# The random variable$$Y = \mathop \smallint \limits_{ - \infty }^\infty W\left( t \right)\phi \left( t \right)dt,$$ where$$\phi \left( t \right)=\;\left\{ {\begin{array}{*{20}{c}} {1;}&{5 \le t \le 7}\\ {0;}&{Otherwise} \end{array}} \right\}$$ And W(t) is a real white Gaussian noise process with two-sided power spectral density SW (f) = 3 W/Hz, for all f. The variance of Y is:

## Random Processes MCQ Question 1 Detailed Solution

Concept:

The mean or the expected value of a Gaussian random variable is 0.

The variance of a random variable 'X' is defined as:

$$E\left[ {{X^2}} \right] = E\left[ {X \cdot X} \right]$$

Also, the autocorrelation function and the Energy spectrum density form Fourier transform pairs, i.e.

$${R_w}\left( \tau \right)\mathop \leftrightarrow \limits^{FT} {S_w}\left( f \right)$$

Application:

Sw(f) = 3 Watt/Hz

Since the power spectral density is constant, the autocorrelation function will be an impulse function, i.e.

$${S_w}\left( f \right)\;\mathop \leftrightarrow \limits^{IFT} {R_w}\left( \tau \right)$$

∴ Rw(τ) = 3 δ (τ)

Also, Rw (t2 – t1) = 3δ (t2 – t1)     …1)

Now,

$$E\left[ Y \right] = E\left[ {\mathop \smallint \nolimits_{ - \infty }^\infty w\left( t \right) \cdot \phi \left( t \right)dt} \right]$$

$$E\left[ Y \right] = \mathop \smallint \nolimits_{ - \infty }^\infty E\left[ {w\left( t \right)} \right]\phi \left( t \right)\;dt$$

Since W(t) is a real white Gaussian noise process:

E[w(t)] = 0

∴ E[Y] = 0

Now, the variance of Y  can be written as:

$$E\left[ {{Y^2}} \right] = E\left[ {Y \cdot Y} \right]$$

Putting on the respective functions, we get:

$$= E\left[ {\mathop \smallint \nolimits_{-\infty} ^{ \infty } w\left( {{t_1}} \right)\phi \left( {{t_1}} \right)d{t_1}\mathop \smallint \nolimits_{ - \infty }^\infty w\left( {{t_2}} \right)\phi \left( {{t_2}} \right)d{t_2}} \right]$$

$$= E\left[ {\mathop \smallint \nolimits_{ - \infty }^\infty \mathop \smallint \nolimits_{ - \infty }^\infty w\left( {{t_1}} \right)w\left( {{t_2}} \right)\phi \left( {{t_1}} \right)\phi \left( {{t_2}} \right)d{t_1}d{t_2}} \right]$$

$$= \mathop \smallint \nolimits_{ - \infty }^\infty \mathop \smallint \nolimits_{ - \infty }^\infty E\left[ {w\left( {{t_1}} \right)w\left( {{t_2}} \right)} \right] \cdot \phi \left( {{t_1}} \right)\phi \left( {{t_2}} \right)d{t_1}d{t_2}$$

$$= \mathop \smallint \nolimits_{ - \infty }^\infty \mathop \smallint \nolimits_{ - \infty }^\infty {R_w}\left( {{t_2} - {t_1}} \right) \cdot \phi \left( {{t_1}} \right)\phi \left( {{t_2}} \right)d{t_1}d{t_2}$$

From Equation (1), we can write:

$$E\left[ {{Y^2}} \right] = \mathop \smallint \nolimits_{ - \infty }^\infty \mathop \smallint \nolimits_{ - \infty }^\infty 3\delta \left( {{t_2} - {t_1}} \right)\phi \left( {{t_1}} \right)\phi \left( {{t_2}} \right)d{t_1}d{t_2}$$

The above integration exists, provided:

t1 = t2 = t

$$E\left[ {{Y^2}} \right] = \mathop \smallint \nolimits_{ - \infty }^\infty \mathop \smallint \nolimits_{ - \infty }^\infty 3\delta \left( 0 \right){\phi ^2}\left( t \right) \cdot dt$$

$$= 3 \times 1 \times \mathop \smallint \nolimits_5^7 {\phi ^2}\left( t \right) \cdot dt$$

