# Communication Systems

 Question 1
X is a random variable with uniform probability density function in the interval [-2,10]. For Y = 2X-6, the conditional probability $P(Y \leq 7|X \geq 5)$ (rounded off to three decimal places) is _____.
 A 0.1 B 0.3 C 0.6 D 0.8
GATE EC 2020      Random Signals and Noise
Question 1 Explanation:
$x$ follows uniform distribution over [-2, 10]
\begin{aligned} \therefore \; f(x)&=\frac{1}{b-a}=\frac{1}{10-(-2)}=\frac{1}{12}\\ \text{Given: } y &= 2x - 6\\ \Rightarrow \; x&=\frac{y+6}{2}\\ \text{For } y = 7 \\ x&=\frac{7+6}{2}=\frac{13}{2}=6.5\\ P\left [ \frac{y\lt7}{x\gt 5} \right ]&=P\left [ \frac{x\lt6.5}{x\gt 5} \right ] \\ &=\frac{P\left [ x\gt 5\, and\, x\lt6.5 \right ]}{P\left [ x\gt 5 \right ]} \\ &=\frac{P\left [ 5\lt x \lt 6.5 \right ]}{P[x\gt 5]} \\ &=\frac{\int_{5}^{6.5}f(x)dx}{\int_{5}^{10}F(x)dx} \\ &=\frac{\int_{5}^{6.5}\frac{1}{12}dx}{\int_{5}^{10}\frac{1}{12}dx} \\ &=\frac{(x)_{6.5}^{5}}{(x)_{5}^{10}}=\frac{1.5}{5}=0.3\end{aligned}
 Question 2
In a digital communication system, a symbol S randomly chosen from the set ($s_1, s_2, s_3, s_4$) is transmitted. It is given that $s_1=-3, s_2=-1, s_3=+1$ and $s_4=+2$. The received symbol is Y = S + W. W is a zero-mean unit-variance Gaussian random variable and is independent of S. $P_i$ is the conditional probability of symbol error for the maximum likelihood (ML) decoding when the transmitted symbol $S=s_i$. The index i for which the conditional symbol error probability $P_i$ is the highest is ______.
 A 1 B 2 C 3 D 4
GATE EC 2020      Random Signals and Noise
Question 2 Explanation:
Since the noise variable is Gaussian with zero mean and ML decoding is used, the decision boundary between two adjacent signal points will be their arithmetic mean. In the following graphs, the shaded area indicates the conditional probability of decoding a symbol correctly when it is transmitted.

By comparing the above graphs, we can conclude that $P_{3}$ is larger among the four. End of Solution
 Question 3
$S_{PM}(t)\; and \; S_{FM}(t)$ as defined below, are the phase modulated and the frequency modulated waveforms, respectively, corresponding to the message signal m(t) shown in the figure.

$S_{PM}(t)= \cos (1000 \pi t+K_p m(t))$
and $S_{FM}(t)= \cos (1000 \pi t+K_f \int_{-\infty }^{t} m(\tau )d\tau )$

where $K_p$ is the phase deviation constant in radians/volt and $K_f$ is the frequency deviation constant in radians/second/volt. If the highest instantaneous frequencies of $S_{PM}(t)\; and \; S_{FM}(t)$ are same, then the value of the ratio $\frac{K_p}{K_f}$ is ______ seconds.
 A 0.5 B 1 C 2 D 2.5
GATE EC 2020      Analog Communication Systems
Question 3 Explanation:
\begin{aligned}S(t)_{pm}&=A_c \cos [2\pi f_{c}t+k_pm(t)]\\ S(t)_{Fm}&=A_c \cos [2\pi f_{c}t+k_{f}\int_{0}^{\infty }m(t)dt]\end{aligned}
Instantaneous frequency are equal
\begin{aligned}f_{i}&=\frac{1}{2\pi }\frac{\mathrm{d} }{\mathrm{d} t}\theta (t)\\f_{iPM}&=f_{c}+\frac{K_p}{2\pi }\frac{\mathrm{d} }{\mathrm{d} t} m(t)\\ f_{iFM}&=f_{c}+\frac{K_f}{2\pi }m(t)\\ \text{Given that, } (f_{iPM})_{max}&=(f_{iFM})_{max} \\ f_{c}+\frac{k_P}{2\pi}[\frac{\mathrm{d} }{\mathrm{d} t}m(t)]_{max}&=f_{c}+\frac{K_{f}}{2\pi }[m(t)]_{max}\\ \left.\begin{matrix} \frac{k_P}{2\pi}\frac{\mathrm{d} }{\mathrm{d} t}m(t) \end{matrix}\right|_{max} &= \left. \begin{matrix} \frac{K_{f}}{2\pi }[m(t)] \end{matrix}\right|_{max}\\ \left.\begin{matrix} \frac{\mathrm{d} }{\mathrm{d} t}m(t) \end{matrix}\right|_{max} &=5, \left.\begin{matrix} m(t) \end{matrix}\right|_{max}=10 \\ 5k_{p}&=10k_{f} \\ \frac{k_{p}}{k_{f}}&=2\end{aligned}
 Question 4
For the modulated signal $x(t)= m(t) \cos(2 \pi f_c t)$, the message signal $m(t) = 4 \cos (1000 \pi t)$ and the carrier frequency $f_c$ is 1 MHz. The signal $x(t)$ is passed through a demodulator, as shown in the figure below. The output $y(t)$ of the demodulator is
 A $\cos (460 \pi t)$ B $\cos (920 \pi t)$ C $\cos (1000 \pi t)$ D $\cos (540 \pi t)$
GATE EC 2020      Analog Communication Systems
Question 4 Explanation:
Output of multiplier
$=x(t)\cos 2\pi (f_{c}+40)t=m(t)\cos 2\pi f_{c}t\cdot \cos2\pi (f_{c}+40)t$
$=\frac{m(t)}{2}[\cos 2\pi (2f_{c}+40)t+\cos 2\pi (40)t]$
Given, $m(t)=4\cos 1000\pi t$
So, output of multiplier
$=2\cos 2\pi (500)t[\cos 2\pi (2f_{c}+40)t+\cos 2\pi (40)t]$
$=\cos 2\pi (2f_{c}+540)t+\cos 2\pi (2f_{c}-460)t+\cos 2\pi (540)t+\cos 2\pi (460)t$
Output of Low pass filter
$=\cos [2\pi (460)]t$
$=\cos 920\pi t$
 Question 5
The random variable

