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 commonplace to take the autocorrelation of these time-series before calculating the Fourier power spectrum.
I have read that in molecular dynamics this practice can be due to the relationship between Fermi's golden rule for quantum mechanical state transitions and the autocorrelation function of say the dipole moment-time vector for infrared spectral calculations of molecules. I have also read that taking the autocorrelation removes the dependence on the initial conditions of the simulation.
a) My current understanding is that the autocorrelation function still contains all the same frequency components in the original time-series. Is this true?
b) If so and the signal is infinitely repeating, do the Fourier power spectra of the time-series and it's autocorrelation become mathematically identical?
c) If not, in what ways are they expected to differ?
d) Are there any other reasons that favor using the autocorrelation in place of the original time-series when computing the power spectrum?
Thanks,
Magnus.
Thanks Nazanin. For anyone who's curious about this topic, I also asked this question in a signal processing group and I've coped out the answer I recieved there below. I've done my best to reformat the LaTeX formulas into plain text, but in case they haven't copied well or are easier to read in the original, the link is here: https://dsp.stackexchange.com/questions/66357/does-a-signal-and-its-autocorrelation-ever-have-the-same-fourier-spectra
(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)=Wsinc(Wt) whose Fourier transform is X(f)=rect(fW) 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 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)=∑n=−∞∞cnexp(j2πnf0t), then the periodic autocorrelation function has Fourier series ∑n=−∞∞|cn|2exp(j2πnf0t). Fourier series are represented by impulse trains in the frequency domain:
X(f)=∑n=−∞∞cnδ(f−nf0),
A(f)=∑n=−∞∞cn2δ(f−nf0),
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
cnδ(f−nf0)c∗mδ∗(f−mf0)= { |cn|2δ(f−nf0), if m=n}
{0, if m≠n}
Note that we are assuming that |δ(f−nf0)|2=δ(f−nf0) and δ(f−nf0)δ∗(f−mf0)=0 for m≠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 |cn|2=cn for all n. Note that this means that for each n, cn 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 a0+∑∞n=1an2cos(2πf0t), ai∈{0,1}, i=0,1,2,…
- Badges
- Science topic
- Similar topics
- Philosophy
- Philosophy Of Science
- Reasoning