How to properly use the ifft function ?
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Hello, I'm trying to apply the ifft function to the Fourier transform of a real signal S I know, so as to get S in the end. (basically ifft(ffft(S)) = S).
My fft was made with a number N of points ( fft(S,N) ) : N = 2^nextpow2(Ns), where Ns is the number of points in S ( Ns = length(S) ).
I don't know what operations (symmetry, normalization, scaling ...) I have to do before applying ifft but for the moment I don't get S. And the signal I get has a non-neglegtable imaginary part.
Thank you for your help, Léo.
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Answers (4)
Wayne King
on 20 Oct 2011
x = randn(1000,1);
xdft = fft(x);
y = ifft(xdft,'symmetric');
max(abs(x-y))
The 'symmetric' option enforces that the imaginary part should be zero in the inverse DFT.
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Daniel Shub
on 20 Oct 2011
Working from Wayne's answer:
x = randn(1000,1);
xdft = fft(x, pow2(nextpow2(length(x))));
y = ifft(xdft);
isreal(y)
I don't need to use the symmetric option and I get a real y. I cannot do x-y, because y is not the same length as x.
Can you show your code Leonard?
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Walter Roberson
on 20 Oct 2011
It is not uncommon for ifft(fft(S)) to have an imaginary part, due to round-off error.
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Daniel Shub
on 20 Oct 2011
Really? Is it because the fft of a real signal results in a non-symmetric complex signal, the ifft of a symmetric signal results in a non-real signal, or both? I understand there is round-off error, I am surprised it is asymmetric and that the FFT algorithm doesn't prevent this. Can you provide an "x" that will demonstrate this?
Temur Dzigrashvili
on 11 Mar 2022
Is it possible to approximate a complex-valued function of real variable in Matlab.
Values of function (complex numbers) and arguments (real numbers) are given.
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