# Digital Communication Systems

 Question 1
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   Communication Systems
Question 1 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 2
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   Communication Systems
Question 2 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 3
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   Communication Systems
Question 3 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 4
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   Communication Systems
Question 4 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}
 Question 5
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   Communication Systems
Question 5 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 $=2 \times \frac{1}{20}=0.10$

There are 5 questions to complete.