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Dividing fourier transform by factor n

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Hello everybody
I have a problem with the fast fourier transform. I would like to extract the low and high frequency power of a signal. I have seen the following example (actually it is Python):
Y = np.fft.fft(signal)/n #Calculate FFT
Y = Y[range(n/2)] #Return one side of the FFT
lf = np.trapz(abs(Y[(frq>=0.04) & (frq<=0.15)]))
hf = np.trapz(abs(Y[(frq>=0.16) & (frq<=0.5)])) #Do the same for 0.16-0.5Hz (HF)
signal is just a an array of values (Let's say 10000 values).
The problem here is that I don't see what n should be. Should n just be the length of signal?

Accepted Answer

Star Strider
Star Strider on 12 Nov 2017
‘Should n just be the length of signal?’
Yes. (Note that it is the length of the original signal, not of a signal zero-padded to have a length of a power-of-2.)
Sepp on 12 Nov 2017
What do you mean with zero-padded?
Star Strider
Star Strider on 12 Nov 2017
Zero-padding a vector to create an ‘augmented’ vector of length equal to the next highest power of 2 is sometimes desired because (1) it increases the efficiency of the fft calculation, and (2) increases the frequency resolution of the transformed vector.
The point is that all the energy in the vector are in the first ‘n’ samples, and that is the correct normalising value.

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