$Y=\int_{-\infty }^{\infty }W(t)\phi (t) dt$
where $\phi (t) =\left\{\begin{matrix} 1; & 5\leq t\leq 7\\ 0; & otherwise \end{matrix}\right.$

and W(t) is a real white Gaussian noise process with two-sided power spectral density $S_W(f) = 3$ W/Hz, for all f. The variance of Y is _________.
 A 3 B 4 C 6 D 8
GATE EC 2020      Random Signals and Noise
Question 5 Explanation:
\begin{aligned} \text{Given: }\phi (t) &=\left\{\begin{matrix} 1 & 5\leq t\leq 7\\ 0 & otherwise \end{matrix}\right. \\ s_{\omega }(f)& =3 \text{ Watts/Hz} \\ R_{\omega }(\tau )&=3\delta (\tau )=3\delta (t_{1}-t_{2}) \\ Var[y]&=E[y^{2}]-{E[y]}^{2} \\ {E[W(t)]}^{2}&=\text{DC power} \\ &=\text{Area under PSD at }f=0 \\ {E[W(t)]}^{2}&=0 \\ E[W(t)]&=0 \\ y&=\int_{-\infty }^{\infty }W(t)\phi (t)dt \\ E[y]&=\int_{-\infty }^{\infty }E[W(t)]\phi (t)dt \\ y&=\int_{-\infty }^{\infty }W(t)\phi (t)dt\rightarrow E[y^{2}]\\ &=S_{\omega }(f)\cdot Energy[\phi (t)]\\ & =3 \times 2=6\\ \text{Var}[y]&=6-0=6\end{aligned}

Detailed Explanations for:
\begin{aligned}y&=\int_{-\infty }^{\infty }W(t)\phi (t)dt \\ E(y^{2})&=E[y \cdot y] \\ &=E[\int_{-\infty }^{\infty }W(t_{1})\phi (t_{1})dt_{1}\int_{-\infty }^{\infty }W(t_{2})\phi (t_{2})dt_{2}] \\ &=E[\int_{-\infty }^{\infty }\int_{-\infty }^{\infty }W(t_{1})W(t_{2})\phi (t_{1})\phi (t_{2})dt_{1}\cdot dt_{2}] \\ &=\int_{-\infty }^{\infty }\int_{-\infty }^{\infty }E[W(t_{1})W(t_{2}) \cdot \phi (t_{1})\phi (t_{2})dt_{1} dt_{2}] \\ &=\int_{-\infty }^{\infty }\int_{-\infty }^{\infty }R_{W}(t_{1}-t_{2})\phi (t_{1})\phi (t_{2})dt_{1}dt_{2} \\ &=\int_{-\infty }^{\infty }\int_{-\infty }^{\infty }3\delta (t_{1}-t_{2})\phi (t_{1})\phi (t_{2})dt_{1}dt_{2}\end{aligned}

