GATE ECE · Communication Systems

Random Signals And Noise – GATE ECE Communication SystemsPractice Questions & PYQs

Generate GATE-level questions on Random Processes and Noise. Focus on: 1. Random Processes: Autocorrelation, Power Spectral Density, and WSS properties. 2. Filtered Random Processes: Response of LTI systems to random inputs. 3. White Noise and Gaussian Noise characteristics. 4. Random Variables: PDFs, Expectations, and Variance calculations.

73 questions · 20 PYQs · 0 AI practice · GATE ECE 2027

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Q1GATE 2025MCQ

Consider an additive white Gaussian noise (AWGN) channel with bandwidth $W$ and noise power spectral density $\frac{N_0}{2}$ . Let $P_{\text{av}}$ denote the average transmit power constraint. Which one of the following plots illustrates the dependence of the channel capacity $C$ on the bandwidth $W$ (keeping $P_{\text{av}}$ and $N_0$ fixed)?

🚀 Solve in Practice Mode

📖 Explanation

For an AWGN channel, the channel capacity is given by the Shannon?Hartley theorem: $C = W \log_2\left(1 + \frac{P_{\text{av}}}{W N_0}\right) \ \text{bits/sec}$ Let?s analyze this expression: $W$ is in both the linear multiplier and inside the denominator of the SNR. As $W \rightarrow 0$ , capacity $C \rightarrow 0$ because bandwidth is 0. As $W \rightarrow \infty$ , $\frac{P_{\text{av}}}{W N_0} \rightarrow 0$ , and we can use $\log_2(1 + x) \approx x/\ln 2$ when $x \ll 1$ . So, $C \approx W \cdot \frac{P_{\text{av}}}{W N_0 \ln 2} = \frac{P_{\text{av}}}{N_0 \ln 2}$ That is, capacity approaches a finite limit. Conclusion: Capacity increases with $W$ , but saturates to a constant as $W$ becomes large. This corresponds to a concave increasing curve that flattens out, which matches Plot (A).

Q2GATE 2025MCQ

Consider a real-valued random process $\mathrm{f}(\mathrm{t})=\sum_{\mathrm{n}=1}^{\mathrm{N}} \mathrm{a}_{\mathrm{n}} \mathrm{p}(\mathrm{t}-\mathrm{nT})$ , where $\mathrm{T}>0$ and N is a positive integer. Here, $p(t)=1$ for $t \in[0,0.5 T]$ and 0 otherwise. The coefficients $\mathrm{a}_{\mathrm{n}}$ are pairwise independent, zero-mean unit-variance random variables. Read the following statements about the random process and choose the correct option. (i) The mean of the process $f(t)$ is independent of time t . (ii) The autocorrelation function $\mathrm{E}[\mathrm{f}(\mathrm{t}) \mathrm{f}(\mathrm{t}+\tau)]$ is independent of time t for all $\tau$ . (Here, $\mathrm{E}[\cdot]$ is the expectation operation.)

🚀 Solve in Practice Mode

📖 Explanation

We are given: $f(t) = \sum_{n=1}^{N} a_n , p(t - nT)$ $a_n$ : zero-mean, unit-variance, pairwise independent $p(t) = 1$ for $t \in [0,\ 0.5T]$ , and 0 otherwise (i) Mean of $f(t)$ :We use linearity of expectation: $E[f(t)] = \sum_{n=1}^{N} E[a_n] \cdot p(t - nT)$ But $E[a_n] = 0$ for all $n$ , so: $E[f(t)] = 0$ for all $t$ $\Rightarrow$ Mean is zero and time-independent So, (i) is TRUE (ii) Autocorrelation function $R_f(t, \tau) = E[f(t)f(t + \tau)]$ : Expand: $f(t) = \sum_{n=1}^{N} a_n p(t - nT)$ $f(t + \tau) = \sum_{m=1}^{N} a_m p(t + \tau - mT)$ So: $E[f(t)f(t + \tau)] = E\left[ \sum_{n=1}^{N} \sum_{m=1}^{N} a_n a_m , p(t - nT) p(t + \tau - mT) \right]$ Since $a_n$ are zero-mean and pairwise independent: $E[a_n a_m] = 0$ for $n \neq m$ , and $E\$ [a_n^2]$ = 1 $So only diagonal terms survive:$ E[f(t)f(t + \tau)] = \sum_{n=1}^{N} p(t - nT) \cdot p(t + \tau - nT) $Note:$ p(t - nT) \cdot p(t + \tau - nT) $depends on$ t $and$ \tau $, so the sum depends on$ t\Rightarrow$ Autocorrelation is not time-invariant So, (ii) is FALSE

