# Communication Systems

 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      Random Signals and Noise
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 a channel over which either symbol $x_A$ or symbol $x_B$ is transmitted. Let the output of the channel Y be the input to a maximum likelihood (ML) detector at the receiver. The conditional probability density functions for Y given $x_A$and $x_B$ are:

$f_{Y|x_A}(y)=e^{-(y+1)}u(y+1),$
$f_{Y|x_B}(y)=e^{(y-1)}(1-u(y-1)),$

where $u(\cdot )$ is the standard unit step function. The probability of symbol error for this system is _________ (rounded off to two decimal places).
 A 0.23 B 0.56 C 0.74 D 0.65
GATE EC 2022      Digital Communication Systems
Question 2 Explanation:

ML decoding $\rightarrow \; f_{Y/X_A}(y)\underset{\overset{ \lt }{X_B}}{\overset{X_A}{ \gt }} f_{Y/X_B}(y)$
$i.e. \;\; f_{Y/X_A}(y) \gt f_{Y/X_B}(y) \rightarrow \text{ Decision favor of }X_A$
$i.e. \;\; f_{Y/X_A}(y) \lt f_{Y/X_B}(y) \rightarrow \text{ Decision favor of }X_B$
For $-1 \lt y \lt 0$ and $1 \lt y \lt \infty \rightarrow f_{Y/X_A}(y) \gt f_{Y/X_B}(y)$
For above internal decision in favor of $X_A$
For $-\infty \lt y \lt -1$ and $0\lt y \lt 1 \rightarrow f_{Y/X_B}(y) \gt f_{Y/X_A}(y)$
For above internal decision in favor of $X_B$
$P_e=P(X_A)\cdot P_{eX_A}+P(X_B)\cdot P_{eX_A}$
$P_{eX_A}\rightarrow$ Probability of error $X_A$ transmitted
$P_{eX_B}\rightarrow$ Probability of error $X_B$ transmitted
$P_{eX_A}=\int_{0}^{1} f_{Y/X_A}(y)dy=\int_{0}^{1}e^{-(y+1)}u(y+1)dy= \int_{0}^{1}e^{-(y+1)} dy=e^{-1}-e^{-2}=0.23$
$P_{eX_B}=\int_{0}^{1} f_{Y/X_B}(y)dy=\int_{0}^{1}e^{(y-1)} [1-u(y-1)]dy= \int_{0}^{1}e^{(y-1)} dy=e^{-1}-e^{-2}=0.23$
$P_e=P(X_A) \times 0.23+P(X_B) \times 0.23 =0.23[P(X_A)+P(X_B)]=0.23$
 Question 3
Consider communication over a memoryless binary symmetric channel using a (7, 4) Hamming code. Each transmitted bit is received correctly with probability $(1-\epsilon )$, and flipped with probability $\epsilon$. For each codeword transmission, the receiver performs minimum Hamming distance decoding, and correctly decodes the message bits if and only if the channel introduces at most one bit error.
For $\epsilon =0.1$, the probability that a transmitted codeword is decoded correctly is _________ (rounded off to two decimal places).
 A 0.25 B 0.65 C 0.85 D 0.94
GATE EC 2022      Information Theory and Coding
Question 3 Explanation:
Here (7, 4) Hamming code is given
P(0/1) = P(1/0) (due to bindary symmetry channel) = 0.1
When n bits are transmitted then probability of getting error in $\gamma$ bits $= ^nC_rP^r(1-p)^{n-r}$
P:Bit error probability
$P_c=C_0(0.1)^0[1-0.1]^{7-0}+^7C_1(0.1)(1-0.1)^{7-1}=(0.9)^7+7 \times 0.1 \times (0.9)^6=0.85$
$P_C$ : Prob of all most one bit error.
 Question 4
The transition diagram of a discrete memoryless channel with three input symbols and three output symbols is shown in the figure. The transition probabilities are as marked.
The parameter $\alpha$ lies in the interval [0.25, 1]. The value of $\alpha$ for which the capacity of this channel is maximized, is ________ (rounded off to two decimal places).

 A -1 B -2 C 2 D 1
GATE EC 2022      Information Theory and Coding
Question 4 Explanation:

Channel capacity,
$C_s=Max[I(X:Y)]$
$I[(X:Y)]=H(Y)-H\left ( \frac{Y}{X} \right )$
Where,
$H\left ( \frac{Y}{X} \right )=-\sum_{i=1}^{3}\sum_{i=1}^{3}P(x_i,y_i) \log_2P\left ( \frac{y_i}{x_i} \right )$
$P\left ( \frac{Y}{X} \right )=\begin{matrix} &y_1 & y_2 & y_3\\ x_1 &1-\alpha & \alpha & 0\\ x_2 &0 &1-\alpha &\alpha \\ x_3 & \alpha & 0 & 1-\alpha \end{matrix}$
For simplication convenience, let $[P(X)]=[1\;\;0\;\;0]$
\begin{aligned} [P(X,Y)]&=[P(X)]_d\cdot \left [ P\left ( \frac{Y}{X} \right ) \right ]\\ [P(X,Y)]&=\begin{bmatrix} 1-\alpha & \alpha &0 \\ 0& 0&0 \\ 0 & 0 &0 \end{bmatrix}\\ H\left ( \frac{Y}{X} \right )&=-((1-\alpha ) \log_2 (1-\alpha )+\alpha \log _2 \alpha )\\ I(X:Y)&=H(Y)+(1-\alpha ) \log_2 (1-\alpha )+ \log _2 \alpha \\ C_s&=MAX\left \{ (X:Y) \right \}\\ &=MAX\left \{ H(Y)+(1-\alpha ) \log_2 (1-\alpha )+ \log _2 \alpha \right \}\\ &=log_2 3+(1-\alpha ) \log_2 (1-\alpha )+\alpha \log _2 \alpha \end{aligned}
Plot of $(1-\alpha ) \log_2 (1-\alpha )+\alpha \log _2 \alpha$

