Finding the autocorrelation function

[Infinity2]

how would I go about solving this question:
An input random process X(t) is real-valued Gaussian white noise process with spectral density No/2. It is filtered by a LTI causal filer with impulse response h(t) =exp(-t/T) 0<=t<infinity. The filter output V(t) is wide sense stationary

What is the autocorrelation function for V(t)?

 

[haadi20]

Just by looking at it...i think you should find the Power Spectral Density of the output and then the inverse Fourier transform of the PSD will give you the auto-correlation function...

1. Find V(f)= X(f)*H(f)
2. Find PSD of V(f)
3. Take Inverse Fourier Transform of PSD

Hope this helps...correct me if i am wrong...

 

[pangpangfish]

Actually your first step is conceptually incorrect. Since the input signal X(t) is a random process, you can't apply Fourier transform to X(t) because FT is reserved for aperiodic deterministic signals (or energy signals). However, X(t) is a random process here.

The correct steps should be as follows:

(1) Find the PSD for the output using S_v(f)=|H(f)|^2 S_x(f), where S_x(f) is the input signal's PSD and it is given as N_0/2.

(2) Take the inverse Fourier transform of S_v(f) and it will get you the autocorrelation function for the output random process V(t).

 

[zorro]

Hi Infinity2,

The autocorrelation of v(t) is No/2 times h(t)*h(-t), where the symbol “*” stands for convolution.
You can see in books on signal and systems, stochastic processes, etc. the relationships between inputs and outputs of LTI systems.
So, you have to solve the convolution integral. If I’m not wrong, the solution is T/2.exp(-|t|/T) .
Regards

Z

 

[purnapragna]

If we can look what Mr Z has said in frequency domain out put is nothing but

Sy(jω)=(|H(jω)|^2) × Sx(jω)

where Sy(jω) is PSD of the output which is nothing bu tFT of the ACF which u

needed, Sx(jω) is the PSD of the input which is constant which is (No/2) and H(jω)

is the impulse response of the system.

So for our case

H(jω) = 1/(a+jω). where a = 1/T

thus

|H(jω)|^2 = 1/(a^2+ω^2)

Thus

Sy(jω) = (No/2) × 1/((a^2+ω^2)

Thus Ry(τ) = No/2 × IFT(1/((a^2+ω^2))

which is nothing but given by Mr Z as No/2 exp(-|t|/T)

Added after 5 minutes:

hey i m sorry that i missed a constant factor which is NoT/4 and not just No/2

sorry for that

regards

sri hari