= 3 × 2 = 6

# One of the following types of noise assumes importance at high frequncies:

1. Thermal noise
2. Shot noise
3. Transit-time noise
4. Flicker noise

Option 3 : Transit-time noise

## Random Processes MCQ Question 2 Detailed Solution

High- Frequency Noise:

• These noises are also known as TRANSIT- TIME Noise
• They are observed in the semiconductor devices when the transit time of a charge carrier while crossing a junction is compared with the time period of that signal

Various types of internal noise are:

Shot Noise:

• These Noises are generally arising in the active devices due to the random behavior of Charge particles or carries
• In the case of the electron tube, shot Noise is produced due to the random emission of electron from cathodes

Partition Noise:

• When a circuit is to divide in between two or more paths then the noise generated is known as Partition noise
• The reason for the generation is a random fluctuation in the division

Low- Frequency Noise:

• They are also known as FLICKER NOISE
• These types of noise are generally observed at a frequency range below few kHz
• Power spectral density of these noise increases with the decrease in frequency
• That's why the name is given Low- Frequency Noise

Thermal Noise:

• Thermal Noise is random and often referred to as White Noise or Johnson Noise
• Thermal noise is generally observed in the resistor or the sensitive resistive components of a complex impedance due to the random and rapid movement of molecules or atoms or electrons

# The response of a Gaussian random process applied to a stable linear system is:1. A Gaussian random process2. Not a Gaussian random process3. Completely specified by its mean and auto-covariance functionsWhich of the above statements is/are correct?

1. 1 only
2. 2 only
3. 2 and 3
4. 1 and 3

Option 4 : 1 and 3

## Random Processes MCQ Question 3 Detailed Solution

An important property of Gaussian random process is that their Probability density function is completely determined by their mean and covariance, i.e.

$${f_X}\left( X \right) = \frac{1}{{\sqrt {2\pi {\sigma ^2}} }}{e^{ - \frac{{{{\left( {x - \mu } \right)}^2}}}{{2{\sigma ^2}}}}}$$

Where μ = mean

σ = Standard deviation.

Fourier analysis states that any Linear system in the frequency spectrum can be represented as the sum of sinusoids, i.e.

$$H\left( \omega \right) = \mathop \sum \limits_{n \to - \infty }^{n \to \infty } sin\left( {n{\omega _0}} \right)$$

Now, If the Input (x(t)) is Gaussian, then the output y(t) in the frequency spectrum will be:

Y(ω) = X(ω) × H(ω)

For a sinusoid input, the output will be varied in Magnitude and phase, but nature will remain the same according to the input.

So, the output will be Gaussian

Observations:

1) The output phase and magnitude will be varied, but nature will remain the same as that of the input.

2) So, if the input is Gaussian, the output will also be Gaussian.

# The noise caused by random variations in the arrival of electrons or holes at the output electrode of an amplifying device

1. white noise
2. flicker
3. Shot Noise
4. Transmit time noise

Option 3 : Shot Noise

## Random Processes MCQ Question 4 Detailed Solution

Shot noise:

• It is a type of noise that appears in all amplifying & active devices.
• It is associated with a current flow
• The noise caused by random variations in the arrival of electrons or holes at the output electrode of an amplifying device is called shot noise.
• In semiconductor devices, shot Noise is due to the random variation in the diffusion of minority carriers
• Current is a flow of carriers each of which carriers a finite amount of charge.
• For diodes, the exact expression for RMS shot noise $$I_n^2 = 2\;{I_{dC}}qB$$

• At low frequencies, the noise varies approximately as 1/f, and is called an excess, or flicker noise. It is caused by the recombination and generation of carriers on the surface of the crystal.
• In intermediate frequencies, the noise is independent of frequency. This white noise is caused by the bulk resistance of the semiconductor material and the statistical variation of the currents (shot noise).
• The third region is characterized by an increase of the noise figure with frequency and is essentially caused by a decrease in power gain with frequency.