Above Integration Exists Provided
\begin{aligned}t_{1}&=t_{2}=t\\ &=\int_{-\infty }^{\infty }3\delta (0)\phi (t)\phi (t)dt \, dt\\ &=3 \int_{-\infty }^{\infty } \delta (0) dt + \int_{-\infty }^{\infty } \phi ^2 (t) dt\\&=2 times 1 times \text{Energy}[\phi (t)] \\E[y^{2}]&=6\end{aligned}
 Question 6
A binary random variable X takes the value +2 or -2. The probability $P(X=+2)= \alpha$. The value of $\alpha$ (rounded off to one decimal place), for which the entropy of X is maximum, is __________
 A 0.2 B 0.5 C 0.7 D 0.9
GATE EC 2020      Random Signals and Noise
Question 6 Explanation:
Given that, $P(X=2)=\alpha$
Entropy Will be Maximum;
Provided Probabilities are equal.
i.e. $P(X=2)=P(X=-2)=\alpha =\frac{1}{2}$
$\alpha =\frac{1}{2}=0.5$
 Question 7
A digital communication system transmits a block of N bits. The probability of error in decoding a bit is $\alpha$. The error event of each bit is independent of the error events of the other bits. The received block is declared erroneous if at least one of the its bits is decoded wrongly. The probability that the received block is erroneous is
 A $N(1-\alpha )$ B $\alpha ^n$ C $1-\alpha ^n$ D $1-(1-\alpha )^n$
GATE EC 2020      Information Theory and Coding
Question 7 Explanation:
Probability of error in decoding single bit=$\alpha$
Then probability of no error will be $1- \alpha$.
Total N-bits transmitted, so that probability of no error in received block
=$(1-\alpha )(1-\alpha )...N \, \text{ Times}$
$=(1-\alpha) ^{N}$
The Probability of received block is erroneous is $=1-(1-\alpha) ^{N}$
 Question 8
A random variable X takes values -1 and +1 with probabilities 0.2 and 0.8, respectively. It is transmitted across a channel which adds noise N, so that the random variable at the channel output is Y=X+N. The noise N is independent of X, and is uniformly distributed over the interval [-2,2]. The receiver makes a decision

$\hat{X}=\left\{\begin{matrix} -1 & if & Y\leq \theta \\ +1& if & Y> \theta \end{matrix}\right.$

where the threshold $\theta \in [-1,1]$ is chosen so as to minimize the probability of error $Pr[\hat{X}\neq X]$. The minimum probability of error, rounded off to 1 decimal place, is ______.
 A 0.1 B 0.2 C 0.3 D 0.8
GATE EC 2019      Digital Communication Systems
Question 8 Explanation:
MAP criteria should be used to minimise the probability of error.
$\begin{array}{c} f_{Y}(y \mid+1) P(+1) \sum_{-1}^{+1} f_{Y}(y \mid-1) F(-1) \\ P(+1)=0.80 \text { and } P(-1)=0.20 \end{array}$

$P_{\theta(m \text { in })}=$ Shaded area $=2 \times \frac{1}{20}=0.10$
 Question 9
A voice signal m(t) is in the frequency range 5 kHz to 15 kHz. The signal is amplitude-modulated to generate an AM signal $f(t)=A(1+m(t))cos 2\pi f_c t$, where $f_c$=600kHz. The AM signal f(t) is to be digitized and archived. This is done by first sampling f(t) at 1.2 times the Nyquist frequency, and then quantizing each sample using a 256-level quantizer. Finally, each quantized sample is binary coded using K bits, where K is the minimum number of bits required for the encoding. The rate, in Megabits per second (rounded off to 2 decimal places), of the resulting stream of coded bits is _______ Mbps.
 A 9.25 B 11.81 C 25.36 D 14.25
GATE EC 2019      Digital Communication Systems
Question 9 Explanation:

\begin{aligned} \text { Nyquist rate } &=2 \times 615 \mathrm{kHz} \\ &=1230 \mathrm{kHz}=1.23 \mathrm{MHz} \\ f_{s} &=1.2 \times 1.23=1.476 \mathrm{MHz} \\ \text { Bits/sample }, n &=\log _{2}(256)=8 \\ \text { So, } \quad A_{D} &=n f_{s}=8 \times 1.476 \\ &=11.808 \simeq 11.81 \mathrm{Mbps} \end{aligned}
 Question 10
Let a random process Y(t) be described as Y(t)=h(t)*X(t)+Z(t), where X(t) is a white noise process with power spectral density $S_X(f)=5$ W/Hz. The filter h(t) has a magnitude response given by |H(f)|=0.5 for $-5\leq f\leq 5$, and zero elsewhere. Z(t) is a stationary random process, uncorrelated with X(t), with power spectral density as shown in the figure. The power in Y(t), in watts, is equal to ________ W (rounded off to two decimal places).
 A 17.5 B 12.24 C 18.88 D 24.36
GATE EC 2019      Random Signals and Noise
Question 10 Explanation:
\begin{aligned} \text { Let, } \quad X_{1}(t) &=h(t) * X(t) \\ S_{X 1}(f) &=|H(f)|^{2} S_{X}(f) \end{aligned}

Given that, Z(t) and X(t) are uncorrelated
\begin{aligned} & \text{So,}\quad S_{y}(f)=S_{X 1}(f)+S_{z}(f)\\ & \text{Power in }\mathrm{y}(t) \\ P_{Y}&=\left[\text { Area under } S_{X 1}(f)\right]+\left[\text { Area under } S_{z}(f)\right] \\ &=(10 \times 1.25)+(5 \times 1)=17.5 \mathrm{W} \end{aligned}

There are 10 questions to complete.