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Fast Fourier Transform of a real function

Asked by amin rasoulof on 7 Jul 2015
Latest activity Answered by Walter Roberson
on 8 Jul 2015
Hi! I got a question about the FFT properties; why FFT of a real function is not a real function as it is a property of Fourier transform? For example FFT of a gaussian function should be a gaussian, however, abs(fft(gaussian)) is a gaussian. thanks

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Answer by Walter Roberson
on 8 Jul 2015

That is not a property of the Fourier transform. The fft of a real function is generally complex and is always hermitian. The complex part is absent only if everything is in phase (e.g., the sum of sines)

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