# The relation between power spectral density S(f) and the total average power over 1Ω is:

1. P = S(f) × B/2
2. $$P = \mathop \smallint \nolimits_{ - \infty }^\infty S\left( f \right)df$$
3. $$P = \mathop \smallint \nolimits_{ - \infty }^\infty S\left( f \right)\exp \left( {j\omega t} \right)df$$
4. P = S(f) × B

Option 2 : $$P = \mathop \smallint \nolimits_{ - \infty }^\infty S\left( f \right)df$$

## Random Processes MCQ Question 5 Detailed Solution

Analysis:

The power spectral density (PSD) of a WSS process is defined as the Fourier transform of its Autocorrelation function. It is given as:

$${S_X}\left( f \right) = \mathop \smallint \nolimits_{ - \infty }^\infty {R_x}\left( \tau \right){e^{ - j2\pi f\tau }} \cdot d\tau$$

Note:

The Autocorrelation can be obtained from the PSD through inverse transform,

$${R_x}\left( \tau \right) = \mathop \smallint \nolimits_{ - \infty }^\infty {S_X}\left( f \right){e^{j2\pi f\tau }}df$$

→ RX (0) = ∫ SX (f) ⋅ df = Power of x(t)

→ SX (f) is real and SX (f) ≥ 0

→ SX (f) = SX (-f)

# The autocorrelation function of a band-limited (B) white noise process is given by

1. R(τ) = B × N0 In (2Bτ)
2. R(τ) = B × N0 × τ × sin (2B)
3. R(τ) = B × N0 log (2Bτ)
4. R(τ) = B × N0 sinc (2Bτ)

Option 4 : R(τ) = B × N0 sinc (2Bτ)

## Random Processes MCQ Question 6 Detailed Solution

Concept:

PSD is the Fourier transform of the autocorrelation function, i.e.

$$R_{nm}(\tau) \leftrightarrow S_{nm}(f)$$

Since white noise is uniformly distributed, the spectral density will be represented as:

Now, using the Fourier transform pair,

Using the duality property of Fourier transform, we can write:

$$2{\rm{asinc}}\left( {{\rm{t}}.2{\rm{a}}} \right)\mathop \leftrightarrow \limits^{{\rm{FT}}} {\rm{rect}}\left( {\frac{{\rm{f}}}{{2{\rm{a}}}}} \right)$$

Replacing, $${\rm{a}} = {\rm{B}}$$

$$2{\rm{Bsinc}}\left( {{\rm{\tau }}.2{\rm{B}}} \right)\mathop \leftrightarrow \limits^{{\rm{FT}}} {\rm{rect}}\left( {\frac{{\rm{f}}}{{2{\rm{B}}}}} \right)$$

Multiplying both sides by $$\frac{{{{\rm{N}}_{\rm{o}}}}}{2}$$, we get:

$${{\rm{N}}_{\rm{o}}}{\rm{Bsinc}}\left( {2{\rm{B\tau }}} \right)\mathop \leftrightarrow \limits^{{\rm{FT}}} \frac{{{{\rm{N}}_0}}}{2}{\rm{rect}}\left( {\frac{{\rm{f}}}{{2{\rm{B}}}}} \right)$$

∴ The autocorrelation function is:

R(τ) = B × N 0 sinc (2Bτ)

# If the autocorrelation function of a random process X(t) is R(τ) its power spectral density is:

1. S(f) = In R(τ)
2. S(f) = exp (R(τ))
3. S(f) = R(τ)*R(τ)
4. $$S\left( f \right) = \mathop \smallint \nolimits_{ - \infty }^\infty R\left( \tau \right)\exp \left\{ { - \left( {j 2\pi f \tau } \right)} \right\}d\tau$$

Option 4 : $$S\left( f \right) = \mathop \smallint \nolimits_{ - \infty }^\infty R\left( \tau \right)\exp \left\{ { - \left( {j 2\pi f \tau } \right)} \right\}d\tau$$

## Random Processes MCQ Question 7 Detailed Solution

Analysis:

The power spectral density (PSD) of a WSS process is defined as the Fourier transform of its Autocorrelation function.

