IN THE EC SYLLABUS

Random processes, power spectral density and noise

A random process is a signal whose value at each time is a random variable, so it is described by averages rather than by a single waveform. This topic covers the autocorrelation function and its properties, the power spectral density as its Fourier transform, how a linear time-invariant system reshapes the PSD by $∣ H (f) ∣^{2}$ , and the properties of white and Gaussian noise. Together they give the output noise power of a filter or a receiver.

6 previous-year questions · 6 solved · 6 practice questions

Random variables and random processes: the basics

A random process $X (t)$ is a collection of random variables, one for each time $t$ . Fix $t$ and you get a random variable; fix an outcome of the experiment and you get one waveform, called a sample function. The set of all sample functions is the ensemble.

MEAN AND VARIANCE OF A RANDOM VARIABLE

μ_{X} = E [X], σ_{X}^{2} = E [(X - μ_{X})^{2}] = E [X^{2}] - μ_{X}^{2}

Holds for any random variable. $E [X^{2}]$ is the mean-square value (average power for a signal).

GAUSSIAN DENSITY

f_{X} (x) = \frac{1}{2 π σ} exp [- \frac{( x - μ ) ^{2}}{2 σ ^{2}}]

Fully described by mean $μ$ and variance $σ^{2}$ .

GAUSSIAN TAIL PROBABILITY

P (X > x) = Q (\frac{x - μ}{σ}), Q (x) = \frac{1}{2} erfc (\frac{x}{2})

Convention used in these notes: $Q (x) = \frac{1}{2} erfc (x / 2)$ . Also $Q (- x) = 1 - Q (x)$ and $Q (0) = 0.5$ .

Ensemble average is taken across the ensemble at a fixed time, for example $E [X (t)]$ . Time average is taken along one sample function over all time. A stationary process is ergodic if its time averages equal its ensemble averages, so one long record is enough to find its mean and power: the time-averaged DC value equals $μ_{X}$ and the time-averaged power equals $R_{X} (0)$ . An ergodic process must be stationary, but a stationary process need not be ergodic. For example, $X (t) = A$ , with $A$ a random variable fixed for all time, is stationary, yet the time average of each sample function is the value $A$ happened to take, not $E [A]$ .

TIME AVERAGE OF A SAMPLE FUNCTION

⟨ x (t)⟩ = T \to \infty lim \frac{1}{2 T} \int_{- T}^{T} x (t) d t, ⟨ x^{2} (t)⟩ = T \to \infty lim \frac{1}{2 T} \int_{- T}^{T} x^{2} (t) d t

For an ergodic process, $⟨ x (t)⟩ = μ_{X}$ and $⟨ x^{2} (t)⟩ = R_{X} (0)$ .

Type	Condition	Remark
Strict-sense stationary (SSS)	All joint statistics unchanged by a time shift	Strongest condition
Wide-sense stationary (WSS)	$E [X (t)]$ is constant and $E [X (t + τ) X (t)]$ depends only on $τ$	Only mean and autocorrelation matter
Relation	SSS implies WSS (if second moments exist)	WSS does not imply SSS, except for Gaussian processes

Autocorrelation function

AUTOCORRELATION OF A WSS PROCESS (REAL)

R_{X} (τ) = E [X (t + τ) X (t)]

Depends only on the lag $τ$ when $X (t)$ is WSS. The process is real-valued here.

Properties of $R_X(\tau)$ for a real WSS process

$R_{X} (0) = E [X^{2} (t)] \geq 0$ is the average power (mean-square value).
$R_{X} (τ) = R_{X} (- τ)$ : it is an even function.
$∣ R_{X} (τ) ∣ \leq R_{X} (0)$ : the maximum is at $τ = 0$ .
If $X (t + τ)$ and $X (t)$ become uncorrelated as $τ \to \infty$ (no periodic component and no random DC level), then $R_{X} (\infty) = μ_{X}^{2}$ , which is the DC power.
The variance is always $σ_{X}^{2} = R_{X} (0) - μ_{X}^{2}$ . It equals $R_{X} (0) - R_{X} (\infty)$ only under the condition in the previous line.
If $X (t)$ contains a periodic component of period $T_{0}$ , $R_{X} (τ)$ contains a periodic component of the same period and does not settle to a constant as $τ \to \infty$ .
The Fourier transform of $R_{X} (τ)$ must be non-negative everywhere, so not every even function is a valid autocorrelation.

