What is the difference between the PSD of a deterministic and stochastic signal?

Question

I am learning about stochastic processes and I don't get one thing:

What is the advantage of calculating the PSD of a signal using the Wiener-Khinchin theorem $Phi(omega) =mathcal{F}{R_{xx}}$ and calculating its PSD with $Phi(omega) =X(omega) X^*(omega)$.

I mean, if I record some random signal and I want to get its spectrum then I do $mathcal{F}{x(t)}$.

What's wrong about using this way? 
Why should I deal with autocorrelation and Wiener-Khinchin-Theorem?

I've created in MATLAB a cosine with random amplitude, so in my understanding this is a stochastic signal

t = 0:0.1:1000
x = rand() * cos(t) 
plot(abs(fft(xcorr(x,x))))
plot(abs(fft(x)).^2)

Both plots seem to be very equal, I can't see the difference.

Jeff · Answer

You need to be very careful about the differences between a "Power Spectral Density," a "Power Spectrum", a "Linear Spectral Density", and a "Linear Spectrum." Simply taking the Fourier transform of a signal does not give you power estimates, it estimates the complex amplitude of the signal at each frequency. Read Heinzel et al "Spectrum and spectral density estimation by the Discrete Fourier transform (DFT), including a comprehensive list of window functions and some new at-top windows" for a better explanation.

Answered by Jeff on November 30, 2021

Laurent Duval · Answer

Generally, one cannot compute the Fourier transform of any random process, because of integrability reasons. The PSD is one way to define it, based on autocorrelation.

And now, this mathematically sound definition of a periodogram is consistent in some way: if you have a realization of a random process, on a limited time interval, then its Fourier transform can be computed, and is consistent with the PSD, which you'd expect.

What is the difference between the PSD of a deterministic and stochastic signal?

2 Answers

Add your own answers!

Ask a Question