Question 1 |

X is a random variable with uniform probability density function in the interval [-2,10]. For Y = 2X-6, the conditional probability P(Y \leq 7|X \geq 5) (rounded off to three decimal places) is _____.

0.1 | |

0.3 | |

0.6 | |

0.8 |

Question 1 Explanation:

x follows uniform distribution over [-2, 10]

\begin{aligned} \therefore \; f(x)&=\frac{1}{b-a}=\frac{1}{10-(-2)}=\frac{1}{12}\\ \text{Given: } y &= 2x - 6\\ \Rightarrow \; x&=\frac{y+6}{2}\\ \text{For } y = 7 \\ x&=\frac{7+6}{2}=\frac{13}{2}=6.5\\ P\left [ \frac{y\lt7}{x\gt 5} \right ]&=P\left [ \frac{x\lt6.5}{x\gt 5} \right ] \\ &=\frac{P\left [ x\gt 5\, and\, x\lt6.5 \right ]}{P\left [ x\gt 5 \right ]} \\ &=\frac{P\left [ 5\lt x \lt 6.5 \right ]}{P[x\gt 5]} \\ &=\frac{\int_{5}^{6.5}f(x)dx}{\int_{5}^{10}F(x)dx} \\ &=\frac{\int_{5}^{6.5}\frac{1}{12}dx}{\int_{5}^{10}\frac{1}{12}dx} \\ &=\frac{(x)_{6.5}^{5}}{(x)_{5}^{10}}=\frac{1.5}{5}=0.3\end{aligned}

\begin{aligned} \therefore \; f(x)&=\frac{1}{b-a}=\frac{1}{10-(-2)}=\frac{1}{12}\\ \text{Given: } y &= 2x - 6\\ \Rightarrow \; x&=\frac{y+6}{2}\\ \text{For } y = 7 \\ x&=\frac{7+6}{2}=\frac{13}{2}=6.5\\ P\left [ \frac{y\lt7}{x\gt 5} \right ]&=P\left [ \frac{x\lt6.5}{x\gt 5} \right ] \\ &=\frac{P\left [ x\gt 5\, and\, x\lt6.5 \right ]}{P\left [ x\gt 5 \right ]} \\ &=\frac{P\left [ 5\lt x \lt 6.5 \right ]}{P[x\gt 5]} \\ &=\frac{\int_{5}^{6.5}f(x)dx}{\int_{5}^{10}F(x)dx} \\ &=\frac{\int_{5}^{6.5}\frac{1}{12}dx}{\int_{5}^{10}\frac{1}{12}dx} \\ &=\frac{(x)_{6.5}^{5}}{(x)_{5}^{10}}=\frac{1.5}{5}=0.3\end{aligned}

Question 2 |

In a digital communication system, a symbol S randomly chosen from the set (s_1, s_2, s_3, s_4)
is transmitted. It is given that s_1=-3, s_2=-1, s_3=+1 and s_4=+2. The received
symbol is Y = S + W. W is a zero-mean unit-variance Gaussian random variable and
is independent of S. P_i is the conditional probability of symbol error for the maximum
likelihood (ML) decoding when the transmitted symbol S=s_i. The index i for which the
conditional symbol error probability P_i is the highest is ______.

1 | |

2 | |

3 | |

4 |

Question 2 Explanation:

Since the noise variable is Gaussian with zero mean and ML decoding is used, the
decision boundary between two adjacent signal points will be their arithmetic mean.
In the following graphs, the shaded area indicates the conditional probability of decoding
a symbol correctly when it is transmitted.

By comparing the above graphs, we can conclude that P_{3} is larger among the four. End of Solution

By comparing the above graphs, we can conclude that P_{3} is larger among the four. End of Solution

Question 3 |

S_{PM}(t)\; and \; S_{FM}(t) as defined below, are the phase modulated and the frequency modulated
waveforms, respectively, corresponding to the message signal m(t) shown in the figure.

S_{PM}(t)= \cos (1000 \pi t+K_p m(t))

and S_{FM}(t)= \cos (1000 \pi t+K_f \int_{-\infty }^{t} m(\tau )d\tau )

where K_p is the phase deviation constant in radians/volt and K_f is the frequency deviation constant in radians/second/volt. If the highest instantaneous frequencies of S_{PM}(t)\; and \; S_{FM}(t) are same, then the value of the ratio \frac{K_p}{K_f} is ______ seconds.

S_{PM}(t)= \cos (1000 \pi t+K_p m(t))

and S_{FM}(t)= \cos (1000 \pi t+K_f \int_{-\infty }^{t} m(\tau )d\tau )

where K_p is the phase deviation constant in radians/volt and K_f is the frequency deviation constant in radians/second/volt. If the highest instantaneous frequencies of S_{PM}(t)\; and \; S_{FM}(t) are same, then the value of the ratio \frac{K_p}{K_f} is ______ seconds.

