Test 1 - Communication Systems | Electronics and Communication (ECE)

Description: Topic wise test for Communication Systems of Electronics and Communication (ECE)
Number of Questions: 25
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Tags: Communications
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Consider the amplitude modulated (AM) signal Ac cos$\omega_0$t + 2cos$\omega_m$t cos$\omega_0$t. For demodulating the signal using envelope detector, the minimum value of Ac should be

  1. 2

  2. 1

  3. 0.5

  4. 0


Correct Option: A
Explanation:

The input to a coherent detector is DSB-SC signal plus noise. The noise at the detector output is

  1. the in-phase component

  2. the quadrature component

  3. zero

  4. the envelope


Correct Option: A
Explanation:

The input is a coherent detector is DSB - SC signal plus noise. The noise at the detector output is the in-phase component as the quadrature component $n_0(t)$ of the noise $n(t)$ is completely rejected by the detector.

In a PCM system, if the code word length is increased from 6 to 8 bits, the signal to quantisation noise ratio improves by the factor

  1. $\dfrac{8}{6}$

  2. 12

  3. 16

  4. 8


Correct Option: C
Explanation:

The power spectral density of a real process X(t) for positive frequencies is shown below. The values of E[X2- (t)] and | E [X (t)] |, respectively are

  1. 6000/$\pi$ and 0

  2. 6400/$\pi$ and 0

  3. 6400/$\pi$ and 20/($\pi$$\sqrt 2$)

  4. 6000/$\pi$ and 20/($\pi$$\sqrt 2$)


Correct Option: B
Explanation:

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The probability density function (pdf) of random variable is as shown below:


Correct Option: A
Explanation:

CDF is the integration of PDF. Plot in option (1) is the integration of plot given in question.

A source generates three symbols with probabilities 0.25, 0.25, 0.50 at a rate of 3000 symbols per second. Assuming independent generation of symbols, the most efficient source encoder would have average bit rate as

  1. 6000 bits/sec

  2. 4500 bits/sec

  3. 3000 bits/sec

  4. 1500 bits/sec


Correct Option: B
Explanation:

Suppose that the modulating signal is m(t) = 2cos (2$\pi$fmt) and the carrier signal is xC(t) = AC cos(2$\pi$fCt), which one of the following is a conventional AM signal without over-modulation?

  1. x(t) = Acm(t) cos(2$\pi$fct)

  2. x(t) = Ac[1 + m(t)]cos(2$\pi$fct)

  3. x(t) = Ac cos(2$\pi$fct) + $\dfrac{A_0}{4}$m(t) cos (2$\pi$fCt)

  4. x(t) = Ac cos(2$\pi$fmt) cos(2$\pi$fct) + Ac sin(2$\pi$fmt) sin(2$\pi$fct)


Correct Option: C
Explanation:

The raised cosine pulse p(t) is used for zero ISI in digital communications. The expression for p(t) with unity roll - off factor is given by p(t) =$\dfrac{sin \ 4\pi Wt}{4\pi Wt(1 - 16 w^2t^2)}$ The value of p(t) at t = $\dfrac{1}{4W}$is

    • 0.5
  1. 0

  2. 0.5

  3. $\infty$


Correct Option: C
Explanation:

A BPSK scheme operating over an AWGN channel with noise power spectral density of N0/2 uses equiprobable signals s1(t) = $\dfrac{2E}{T}$sin ($\omega_0 t$) and s2 (t) = -$\dfrac{2E}{T}$ sin ($\omega_0 t$) over the symbol interval (0,T). If the local oscillator in a coherent receiver is ahead in phase by 450 with respect to the received signal, the probability of error in the resulting system is

  1. Q$\left( \sqrt{ \dfrac{2E}{N_0} } \right) $

  2. Q$\left( \sqrt{ \dfrac{E}{N_0} } \right) $

  3. Q$\left( \sqrt{ \dfrac{E}{2N_0} } \right) $

  4. Q$\left( \sqrt{ \dfrac{E}{4N_0} } \right) $


Correct Option: B
Explanation:

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The diagonal clipping in Amplitude Demodulation (using envelope detector) can be avoided if RC time-constant of the envelope detector satisfies the following condition, (here W is message bandwidth and c w is carrier frequency both in rad/sec)

  1. RC < $\dfrac{1}{W}$

  2. RC > $\dfrac{1}{W}$

  3. RC < $\dfrac{1}{\omega_c}$

  4. RC > $\dfrac{1}{\omega_c}$


Correct Option: A
Explanation:

Consider a baseband binary PAM receiver shown below. The additive channel noise n(t) is whit with power spectral density SN(f)=N0/2=10-20 W/Hz. The low-pass filter is ideal with unity gain and cutoff frequency 1MHz. Let Yk represent the random variable y(tk). Yk=Nk if transmitted bit bk=0 Yk=a+Nk if transmitted bit bk=1 Where Nk represents the noise sample value. The noise sample has a probability density function, PNk(n)=0.5لe-ل|n| (This has mean zero and variance 2/ل2). Assume transmitted bits to be equiprobable and threshold z is set to a/2=10-6V.

The probability of bit error is

  1. 0.5xe-3.5

  2. 0.5xe-5

  3. 0.5xe-7

  4. 0.5xe-10


Correct Option: D
Explanation:

In a baseband communications link, frequencies upto 3500 Hz are used for signaling. Using a raised cosine pulse with 75% excess bandwidth and for no inter - symbol interference, the maximum possible signaling rate in symbols per second is

  1. 1750

  2. 2625

  3. 4000

  4. 5250


Correct Option: C

Consider a baseband binary PAM receiver shown below. The additive channel noise n(t) is whit with power spectral density SN(f)=N0/2=10-20 W/Hz. The low-pass filter is ideal with unity gain and cutoff frequency 1MHz. Let Yk represent the random variable y(tk). Yk=Nk if transmitted bit bk=0 Yk=a+Nk if transmitted bit bk=1 Where Nk represents the noise sample value. The noise sample has a probability density function, PNk(n)=0.5لe-ل|n| (This has mean zero and variance 2/ل2). Assume transmitted bits to be equiprobable and threshold z is set to a/2=10-6V.