RANDOM-PHASE SINUSOID

X (t) = A cos (2 π f_{0} t + Θ), Θ \sim U (0, 2 π) \Rightarrow R_{X} (τ) = \frac{A ^{2}}{2} cos (2 π f_{0} τ)

$A$ and $f_{0}$ are constants. A WSS process with zero mean and power $A^{2} /2$ . If $Θ$ is uniform over only $(0, π)$ , the mean depends on $t$ and the process is not WSS.

EXAM TIP

To read a mean and variance from a given $R_{X} (τ)$ that has no periodic part and settles to a constant: that constant is $μ_{X}^{2}$ , and $R_{X} (0)$ is the total power $E [X^{2}]$ . The variance is the difference of the two. Only $μ_{X}^{2}$ is found this way, so $R_{X} (τ)$ does not fix the sign of the mean.

Power spectral density and the Wiener-Khinchin theorem

The power spectral density $S_{X} (f)$ shows how the average power of a WSS process is spread over frequency, with $f$ in Hz. It is the Fourier transform of the autocorrelation function. Power is normalised to a 1 Ω resistance throughout.

WIENER-KHINCHIN THEOREM

S_{X} (f) = \int_{- \infty}^{\infty} R_{X} (τ) e^{- j 2 π f τ} d τ, R_{X} (τ) = \int_{- \infty}^{\infty} S_{X} (f) e^{j 2 π f τ} df

Applies to a WSS process. $S_{X} (f)$ is two-sided, so it covers negative and positive frequencies.

POWER FROM THE PSD

P = E [X^{2}] = R_{X} (0) = \int_{- \infty}^{\infty} S_{X} (f) df

Integrate over both sides of the spectrum. A nonzero mean contributes an impulse $μ_{X}^{2} δ (f)$ at $f = 0$ .

Properties of $S_X(f)$

$S_{X} (f) \geq 0$ for all $f$ .
$S_{X} (f) = S_{X} (- f)$ for a real process.
$S_{X} (f)$ is real.
The units are watts per hertz (power normalised to 1 Ω).

In the table below, $sinc (x) = \frac{sin π x}{π x}$ .

$R_{X} (τ)$	$S_{X} (f)$
$δ (τ)$	$1$
$K$ (a constant)	$K δ (f)$
$e^{- a ∣ τ ∣}, a > 0$	$\frac{2 a}{a ^{2} + 4 π ^{2} f ^{2}}$
$cos (2 π f_{0} τ)$	$\frac{1}{2} [δ (f - f_{0}) + δ (f + f_{0})]$
$2 B sinc (2 B τ)$	$1$ for $∣ f ∣ \leq B$ , else $0$
$1 - \frac{∣ τ ∣}{T}$ for $∣ τ ∣ \leq T$ , else $0$	$T sinc^{2} (f T)$

Random processes through LTI systems

Let a WSS input $X (t)$ , applied for all time (steady state), drive a stable LTI system with real impulse response $h (t)$ and frequency response $H (f)$ . The output $Y (t) = h (t) * X (t)$ is also WSS, and a Gaussian input gives a Gaussian output.

OUTPUT MEAN

μ_{Y} = μ_{X} H (0)

$H (0)$ is the DC gain.

OUTPUT PSD

S_{Y} (f) = ∣ H (f) ∣^{2} S_{X} (f)

Only the magnitude of $H (f)$ matters; phase has no effect on the output PSD.

OUTPUT AUTOCORRELATION

R_{Y} (τ) = h (τ) * h (- τ) * R_{X} (τ)

The Fourier transform of this is $∣ H (f) ∣^{2} S_{X} (f)$ for real $h (t)$ .

OUTPUT POWER

P_{Y} = \int_{- \infty}^{\infty} ∣ H (f) ∣^{2} S_{X} (f) df

Total output mean-square value, including any DC part.

INPUT-OUTPUT CROSS-CORRELATION

R_{Y X} (τ) = h (τ) * R_{X} (τ), S_{Y X} (f) = H (f) S_{X} (f)

Defined as $R_{Y X} (τ) = E [Y (t + τ) X (t)]$ . For the other order, $R_{X Y} (τ) = h (- τ) * R_{X} (τ)$ and $S_{X Y} (f) = H^{*} (f) S_{X} (f)$ .