0.5 | |

1 | |

2 | |

2.5 |

Question 3 Explanation:

\begin{aligned}S(t)_{pm}&=A_c \cos [2\pi f_{c}t+k_pm(t)]\\ S(t)_{Fm}&=A_c \cos [2\pi f_{c}t+k_{f}\int_{0}^{\infty }m(t)dt]\end{aligned}

Instantaneous frequency are equal

\begin{aligned}f_{i}&=\frac{1}{2\pi }\frac{\mathrm{d} }{\mathrm{d} t}\theta (t)\\f_{iPM}&=f_{c}+\frac{K_p}{2\pi }\frac{\mathrm{d} }{\mathrm{d} t} m(t)\\ f_{iFM}&=f_{c}+\frac{K_f}{2\pi }m(t)\\ \text{Given that, } (f_{iPM})_{max}&=(f_{iFM})_{max} \\ f_{c}+\frac{k_P}{2\pi}[\frac{\mathrm{d} }{\mathrm{d} t}m(t)]_{max}&=f_{c}+\frac{K_{f}}{2\pi }[m(t)]_{max}\\ \left.\begin{matrix} \frac{k_P}{2\pi}\frac{\mathrm{d} }{\mathrm{d} t}m(t) \end{matrix}\right|_{max} &= \left. \begin{matrix} \frac{K_{f}}{2\pi }[m(t)] \end{matrix}\right|_{max}\\ \left.\begin{matrix} \frac{\mathrm{d} }{\mathrm{d} t}m(t) \end{matrix}\right|_{max} &=5, \left.\begin{matrix} m(t) \end{matrix}\right|_{max}=10 \\ 5k_{p}&=10k_{f} \\ \frac{k_{p}}{k_{f}}&=2\end{aligned}

Instantaneous frequency are equal

\begin{aligned}f_{i}&=\frac{1}{2\pi }\frac{\mathrm{d} }{\mathrm{d} t}\theta (t)\\f_{iPM}&=f_{c}+\frac{K_p}{2\pi }\frac{\mathrm{d} }{\mathrm{d} t} m(t)\\ f_{iFM}&=f_{c}+\frac{K_f}{2\pi }m(t)\\ \text{Given that, } (f_{iPM})_{max}&=(f_{iFM})_{max} \\ f_{c}+\frac{k_P}{2\pi}[\frac{\mathrm{d} }{\mathrm{d} t}m(t)]_{max}&=f_{c}+\frac{K_{f}}{2\pi }[m(t)]_{max}\\ \left.\begin{matrix} \frac{k_P}{2\pi}\frac{\mathrm{d} }{\mathrm{d} t}m(t) \end{matrix}\right|_{max} &= \left. \begin{matrix} \frac{K_{f}}{2\pi }[m(t)] \end{matrix}\right|_{max}\\ \left.\begin{matrix} \frac{\mathrm{d} }{\mathrm{d} t}m(t) \end{matrix}\right|_{max} &=5, \left.\begin{matrix} m(t) \end{matrix}\right|_{max}=10 \\ 5k_{p}&=10k_{f} \\ \frac{k_{p}}{k_{f}}&=2\end{aligned}

Question 4 |

For the modulated signal x(t)= m(t) \cos(2 \pi f_c t), the message signal m(t) = 4 \cos (1000 \pi t) and the carrier frequency f_c is 1 MHz. The signal x(t) is passed
through a demodulator, as shown in the figure below. The output y(t) of the
demodulator is

\cos (460 \pi t) | |

\cos (920 \pi t) | |

\cos (1000 \pi t) | |

\cos (540 \pi t) |

Question 4 Explanation:

Output of multiplier

=x(t)\cos 2\pi (f_{c}+40)t=m(t)\cos 2\pi f_{c}t\cdot \cos2\pi (f_{c}+40)t

=\frac{m(t)}{2}[\cos 2\pi (2f_{c}+40)t+\cos 2\pi (40)t]

Given, m(t)=4\cos 1000\pi t

So, output of multiplier

=2\cos 2\pi (500)t[\cos 2\pi (2f_{c}+40)t+\cos 2\pi (40)t]

=\cos 2\pi (2f_{c}+540)t+\cos 2\pi (2f_{c}-460)t+\cos 2\pi (540)t+\cos 2\pi (460)t

Output of Low pass filter

=\cos [2\pi (460)]t

=\cos 920\pi t

=x(t)\cos 2\pi (f_{c}+40)t=m(t)\cos 2\pi f_{c}t\cdot \cos2\pi (f_{c}+40)t

=\frac{m(t)}{2}[\cos 2\pi (2f_{c}+40)t+\cos 2\pi (40)t]

