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

Derivation:
The cross-correlation function between the input and output processes is given by:
R_xy (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)

For a random process with mean μ_x 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

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)]

Statement 1 is correct.

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

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 P_SD is related to the input PSD as:
Sy(f) = |H(f)|² S_N(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:

1. Autocorrelation function for a process Y(t) is given by R_x(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 R_x(τ) = E[Y(t) Y(t + τ)]
R_Y(τ) = E[[(X(t) - X(t - T₀)][X (t + τ) = X(t + τ - T₀)]]

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

The cross-correlation function between the input and output processes is given by:
R_xy (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:
ϕ_x[m] = ϕ_xx [m] * h[⋅]

Since x(n) * δ(n) = x(n)

PSD_o = PSD_i |H(f)|²
Calculations:
h(t) = e^-t u(t)

To find Auto correlation of output
R_y(z) = IFT {S_y (b)}

For a system with given input power spectral density, and impulse response, the output spectral density is given by:
S_Y(ω) = S_X(ω) H(ω)²
S_X(ω) = Input spectral density
S_Y(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:
S_Y(ω) = S_Y1(ω) H(ω)²