SUM OF TWO PROCESSES

Z (t) = X (t) + Y (t) : R_{Z} (τ) = R_{X} (τ) + R_{Y} (τ), S_{Z} (f) = S_{X} (f) + S_{Y} (f)

Holds when $X (t)$ and $Y (t)$ are jointly WSS and uncorrelated, with at least one of them zero-mean (for example, a signal plus independent zero-mean noise). Otherwise add the cross terms $R_{X Y} (τ) + R_{Y X} (τ)$ .

MODULATION BY A RANDOM-PHASE CARRIER

Y (t) = X (t) cos (2 π f_{c} t + Θ) : R_{Y} (τ) = \frac{1}{2} R_{X} (τ) cos (2 π f_{c} τ), S_{Y} (f) = \frac{1}{4} [S_{X} (f - f_{c}) + S_{X} (f + f_{c})]

$X (t)$ is WSS and $Θ \sim U (0, 2 π)$ is independent of $X (t)$ . The output power is $P_{Y} = R_{X} (0) /2$ .

DIFFERENTIATOR

Y (t) = \frac{d X ( t )}{d t} : H (f) = j 2 π f, S_{Y} (f) = 4 π^{2} f^{2} S_{X} (f), R_{Y} (τ) = - \frac{d ^{2} R _{X} ( τ )}{d τ ^{2}}

Follows from $S_{Y} (f) = ∣ H (f) ∣^{2} S_{X} (f)$ .

EXAM TIP

Procedure for a filtering question: (1) write $S_{X} (f)$ ; (2) find $∣ H (f) ∣^{2}$ ; (3) multiply to get $S_{Y} (f)$ ; (4) integrate for power, or inverse-transform for $R_{Y} (τ)$ . Remember to integrate over negative frequencies too, or use a factor of 2 on the positive side.

White noise and noise bandwidth

Convention: white noise is zero-mean with a flat two-sided PSD $N_{0} /2$ (W/Hz) for all $f$ , with spectra in Hz. $N_{0}$ is then the one-sided PSD, defined for $f \geq 0$ only.

WHITE NOISE

S_{W} (f) = \frac{N _{0}}{2}, R_{W} (τ) = \frac{N _{0}}{2} δ (τ)

Samples at different instants are uncorrelated (independent if the noise is also Gaussian). Total power is infinite, so ideal white noise is a model that is used only after some filtering.

WHITE NOISE THROUGH AN IDEAL LOW-PASS OR BAND-PASS FILTER

P_{out} = \frac{N _{0}}{2} \cdot ∣ H_{0} ∣^{2} \cdot 2 B = N_{0} ∣ H_{0} ∣^{2} B

$H_{0}$ is the passband gain and $B$ is the one-sided bandwidth in Hz (for a low-pass filter, $∣ f ∣ \leq B$ ; for a band-pass filter, total width $B$ centred on $\pm f_{c}$ ).

AUTOCORRELATION OF WHITE NOISE AFTER AN IDEAL LOW-PASS FILTER

R_{Y} (τ) = N_{0} B sinc (2 B τ)

Unity gain, passband $∣ f ∣ \leq B$ , $sinc (x) = \frac{sin π x}{π x}$ . $R_{Y} (τ)$ is zero at $τ = k / (2 B)$ , $k \neq = 0$ , so samples taken $1/ (2 B)$ apart are uncorrelated (independent if the noise is Gaussian).

NOISE-EQUIVALENT BANDWIDTH

B_{N} = \frac{\int _{0}^{\infty} ∣ H ( f ) ∣ ^{2} df}{∣ H ( f ) ∣ _{m a x}^{2}}

Width of an ideal filter of the same peak gain that passes the same white-noise power: $P_{out} = N_{0} ∣ H ∣_{m a x}^{2} B_{N}$ . The peak is at $f = 0$ for a low-pass filter and at the centre frequency for a band-pass filter; $f$ in Hz.

RC LOW-PASS FILTER WITH WHITE NOISE INPUT

∣ H (f) ∣^{2} = \frac{1}{1 + ( 2 π f R C ) ^{2}}, R_{Y} (τ) = \frac{N _{0}}{4 R C} e^{- ∣ τ ∣/ R C}, P_{Y} = \frac{N _{0}}{4 R C}

Here $B_{N} = \frac{π}{2} f_{3 dB} = \frac{1}{4 R C}$ with $f_{3 dB} = \frac{1}{2 π R C}$ . The output is coloured noise.