Given, m(t)=4\cos 1000\pi t

So, output of multiplier

=2\cos 2\pi (500)t[\cos 2\pi (2f_{c}+40)t+\cos 2\pi (40)t]

=\cos 2\pi (2f_{c}+540)t+\cos 2\pi (2f_{c}-460)t+\cos 2\pi (540)t+\cos 2\pi (460)t

Output of Low pass filter

=\cos [2\pi (460)]t

=\cos 920\pi t

Question 5 |

The random variable

Y=\int_{-\infty }^{\infty }W(t)\phi (t) dt

where \phi (t) =\left\{\begin{matrix} 1; & 5\leq t\leq 7\\ 0; & otherwise \end{matrix}\right.

and W(t) is a real white Gaussian noise process with two-sided power spectral density S_W(f) = 3 W/Hz, for all f. The variance of Y is _________.

Y=\int_{-\infty }^{\infty }W(t)\phi (t) dt

where \phi (t) =\left\{\begin{matrix} 1; & 5\leq t\leq 7\\ 0; & otherwise \end{matrix}\right.

and W(t) is a real white Gaussian noise process with two-sided power spectral density S_W(f) = 3 W/Hz, for all f. The variance of Y is _________.

3 | |

4 | |

6 | |

8 |

Question 5 Explanation:

\begin{aligned} \text{Given: }\phi (t) &=\left\{\begin{matrix} 1 & 5\leq t\leq 7\\ 0 & otherwise \end{matrix}\right. \\ s_{\omega }(f)& =3 \text{ Watts/Hz} \\ R_{\omega }(\tau )&=3\delta (\tau )=3\delta (t_{1}-t_{2}) \\ Var[y]&=E[y^{2}]-{E[y]}^{2} \\ {E[W(t)]}^{2}&=\text{DC power} \\ &=\text{Area under PSD at }f=0 \\ {E[W(t)]}^{2}&=0 \\ E[W(t)]&=0 \\ y&=\int_{-\infty }^{\infty }W(t)\phi (t)dt \\ E[y]&=\int_{-\infty }^{\infty }E[W(t)]\phi (t)dt \\ y&=\int_{-\infty }^{\infty }W(t)\phi (t)dt\rightarrow E[y^{2}]\\ &=S_{\omega }(f)\cdot Energy[\phi (t)]\\ & =3 \times 2=6\\ \text{Var}[y]&=6-0=6\end{aligned}

Detailed Explanations for:

\begin{aligned}y&=\int_{-\infty }^{\infty }W(t)\phi (t)dt \\ E(y^{2})&=E[y \cdot y] \\ &=E[\int_{-\infty }^{\infty }W(t_{1})\phi (t_{1})dt_{1}\int_{-\infty }^{\infty }W(t_{2})\phi (t_{2})dt_{2}] \\ &=E[\int_{-\infty }^{\infty }\int_{-\infty }^{\infty }W(t_{1})W(t_{2})\phi (t_{1})\phi (t_{2})dt_{1}\cdot dt_{2}] \\ &=\int_{-\infty }^{\infty }\int_{-\infty }^{\infty }E[W(t_{1})W(t_{2}) \cdot \phi (t_{1})\phi (t_{2})dt_{1} dt_{2}] \\ &=\int_{-\infty }^{\infty }\int_{-\infty }^{\infty }R_{W}(t_{1}-t_{2})\phi (t_{1})\phi (t_{2})dt_{1}dt_{2} \\ &=\int_{-\infty }^{\infty }\int_{-\infty }^{\infty }3\delta (t_{1}-t_{2})\phi (t_{1})\phi (t_{2})dt_{1}dt_{2}\end{aligned}

Above Integration Exists Provided

\begin{aligned}t_{1}&=t_{2}=t\\ &=\int_{-\infty }^{\infty }3\delta (0)\phi (t)\phi (t)dt \, dt\\ &=3 \int_{-\infty }^{\infty } \delta (0) dt + \int_{-\infty }^{\infty } \phi ^2 (t) dt\\&=2 times 1 times \text{Energy}[\phi (t)] \\E[y^{2}]&=6\end{aligned}

