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).
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).
1.16 | |
1.85 | |
2.21 | |
3.63 |
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
\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).
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 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).
For \epsilon =0.1 , the probability that a transmitted codeword is decoded correctly is _________ (rounded off to two decimal places).
0.25 | |
0.65 | |
0.85 | |
0.94 |
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.
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).
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).
-1 | |
-2 | |
2 | |
1 |
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 ________.
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 ________.
R=1k\Omega ,C=0.1\mu F | |
R=2k\Omega ,C=1\mu F | |
R=1k\Omega ,C=2\mu F | |
R=2k\Omega ,C=0.5\mu F |
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
\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).
2.2 | |
3 | |
4.8 | |
5.3 |
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?
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 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?
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?
Statement I is correct, Statement II is correct | |
Statement I is correct, Statement II is incorrect | |
Statement I is incorrect, Statement II is correct | |
Statement I is incorrect, Statement II is incorrect |
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.
0.2 | |
0.01 | |
0.04 | |
0.4 |
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 ______
38.25 | |
64.2 | |
12.45 | |
30.72 |
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.