My understanding is that Fourier power spectra give the frequencies and corresponding intensities/amplitudes contained within a signal. In certain techniques, such as molecular dynamics, it is common to take the autocorrelation of a time-series before calculating the Fourier spectrum.

I have read that in molecular dynamics this practice can be due to the appearance of autocorrelations in quantum mechanics, and I have also read that taking the autocorrelation removes the dependence on the initial conditions.

a) Does the autocorrelation contain the same frequency components as the original signal?

b) Are there any conditions under which the Fourier power spectrum of the signal and it's autocorrelation become mathematically identical? eg. infinite signal length, perhaps.

c) What are the factors that cause these two spectra to differ?

d) Are there any other reasons to favor using the autocorrelation in place of the original signal when computing the power spectrum?



2 Answers 2


(a). Yes, a signal and its autocorrelation have the same frequencies. If $X(f)$ is the Fourier transform of the signal and $A(f)$ the Fourier transform of its autocorrelation, then $A(f) = |X(f)|^2$. So, in general, a signal and its autocorrelation don’t have the same Fourier transform, but there is at least one exceptional case: $x(t) = W \operatorname{sinc}(Wt)$ whose Fourier transform is $X(f) = \operatorname{rect}\left(\frac fW\right)$ and so $A(f) = |X(f)|^2 = X(f)$. Those who wish to generalize this notion to signals whose Fourier spectra are the sum of non-overlapping $\operatorname{rect}$ functions should write out the details for themselves.

All of the above is for finite-energy signals.

If one is thinking of periodic finite-power signals with Fourier series $x(t) = \sum_{n=-\infty}^\infty c_n \exp(j2\pi nf_0t)$, then the periodic autocorrelation function has Fourier series $\sum_{n=-\infty}^\infty |c_n|^2 \exp(j2\pi nf_0t)$. Fourier series are represented by impulse trains in the frequency domain: \begin{align} X(f) &= \sum_{n=-\infty}^\infty c_n \delta (f-nf_0),\\ A(f) &= \sum_{n=-\infty}^\infty |c_n|^2 \delta (f-nf_0) \end{align} and if we wish to continue to use the formula $A(f) = |X(f)|^2$ with the spectra of these periodic signals, then we must assume that $$c_n \delta(f- nf_0)c_m^* \delta^*(f- mf_0) = \begin{cases} |c_n|^2 \delta(f- nf_0), & \text{if}~m=n,\\ 0, &\text{if}~m\neq n. \end{cases}$$ Note that we are assuming that $|\delta(f-nf_0)|^2 = \delta(f-nf_0)$ and $\delta(f- nf_0) \delta^*(f- mf_0)=0$ for $m\neq n$, but if we are willing to swallow this bald and unconvincing narrative, then everything is hunky-dory. Once again, we see that the signal and its autocorrelation have the same frequencies but that their Fourier transforms are generally not equal unless it so happens that $|c_n|^2 = c_n$ for all $n$. Note that this means that for each $n$, $c_n$ must equal $1$ or $0$. Thus, the only real-valued periodic signals whose Fourier transforms equal the Fourier transform of their autocorrelation function are of the form $a_0 + \sum_{n=1}^\infty a_n 2\cos(2\pi f_0 t),~~ a_i \in \{0,1\}, i = 0, 1, 2, \ldots$

  • $\begingroup$ Thank you. If I've understood, that would mean that in general the square root of Fourier transform of the autocorrelation would be mathematically identical to the absolute Fourier transform of a general signal. Are there any limiting cases where this wouldn't be true? $\endgroup$
    – Magnus
    Apr 12, 2020 at 5:29
  • $\begingroup$ I have tried implementing this relationship in a LabVIEW test environment using sine waves, and I'm struggling to get a matching set of signals. Certainly their Fourier Transforms alone are not related in this way, at least when calculated using the fast Fourier transform algorithm implemented, as the autocorrelation Fourier transform has both negative and positive amplitude. Is there another resource you could recommend for further reading on this relationship? $\endgroup$
    – Magnus
    Apr 12, 2020 at 8:41
  • $\begingroup$ Sine waves are a particular exception to the general result $A(f)=|X(f)|^2$ because $X(f)$ is a pair of impulses and the "square" of an unit-amplitude impulse is the unit-amplitude impulse itself while the "square" of $a\delta(f-f_0)$ is $a^2\delta(f-f_0)$. FFTs as substitutes for continuous-time Fourier transforms are even worse. $\endgroup$ Apr 12, 2020 at 17:00
  • $\begingroup$ In the case of a signal built out of either a sine wave, multiple sine waves, or waves that are almost sine waves, is there a formulation which relates the signal and autocorrelation Fourier transforms to each other? Presumably they still both contain the same frequencies. Also, does the fact that the signal is of a finite length change these relationships in general? I'm intending to look at molecular vibrations, which tend to have a large component of harmonic motion. $\endgroup$
    – Magnus
    Apr 12, 2020 at 22:02
  • $\begingroup$ I'm pretty sure that the equation $\delta^2(x)=\delta(x)$ is incorrect. $\endgroup$
    – Matt L.
    Apr 13, 2020 at 19:19

A simple example would be a $\delta(t)$ function. Autocorrelation of a delta function is a delta function. So, Fourier transform of the signal and it's autocorrelation is same in this case.


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