Question about PSD calculation using FFT

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Chris Hoer 2016-6-5

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https://ww2.mathworks.cn/matlabcentral/answers/287783-question-about-psd-calculation-using-fft

评论： Francesca 2022-10-18

采纳的回答： Jeremy

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Hello everyone,

a short question about the calculation of the PSD using the FFT. I used the example code provided:

rng default
Fs = 1000;
t = 0:1/Fs:1-1/Fs;
x = cos(2*pi*100*t) + randn(size(t));
N = length(x);
xdft = fft(x);
xdft = xdft(1:N/2+1);
psdx = (1/(Fs*N)) * abs(xdft).^2;
psdx(2:end-1) = 2*psdx(2:end-1);
freq = 0:Fs/length(x):Fs/2;

I understand each step except this one:

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

Why do we double entry except the first and the last? Is it about the symmetry of the FFT?

Thanks ahead and best regards, Chris

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Francesca 2022-10-18

Hi, i'm working on a similar work: can I ask you what this part means: PSDx = (1/(Fs*N)) * abs(DFTx).^2?

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Answer 1

Jeremy 2016-6-14

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The fft results in a spectra that includes positive and negative frequencies. Don't ask me to explain what a negative frequency really means but the negative frequency esults are just the conjugates of the positive frequency results. They are not needed for the PSD calculation which is why they are rmeoved by the following line:

xdft = xdft(1:N/2+1);

You have to account for the enrgy lost when this portion of the spectra is removed so we multiply the rest by 2. The 0Hz and Nyquist frequency results don't have an imaginary portion and are not included in the negative frequency portion so they are not muiltiplied by 2.