The frequency axis for the PSD of a signal

Suppose I have the signal x that contains N samples, and I take the FFT of this signal and rearrange the frequency components to have the 0 frequency componenet to the center

X = fftshift(fft(x));

and then calculate the PSD of the signal as

psd = (1/(fs*N))*abs(X).^2;

where fs is the sampling frequency, what is the exact frequency axis that corresponds to psd in this case? If I defined it as

freq = -fs/2:fs/N:fs/2;

it would be 1 element greater than the size of psd, but if I define it as

freq = -fs/2+fs/N:fs/N:fs/2;

it will be of the same size as psd.

Is the second definition accurate? If yes, why the negative frequencies don't extend to -fs/2 as it's the case for the positive frequencies?

Answer 1

[Abhaya]

Hi Mawe,

I understand you want to find the frequency axis for the Power Spectral Density (PSD) of a signal, spanning from ‘-fs/2’ to ‘fs/2’.

The following line of the given code returns ‘N+1’ points, which includes both the endpoints.

freq = -fs/2:fs/N:fs/2

Also, the second definition in the code provides ‘N’ points but does not maintain zero symmetry.

freq = -fs/2+fs/N:fs/N:fs/2

To achieve all frequency bins symmetrically around zero, you can use the code given below:

freq = -fs/2 : fs/N : fs/2 - fs/N;

If you want to include both the endpoints in frequency vector, you can use ‘linspace’ function of MATLAB.

freq= linspace(-fs/2,fs/2,N);

For better understanding, please follow the MATLAB documentations given below.

https://www.mathworks.com/help/releases/R2019b/matlab/ref/colon.html

https://www.mathworks.com/help/releases/R2019b/matlab/ref/linspace.html

Hope this solves your query.