It is given as:

$${S_X}\left( f \right) = \mathop \smallint \nolimits_{ - \infty }^\infty {R_x}\left( \tau \right){e^{ - j2\pi f\tau }} \cdot d\tau$$

Note:

The Autocorrelation can be obtained from the PSD through inverse transform as:

$${R_x}\left( \tau \right) = \mathop \smallint \nolimits_{ - \infty }^\infty {S_X}\left( f \right){e^{j2\pi f\tau }}df$$

SX (f) is real and SX (f) ≥ 0

SX (f) = SX (-f)

RX (0) = ∫ SX (f) ⋅ df = Power of x(t)

# Spectral density of white noise:

1. is constant
2. Varies with frequency
3. varies with bandwidth
4. varies with amplitude of the signal

Option 1 : is constant

## Random Processes MCQ Question 8 Detailed Solution

White noise is that signal whose frequency spectrum is uniform i.e. it has flat spectral density.

The power spectral density (PSD) of white noise is uniform throughout the frequency spectrum as shown:

# Cross-correlation function provides:

1. Information about the structure of only one signal
2. Information about the behavior of only one signal in the time domain
3. The measure of dissimilarities between two signals
4. The measure of similarities between two signals

Option 4 : The measure of similarities between two signals

## Random Processes MCQ Question 9 Detailed Solution

The correlation function measures the similarity between two random signals, i.e. how often the two random signals have similar values.

Depending upon the signals we are correlating, there are two kinds of correlation.

Autocorrelation:

This is a type of correlation in which the given signal is correlated with itself mathematically, this is defined as:

$${R_{xx}}\left( y \right) = \;\mathop \smallint \limits_{ - \infty }^\infty x\left( t \right)\;{x^*}\;\left( {t - \tau } \right)dt$$

Cross-correlation:

This is a type of correlation, in which the signal in hand is correlated with another signal to understand how much they resemble each other 'or' how much similar they are to each other.

Mathematically, this is defined as:

$${R_{xy}}\left( \tau \right) = \;\mathop \smallint \limits_{ - \infty }^\infty x\left( t \right)y\left( {t - \tau } \right)d\tau$$

# The autocorrelation of a wide-sense stationary random process is given by:  e-2|τ| .The peak value of the spectral density is

1. 2
2. 1
3. e-1/2
4. e

Option 2 : 1

## Random Processes MCQ Question 10 Detailed Solution

Concept:

The power spectral density is basically the Fourier transform of the autocorrelation function of the power signal, i.e.

$${S_x}\left( f \right) = F.T.\left\{ {{R_x}\left( \tau \right)} \right\}$$

Analysis:

Given, the autocorrelation function of the random signal X(t) as:

$${R_X}\left( \tau \right) = {e^{ - 2\left| \tau \right|}}$$

So, its power spectral density is obtained as:

$${S_X}\left( f \right) = {\cal F}\left\{ {{R_X}\left( \tau \right)} \right\}$$

$$= \mathop \smallint \limits_{ - \infty }^\infty {R_X}\left( \tau \right){e^{ - j2\pi f\tau }}d\tau$$

$$= \mathop \smallint \limits_{ - \infty }^0 {e^{2\tau }}{e^{ - j2\pi f\tau }}d\tau + \mathop \smallint \limits_0^\infty {e^{ - 2\tau }}{e^{ - j2\pi f\tau }}d\tau$$

$$= \frac{1}{{2\left( {1 - j\pi f} \right)}} + \frac{1}{{2\left( {1 + j\pi f} \right)}}$$

$$= \frac{1}{{{1} + {\pi ^2}{f^2}}}$$

The peak value of PSD is at f = 0,

SX(0) = 1

# Following are few statements regarding noise in communication system:a) Atmospheric noise, shot noise, solar noise are examples of external noise sourcesb) Noise temperature is useful in dealing with UHF noisec) Thermal agitation is the only source of noise in receiverChoose the correct answer:

