What is the power magnitude in FFT?

Hi,

I'm computing the fft with the following code:

%********** Més dades per la FFT ***********************
        Fs = 99840;             %ho trec del PicoScope
        m =length (tensio);     %calculo la longitud de les mostres de tensió
        n = pow2(nextpow2(m));  %calculo el número de punts a representar

          %********** càlcul de la FFT ***********************
          y = fft (tensio, n);    %DFT (discret fourier transform)
          f = (0:n-1)*Fs/n;       %rang de freqüències
          power = y.*conj(y)/n;   %calcula la potència de la DFT
          freqNyquist = Fs/2;

          %********** càlcul de la FFT centrada ***********************
          y0=fftshift(y);             %centro la FFT d'y
          f0=(-n/2:n/2-1)*Fs/n;       %marco el rang de les freqüències
          power0 = y0.*conj(y0)/n; %calculo la potència corresponent a la DFT centrada

What is the 'power' magnitude? dB? V? The input signal 'y' is in Volts.

Thanks,

Roser

Answer 1

[Wayne King]

Hi Roser, "power" is basicially magnitude squared. If you are trying to construct a power spectral density estimate, the units are in V^2/Hz.

Then it is customary to take the logarithm of that because the log is a variance stabilizing transformation for the power spectral density.

The periodogram is a nonparametric estimator of the power spectral density which you can implement using fft() as follows.

 Fs = 1000;
 t = 0:1/Fs:1-1/Fs;
 x = cos(2*pi*100*t)+randn(size(t));
 xdft = fft(x);
 % only use one side because signal is real-valued
 xdft = xdft(1:length(x)/2+1);
 per = 1/(Fs*length(x))*abs(xdft).^2;
 freq = 0:Fs/length(x):Fs/2;
 plot(freq,10*log10(per))
 xlabel('Hz'); ylabel('dB')

MATLAB additionally scales all the frequencies except 0 and the Nyquist by 2 in a one-sided PSD estimate, so if you want to demonstrate agreemen with:

    [Pxx,F] = periodogram(x,rectwin(length(x)),length(x),Fs);

Add the line

 per(2:end-1) = 2*per(2:end-1);