Q3GATE 2025MCQ

A source transmits symbol $S$ that takes values uniformly at random from the set $\{-2,0,2\}$ . The receiver obtains $Y=S+N$ , where $N$ is a zero-mean Gaussian random variable independent of $S$ . The receiver uses the maximum likelihood decoder to estimate the transmitted symbol $S$ . Suppose the probability of symbol estimation error $\mathrm{P}_{\mathrm{e}}$ is expressed as follows: $\mathrm{P}_{\mathrm{e}}=\alpha \mathrm{P}(\mathrm{N}>1)$ , where $\mathrm{P}(\mathrm{N}>1)$ denotes the probability that $N$ exceeds 1. What is the value of $\alpha$ ?

🚀 Solve in Practice Mode

📖 Explanation

Given: $S \in {-2,\ 0,\ 2}$ uniformly at random $Y = S + N$ , where $N \sim \mathcal{N}(0,\ \sigma^2)$ ML decoder chooses the symbol closest to $Y$ , so decision boundaries are at $Y = -1$ and $Y = 1$ Case 1: $S = -2$ $Y = -2 + N$ Error occurs when $Y > -1 \Rightarrow N > 1$ So, $P_e|S = -2 = P(N > 1)$ Case 2: $S = 0$ $Y = N$ Error occurs when $|N| > 1$ So, $P_e|S = 0 = P(N < -1) + P(N > 1) = 2P(N > 1)$ Case 3: $S = 2$ $Y = 2 + N$ Error occurs when $Y < 1 \Rightarrow N < -1$ So, $P_e|S = 2 = P(N < -1) = P(N > 1)$ Average probability of error: $P_e = \frac{1}{3} \left(P(N > 1) + 2P(N > 1) + P(N > 1)\right)$ $P_e = \frac{4}{3} \cdot P(N > 1)$ So, $\alpha = \frac{4}{3}$

Q4GATE 2024NAT

Let $X(t)=A \cos \left(2 \pi f_{0} t+\theta\right)$ be a random process, where amplitude $A$ and phase $\theta$ are independent of each other, and are uniformly distributed in the intervals $[-2,2]$ and $[0,2 \pi]$ , respectively. $X(t)$ is fed to an 8-bit uniform mid-rise type quantizer. Given that the autocorrelation of $X(t)$ is $R_{x}(\tau)=\frac{2}{3} \cos \left(2 \pi f_{0} \tau\right)$ , the signal to quantization noise ratio (in $\mathrm{dB}$ , rounded off to two decimal places) at the output of the quantizer is ______

🚀 Solve in Practice Mode

📖 Explanation

\begin{aligned} S & =R_{x}(0)=\frac{2}{3} \\ N_{q} & =\frac{\Delta^{2}}{12} \\ \Delta & =\frac{2-(-2)}{2^{8}}=\frac{4}{2^{8}} \end{aligned}

\begin{aligned} N_{q} & =\frac{4^{2}}{\frac{2^{16}}{12}}=\frac{2^{4}}{12 \times 2^{16}}=\frac{1}{12 \times 2^{12}} \\ \frac{S}{N_{a}} & =\frac{2}{3} \times \frac{12 \times 2^{12}}{1}=32,768 \\ \left(\frac{S}{N_{a}}\right)_{d B} & =10 \log 32768=45.15 \mathrm{~dB} \end{aligned}

Explanation Figure 1

Q5GATE 2024MCQ

A White Gaussian noise $w(t)$ with zero mean and power spectral density $\frac{N_{0}}{2}$ , when applied to a first-order $R C$ low pass filter produces an output $n(t)$ . At a particular time $t=t_{k}$ , the variance of the random variable $n\left(t_{k}\right)$ is _____