Detailed Explanations for:

\begin{aligned}y&=\int_{-\infty }^{\infty }W(t)\phi (t)dt \\ E(y^{2})&=E[y \cdot y] \\ &=E[\int_{-\infty }^{\infty }W(t_{1})\phi (t_{1})dt_{1}\int_{-\infty }^{\infty }W(t_{2})\phi (t_{2})dt_{2}] \\ &=E[\int_{-\infty }^{\infty }\int_{-\infty }^{\infty }W(t_{1})W(t_{2})\phi (t_{1})\phi (t_{2})dt_{1}\cdot dt_{2}] \\ &=\int_{-\infty }^{\infty }\int_{-\infty }^{\infty }E[W(t_{1})W(t_{2}) \cdot \phi (t_{1})\phi (t_{2})dt_{1} dt_{2}] \\ &=\int_{-\infty }^{\infty }\int_{-\infty }^{\infty }R_{W}(t_{1}-t_{2})\phi (t_{1})\phi (t_{2})dt_{1}dt_{2} \\ &=\int_{-\infty }^{\infty }\int_{-\infty }^{\infty }3\delta (t_{1}-t_{2})\phi (t_{1})\phi (t_{2})dt_{1}dt_{2}\end{aligned}

Above Integration Exists Provided

\begin{aligned}t_{1}&=t_{2}=t\\ &=\int_{-\infty }^{\infty }3\delta (0)\phi (t)\phi (t)dt \, dt\\ &=3 \int_{-\infty }^{\infty } \delta (0) dt + \int_{-\infty }^{\infty } \phi ^2 (t) dt\\&=2 times 1 times \text{Energy}[\phi (t)] \\E[y^{2}]&=6\end{aligned}

Question 6 |

A binary random variable X takes the value +2 or -2. The probability P(X=+2)= \alpha.
The value of \alpha (rounded off to one decimal place), for which the entropy of X is maximum,
is __________

0.2 | |

0.5 | |

0.7 | |

0.9 |

Question 6 Explanation:

Given that, P(X=2)=\alpha

Entropy Will be Maximum;

Provided Probabilities are equal.

i.e. P(X=2)=P(X=-2)=\alpha =\frac{1}{2}

\alpha =\frac{1}{2}=0.5

Entropy Will be Maximum;

Provided Probabilities are equal.

i.e. P(X=2)=P(X=-2)=\alpha =\frac{1}{2}

\alpha =\frac{1}{2}=0.5

Question 7 |

A digital communication system transmits a block of N bits. The probability of error in
decoding a bit is \alpha. The error event of each bit is independent of the error events of
the other bits. The received block is declared erroneous if at least one of the its bits
is decoded wrongly. The probability that the received block is erroneous is

N(1-\alpha ) | |

\alpha ^n | |

1-\alpha ^n | |

1-(1-\alpha )^n |

Question 7 Explanation:

Probability of error in decoding single bit=\alpha

Then probability of no error will be 1- \alpha.

Total N-bits transmitted, so that probability of no error in received block

=(1-\alpha )(1-\alpha )...N \, \text{ Times}

=(1-\alpha) ^{N}

The Probability of received block is erroneous is =1-(1-\alpha) ^{N}

Then probability of no error will be 1- \alpha.

Total N-bits transmitted, so that probability of no error in received block

=(1-\alpha )(1-\alpha )...N \, \text{ Times}

=(1-\alpha) ^{N}

The Probability of received block is erroneous is =1-(1-\alpha) ^{N}

Question 8 |

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 8 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

Question 9 |

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.

9.25 | |

11.81 | |

25.36 | |

14.25 |

Question 9 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 10 |

Let a random process Y(t) be described as Y(t)=h(t)*X(t)+Z(t), where X(t) is a white noise process with power spectral density S_X(f)=5 W/Hz. The filter h(t) has a magnitude response given by |H(f)|=0.5 for -5\leq f\leq 5, and zero elsewhere. Z(t) is a stationary random process, uncorrelated with X(t), with power spectral density as shown in the figure. The power in Y(t), in watts, is equal to ________ W (rounded off to two decimal places).

17.50 | |

12.24 | |

18.88 | |

24.36 |

Question 10 Explanation:

\begin{aligned} \text { Let, } \quad X_{1}(t) &=h(t) * X(t) \\ S_{X 1}(f) &=|H(f)|^{2} S_{X}(f) \end{aligned}

Given that, Z(t) and X(t) are uncorrelated

\begin{aligned} & \text{So,}\quad S_{y}(f)=S_{X 1}(f)+S_{z}(f)\\ & \text{Power in }\mathrm{y}(t) \\ P_{Y}&=\left[\text { Area under } S_{X 1}(f)\right]+\left[\text { Area under } S_{z}(f)\right] \\ &=(10 \times 1.25)+(5 \times 1)=17.5 \mathrm{W} \end{aligned}

Given that, Z(t) and X(t) are uncorrelated

\begin{aligned} & \text{So,}\quad S_{y}(f)=S_{X 1}(f)+S_{z}(f)\\ & \text{Power in }\mathrm{y}(t) \\ P_{Y}&=\left[\text { Area under } S_{X 1}(f)\right]+\left[\text { Area under } S_{z}(f)\right] \\ &=(10 \times 1.25)+(5 \times 1)=17.5 \mathrm{W} \end{aligned}

There are 10 questions to complete.