The frequency axis for the PSD of a signal

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MAWE
MAWE 2022-11-4
回答： Abhaya 2024-10-11
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?

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回答（1 个）

Abhaya
Abhaya 2024-10-11
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.

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