Calculate PSD using FFT

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

[Noel Khan]

Do you have the pwelch implementation? I would like to see how you got your estimate if you dont mind :)

Answer 1

[Frantz Bouchereau]

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

[Aditya]

This page appears to have a conceptual error in the first example. Amplitudes must be multiplied by two before taking one-sided PSD (as pointed out in reply by Chris Schwarz on 27 Aug 2020). Can the documentation be fixed?

[Frantz Bouchereau]

Aditya, we do correct for the one-sided power. See the section of the example where the one-sided power correction is done:

"Obtain the periodogram using fft. The signal is real-valued and has even length. Because the signal is real-valued, you only need power estimates for the positive or negative frequencies. In order to conserve the total power, multiply all frequencies that occur in both sets — the positive and negative frequencies — by a factor of 2. Zero frequency (DC) and the Nyquist frequency do not occur twice. Plot the result."

Answer 2

[Wayne King]

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).

Answer 3

[Wayne King]

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.

[Wayne King]

If you want to use a window, like Hamming, etc. You can do:

plot(psd(spectrum.periodogram('Hamming'),x,'Fs',Fs,'NFFT',length(x)));

[Chappi]

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

Answer 4

[Peppe]

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;

[Peppe]

Shouldnt I multiply the hanning windows with my time vector?

Answer 5

[Chris Schwarz]

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;

[PK]

psdest = abs(xdft/length(x)).^2/(2*df); % modified

why '2*df'? For the zero frequency also 2*df?

Thanks

[karinkan]

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;