The value of the parameter ل (in V-1) is

  1. 1010

  2. 107

  3. 1.414$\times$10-10

  4. 2$\times$10-20


Correct Option: B
Explanation:

Let x(t) = 2 cos (800$\pi$t) + cos (1400$\pi$t). x(t) is sampled with the rectangular pulse train shown in figure. The only spectral components (in kHz) present in the sampled signal in the frequency range 2.5 kHz to 3.5 kHz are

  1. 2.7, 3.4

  2. 3.3, 3.6

  3. 2.6, 2.7, 3.3, 3.4, 3.6

  4. 2.7, 3.3


Correct Option: D
Explanation:

Choose the correct one from among the alternatives A, B, C, D after matching an item in Group 1 with the most appropriate item in Group 2.||| |---|---| | Group 1| Group 2| | P Ring modulator| 1 Clock recovery| | Q VCO| 2 Demodulation of FM| | R Foster-Seely discriminator| 3 Frequency conversion| | S Mixer| 4 Summing the two inputs| | | 5 Generation of FM| | | 6 Generation of DSB-Sc|

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

  2. P - 6, Q - 5, R - 2, S - 3

  3. P - 6, Q - 1, R - 3, S - 2

  4. P - 5, Q - 6, R - 1, S - 3


Correct Option: B
Explanation:

A memory less source emits n symbols each with a probability p. The entropy of the source as a function of n

  1. increases as log n

  2. decreases as log $\dfrac{1}{n}$

  3. increases as n

  4. increases as n log n


Correct Option: A
Explanation:

In a GSM system, 8 channels can co-exist in 200 KHz bandwidth using TDMA. A GSM based cellular operator is allocated 5 MHz bandwidth. Assuming a frequency reuse factor of $\dfrac{1}{5}$ i.e a five-cell repeat pattern, the maximum number of simultaneous channels that can exist in one cell is

  1. 200

  2. 40

  3. 25

  4. 5


Correct Option: B
Explanation:

Let g (t) = p (t) * p (t), where * denotes convolution and p (t) = u (t) − u (t − 1) with u (t) being the unit step function

The impulse response of filter matched to the signal s (t) = g (t) − $\delta$ (t − 2) * g (t) is given as:

  1. s (1 − t)

  2. − s (1 − t)

  3. − s (t)

  4. s (t)


Correct Option: A
Explanation:

If R($\tau$) is the autocorrelation function of a real, wide-sense stationary random process, then which of the following is NOT true?

  1. R($\tau$) = R(-$\tau$)

  2. |R($\tau$)| $\le$ R(0)

  3. R($\tau$) = –R(–$\tau$)

  4. The mean square value of the process is R(0)


Correct Option: C
Explanation:

Autocorrelation is an even function.

A low-pass filter having a frequency response H (j$\omega$) = A ($\omega$)$e^{j \phi (\omega)}$ does not produce any phase distortion if

  1. A($\omega$) = C$\omega$2, $\phi$($\omega$) = k$\omega$3

  2. A($\omega$) = C$\omega$2, $\phi$($\omega$) = k$\omega$

  3. A($\omega$) = C$\omega$2, $\phi$($\omega$) = k$\omega$2

  4. A($\omega$) = C$\omega$2, $\phi$($\omega$) = k$\omega$-1


Correct Option: B
Explanation:

A LPE will not produce phase distortion if phase varies linearly with frequecy.

$\phi (\omega) \propto \omega \\ i.e \qquad \phi (\omega) = k \omega $

Let g (t) = p (t) * p (t), where * denotes convolution and p (t) = u (t) − u (t − 1) with u (t) being the unit step function.

An Amplitude Modulated signal is given as xAM (t) = 100 (p (t) + 0.5 g (t)) cos $\omega_0 t$ in the interval 0 $\le$ t $\le$ 1. One set of possible values of the modulating signal and modulation index would be

  1. t, 0.5

  2. t, 1.0

  3. t, 2.0

  4. 2t, 0.5


Correct Option: A
Explanation:

An AM signal is detected using an envelop detector. The carrier frequency and modulating signal frequency are 1 MHz and 2 kHz respectively. An appropriate value for the time constant of the envelop detector is

  1. 500 $\mu$sec

  2. 20 $\mu$sec

  3. 0.2 $\mu$sec

  4. 1 $\mu$sec


Correct Option: B
Explanation:

Consider two independent random variables X and Y with identical distributions. The variables X and Y take values 0, 1 and 2 with probabilities $\dfrac{1}{2}. \dfrac{1}{4}$ and $\dfrac{1}{4}$ respectively. What is the conditional probability P (X + Y = 2, X - Y = 0)?

  1. 0

  2. 1/16

  3. 1/6

  4. 1


Correct Option: C
Explanation:

For a message signal m(t) = cos $2\pi f_m t$ and carrier of frequency fc, which of the following represents a single side band (SSB) signal?

  1. cos $2\pi f_m t$ cos $2\pi f_o t$

  2. cos $2\pi f_o t$

  3. cos $[ (2\pi (f_o + f_m) ) t]$

  4. [1 + cos $2\pi f_m t$ cos $2\pi f_o t$ ]


Correct Option: C
Explanation:

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