1. Only b)
2. a) and b)
3. b) and c)
4. a) and c)

Option 1 : Only b)

## Random Processes MCQ Question 11 Detailed Solution

Statement a:

Noise in communication systems is generally divided into two categories-

i) External noise like atmospheric noise, industrial noise solar noise, and cosmic noise.

ii) Internal noise like shot noise, thermal noise, low and high-frequency noise. (Statement a is false)

Statement b:

Noise temperature is one way of expressing the level of available noise power introduced by a component or source. The power spectral density of the noise is expressed in terms of the temperature (in kelvins) that would produce that level of Johnson–Nyquist noise.

It is useful in dealing with UHF. (Statement b is True)

Statement c:

Internal and external noises are the main source of noise in the receiver (Statetemt c is false)

Note:

Thermal Agitation:

• It is also called Johnson-Nyquist noise
• It is distributed across the entire frequency spectrum range so it is also called as white noise
• Thermal noise is generally observed in the passive component like R, L & C due to the random movement of charge carriers electrons.
• Random noise power is

Pn = KTB

Where,

K = Boltz man's constant

T = temperature

B = Bandwidth over which noise is measured

# A telephone exchange has 9000 subscribers. If the number of calls originating at peak time is 10,000 in one hour, the calling rate is

1. 0.9
2. 10/9
3. 0.81
4. 0.1

Option 2 : 10/9

## Random Processes MCQ Question 12 Detailed Solution

Calling rate is the number of calls per subscriber.

Given that, the number of subscribers = 9000

Number of calls = 10,000

Calling rate = 10,000/9000 = 10/9

# Let (𝑡) be a wide sense stationary (WSS) random process with power spectral density 𝑆𝑋(𝑓). If Y(t) is the process defined as 𝑌(𝑡) = 𝑋(2𝑡 − 1), the power spectral density 𝑆𝑌(𝑓) is

1. $${S_Y}\left( f \right) = \frac{1}{2}{S_X}\left( {\frac{f}{2}} \right){e^{ - j\pi f}}$$
2. $${S_Y}\left( f \right) = \frac{1}{2}{S_X}\left( {\frac{f}{2}} \right){e^{ - j\pi f/2}}$$
3. $${S_Y}\left( f \right) = \frac{1}{2}{S_X}\left( {\frac{f}{2}} \right)$$
4. $${S_Y}\left( f \right) = \frac{1}{2}{S_X}\left( {\frac{f}{2}} \right){e^{ - j2\pi f}}$$

Option 3 : $${S_Y}\left( f \right) = \frac{1}{2}{S_X}\left( {\frac{f}{2}} \right)$$

## Random Processes MCQ Question 13 Detailed Solution

Concept:

Power spectral density function (PSD) is the Fourier transform of the Autocorrelation function of the WSS process.

$$h\left( t \right)\mathop \to \limits^{FT} H\left( f \right)$$

$$h\left( {at} \right)\mathop \to \limits^{FT} \frac{1}{a}H\left( {\frac{f}{a}} \right)$$

The shifting in the time domain does not change the PSD.

Calculation:

Let the autocorrelation function be denoted as R(τ).

Therefore,

$${R_x}\left( \tau \right)\mathop \to \limits^{FT} {S_X}\left( f \right)$$

Then,

$$Y\left( t \right) = X\left( {2t - 1} \right)\mathop \to \limits^{Autocorrelation} {R_y}\left( \tau \right) = {R_x}\left( {2\tau } \right)$$

$${R_x}\left( {2\tau } \right)\mathop \to \limits^{FT} \frac{1}{2}{S_x}\left( {\frac{f}{2}} \right)$$

$${R_y}\left( \tau \right)\mathop \to \limits^{FT} {S_y}\left( f \right) = \frac{1}{2}{S_x}\left( {\frac{f}{2}} \right)$$

# A random process X(t) is called ‘white noise’ if the power spectral density is equal to

1. $$\frac{\eta }{8}$$
2. $$\frac{\eta }{2}$$
3. $$\frac{3\eta }{4}$$
4. η

Option 2 : $$\frac{\eta }{2}$$

## Random Processes MCQ Question 14 Detailed Solution

Concept:

The power spectral density is basically the Fourier transform of the autocorrelation function of the power signal, i.e.

$${S_x}\left( f \right) = F.T.\left\{ {{R_x}\left( \tau \right)} \right\}$$

Also, the inverse Fourier transform of a constant function is a unit impulse.

