Units of frequency when using FFT in NumPy

Question

I am using the FFT function in NumPy to do some signal processing. I have array called signal which has one data point for each hour and has a total of 576 data points. I use the following code on signal to look at its fourier transform.

t = len(signal)
ft = fft(signal,n=t)
mgft=abs(ft)
plot(mgft[0:t/2+1])

I see two peaks but I am unsure as to what the units of the x axis are i.e., how they map onto hours? Any help would be appreciated.

Answer 1

Given sampling rate FSample and transform blocksize N, you can calculate the frequency resolution deltaF, sampling interval deltaT, and total capture time capT using the relationships:

deltaT = 1/FSample = capT/N
deltaF = 1/capT = FSample/N

Keep in mind also that the FFT returns value from 0 to FSample, or equivalently -FSample/2 to FSample/2. In your plot, you're already dropping the -FSample/2 to 0 part. NumPy includes a helper function to calculate all this for you: fftfreq.

For your values of deltaT = 1 hour and N = 576, you get deltaF = 0.001736 cycles/hour = 0.04167 cycles/day, from -0.5 cycles/hour to 0.5 cycles/hour. So if you have a magnitude peak at, say, bin 48 (and bin 528), that corresponds to a frequency component at 48*deltaF = 0.0833 cycles/hour = 2 cycles/day.

In general, you should apply a window function to your time domain data before calculating the FFT, to reduce spectral leakage. The Hann window is almost never a bad choice. You can also use the rfft function to skip the -FSample/2, 0 part of the output. So then, your code would be:

ft = np.fft.rfft(signal*np.hanning(len(signal)))
mgft = abs(ft)
xVals = np.fft.fftfreq(len(signal), d=1.0) # in hours, or d=1.0/24 in days
plot(xVals[:len(mgft)], mgft)