# Communication Systems

 Question 1
The frequency of occurrence of 8 symbols $(a-h)$ is shown in the table below. A symbol is chosen and it is determined by asking a series of "yes/no" questions which are assumed to be truthfully answered. The average number of questions when asked in the most efficient sequence, to determine the chosen symbol, is ____
(rounded off to two decimal places).
$\begin{array}{|c|c|c|c|c|c|c|c|c|} \hline Symbols & a & b & c & d & e & f & g & h \\ \hline Frequency& \frac{1}{2} & \frac{1}{4} & \frac{1}{8} & \frac{1}{16} & \frac{1}{32} & \frac{1}{64} & \frac{1}{128} & \frac{1}{128} \\ of \; occurrence& &&&&&&& \\ \hline \end{array}$
 A 0.25 B 1.98 C 2.48 D 3.36
GATE EC 2023      Information Theory and Coding
Question 1 Explanation:
The average number of questions when asked in the most efficient sequence, to determine the chosen symbol $=\min$ possible number of questions per symbol $(H)$
$H=\sum_{i} P_{x}\left(x_{i}\right) \log _{2} \frac{1}{P_{x}\left(x_{i}\right)} =\frac{1}{2} \log _{2} 2+\frac{1}{4} \log _{2} 4+\frac{1}{8} \log _{2} 8+\frac{1}{16} \log _{2} 16+\frac{1}{32} \log _{2} 32+\frac{1}{64} \log _{2} 64+2 \times \frac{1}{128} \log _{2} 128 =1.984 \frac{\text { Questions }}{\text { Symbol }}$
 Question 2
Let a frequency modulated (FM) signal $x(t)=A \cos \left(\omega_{c} t+k_{f} \int_{-\infty}^{t} m(\lambda) d \lambda\right)$, where $m(t)$ is a message signal of bandwidth $W$. It is passed through a non-linear system with output $y(t)=2 x(t)+5(x(t))^{2}$. Let $B_{T}$ denote the FM bandwidth. The minimum value of $\omega_{c}$ required to recover $x(t)$ from $y(t)$ is
 A $B_{T}+W$ B $\frac{3}{2} B_{T}$ C $2 B_{T}+W$ D $\frac{5}{2} B_{T}$
GATE EC 2023      Analog Communication Systems
Question 2 Explanation:
$x(t)=A \cos \left[\omega_{c} t+K_{f} \int_{-\infty}^{t} m(\lambda) d \lambda\right]$
B.W. $[x(t)] \rightarrow B T=2[\Delta f+\omega]$ $\left.x^{2}(t) \rightarrow \begin{array}{c} \Delta f^{\prime}=2 \Delta f \\ \omega_{c}^{\prime}=2 \omega_{c} \end{array}\right\}$ $\mathrm{BW}\left[x^{2}(t)\right]=2\left[\Delta f^{\prime}+\omega\right]$ $=2[2 \Delta f+\omega]=B T+2 \Delta f$
$y(t)=2x(t)+5x^2(t)$ To recover $x(t) \rightarrow \quad 2 \omega_{C}-\frac{B_{T}}{2}-\Delta f \gt \omega_{C}+\frac{B_{T}}{2}$
$\omega_{c} \gt \Delta f+B T$
$\omega_{c} \gt \Delta f+2 \Delta f+2 \omega$
$\omega_{c} \gt 3 \Delta f+2 \omega$
$\omega_{c} \gt \frac{3}{2}\{2[\Delta f+\omega]\}-\omega$
$\omega_{c} \gt \frac{3}{2} B_{T}-\omega$
Compared to FM BW, message BW is very small. So, that it can be ignored.
\begin{aligned} \omega_{C} & \gt \frac{3}{2} B_{T} \\ \left[\omega_{C}\right]_{min } & =\frac{3}{2} B_{T} \end{aligned}

 Question 3
The signal-to-noise ratio (SNR) of an ADC with a full-scale sinusoidal input is given to be $61.96 \mathrm{~dB}$. The resolution of the ADC is _____ bits. (rounded off to the nearest integer).
 A 2 B 8 C 10 D 12
GATE EC 2023      Analog Communication Systems
Question 3 Explanation:
We know that for sinusoidal input, the signal to noise ratio (SNR) is given as,
\begin{aligned} \mathrm{SNR} & =1.76+6.02 \mathrm{n} \mathrm{dB} \\ 61.96 \mathrm{~dB} & =1.76+6.02 \mathrm{n} \mathrm{dB} \\ 6.02 \mathrm{n} & =61.96-1.76 \\ n & =10 \text { bits } \end{aligned}
 Question 4
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      Random Signals and Noise
Question 4 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 5
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 5 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
$\alpha -\beta =\frac{1}{2}-\left ( -\frac{2}{3} \right )=\frac{7}{6}=1.167$

There are 5 questions to complete.