Random Signals and Noise

Question 1
Consider a real valued source whose samples are independent and identically distributed random variables with the probability density function, f(x), as shown in the figure.

Consider a 1 bit quantizer that maps positive samples to value \alpha and others to value \beta . If \alpha ^* and \beta ^* are the respective choices for \alpha and \beta that minimize the mean square quantization error, then (\alpha ^*- \beta ^*)= _________ (rounded off to two decimal places).
A
1.16
B
1.85
C
2.21
D
3.63
GATE EC 2022   Communication Systems
Question 1 Explanation: 
\frac{1}{2} \times K \times 2+1 \times K=1\Rightarrow K=0.5
\begin{aligned} f_X(x)&=mx+C \\ &=0.25+C \;\;;\;(-2\leq x\leq 0) \\ when \; x &=-2 \Rightarrow f_X(x)=0\\ 0&=0.25x-2+C \\ C&=0.5 \\ f_X(x)&= \frac{1}{4}x+\frac{1}{2}=-2\leq x\leq 0\\ f_X(x)&=0.5;\;\;0\leq x\leq 1 \\ x_q&=\alpha ;\;\; for\; 0\leq x\leq 1 \\; x_q&= \beta ;\;\; for\; -2\leq x\leq 0 \\; \end{aligned}
Again, MSQ[Q_e]=E\left [ (X-X_q)^2 \right ]
Quantization noise power =N_o
=MSQ[Q_e]=\int (X-X_a)^2 f_X(x)dx
for -2 \leq x\leq 0\Rightarrow N_Q=\int_{-2}^{0}(x-\beta )^2 \times \left ( \frac{1}{4}x+\frac{1}{2} \right )dx
=\int_{-2}^{0}[x^2+\beta ^2-2x\beta ] \left [ \frac{x}{4}+\frac{1}{2} \right ]dx
\Rightarrow N_Q=\frac{\beta ^2}{2}+\frac{2}{3}\beta -\frac{1}{3}
N_Q to be minimum:
\frac{dN_Q}{d\beta }=0
\Rightarrow \frac{1}{2} \times 2\beta +\frac{2}{3}=0
\beta =-\frac{2}{3}
for 0\leq x\leq 1
\Rightarrow N_Q=\int_{0}^{1}(x-\alpha )^2 \times \frac{1}{2}dx =\frac{1}{6} [(1-\alpha )^3+\alpha ^3]
Similarly for '\alpha '
\frac{dN_Q}{d\alpha }=0
\Rightarrow \;\frac{1}{6} [3(1-\alpha )^2(-1)+3\alpha ^2]=0
\alpha =1/2
For N_q to be minimum
\alpha -\beta =\frac{1}{2}-\left ( -\frac{2}{3} \right )=\frac{7}{6}=1.167
Question 2
Consider an FM broadcast that employs the pre-emphasis filter with frequency response
H_{pe}(\omega )=1+\frac{j\omega }{\omega _0}
where \omega _0=10^4 rad/sec. For the network shown in the figure to act as a corresponding de-emphasis filter, the appropriate pair(s) of (R,C) values is/are ________.

A
R=1k\Omega ,C=0.1\mu F
B
R=2k\Omega ,C=1\mu F
C
R=1k\Omega ,C=2\mu F
D
R=2k\Omega ,C=0.5\mu F
GATE EC 2022   Communication Systems
Question 2 Explanation: 
\begin{aligned} H_{pe}(f) &=\frac{1}{H_{de}(f)} \\ \Rightarrow |H_{pe}(f)|^2&= \frac{1}{|H_{de}(f)|^2} \;\;\;...(i)\\ H_{Pe}(\omega )&=1+j\frac{\omega }{\omega _0}\;\;where\; \omega _0=10^4 \\ |H_{Pe}(\omega )|&= \sqrt{1+(\omega /\omega _0)^2}\\ |H_{Pe}(\omega )|^2 &=1+ (\omega /\omega _0)^2 \;\;...(ii)\\ H_{de}(\omega ) &=\frac{1}{1+j\omega RC} \\ |H_{de}(\omega )|^2 &=\frac{1}{1+(j\omega RC)^2} \;\;...(iii)\\ from \; (i)&,(ii),(iii) \\ \omega _0&= \frac{1}{RC}=10^4 \end{aligned}
\Rightarrow R=1k\Omega ,C=0.1\mu F satisfies only
Question 3
The frequency response H(f) of a linear time-invariant system has magnitude as shown in the figure.

Statement I: The system is necessarily a pure delay system for inputs which are bandlimited to -\alpha \leq f\leq \alpha .
Statement II: For any wide-sense stationary input process with power spectral density S_X(f) , the output power spectral density S_Y(f) obeys S_Y(f)=S_X(f) for -\alpha \leq f\leq \alpha .

