Question 1 |
For a real signal, which of the following is/are valid power spectral density/densities?
A | |
B | |
C | |
D |
Question 1 Explanation:
(i) S_{x}(\omega) \geq 0
(ii) S_{x}(\omega) is even function
Hence, options (A) and (B) are valid power spectral densities.
(ii) S_{x}(\omega) is even function
Hence, options (A) and (B) are valid power spectral densities.
Question 2 |
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 2 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 3 |
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 3 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 4 |
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 4 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 5 |
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 5 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}
There are 5 questions to complete.