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I have a piece of code that gets the FFT of a part of the signal and I'm now trying to get the PSD,

Fs = 44100; 
cj = sqrt(-1);
 %T=.6; 
dt = 1/Fs; 
left=test(:,1);
 right=test(:,2); 
time = 45; 
interval =.636;
 w_range = time*Fs: (time+interval)*Fs-1;
 I = left(w_range);
 Q = right(w_range); 
n = interval * Fs; 
f = -Fs/2:Fs/n:Fs/2-Fs/n;

s = I+cj.*Q;

% Smooth the signal ss = smooth(s,201);

sf = (fftshift(fft(ss(1:n)))); %FFT of signal

figure(1) plot(f,((20*log10((abs(sf))./max(abs(sf))))))

From my understanding, in order to get the PSD I just need to raise sf to the power of 2, or is there anything more I need to perform?

Sam
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fragkos
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  • I guess it should be `sf.*conj(sf)` or `(abs(sf)).^2`? – Yvon Jul 27 '14 at 18:20
  • What I meant is to calculate PSD from absolute-squaring the frequency spectrum. Another approach is using Wiener-Khinchine theorem which enables you calculate PSD by Fourier-transforming autocorrelation function. I will say sorry if I have provided any irrelevant information, since I didn't look up textbook for PSD. – Yvon Jul 27 '14 at 18:45

2 Answers2

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Technically yes, you can obtain the power-spectral density (PSD) of a periodic signal by taking the squared-magnitude of its FFT. Note that if you are going to plot it on a logarithmic decibel scale, there is really no difference between 20*log10(abs(sf)) or 10*log10(abs(sf).^2).

There is however generally more to it in the sense that the PSD estimate computed in this way tends to have a fairly large variance. There are a number of techniques which can be used to improve the estimate. A simple one consists of applying a window to sections of data, perform the FFT, then averaging the resulting PSDs (i.e. averaging the squared-magnitudes).

SleuthEye
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You are perfectly right. You just have to built the square of the absolute values.

Heinz M.
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