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).
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).
0.23 | |
0.56 | |
0.74 | |
0.65 |
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).
2.2 | |
3 | |
4.8 | |
5.3 |
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?
H(X)\leq \log _2 K bits | |
H(X)\leq H(2X) | |
H(X)\leq H(X^2) | |
H(X)\leq H(2^X) |
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 ______
38.25 | |
64.2 | |
12.45 | |
30.72 |
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 ______.
\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 ______.
0.1 | |
0.2 | |
0.3 | |
0.8 |
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
\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.