rate of R is

  1. R = 32 kbps, SNRq = 25.8 dB

  2. R = 64 kbps, SNRq = 49.8 dB

  3. R = 64 kbps, SNRq = 55.8 dB

  4. R = 32 kbps, SNRq = 49.8 dB


Correct Option: B
Explanation:

Consider the following Amplitude Modulated (AM) signal, where fm < B : $X_{AM}(t) = 10 (1+0.5 sin 2\pi f_m t) cos2\pi f_ct$

The average side band power for the AM signal given above is:

  1. 25

  2. 12.5

  3. 6.25

  4. 3.125


Correct Option: C
Explanation:

Consider the following Amplitude Modulated (AM) signal, where fm < B:

$X_{AM}(t) = 10 (1+0.5 sin 2\pi f_m t) cos2\pi f_ct$

The AM signal gets added to a noise with Power Spectral Density Sn (f) given in the figure below. The ratio of average sideband power to mean noise power would be

  1. $\dfrac{25}{8 N_0 B}$

  2. $\dfrac{25}{4 N_0 B}$

  3. $\dfrac{25}{2 N_0 B}$

  4. $\dfrac{25}{N_0 B}$


Correct Option: B
Explanation:

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