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I find myself in the position of having to lower the FFT resolution. Basically I have a signal of length M and I would like to make an FFT with N<M frequency bins. I cannot simply make several FFT's of length N and average them together because I need to preserve the phase of the frequency bin, since I need them for reconstructing a part of the signal. I just do not need a frequency resolution of M.

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    $\begingroup$ Why not use the DTFT? It can be implemented in five lines of code and allows you to sample at any frequencies you want. As long as M is not too large, execution speed won't be a problem on any modern computer. $\endgroup$
    – MBaz
    Commented May 3, 2022 at 14:19

1 Answer 1

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As @MBaz says in the comments, just do:

M = 1024;
N = 128;
x = randn(1,M);

X = fft(x);

XX = zeros(1,N);

for k=0:(N-1)
    for t=0:(M-1)
        XX(k+1) = x(t+1)*exp(-1j*2*pi*t*k/N) + XX(k+1);
    end
end

plot([0:M-1]/M,abs(X));
hold on;
plot([0:N-1]/N,abs(XX),'r.')

And you get:

Example of subsampling in the frequency domain.

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  • $\begingroup$ In this case, how would you calculate the inverse DTFT as it involves a continuous integral of the frequency? $\endgroup$
    – ecook
    Commented May 4, 2022 at 1:10
  • $\begingroup$ @ecook The inverse really won't have enough information to fully reconstitute the original signal. $\endgroup$
    – Peter K.
    Commented May 4, 2022 at 1:17
  • $\begingroup$ It's impossible to recover the time domain signal unless the frequency axis is densely sampled ($N>>M$ in the code)? If the frequency axis is oversampled, do we have to evaluate the IDTFT integral to find the original signal? Are there any fast methods? $\endgroup$
    – ecook
    Commented May 4, 2022 at 1:24

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