Application:

The power spectral density of white noise is given by:

$$\rm{S_{X}\left(f\right)=\frac{\eta}{2}}$$ for all frequency 'f', i.e.

Now auto-correlation is inverse Fourier transform (IFT) of power spectral density function.

$$\rm{R_x\left(\tau\right)\mathop \to \limits^{IFT}S_X\left(f\right)}$$

The inverse Fourier transform of the power spectrum of white noise will be an impulse as shown:

$$\rm{R_x\left(\tau\right)=\frac{\eta}{2}\delta\left(\tau\right)}$$

# 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 SX(f) = 5 W/Hz. The filter h(t) has a magnitude response given by |H(f)| = 0.5 for −5 ≤ f ≤ 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).

## Random Processes MCQ Question 15 Detailed Solution

Concept: Average power = Area bounded between the Power Spectral Density - frequency curve.

Calculation:

Given:

y(t) = h(t) * x(t) + z(t)

Power in y(t) = Power [h(t) * x(t)] + Power [z(t)]

H(f) will determine the band width of x(t).

h(t) * x(t) is bandlimited to – 5 Hz to + 5 Hz

$$Power\left[ {h\left( t \right)*x\left( t \right)} \right] = \mathop \smallint \limits_{ - \infty }^{ + \infty } {\left| {H\left( f \right)} \right|^2}5 \times \left( f \right)$$

$$= \mathop \smallint \limits_{ - 5}^{ + 5} {\left( {0.5} \right)^2}5df$$

$$\Rightarrow \mathop \smallint \limits_{ - 5}^{ + 5} \left( {0.25} \right)\left( {0.5} \right)df$$

= 10(1.25) = 12.5 W

power [z (t)] = Area under power spectral density

$${P_z}\left( t \right) = \left( {\frac{1}{2}} \right)\left( {10} \right)\left( 1 \right) = 5W$$

Total power = (5 + 12.5) W

= 17.5 W

# Consider a random process $$\rm X(t) = 3V(t) − 8$$, where (𝑡) is a zero mean stationary random process with autocorrelation $$\rm R_{V}(\tau)=4e^{-5\left|\tau\right|}$$. The power in $$\rm X(t)$$ is ________

## Random Processes MCQ Question 16 Detailed Solution

Concept:

ACF is defined as:

Rx(τ) = E[x(t) x(t + τ)] = E[x(t) x(t - τ)]

Properties of ACF:

1. Rx(-τ) = Rx (τ)

2. Rx (0) = E [x2(t)] = Power of x(t)

Calculation:

Given:

x(t) = 3 V(t) – 8,  Rv(τ) = 4e-5|τ|

E [V(t)] = 0

We know that,

Power of x(t) = E[x2(t)]

= E[9V2(t) + 64 – 48 E[V(t)]]

= 9E [V2(t)] + 64 – 48 E[V(t)]

Now,

E[V2(t)] = Rv(0) = 4

So,

Power of x(t) = (9 × 4) + 64 – 0

Power of x(t) = 100

# The Fourier transform of a Gaussian time pulse is

1. Uniform
2. A pair of impulses
3. Gaussian
4. Rayleigh

Option 3 : Gaussian

## Random Processes MCQ Question 17 Detailed Solution

The normalized Gaussian pulse is given as:

$$f\left( t \right) = K{e^{ - a{t^2}}}$$

And its Fourier transform is given as:

$$F.T.\;\left\{ {f\left( t \right)} \right\} = \sqrt {\frac{\pi }{a}} \;{e^{ - \frac{{{\pi ^2}{t^2}}}{a}}}$$

$$= \sqrt {\frac{\pi }{a}} \;{e^{ - {{\left( {\frac{\pi }{a}} \right)}^{\pi {t^2}}}}}$$

$$= \sqrt A \;{e^{ - A\pi {t^2}}}$$

$$= B{e^{ - C{t^2}}}$$

Where, A, B, C, K all are constant values.

