# For binary (spike train) signals, take FFT of signal or autocorrelation of signal?

I want to characterize a binary time-series signal x (derived from neuron action potential data) in the frequency domain.

Should I use the FFT of the original signal x, or the power spectrum (FFT of the autocorrelation of x, which is defined as $k(n)=E[x(t)\cdot x(t-n)]$ )?

• What is your ultimate objective with "characterizing" the data? Definitely read Spikes when you have a chance. The short answer from there is that it depends on what you are looking to get out of the signal. If you're looking to drive a device or something of that nature, the FFT is going to be computationally burdensome, and may not give you sufficient information. – Chuck Sherrington May 28 '14 at 21:43
• As an aside, for papers to look at, George Gertstein is the guru of the analysis of spike trains. – Chuck Sherrington May 28 '14 at 21:43
• I'm using "information" as a loaded term. I was referring to both measures that you are talking about, sorry if that wasn't clear. Neither is going to tell you anything meaningful. If you're locking to a specific event (sensory input or specific motor output - e.g., an EMG of a muscle) the frequency variation will tell you something, but a "free form" recording really isn't. You'd be better off examining local field potentials (most extracellular recording systems can separate these out in the spike sorting processing) if you are trying to tease out frequency changes. – Chuck Sherrington May 29 '14 at 20:02
• Otherwise, this is really just an exercise in signal processing, to be frank. – Chuck Sherrington May 29 '14 at 20:03
• (not trying to be an overbearing curmudgeon here, just offering some constructive criticism. I'm happy to see a methods question on here!) – Chuck Sherrington May 29 '14 at 21:00

As this answer explains, the Fourier transform of the autocorrelation of $x$ is the same as the magnitude squared of the Fourier transform of $x$. The autocorrelation of x can be expressed as $x(t)*x(-t)$, where * means convolution and the Fourier transform of the autocorrelation is then $X(w)X^*(w)=|X(w)^2|$, which of course is the Fourier transform of the original signal $x(t)$.