THERMAL NOISE OF A RESISTOR

\overline{v_{n}^{2}} = 4 k T R B, P_{avail} = k T B, \frac{N _{0}}{2} = \frac{k T}{2}

$k = 1.38 \times 1 0^{- 23}$ J/K, $T$ in kelvin, $B$ in Hz. $\overline{v_{n}^{2}}$ is the open-circuit mean-square voltage of a resistor $R$ , and $P_{avail}$ is the power it delivers to a matched load. Thermal noise is Gaussian and is white for frequencies well below $k T / h$ (about $6 \times 1 0^{12}$ Hz at 290 K).

EXAM TIP

The factor-of-two trap: with a two-sided PSD $N_{0} /2$ , the power in a bandwidth $B$ (Hz, positive side) is $N_{0} B$ , not $N_{0} B /2$ . Always check whether a question gives $N_{0} /2$ or $N_{0}$ .

Gaussian processes and narrowband noise

A process is Gaussian if every set of samples is jointly Gaussian. Such a process is completely described by its mean and autocorrelation. Therefore a WSS Gaussian process is also strict-sense stationary, and any linear operation (filtering by a stable LTI system, sampling, adding) on it gives another Gaussian result. Uncorrelated jointly Gaussian samples are independent.

Statement	True for	Remark
Uncorrelated implies independent	Jointly Gaussian variables only	False in general
WSS implies SSS	Gaussian processes	False in general
Output is Gaussian	Linear filtering of Gaussian input	Holds for any stable LTI filter

Noise at a receiver, after a band-pass filter of width $B$ around a centre frequency $f_{c}$ (with $f_{c} > B /2$ ), is narrowband noise. It can be written in terms of in-phase and quadrature components.

NARROWBAND NOISE REPRESENTATION

n (t) = n_{c} (t) cos (2 π f_{c} t) - n_{s} (t) sin (2 π f_{c} t)

$n (t)$ is zero-mean, WSS and band-pass. $n_{c} (t)$ and $n_{s} (t)$ are low-pass, zero-mean and jointly WSS; if $n (t)$ is Gaussian, they are jointly Gaussian.

POWERS OF THE COMPONENTS

E [n^{2}] = E [n_{c}^{2}] = E [n_{s}^{2}] = \int_{- \infty}^{\infty} S_{n} (f) df

For white noise $N_{0} /2$ through an ideal band-pass filter of width $B$ (gain 1): each of the three powers equals $N_{0} B$ .

SPECTRUM OF THE COMPONENTS

S_{n_{c}} (f) = S_{n_{s}} (f) = {S_{n} (f - f_{c}) + S_{n} (f + f_{c}), 0, ∣ f ∣ \leq B /2 otherwise

Holds for any band-pass spectrum confined to $f_{c} - B /2 \leq ∣ f ∣ \leq f_{c} + B /2$ . For white noise $N_{0} /2$ through an ideal band-pass filter of width $B$ (gain 1), $S_{n_{c}} (f) = S_{n_{s}} (f) = N_{0}$ for $∣ f ∣ \leq B /2$ . At the same instant, $n_{c} (t)$ and $n_{s} (t)$ are always uncorrelated (independent if Gaussian); if $S_{n} (f)$ is also symmetric about $\pm f_{c}$ , $n_{c} (t)$ and $n_{s} (t + τ)$ are uncorrelated for every $τ$ .

ENVELOPE AND PHASE OF NARROWBAND GAUSSIAN NOISE

r = n_{c}^{2} + n_{s}^{2}, f_{R} (r) = \frac{r}{σ ^{2}} exp (- \frac{r ^{2}}{2 σ ^{2}}), r \geq 0; ψ \sim U (0, 2 π)

For zero-mean Gaussian narrowband noise with $σ^{2} = E [n^{2}]$ , written as $n (t) = r (t) cos (2 π f_{c} t + ψ (t))$ : the envelope is Rayleigh, and the phase is uniform and independent of the envelope. $P (r > a) = exp (- a^{2} /2 σ^{2})$ and $E [r^{2}] = 2 σ^{2}$ . If a sinusoid of amplitude $A$ is added, the envelope becomes Rician: close to Rayleigh for $A ≪ σ$ , and close to Gaussian about $A$ for $A ≫ σ$ .

PREVIOUS-YEAR QUESTIONS

6 questions from the EC papers

Try each question first, then open the answer. We placed each question under a topic ourselves; the exam paper does not label topics.

