FFT of the average vs average of the FFT

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Gianmarco Venditti 2022-4-13

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编辑： Paul 2022-4-13

Hello everyone,

I've a (probably naive and simple question):

I've a NxM matrix (S): N measures, with M data point

I do the average along N and than compute the FFT =>

A = FFT(mean(S) )

on the other side I do first the FFT of each of the N measure and than I average along N:

B = mean(FFT(S))

Now my A and B are different, and look like that the average and FFT are not abelian operations.

However from my memory using the linearity of the Fourier transform and the Fubini-Tonelli theorem (you can switch sum and integral if every integral is finite) A and B should be the same.

I mean: the fourier transform of the average of a set of signals should be the average of the fourier transforms of eahc signals

.

Am I missing something? Should I expect my A and B to be the same? And if not, why?

Thanks in advance

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Answer 1

Paul 2022-4-13

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https://ww2.mathworks.cn/matlabcentral/answers/1695660-fft-of-the-average-vs-average-of-the-fft#answer_942100

编辑：Paul 2022-4-13

在 MATLAB Online 中打开

For a matrix input, fft() works down the columns. Maybe this:

rng(100);
S = rand(5);
fft(mean(S,1))  % mean for each column, fft of resulting row
ans = 
   2.3586 + 0.0000i  -0.0250 + 0.2397i  -0.1065 - 0.2300i  -0.1065 + 0.2300i  -0.0250 - 0.2397i
mean(fft(S.').',1)  % fft of each row S, mean down the columns
ans = 
   2.3586 + 0.0000i  -0.0250 + 0.2397i  -0.1065 - 0.2300i  -0.1065 + 0.2300i  -0.0250 - 0.2397i