🚀 Solve in Practice Mode

📖 Explanation

\begin{aligned} E[n(t)] & =E[W(t)] \cdot H(0) \\ E[n(t)] & =0 \\ E\left[n^{2}(t)\right]\$ & =\left\{E(n(t)\}^{2}+\operatorname{var}[n(t)]\right. \\ \operatorname{var}[n(t)] & =E\left[n^{2}(t)\right]\$ \end{aligned}

at $t=t_{k}$

\begin{aligned} \left.\operatorname{var}[n(t)]\right|_{t=t_{k}} & =E\left[n^{2}\left(t_{k}\right)\right]\$=E\left[n\left(t_{k}\right) \cdot n\left(t_{k}\right)\right]=R_{n}(0) \\ S_{n}(f) & =S_{W}(f) \cdot|H(f)|^{2} \\ & =\frac{N_{0}}{2} \cdot \frac{1}{1+(\omega R C)^{2}} \\ R_{n}(\tau) & =\operatorname{IFT}\left[S_{n}(f)\right]=\frac{N_{0}}{4 R C} \cdot e^{\frac{-|\tau|}{R C}} \\ R_{n}(0) & =\frac{N_{0}}{4 R C} \end{aligned}

Explanation Figure 1

Q6GATE 2023MSQ

For a real signal, which of the following is/are valid power spectral density/densities?

🚀 Solve in Practice Mode

📖 Explanation

(i) $S_{x}(\omega) \geq 0$ (ii) $S_{x}(\omega)$ is even function Hence, options (A) and (B) are valid power spectral densities.

Q7GATE 2022MCQ

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 ________.

🚀 Solve in Practice Mode

📖 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

Q8GATE 2022NAT

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).

🚀 Solve in Practice Mode

📖 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$

Q9GATE 2022MCQ

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?

🚀 Solve in Practice Mode

📖 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.

Explanation Figure 2

Q10GATE 2021NAT

The autocorrelation function $R_{X}\left ( \tau \right )$ of a wide-sense stationary random process $X(t)$ is shown in the figure. The average power of $X(t)$ is ________________

🚀 Solve in Practice Mode

📖 Explanation

Average power $[X(t)]=R_{x}(0)=2$

Explanation Figure 2

Q11GATE 2021NAT

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.

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📖 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}

Explanation Figure 1

Q12GATE 2021MCQ

Two continuous random variables X and Y are related as $Y=2X+3$ Let $\sigma ^{2}_{X}$ and $\sigma ^{2}_{Y}$ denote the variances of X and Y, respectively. The variances are related as

🚀 Solve in Practice Mode

📖 Explanation

\begin{aligned} Y &=2 X+3 \\ \operatorname{Var}[Y] &=E\left[(Y-\bar{Y})^{2}\right]\$ \\ E[Y] &=\bar{Y}=2 \bar{X}+3 \\ \operatorname{Var}[Y] &=E\left[(2 X+3-2 \bar{X}-3)^{2}\right]\$ \\ &=E\left[4(X-\bar{X})^{2}\right]\$ \\ &=4 \cdot E\left[(X-\bar{X})^{2}\right]\$ \\ \sigma_{Y}^{2} &=4 \cdot \sigma_{X}^{2} \end{aligned}

Q13GATE 2020NAT

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 ______.

🚀 Solve in Practice Mode

📖 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

Explanation Figure 1

Q14GATE 2020NAT

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 __________

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📖 Explanation

Given that, $P(X=2)=\alpha$ Entropy Will be Maximum; Provided Probabilities are equal. i.e. $P(X=2)=P(X=-2)=\alpha =\frac{1}{2}$ $\alpha =\frac{1}{2}=0.5$

Q15GATE 2020NAT

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 _____.

🚀 Solve in Practice Mode

📖 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}

Q16GATE 2020NAT

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 _________.

🚀 Solve in Practice Mode

📖 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}

Q17GATE 2019NAT

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).

🚀 Solve in Practice Mode

📖 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}

Explanation Figure 2

Q18GATE 2018MCQ

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

🚀 Solve in Practice Mode

📖 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}

Q19GATE 2017NAT

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 ____________

🚀 Solve in Practice Mode

📖 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}

Explanation Figure 1

Q20GATE 2017MCQ

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:

🚀 Solve in Practice Mode

📖 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.

Explanation Figure 2

Explanation Figure 3

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