Here, its Fourier transform is also coming a Gaussian pulse shape.

So, the amplitude spectrum is the same as the Gaussian pulse.

# The power spectral density of the stationary noise process N(t), having auto-correlation Ruu(τ) = Ke-3|τ|  is

1. $$\frac{{3K}}{{3 + {\omega ^2}}}$$
2. $$\frac{{3K}}{{3 - {\omega ^2}}}$$
3. $$\frac{{6K}}{{9 + {\omega ^2}}}$$
4. $$\frac{{6K}}{{9 - {\omega ^2}}}$$

Option 3 : $$\frac{{6K}}{{9 + {\omega ^2}}}$$

## Random Processes MCQ Question 18 Detailed Solution

Concept: The Fourier Transform of the Auto-Correlation Ruu(τ) = Spectral density of that process

Analysis:

Given Ruu(τ) = ke-3|τ|

Fourier Transform of the above will be:

$$X\left( \omega \right) = \mathop \smallint \limits_{ - \infty }^\infty x\left( t \right)\;{e^{ - j\omega t}}dt$$

Spectral density will be:

$${S_u}\left( \omega \right) = \mathop \smallint \limits_{ - \infty }^\infty {R_{uu}}\left( \tau \right){e^{ - j\omega \tau }}d\tau$$

$${S_u}\left( \omega \right) = \mathop \smallint \limits_{ - \infty }^\infty k{e^{ - 3\left| \tau \right|}}\;{e^{ - j\omega \tau }}d\tau$$

$${S_u}\left( \omega \right) = k\left[ {\mathop \smallint \limits_{ - \infty }^0 {e^{3\tau }}{e^{ - j\omega \tau }}d\tau + \mathop \smallint \limits_0^\infty {e^{ - 3\tau }}{e^{ - j\omega \tau }}d\tau } \right]$$

$${S_u}\left( \omega \right) = k\left[ {\mathop \smallint \limits_{ - \infty }^0 {e^{\left( {3 - j\omega } \right)\tau }}d\tau + \mathop \smallint \limits_0^\infty {e^{ - \left( {3 + j\omega } \right)\tau }}d\tau } \right]$$

$${S_u}\left( \omega \right) = k\left[ {\left( {\frac{{1.{e^{\left( {3 - j\omega } \right)\tau }}}}{{3 - j\omega }}} \right)_{ - \infty }^0 + \left( {\frac{{ - {e^{ - \left( {3 + j\omega } \right)\tau }}}}{{\left( {3 + j\omega } \right)}}} \right)_0^\infty } \right]$$

$${S_u}\left( \omega \right) = k\left[ {\frac{1}{{3 - j\omega }} + \frac{1}{{3 + j\omega }}} \right]$$

$$= k\left[ {\frac{{3 + j\omega + 3 - j\omega }}{{9 + {\omega ^2}}}} \right]$$

$${S_u}\left( \omega \right) = \frac{{6k}}{{9 + {\omega ^2}}}$$

Tips and Tricks:

Remember the following for saving time in the exam.

$${e^{ - \left| a \right|t}}\mathop \leftrightarrow \limits^{\;F.T\;} \frac{{2a}}{{{a^2} + {\omega ^2}}}$$

$$k{e^{ - 3\left| \tau \right|}}\mathop \leftrightarrow \limits^{\;F.T\;} \;k\left( {\frac{{2 \times 3}}{{9 + {\omega ^2}}}} \right)$$

$$= \frac{{6k}}{{9 + {\omega ^2}}}$$

# If the covariance of two random variable X and Y is μXY, their correlation coefficient ρXY is:

1. In μXY
2. $$ln\;\left( {\frac{{{\mu _{XY}}}}{{{\sigma _X}{\sigma _Y}}}} \right)$$
3. $$\frac{{{\sigma _X}{\sigma _Y}}}{{{\mu _{XY}}}}$$
4. $$\frac{{{\mu _{XY}}}}{{{\sigma _X}{\sigma _Y}}}$$

