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Calculate PSD using FFT

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Hi, The question is to calculate PSD using FFT function in MATLAB. Ive already done it with pwelch command in MATLAB and now it's time to do it with FFT command and compare the results. If I have file named: file2.Mat which contains 3 columns. first column is time, second Force and the third is acceleration. the sampling is 4000Hz and the number of NFFT is ,let us say, 4444. I know that we need to multiply the window with time column. And then what?
Does anybody know how to do it? Ive been work on this for like 3 hours.
regards
  1 Comment
Noel Khan
Noel Khan on 16 Oct 2020
Do you have the pwelch implementation? I would like to see how you got your estimate if you dont mind :)

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Accepted Answer

Frantz Bouchereau
Frantz Bouchereau on 29 Jul 2021
Here is a popular MATLAB doc page that explains the relationship between FFT and true power spectra: Power Spectral Density Estimates Using FFT.

More Answers (4)

Wayne King
Wayne King on 21 Nov 2011
Why don't you just use spectrum.periodogram?
Fs = 1e3;
t = 0:1/Fs:1-1/Fs;
x = cos(2*pi*100*t)+randn(size(t));
plot(psd(spectrum.periodogram,x,'Fs',Fs,'NFFT',length(x)));
If you want to do it simply with fft()
xdft = fft(x);
xdft = xdft(1:length(x)/2+1);
xdft(2:end-1) = 2*xdft(2:end-1);
psdest = 1/(length(x)*Fs)*abs(xdft).^2;
freq = 0:Fs/length(x):Fs/2;
plot(freq,10*log10(psdest));
grid on;
Compare the plots.
  2 Comments
Chappi
Chappi on 17 Dec 2019
Hi, can I ask about how you can apply Hamming window using FFT method? I dont have Signal Processing Tool so I can not use periodogram function. Thank you

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Payam
Payam on 21 Nov 2011
Why xdft = xdft(1:length(x)/2+1); % what does that mean? xdft(2:end-1) = 2*xdft(2:end-1); %? dubbel side? and why?
I dont understand these two lines psdest = 1/(length(x)*Fs)*abs(xdft).^2; freq = 0:Fs/length(x):Fs/2;
  1 Comment
Payam
Payam on 21 Nov 2011
Shouldnt I multiply the hanning windows with my time vector?

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Wayne King
Wayne King on 21 Nov 2011
The PSD is an even function of frequency, so you only need from 0 to the Nyquist, if you want to conserve the total power, you have to multiply all frequencies except 0 and the Nyquist by two if you only keep 1/2 the frequencies. 0 and the Nyquist only occur once in the PSD estimate, all other frequencies occur twice. If you look at the example I gave you, then you see it agrees with the scaling in MATLAB's periodogram.
The answer about multiplying by a window, Hanning, Hamming, Blackman, Tukey. etc. is that it depends. A window reduces the bias in the periodogram, but that comes at the cost of reduced frequency resolution (a broader main lobe).

Chris Schwarz
Chris Schwarz on 27 Aug 2020
An adjustment to Wayne's code that gives an exact match to the periodogram is:
Fs = 1e3;
t = 0:1/Fs:1-1/Fs;
x = cos(2*pi*100*t)+randn(size(t));
plot(psd(spectrum.periodogram,x,'Fs',Fs,'NFFT',length(x)));
xdft = fft(x);
xdft = xdft(1:length(x)/2+1);
xdft(2:end-1) = 2*xdft(2:end-1);
freq = 0:Fs/length(x):Fs/2;
df = diff(freq);
df = df(1);
% psdest = 1/(length(x)*Fs)*abs(xdft).^2; % original
psdest = abs(xdft/length(x)).^2/(2*df); % modified
hold on
plot(freq,10*log10(psdest),'r');
grid on;
  2 Comments
karinkan
karinkan on 7 Mar 2021
close all
Fs = 1e3;
t = 0:1/Fs:1-1/Fs;
x = cos(2*pi*100*t)+randn(size(t));
L =length(x);
NFFT = 2^nextpow2(L);
plot(psd(spectrum.periodogram,x,'Fs',Fs,'NFFT',NFFT));
df = Fs/NFFT;
freq = 0:df:Fs/2;
xdft = fft(x,NFFT);
xdft_s = xdft(1:NFFT/2+1);
amp = abs(xdft_s)/NFFT;
psdest = amp.^2/(df); % original
psdest(2:end-1) = 2*psdest(2:end-1);
hold on
plot(freq,10*log10(psdest),'r');
grid on;

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