Random Signals and Noise

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 _____.
A
0.1
B
0.3
C
0.6
D
0.8
GATE EC 2020   Communication Systems
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}
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 ______.
A
1
B
2
C
3
D
4
GATE EC 2020   Communication Systems
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
Question 3
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 _________.
A
3
B
4
C
6
D
8
GATE EC 2020   Communication Systems
Question 3 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}
Question 4
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 __________
A
0.2
B
0.5
C
0.7
D
0.9
GATE EC 2020   Communication Systems
Question 4 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
Question 5
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).
A
17.50
B
12.24
C
18.88
D
24.36
GATE EC 2019   Communication Systems
Question 5 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}
Question 6
Consider a white Gaussian noise process N(t) with two-sided power spectral density S_{N}(f)=0.5 W/Hz as input to a filter with impulse response 0.5e^{\frac{-t^{2}}{2}} (where t is in seconds) resulting in output Y(t). The power in Y(t) in watts is
A
0.11
B
0.22
C
0.33
D
0.44
GATE EC 2018   Communication Systems
Question 6 Explanation: 
PSD of noise input.
S_{(t)}^{S}(f)=0.50 \mathrm{W} / \mathrm{Hz}
Power of y(t)
\begin{aligned} P_{y} &=\int_{-\infty}^{\infty} S_{N}(f)|H(f)|^{2} d f \\ &=0.50 \int_{-\infty}^{\infty}|H(f)|^{2} d f \\ &=0.50 \int_{-\infty}^{\infty}|h(t)|^{2} d t\\ \text{Given that,}\\ h(t)&=\frac{1}{2} e^{-t^{2} / 2}\\ \text{So}\quad P_{y} &=\frac{1}{2} \int_{-\infty}^{\infty}\left(\frac{1}{2} e^{-t^{2} / 2}\right)^{2} d t \\ &=\frac{1}{8} \int_{-\infty}^{\infty} e^{-t^{2}} d t \\ &=\frac{\sqrt{\pi}}{8}=0.2215 \mathrm{W} \simeq 0.22 \mathrm{W} \end{aligned}
Question 7
Consider the random process X(t)=U+Vt, where U is a zero-mean Gaussian random variable and V is a random variable uniformly distributed between 0 and 2. Assume that U and V are statistically independent. The mean value of the random process at t = 2 is ____________
A
1
B
2
C
3
D
4
GATE EC 2017-SET-2   Communication Systems
Question 7 Explanation: 
\begin{aligned} X(t) &=U+V t \\ \text { At } t=2, \quad X(t) &=X(2)=U+2 V \\ E[X(t)] &=E[U+2 V] \\ &=E[U]+2 E[V]\\ \text{Given that,}\\ E[U]&=0\\ \end{aligned}


Also given that,
V is uniformly distributed between 0 and 2
\begin{aligned} \text{So, } \quad E[V]&=\int_{-\infty}^{\infty} f_{v}(V) d v=\int_{0}^{2}\left(\frac{1}{2}\right) d v=1 \\ \text{So, } \quad E[X(f)]&=0+2(1)=2 \end{aligned}
Question 8
Let X(t) be a wide sense stationary random process with the power spectral density S_{X}(f) as shown in Figure (a), where f is in Hertz (Hz). The random process X(t) is input to an ideal low pass filter with frequency response
H(f)=\left\{\begin{matrix} 1 & |f|\leq 1/2 Hz\\ 0& |f|>1/2 Hz \end{matrix}\right.
As shown in Figure (b). The output of the lowpass filter is Y(t).

Let E be the expectation operator and consider the following statements.
I. E(X(t))=E(Y(t))
II. E(X^{2}(t))=E(Y^{2}(t))
III.E(Y^{2}(t))=2
Select the correct option:
A
only I is true
B
only II and III are true
C
only I and II are true
D
only I and III are true
GATE EC 2017-SET-1   Communication Systems
Question 8 Explanation: 
The given input power spectral density is as follows:


Frequency response of the low pass filter is as follows:


\begin{aligned} E[Y(t)] &=H(0) E[X(t)] \\ H(0) &=1 \\ \text{So,}\quad E[Y(t)] &=E[X(t)]\\ E\left[Y^{2}(t)\right] &\neq E\left[X^{2}(t)\right]\\ \end{aligned}
Since, LPF does not allow total power from input to output.
\begin{aligned} E\left(x^{2}(t)\right]&=\int_{0}^{\infty} S_{X}(f) d f=2 W\\ \text{As }E\left[Y^{2}(t)\right] \neq E\left[X^{2}(t)\right]& \\ E\left[Y^{2}(t)\right] &\neq 2 \end{aligned}
So, only statement - I is correct.
Question 9
A wide sense stationary random process X(t) passes through the LTI system shown in the figure. If the autocorrelation function of X(t) is R_X(\tau), then the autocorrelation function R_Y(\tau) of the output Y(t) is equal to
A
2R_{x}(\tau )+R_{X}(\tau -T_{0})+R_{X}(\tau +T_{0})
B
2R_{x}(\tau )-R_{X}(\tau -T_{0})-R_{X}(\tau +T_{0})
C
2R_{x}(\tau )+2R_{X}(\tau -T_{0})
D
2R_{x}(\tau )-2R_{X}(\tau -T_{0})
GATE EC 2016-SET-3   Communication Systems
Question 9 Explanation: 
\begin{aligned} Y(t)&=X(t)-\left(t-T_{0}\right) \\ R_{N}(\tau)&=E[Y(t+\tau) Y(t)] \\ &=E[(X(t+\tau)-X(t+\tau-T_{0}))(X(t)-X(t-T_{0}))] \\ &=E[X(t+\tau) X(t)-X(t) X(t+\tau-T_{0}) \\ &-X(t+\tau) X(t-T_{0}) +X(t+\tau-T_{0}) X(t-T_{0})] \end{aligned}
since, X(t) is a WSS process,
R_{Y}(\tau)=R_{X}(\tau)-R_{X}\left(\tau-T_{0}\right)-R_{X}\left(\tau+T_{0}\right)+R_{X}(\tau)
=2 R_{X}(\tau)-R_{X}\left(\tau-T_{0}\right)-R_{X}\left(\tau+T_{0}\right)
Question 10
Consider a random process X(t) = 3V(t) - 8, where V(t) is a zero mean stationary random process with autocorrelation R_{v}(\tau )=4e^{-5|\tau |}. The power in X(t) is ________
A
50
B
80
C
100
D
120
GATE EC 2016-SET-2   Communication Systems
Question 10 Explanation: 
\begin{aligned} X(t) &=3 V(t)-8 \quad \text { and } \quad E[V(t)]=0 \\ R_{v}(\tau) &=4 e^{-5|\tau|} \\ \text { Power of } X(t) &=E\left[X^{2}(t)\right] \\ &=E\left[9 V^{2}(t)\right]+64-48 E[V(t)] \\ &=9 E\left[V^{2}(t)\right]+64-48 E[V(t)] \\ E\left[V^{2}(t)\right] &=R_{v}(0)=4 \\ \text { Power of } X(t) &=((9 \times 4)+64) \\ &=100 \end{aligned}
There are 10 questions to complete.
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