Option 4 : $$\frac{{{\mu _{XY}}}}{{{\sigma _X}{\sigma _Y}}}$$

## Random Processes MCQ Question 19 Detailed Solution

Concept:

The correlation coefficient (ρxy) of two random, variable X, and Y is given as:

$${\rho _{xy}} = \frac{{COV\left[ {X,Y} \right]}}{{\sqrt {Var\left( x \right)Var\left( y \right)} }}$$     ---(1)

Where,

COV[X, Y] = E[X,Y] – E[X] E[Y]

Var (x) = σx2

Var (y) = σy2

Analysis:

Given:

COV [X, Y] = μxy

From equation 1),

$${\rho _{xy}} = \frac{{COV\left[ {X,Y} \right]}}{{\sqrt {{V_{ar}}\left( x \right){V_{ar}}\left( y \right)} }}$$

$${\rho _{xy}} = \frac{{{\mu _{xy}}}}{{\sqrt {\sigma _x^2\sigma _y^2} }}$$

$${\rho _{xy}} = \frac{{{\mu _{xy}}}}{{{\sigma _x}{\sigma _y}}}$$

# The power spectral density of a real stationary random process (𝑡) is given by$${S_X}\left( f \right) = \left\{ {\begin{array}{*{20}{c}} {\frac{1}{W},\;\;\;\;\;\left| f \right| \le W}\\ {0,\;\;\;\;\;\left| f \right| > W} \end{array}} \right.\;\;$$ The value of the expectation $$E\left[ {\pi X\left( t \right)X\left( {t - \frac{1}{{4W}}} \right)} \right]$$ is ______

## Random Processes MCQ Question 20 Detailed Solution

Concept:

The Auto correlation Rx(τ) is defined as:

Rx(τ) = E[x(t) x(t + τ)]

Or

Rx(τ) = E[x(t) x(t – τ)]

Also, the Fourier transform of Rx(τ) is the power spectral density, i.e.

Rx(τ) ↔  Sx(f)

Analysis:

$$E\left[ {\pi \;x\left( t \right)\;x\left( {t - \frac{1}{{4\omega }}} \right)} \right]$$

$$= \pi \;E\left[ {x\left( t \right)x\left( {t - \frac{1}{{4\omega }}} \right)} \right]$$

$$Here\;\tau = \frac{1}{{4\omega }}$$ .

The above equation becomes:

$$= \pi \;{R_x}\left( {\frac{1}{{4\omega }}} \right)$$

$${S_x}\left( f \right) = \left\{ {\begin{array}{*{20}{c}} {\frac{1}{\omega }\;\;\;\;\;\;\;\;\left| f \right| \le \omega }\\ {0\;\;\;\;\;\;\;\;\;\;\;\;\left| f \right| > \omega } \end{array}} \right.$$ \

Taking its Inverse Fourier transform, we get:

If x(t) ↔ X(f), then the duality property states that:

X(t) ↔ x(-f)

$${R_x}\left( \tau \right) = \frac{1}{\omega }\left( {2\omega } \right)\sin c\;\left( {2\omega \;\tau } \right)$$

Rx(τ) = 2 sin c (2ωτ)

$$\pi E\left[ {x\left( t \right)x\left( {t - \frac{1}{{4\omega }}} \right)} \right] = \pi {R_x}\left( {\frac{1}{{4\omega }}} \right)$$

$$= \pi 2\sin c\left( {\;2\omega \frac{1}{{4\omega }}} \right)$$

$$= \pi 2\sin c\;\left( {\frac{1}{2}} \right)$$

$$= \frac{{2\pi \sin \frac{\pi }{2}}}{{\frac{\pi }{2}}}$$

$$\pi \;E\left[ {x\left( t \right)\;x\left( {t - \frac{1}{{4\omega }}} \right)} \right] = 4$$