# Random Signals and Noise

 Question 1
For a real signal, which of the following is/are valid power spectral density/densities? A A B B C C D D
GATE EC 2023   Communication Systems
Question 1 Explanation:
(i) $S_{x}(\omega) \geq 0$
(ii) $S_{x}(\omega)$ is even function

Hence, options (A) and (B) are valid power spectral densities.
 Question 2
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 2 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 3
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 3 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 4
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 4 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 5
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 5 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+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}

There are 5 questions to complete.