Which one of the following combinations is true?

A
Statement I is correct, Statement II is correct
B
Statement I is correct, Statement II is incorrect
C
Statement I is incorrect, Statement II is correct
D
Statement I is incorrect, Statement II is incorrect
GATE EC 2022   Communication Systems
Question 3 Explanation: 


For the system to be delay system
\begin{aligned} y(t)&=x(t-t_d) \\ y(F)&=e^{-J\omega t_d} \times F\\ \Rightarrow H(F)&=\frac{Y(F)}{X(F)}=e^{-J\omega t_d} \end{aligned}
Therefore, TF of delay system
Here given system is constant, hence this is not delay system, therefore statement-I is Incorrect
S_y(f)=S_x(f)|H(f)|^2
and |H(f)|=1 (given)
Hence, S_y(f)=S_x(f)\; for \; -\alpha \leq f\leq \alpha
Statement - II is correct.
Question 4
Consider a polar non-return to zero (\text{NRZ}) waveform, using +2\:V and -2\:V for representing binary '1' and '0' respectively, is transmitted in the presence of additive zero-mean white Gaussian noise with variance 0.4\:V^{2}. If the a priori probability of transmission of a binary '1' is 0.4, the optimum threshold voltage for a maximum a posteriori (\text{MAP}) receiver (rounded off to two decimal places) is ______ V.
A
0.2
B
0.01
C
0.04
D
0.4
GATE EC 2021   Communication Systems
Question 4 Explanation: 


\begin{aligned} H_{1}: X &=+2 \mathrm{~V} \\ H_{0}: X &=-2 \mathrm{~V} \\ \text{Var}[N] &=\sigma_{n}^{2}=0.4 \mathrm{~V}^{2} \\ \mathrm{E}[\mathrm{N}] &=0 \\ P(1) &=0.4 \\ \Rightarrow\qquad P(0) &=0.6 \end{aligned}
Opt V_{\text {Th }} by using MAP theorem
\begin{aligned} \frac{V_{T h}\left[a_{1}-a_{2}\right]}{\sigma^{2}}-\frac{a_{1}^{2}-a_{2}^{2}}{2 \sigma^{2}} &=\ln \frac{P(0)}{P(1)} \\ H_{1}: a_{1} &=E[2+N]=E[2]+E[N]=2 \\ H_{0}: a_{2} &=-2 V=E[-2+N]=E[-2]+E[N]=-2 \\ \sigma^{2} &=\text{Var}[Y]=\text{Var}[X+N] \\ &=\text{Var}[X]+\text{Var}[N]=0+0.4=0.4 \\ \frac{V_{T h}[2+2]}{0.4}-\frac{4-4}{2 \times 0.4} &=\ln \frac{0.6}{0.4} \\ V_{T h} &=\frac{0.4}{4} \ln \frac{0.6}{0.4}=0.0405 \\ \text { Opt } V_{\text {Th }} &=0.0405 \text { Volts } \end{aligned}
Question 5
The autocorrelation function R_{X}\left ( \tau \right ) of a wide-sense stationary random process X(t) is shown in the figure.

The average power of X(t) is ________________
A
1
B
2
C
3
D
4
GATE EC 2021   Communication Systems
Question 5 Explanation: 


Average power [X(t)]=R_{x}(0)=2
Question 6
Two continuous random variables X and Y are related as
Y=2X+3
Let \sigma ^{2}_{X} and \sigma ^{2}_{Y} denote the variances of X and Y, respectively. The variances are related as
A
\sigma ^{2}_{Y}=2 \sigma ^{2}_{X}
B
\sigma ^{2}_{Y}=4 \sigma ^{2}_{X}
C
\sigma ^{2}_{Y}=5 \sigma ^{2}_{X}
D
\sigma ^{2}_{Y}=25 \sigma ^{2}_{X}
GATE EC 2021   Communication Systems
Question 6 Explanation: 
\begin{aligned} Y &=2 X+3 \\ \operatorname{Var}[Y] &=E\left[(Y-\bar{Y})^{2}\right] \\ E[Y] &=\bar{Y}=2 \bar{X}+3 \\ \operatorname{Var}[Y] &=E\left[(2 X+3-2 \bar{X}-3)^{2}\right] \\ &=E\left[4(X-\bar{X})^{2}\right] \\ &=4 \cdot E\left[(X-\bar{X})^{2}\right] \\ \sigma_{Y}^{2} &=4 \cdot \sigma_{X}^{2} \end{aligned}
Question 7
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   Communication Systems
Question 7 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 8
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   Communication Systems
Question 8 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 9
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   Communication Systems
Question 9 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 10
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   Communication Systems
Question 10 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
There are 10 questions to complete.