$C_s$ will be maximum at $\alpha =0$ and 1.
Given $\alpha \in [0.25,1]$
Hence, $\alpha =1$ is correct answer.
 Question 5
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      Random Signals and Noise
Question 5 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 6
A symbol stream contains alternate QPSK and 16-QAM symbols. If symbols from this stream are transmitted at the rate of 1 mega-symbols per second, the raw (uncoded) data rate is _______ mega-bits per second (rounded off to one decimal place).
 A 2.2 B 3 C 4.8 D 5.3
GATE EC 2022      Digital Communication Systems
Question 6 Explanation:
\begin{aligned} \text{BIT RATE}&=[\text{SYBOL RATE}] \times \log_{20}M\\ 1.\;\; QPSK, \; M&=4,n=2\\ R_{b1}&=1 \times 2 =2mbps\\ 2.\;\; 16 QAM_1\Rightarrow M=16,n=4\\ R_{b2}&=1 \times 4=4mbps\\ R_b&=\frac{2m+4m}{2}\\ R_b&=3 mbps \end{aligned}
 Question 7
Let $H(X)$ denote the entropy of a discrete random variable $X$ taking $K$ possible distinct real values. Which of the following statements is/are necessarily true?
 A $H(X)\leq \log _2 K bits$ B $H(X)\leq H(2X)$ C $H(X)\leq H(X^2)$ D $H(X)\leq H(2^X)$
GATE EC 2022      Digital Communication Systems
Question 7 Explanation:

$H(X^2)=H(Y)=\Sigma P_Y(Y_i) \log _2 \frac{1}{P_Y(Y_i)}= \frac{1}{2} \log _2 2+\frac{1}{2} \log _2 2=1 bit/symbol$
$H(X)=\frac{1}{4} \log _2 4+\frac{1}{2} \log _2 2+\frac{1}{4} \log _2 4=1.5 bit/symbol$
$H(X) \gt H(X^2)$
Hence option (C) is not correct.
 Question 8
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      Random Signals and Noise
Question 8 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 9
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      Random Signals and Noise
Question 9 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 10
A message signal having peak-to-peak value of $2\:V$, root mean square value of $0.1\:V$ and bandwidth of $\text{5 kHz}$ is sampled and fed to a pulse code modulation $(\text{PCM})$ system that uses a uniform quantizer. The $\text{PCM}$ output is transmitted over a channel that can support a maximum transmission rate of $\text{50 kbps}$. Assuming that the quantization error is uniformly distributed, the maximum signal to quantization noise ratio that can be obtained by the $\text{PCM}$ system (rounded off to two decimal places) is ______
 A 38.25 B 64.2 C 12.45 D 30.72
GATE EC 2021      Digital Communication Systems
Question 10 Explanation:
\begin{aligned} v_{p-p} &=2 \mathrm{~V} \\ \text { Root MSQ }[m(t)] &=0.1 \mathrm{~V} ; \quad f_{m}=5 \mathrm{kHz} \\ \text { Channel capacity, } \quad C &=50 \mathrm{kbps} \\ \text{Max} \frac{S}{N_{Q}}=?\\ & \text{Signal power},\\ S&=\text{MSQ}[m(t)]=(0.1)^{2}=0.01\\ C &\geq R_{b} \Rightarrow 50 \mathrm{kbps} \geq n f_{s}\\ \because \qquad f_{s}&=N R=2 f_{m}=10 \mathrm{kHz}\\ n &\leq 5 \Rightarrow n_{\max }=5\\ N_{Q} &=\frac{\Delta^{2}}{12} \\ \therefore\qquad \Delta &=\frac{V_{p-p}}{2^{n}} \\ \Delta_{\min } &=\frac{V_{p-p}}{2^{7} \max }=\frac{2 V}{2^{5}}=\frac{1}{16} \\ \left(N_{Q}\right)_{\min } &=\frac{\Delta_{\min }^{2}}{12}=3.25 \times 10^{-4} \\ \left(\frac{S}{N Q}\right)_{\max } &=\frac{S}{\left(N_{Q}\right)_{\min }}=\frac{0.01}{3.25 \times 10^{-4}}=30.72 \end{aligned}

There are 10 questions to complete.