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Test: Transmission of Random processes through Linear Systems - Electronics and Communication Engineering (ECE) MCQ


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10 Questions MCQ Test - Test: Transmission of Random processes through Linear Systems

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Test: Transmission of Random processes through Linear Systems - Question 1

Consider an LTI system subjected to a wide-sense stationary input {x(n)}, which is a white noise sequence. The cross-correlation ϕx[m] between the input x(n), and output y(n) is:
Where  and h[⋅] is the impulse response.

Detailed Solution for Test: Transmission of Random processes through Linear Systems - Question 1

Derivation:
The cross-correlation function between the input and output processes is given by:
Rxy (t1, t2) = E{X(t1) Y*(t2)}

∴ The output cross-correlation between the input and the output is the convolution of the autocorrelation function of the input with the conjugate of the impulse response of the system.
Analysis:
Given the autocorrelation of the input wide-sense stationary signal as:

The cross-correlation will be:

Since x(n) * δ(n) = x(n)

Test: Transmission of Random processes through Linear Systems - Question 2

A WSS random process having mean m is applied as input to LTI system having impulse response  . The mean of the output is same as the mean of the input. The value of a is _____

Detailed Solution for Test: Transmission of Random processes through Linear Systems - Question 2

For a random process with mean μx applied to a system with impulse response h(t), the mean of the output random process generated will be:

Calculation:
E[X(t)] = m
E[Y(t)] = m H(0)

Now
m = m/a
∴ a = 1

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Test: Transmission of Random processes through Linear Systems - Question 3

Let X(t) be a wide sense stationary random process with the power spectral density SX(f) as shown in Figure (a), where f is in Hertz (Hz). The random process X(t) is input to an ideal lowpass filter with the frequency response:

This is as shown in Figure (b). The output of the lowpass filter is Y(t).

Let E be the expectation operator. Consider the following statements:
I. E(X(t)) = E(Y(t))
II. E(X2(t)) = E(Y2(t))
III. E(Y2(t)) = 2
Select the correct option:

Detailed Solution for Test: Transmission of Random processes through Linear Systems - Question 3


Application: 
In the given question,
H(0) = 1

2 - 2 e-0.5
Low pass filter does not allow total power to pass from input to output.
Hence,
E[X2(t)] ≠ E[Y2(t)]

Test: Transmission of Random processes through Linear Systems - Question 4

Let X(t) be a wide sense stationary random process with the power spectral density SX(f) as shown in Figure (a), where f is in Hertz (Hz). The random process X(t) is input to an ideal lowpass filter with the frequency response:

This is as shown in Figure (b). The output of the lowpass filter is Y(t).

Let E be the expectation operator. Which of the following statements is/are correct?

Detailed Solution for Test: Transmission of Random processes through Linear Systems - Question 4


Statement 1 is correct.

E[y2(t)] = 2 - 2 e-0.5
Statement 2 is incorrect.

Test: Transmission of Random processes through Linear Systems - Question 5

Consider a white Gaussian noise process N(t) with two-sided power spectral density SN(f) = 0.5 W/Hz as input to a filter with impulse response 0.5e-t2/2 (t is in seconds) resulting in output Y(t). The power in Y(t), in watts, will be:

Detailed Solution for Test: Transmission of Random processes through Linear Systems - Question 5

1. PSD of output = |H(f)|2  PSD of input. 
2. Fourier transform of Gaussian Pulse is also Gaussian in nature i.e 

3. Average Power from PSD can be calculated as 
Application: Given 

The output PSD is related to the input PSD as:
Sy(f) = |H(f)|2 SN(f)    ----(1)
 the Fourier transform is calculated as:

The Fourier transform of h(t) will be:

Substituting this in Equation (1), we get:

The integration of PSD gives power, i.e. the area under the PSD gives power.
Total power P will therefore be:

The area of   is always equal to 1, i.e.

Using the area scaling property, we get:

Applying this property to Equation (1), we can write:

Test: Transmission of Random processes through Linear Systems - Question 6

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 RX(τ), then the autocorrelation function RY(τ) of the output Y(t) is equal to

Detailed Solution for Test: Transmission of Random processes through Linear Systems - Question 6

1. Autocorrelation function for a process Y(t) is given by Rx(τ) = E[Y(t) Y(t + τ)].
2 . For a WSS process Shifting does not affect the autocorrelation function.
Application:
We have Y(t) = X(t) – X(t – T0)
Now Rx(τ) = E[Y(t) Y(t + τ)]
RY(τ) = E[[(X(t) - X(t - T0)][X (t + τ) = X(t + τ - T0)]]

Test: Transmission of Random processes through Linear Systems - Question 7

The two sided PSD of noise signal (PSD for the positive frequencies is shown in below figure) x(t) shown in figure is applied to RC LPF whose time constant is 1 ms.

The output power P out is.

Detailed Solution for Test: Transmission of Random processes through Linear Systems - Question 7

For an RC low pass filter, 
 (being the two sided PSD of x(t))
We have the output power spectral density given by, 

Test: Transmission of Random processes through Linear Systems - Question 8

Consider an LTI system subjected to a wide-sense stationary input {x(n)}, which is a white noise sequence. The cross-correlation ϕx[m] between the input x(n), and output y(n) is:
Where  and h[⋅] is the impulse response.

Detailed Solution for Test: Transmission of Random processes through Linear Systems - Question 8

The cross-correlation function between the input and output processes is given by:
Rxy (t1, t2) = E{X(t1) Y*(t2)}

∴ The output cross-correlation between the input and the output is the convolution of the autocorrelation function of the input with the conjugate of the impulse response of the system.
Analysis:
Given the autocorrelation of the input wide-sense stationary signal as:

The cross-correlation will be:
ϕx[m] = ϕxx [m] * h[⋅]

Since x(n) * δ(n) = x(n)

Test: Transmission of Random processes through Linear Systems - Question 9

Let X(t) be a white Gaussian noise with two sided PSD 
Assume x(t) is input to an LTI system with impulse response h(t) = e-t u(t)
If Y(t) is the output then E[Y2(t)] is ________

Detailed Solution for Test: Transmission of Random processes through Linear Systems - Question 9


PSDo = PSDi |H(f)|2
Calculations:
h(t) = e-t u(t)

To find Auto correlation of output
Ry(z) = IFT {Sy (b)}

Test: Transmission of Random processes through Linear Systems - Question 10

A random noise X(t) having a power spectrum

Is applied to a differentiator that has a transfer function H(ω) = jω. The output is applied to a network for which h(t) = t2 e-7t u(t) 
The power spectrum of Y(t) is

Detailed Solution for Test: Transmission of Random processes through Linear Systems - Question 10

For a system with given input power spectral density, and impulse response, the output spectral density is given by:
SY(ω) = SX(ω) H(ω)2
SX(ω) = Input spectral density
SY(ω) = Output spectral density
H(ω) = Impulse response of the system
Analysis:
Given the power spectrum of input as:

As this passes through the differentiator the output PSD becomes:

The transfer function for the second system is:

So, we obtain the transfer function in the frequency domain as:

Therefore, the final output spectrum is given by:
SY(ω) = SY1(ω) H(ω)2

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