I'm plotting the FFT power-spectrum of a signal in MATLAB. I uploaded the 8000 samples time-series signal in a text file here: http://wikisend.com/download/896484/signal.txt

I'm using the following code:

Fs = 500; % sampling frequency
L = length(y); %number of samples

complex = fft(y)/L; % complex signals
f = 0 : Fs/L : Fs/2; % frequency bins
amplitude = 2*abs(complex(1:L/2+1)); % amplitudes
pow = (amplitude).^2/2; % power

grid on;

And I'm having the following plot: enter image description here

The confusion is, where I work they are using their own modified FFT function and if I run that code I obtain the following plot:

enter image description here

If you check the power in above figures the linear power ratio is 16 between my code and their's. And for other measurement I'm getting the ratio 60, 120 or even 1.

I'm confused and need help if one can plot FFT for me and see if my code and plot correct.

  • $\begingroup$ Why are you bothered by the different ratio? $\endgroup$
    – Moti
    Nov 21, 2015 at 6:41
  • $\begingroup$ my fft result is different than theirs. for the signal i uploaded the difference is 16 linear ratio. i first thought there is some error on my calculstion. but for another signal the ratio is different. im wondering if my fft is correct. could it be they apply window ect? but first i need to know if my fft result is right? $\endgroup$
    – user16307
    Nov 21, 2015 at 10:15
  • $\begingroup$ i dont understand that code there are functions like floor ect. but are you sure my power spectrum is correct? if my fft and their fft is different i dont get how csn it be $\endgroup$
    – user16307
    Nov 21, 2015 at 12:51

1 Answer 1


The difference is in the scaling of the power spectrum. I suppose you will find that the difference in scaling always equals $L/F_s$, which for the numbers in your question is indeed $8000/500=16$.

Take a look at Power Spectral Density Estimates Using FFT for the correct scaling. If you normalize the FFT result by the FFT length, you need to scale the squared magnitude of the normalized FFT by $L/F_s$. Furthermore, you need a factor of $2$ if you throw away the negative frequencies (I see you did that). Note, however, that you shouldn't apply a factor of $2$ for the bins at DC and at Nyquist, because they are not mirrored at negative frequencies.

Here is a simple Matlab code from the above quoted Mathworks page for computing a periodogram-based one-sided power spectrum estimate using the FFT (my comments):

Fs = 1000;                          % sampling frequency (Hz)
N = length(x);                      % even! (=> bin N/2+1 is Nyquist)
xdft = fft(x);
xdft = xdft(1:N/2+1);               % DC to Nyquist (one-sided)
psdx = (1/(Fs*N)) * abs(xdft).^2;   % periodogram scaling
psdx(2:end-1) = 2*psdx(2:end-1);    % scaling for one-sided PSD
                                    % note: no factor 2 at DC and Nyquist
  • $\begingroup$ Thanks a lot, you saved my day! Now both script produces the same results after I scale my code with L/Fs as you suggested. You wrote: "If you normalize the FFT result by the FFT length.." What does that mean? Could you give me a hint? Does it have to be like that always? The thing is if there is no scaling, and one looks at 50Hz x-axis in my plot he will find different power for that freq. at y-axis. Shouldn't that be mentioned? $\endgroup$
    – user16307
    Nov 21, 2015 at 14:18
  • $\begingroup$ @user16307: I meant what you did: fft(y)/L, i.e. normalize the FFT by the length. $\endgroup$
    – Matt L.
    Nov 21, 2015 at 14:21
  • $\begingroup$ oh so my script was wrong, since I did fft(y)/L, but I didn't scale the squared magnitude of the normalized FFT by L/Fs later on? Mine was wrong right? $\endgroup$
    – user16307
    Nov 21, 2015 at 14:25
  • $\begingroup$ @user16307: That's right, you have to scale your result by $L/F_s$. $\endgroup$
    – Matt L.
    Nov 21, 2015 at 18:53

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