See these papers in full: GATE ECE 2025 · GATE ECE 2024 · GATE ECE 2023 · GATE ECE 2020 · GATE ECE 2018

GATE 2025 EC Q.42

GATE 2025 · ECQ.42 · 2 marks · MCQ

Consider a real-valued random process f(t) = sum_{n=1}^{N} a_n p(t - nT), where T > 0 and N is a positive integer. Here, p(t) = 1 for t in [0, 0.5T] and 0 otherwise. The coefficients a_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 E[f(t)f(t+tau)] is independent of time t for all tau. (Here, E[.] is the expectation operation.)

(A)
(i) is TRUE and (ii) is FALSE
(B)
Both (i) and (ii) are TRUE
(C)
Both (i) and (ii) are FALSE
(D)
(i) is FALSE and (ii) is TRUE

Answer and solution

GATE 2025 EC Q.42: answer and solution

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GATE 2024 EC Q.20

GATE 2024 · ECQ.20 · 1 mark · MCQ

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

(A)
$\frac{N _{0}}{4 R C}$
(B)
$\frac{N _{0}}{2 R C}$
(C)
$\frac{N _{0}}{R C}$
(D)
$\frac{2 N _{0}}{R C}$

Answer and solution

GATE 2024 EC Q.20: answer and solution

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GATE 2023 EC Q.30

GATE 2023 · ECQ.30 · 1 mark · MSQ

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

GATE 2023 EC question 30: Options (C) and (D) are plots of S_X(omega) against omega. — Figure from the GATE 2023 EC paper. It is part of the question.

(A)
S_X(omega) = 2/(9 + omega^2)
(B)
S_X(omega) = e^{-omega^2} cos^2(omega)
(C)
Figure — S_X(omega) equal to +1 on a band 0 < omega < omega_1 and equal to -1 on a band -omega_1' < omega < 0 (odd-shaped, taking negative values)
(D)
Figure — S_X(omega) equal to 1 on a band 0 < omega < omega_1 only, zero for omega < 0 (one-sided, not symmetric about omega = 0)

Answer and solution

GATE 2023 EC Q.30: answer and solution

GATE 2023 EC Q.59

GATE 2023 · ECQ.59 · 2 marks · NAT

Let $X (t)$ be a white Gaussian noise with power spectral density $\frac{1}{2}$ W/Hz. If $X (t)$ is input to an LTI system with impulse response $e^{- t} u (t)$ . The average power of the system output is ______ W (rounded off to two decimal places).

Answer and solution

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GATE 2020 EC Q.24

GATE 2020 · ECQ.24 · 1 mark · NAT

The random variable $Y = \int_{- \infty}^{\infty} W (t) ϕ (t) d t$ , where $ϕ (t) = 1$ for $5 \leq t \leq 7$ and $0$ otherwise, 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 ______.

Answer and solution

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GATE 2018 EC Q.28

GATE 2018 · ECQ.28 · 2 marks · MCQ

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 $0.5 e^{- t^{2} /2}$ (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

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PRACTICE

6 questions to test yourself

PRACTICE 11 mark · MCQ

Which of the following functions can be the autocorrelation function of a real wide-sense stationary process? Here $rect (τ) = 1$ for $∣ τ ∣ < \frac{1}{2}$ and $0$ otherwise, and $tri (τ) = 1 - ∣ τ ∣$ for $∣ τ ∣ \leq 1$ and $0$ otherwise.

(A)
$3 e^{- 2∣ τ ∣} - sin τ$
(B)
$τ [2 cos (5 τ) + 3]$
(C)
$4 rect (τ)$
(D)
$5 tri (τ)$

Answer and solution

Answer: (D)

An autocorrelation function must be even, have its maximum at $τ = 0$ , and have a non-negative Fourier transform. Below, $sinc (x) = \frac{sin π x}{π x}$ .
Option A is not even (the $- sin τ$ term is odd), so it fails.
Option B is odd (it is $τ$ times an even function), it is zero at $τ = 0$ but nonzero elsewhere, and it is unbounded, so it fails.
Option C: the transform is $4 sinc (f)$ , which takes negative values (for example for $1 < f < 2$ ), so it fails.
Option D: the transform is $5 sinc^{2} (f) \geq 0$ , and the function is even with its maximum at $0$ . Answer D.

