While I do know the difference between the two, in theory, I am not very sure about why we look at the Fourier transform of the auto-covariance function. What extra information does it give us over and above what we can already get by taking the Fourier transform of the original time-series? Also, when is one advantageous over the other?


1 Answer 1


The interest in looking at the Fourier transform of the autocorrelation (which is usually equivalent to autocoveriance) comes from the Wiener-Khinchin theorem that states

Consider a wide-sense stationary signal $x(t)$. Then the power spectral density $S_x(f)$ is given by

$$S_x(f) = \int_{-\infty}^{\infty}R(\tau)e^{-j2{\pi}f{\tau}}d{\tau}$$

Where $R(\tau)$ is the autocorrelation function of $x(t)$.

There are times where the Fourier transform of a signal does not exist. However, this result still allows us to analyze the signal in the frequency domain and retrieve information by using the autocorrelation and not the signal itself.

This is also helpful when analyzing LTI systems. Given a system $h(t)$ with input $x(t)$ and output $y(t)$, we have a relationship between the power spectral densities given by

$$S_y(f) = S_x|H(f)|^2$$

This aids in estimating properties of the signals as well as the system itself, depending on what information is already known.

In the discrete domain, the DFT can always be calculated for a finite-length signal. The theorem still holds however, and despite being able to "take" the Fourier transform as in the DFT, errors may begin to appear when estimating parameters.

References by OP's request:




  • $\begingroup$ I think the OP is asking about deterministic functions, not random processes. $\endgroup$ Commented Nov 3, 2020 at 21:42
  • $\begingroup$ Is the PSD of the time-series and its auto-covariance/auto-correlation the same? Could you cite some reference which looks at this? $\endgroup$
    – miNiON
    Commented Nov 4, 2020 at 21:22
  • $\begingroup$ @miNiON Without getting into too many details, if the Fourier transform of the signal exists, then the Wiener-Khinchin theorem shows that the PSD's are the same. There are of course situations where this is not true. I'll update my answer with references. $\endgroup$
    – Envidia
    Commented Nov 4, 2020 at 21:33
  • $\begingroup$ Thanks a lot! This is exactly what I was looking for. $\endgroup$
    – miNiON
    Commented Nov 5, 2020 at 22:46
  • $\begingroup$ So the OP liked your very helpful answer and then did not even bother to upvote it. At least I can fix that. $\endgroup$
    – Ed V
    Commented Apr 3, 2021 at 23:30

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.