Basic question about using MATLAB fft: divide fft() by number of data points?
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Hi,
I have a very basic question regarding the use of the MATLAB fft() function.
In the MATLAB documentation on fft http://www.mathworks.com/help/matlab/ref/fft.html, an example was given for calculating the spectrum of a time domain vector y, with length L. The example code is as below for reference.
****************************************
NFFT = 2^nextpow2(L); % Next power of 2 from length of y
Y = fft(y,NFFT)/L;
f = Fs/2*linspace(0,1,NFFT/2+1);
******************************************
My question is, why is the fft() result divided by L here? The actual number of data points used is NFFT, not L. Should the fft() result be divided by NFFT instead?
Thanks!
Xi
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Star Strider
2014-11-11
0 个投票
The nextpow2 function zero-pads the signal so the fft calculation is more efficient. It adds no energy to the signal, so dividing by the length of the original signal is appropriate to normalise the amplitudes of the transformed components.
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