PRACTICE 22 marks · NAT

Zero-mean white noise with two-sided PSD $N_{0} /2 = 1 \times 1 0^{- 9}$ W/Hz is applied to an ideal low-pass filter with gain $2$ in the passband $∣ f ∣ \leq 5$ kHz and zero gain elsewhere. Find the average output noise power in microwatts ( $μ$ W).

Answer and solution

Answer: $40$ (accepted range $\pm 0.1$ )

$S_{Y} (f) = ∣ H (f) ∣^{2} S_{X} (f) = 4 \times 1 0^{- 9}$ W/Hz for $∣ f ∣ \leq 5000$ Hz.
$P_{Y} = \int S_{Y} df = 4 \times 1 0^{- 9} \times (2 \times 5000) = 4 \times 1 0^{- 5}$ W.
This is $40 μ$ W.

PRACTICE 32 marks · NAT

White noise with one-sided PSD $N_{0} = 2 \times 1 0^{- 9}$ W/Hz (two-sided $N_{0} /2$ ) is passed through a first-order RC low-pass filter with unity DC gain and 3 dB frequency $1$ kHz. Using the noise-equivalent bandwidth, find the output noise power in $μ$ W. Take $π = 3.1416$ .

Answer and solution

Answer: $3.14$ (accepted range $\pm 0.03$ )

For an RC low-pass filter, $B_{N} = \frac{π}{2} f_{3 dB} = \frac{π}{2} \times 1000 \approx 1570.8$ Hz.
Output power $= N_{0} ∣ H (0) ∣^{2} B_{N} = 2 \times 1 0^{- 9} \times 1 \times 1570.8 \approx 3.14 \times 1 0^{- 6}$ W.
Answer: $3.14 μ$ W.

PRACTICE 42 marks · MSQ

Let $X (t)$ be a real, zero-mean, wide-sense stationary process with autocorrelation $R_{X} (τ)$ and PSD $S_{X} (f)$ . Which of the following statements are always true?

(A)
$R_{X} (τ) = R_{X} (- τ)$
(B)
$∣ R_{X} (τ) ∣ \leq R_{X} (0)$ for all $τ$
(C)
$S_{X} (f)$ may be negative at some frequencies, provided its total integral is positive
(D)
If $X (t)$ is also Gaussian, it is strict-sense stationary

Answer and solution

Answer: (A), (B), (D)

A is true: a real WSS autocorrelation is even.
B is true: the autocorrelation peaks at zero lag.
C is false: the PSD is non-negative at every frequency.
D is true: a Gaussian process is fixed by its mean and autocorrelation, so WSS implies SSS.
Answer: A, B, D.

PRACTICE 51 mark · MCQ

A zero-mean Gaussian WSS process has $R_{X} (τ) = 4 e^{- ∣ τ ∣}$ . The probability that $X (t_{0}) > 4$ at a given instant $t_{0}$ is

(A)
$Q (1)$
(B)
$Q (2)$
(C)
$Q (4)$
(D)
$Q (0.5)$

Answer and solution

Answer: (B)

Variance $= R_{X} (0) - μ^{2} = 4 - 0 = 4$ , so $σ = 2$ .
$P (X > 4) = Q (\frac{4 - 0}{2}) = Q (2)$ (about $0.0228$ ).
Answer B.

PRACTICE 62 marks · NAT

A zero-mean WSS process $X (t)$ has PSD $S_{X} (f) = 1 0^{- 3}$ W/Hz for $∣ f ∣ \leq 1$ kHz and zero elsewhere. It is multiplied by a carrier to give $Y (t) = X (t) cos (2 π \times 1 0^{4} t + Θ)$ , where $Θ$ is uniform over $(0, 2 π)$ and independent of $X (t)$ . Find the average power of $Y (t)$ in watts.

Answer and solution

Answer: $1$ (accepted range $\pm 0.01$ )

$P_{X} = \int S_{X} (f) df = 1 0^{- 3} \times 2000 = 2$ W.
$S_{Y} (f) = \frac{1}{4} [S_{X} (f - f_{c}) + S_{X} (f + f_{c})]$ , so $P_{Y} = \frac{1}{4} (P_{X} + P_{X}) = P_{X} /2$ .
Equivalently $R_{Y} (0) = \frac{1}{2} R_{X} (0) cos 0 = 1$ W.
Answer: $1$ W.

OFFICIAL GATE 2027 SYLLABUS WORDING

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GATE EC

Random Processes: Autocorrelation and power spectral density, properties of white noise, filtering of random signals through LTI systems

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