Test 1 - Communication Systems | Electronics and Communication (ECE)
Topic wise test for Communication Systems of Electronics and Communication (ECE)
Questions
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
- 2
- 1
- 0.5
- 0
The input to a coherent detector is DSB-SC signal plus noise. The noise at the detector output is
- the in-phase component
- the quadrature component
- zero
- the envelope
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
- $\dfrac{8}{6}$
- 12
- 16
- 8
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
- 6000/$\pi$ and 0
- 6400/$\pi$ and 0
- 6400/$\pi$ and 20/($\pi$$\sqrt 2$)
- 6000/$\pi$ and 20/($\pi$$\sqrt 2$)
The probability density function (pdf) of random variable is as shown below:

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
- 6000 bits/sec
- 4500 bits/sec
- 3000 bits/sec
- 1500 bits/sec
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?
- x(t) = Acm(t) cos(2$\pi$fct)
- x(t) = Ac[1 + m(t)]cos(2$\pi$fct)
- x(t) = Ac cos(2$\pi$fct) + $\dfrac{A_0}{4}$m(t) cos (2$\pi$fCt)
- x(t) = Ac cos(2$\pi$fmt) cos(2$\pi$fct) + Ac sin(2$\pi$fmt) sin(2$\pi$fct)
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
- 0
- 0.5
- $\infty$
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
- Q$\left( \sqrt{ \dfrac{2E}{N_0} } \right) $
- Q$\left( \sqrt{ \dfrac{E}{N_0} } \right) $
- Q$\left( \sqrt{ \dfrac{E}{2N_0} } \right) $
- Q$\left( \sqrt{ \dfrac{E}{4N_0} } \right) $
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)
- RC < $\dfrac{1}{W}$
- RC > $\dfrac{1}{W}$
- RC < $\dfrac{1}{\omega_c}$
- RC > $\dfrac{1}{\omega_c}$
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 probability of bit error is
- 0.5xe-3.5
- 0.5xe-5
- 0.5xe-7
- 0.5xe-10
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
- 1750
- 2625
- 4000
- 5250
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
- 1010
- 107
- 1.414$\times$10-10
- 2$\times$10-20
Let x(t) = 2 cos (800$\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

- 2.7, 3.4
- 3.3, 3.6
- 2.6, 2.7, 3.3, 3.4, 3.6
- 2.7, 3.3
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|
- P - 1, Q - 3, R - 2, S - 4
- P - 6, Q - 5, R - 2, S - 3
- P - 6, Q - 1, R - 3, S - 2
- P - 5, Q - 6, R - 1, S - 3
A memory less source emits n symbols each with a probability p. The entropy of the source as a function of n
- increases as log n
- decreases as log $\dfrac{1}{n}$
- increases as n
- increases as n log n
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
- 200
- 40
- 25
- 5
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:
- s (1 − t)
- − s (1 − t)
- − s (t)
- s (t)
If R($\tau$) is the autocorrelation function of a real, wide-sense stationary random process, then which of the following is NOT true?
- R($\tau$) = R(-$\tau$)
- |R($\tau$)| $\le$ R(0)
- R($\tau$) = –R(–$\tau$)
- The mean square value of the process is R(0)
A low-pass filter having a frequency response H (j$\omega$) = A ($\omega$)$e^{j \phi (\omega)}$ does not produce any phase distortion if
- A($\omega$) = C$\omega$2, $\phi$($\omega$) = k$\omega$3
- A($\omega$) = C$\omega$2, $\phi$($\omega$) = k$\omega$
- A($\omega$) = C$\omega$2, $\phi$($\omega$) = k$\omega$2
- A($\omega$) = C$\omega$2, $\phi$($\omega$) = k$\omega$-1
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
- t, 0.5
- t, 1.0
- t, 2.0
- 2t, 0.5
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
- 500 $\mu$sec
- 20 $\mu$sec
- 0.2 $\mu$sec
- 1 $\mu$sec
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)?
- 0
- 1/16
- 1/6
- 1
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?
- cos $2\pi f_m t$ cos $2\pi f_o t$
- cos $2\pi f_o t$
- cos $[ (2\pi (f_o + f_m) ) t]$
- [1 + cos $2\pi f_m t$ cos $2\pi f_o t$ ]
















