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
Question 6
A voice signal m(t) is in the frequency range 5 kHz to 15 kHz. The signal is amplitude-modulated to generate an AM signal f(t)=A(1+m(t))cos 2\pi f_c t, where f_c=600kHz. The AM signal f(t) is to be digitized and archived. This is done by first sampling f(t) at 1.2 times the Nyquist frequency, and then quantizing each sample using a 256-level quantizer. Finally, each quantized sample is binary coded using K bits, where K is the minimum number of bits required for the encoding. The rate, in Megabits per second (rounded off to 2 decimal places), of the resulting stream of coded bits is _______ Mbps.
A
9.25
B
11.81
C
25.36
D
14.25
GATE EC 2019   Communication Systems
Question 6 Explanation: 


\begin{aligned} \text { Nyquist rate } &=2 \times 615 \mathrm{kHz} \\ &=1230 \mathrm{kHz}=1.23 \mathrm{MHz} \\ f_{s} &=1.2 \times 1.23=1.476 \mathrm{MHz} \\ \text { Bits/sample }, n &=\log _{2}(256)=8 \\ \text { So, } \quad A_{D} &=n f_{s}=8 \times 1.476 \\ &=11.808 \simeq 11.81 \mathrm{Mbps} \end{aligned}
Question 7
A single bit, equally likely to be 0 and 1, is to be sent across an additive white Gaussian noise (AWGN) channel with power spectral density N_0/2. Binary signaling, with 0\rightarrow p(t) and 1\rightarrow q(t), is used for the transmission, along with an optimal receiver that minimizes the bit-error probability.

Let \varphi _1(t),\varphi _2(t) form an orthonormal signal set.
If we choose p(t)=\varphi _1(t)\; and \; q(t)=-\varphi _2(t), we would obtain a certain bit-error probability P_b.
If we keep p(t)=\varphi _1(t), but take q(t)=\sqrt{E}\varphi _2(t), for what value of E would we obtain the same bit-error probability P_b?
A
0
B
1
C
2
D
3
GATE EC 2019   Communication Systems
Question 7 Explanation: 
\text { When } p(t)=\phi_{1}(t) \text { and } q(t)=-\phi_{1}(t)


d_{\min }=2
\text { When } p(t)=\phi_{1}(t) \text { and } q(t)=-\sqrt{E}\phi_{2}(t)
d_{\min }=\sqrt{(\sqrt{E})^{2}+1}=\sqrt{E+1}


To obtain same bit-error probability, \sigma_{\text {min }} should be same.
\begin{aligned} \text { So, } \quad \sqrt{E+1} &=2 \\ E &=3 \end{aligned}
Question 8
A random variable X takes values -0.5 and 0.5 with probabilities 1/4 and 3/4, respectively. The noisy observation of X is Y = X + Z, where Z has uniform probability density over the interval (-1, 1). X and Z are independent. If the MAP rule based detector outputs \hat{X} as

\hat{X}=\left\{\begin{matrix} -0.5 \; \; \; Y \lt \alpha \\ 0.5 \; \; \; Y \geq \alpha \end{matrix}\right.

then the value of \alpha (accurate to two decimal places) is _______.
A
-0.5
B
-1
C
1
D
0.5
GATE EC 2018   Communication Systems
Question 8 Explanation: 


\begin{array}{c} P\left(x_{0}\right)=\frac{1}{4} ; P\left(x_{1}\right)=\frac{3}{4} \\ \text { MAP criteria, } f_{Y}\left(y \mid x_{0}\right) P\left(x_{0}\right) \overset{x_{2}}{\underset{x_{1}}{\gtrless}}\left(y \mid x_{1}\right) P\left(x_{1}\right) \end{array}


\text{So}\quad \alpha=-0.5
Question 9
A binary source generates symbols X \in \{-1,1\} which are transmitted over a noisy channel. The probability of transmitting X= 1 is 0.5. Input to the threshold detector is R= X + N. The probability density function f_{N}(n) of the noise N is shown below.
If the detection threshold is zero, then the probability of error (correct to two decimal places) is __________.
A
0.12
B
0.2
C
0.8
D
1
GATE EC 2018   Communication Systems
Question 9 Explanation: 
Let s_{0} and s_{1} be the transmitted symbols representing the transmitted value {-1 , 1} respectively and let r_{0} and r_{1} be the received symbols.

Probability of error,
\begin{aligned} P_{e} &=P\left(s_{1}\right) P\left(r_{0} \mid s_{1}\right)+P\left(s_{0}\right) P\left(r_{1} \mid s_{0}\right) \\ P\left(r_{0} \mid s_{1}\right) &=P\left(r_{1} \mid s_{0}\right)=\frac{1}{2} \times 1 \times \frac{1}{4}=\frac{1}{8}\\ \text{Given that,}\\ P\left(s_{0}\right)&=P\left(s_{1}\right)=\frac{1}{2}\\ \text{So}\quad P_{e}&=\frac{1}{2}\left(\frac{1}{8}+\frac{1}{8}\right)=\frac{1}{8}=0.125 \end{aligned}
Question 10
A sinusoidal message signal is converted to a PCM signal using a uniform quantizer. The required signal-to-quantization noise ratio (SQNR) at the output of the quantizer is 40dB. The minimum number of bits per sample needed to achieve the desired SQNR is _______
A
5
B
6
C
7
D
8
GATE EC 2017-SET-2   Communication Systems
Question 10 Explanation: 
For sinusoidal input is applied to a PCM system,
\mathrm{SQNR}=6 n+1.8
n= number of bits per sample
Given that, the required \mathrm{SQNR}=40 \mathrm{dB}
\begin{aligned} \mathrm{so}, 6 n+1.8 &\geq 40 \\ n &\geq \frac{38.2}{6}\\ n_{\min }&=7 \\ \because \; &n \text{ is always an integer value} \end{aligned}
There are 10 questions to complete.