Ooura inverse FFT differences from Matlab ifft

Question

Today while struggling to make my C++ code (using Ooura FFT library) to give the same results as Matlab does, I finally found a problem and a solution.

In Matlab calculating reverse coefficients for a grid of amplitude-frequency response is done in a following way (code fragment from Matlab's internal fir2() function):

%H contains 8192 points of AFR data
Hconj = [H conj(H(npt-1:-1:2))];   % Fourier transform of real series
ht = real(ifft(Hconj));            % Symmetric real series

as a result we get back 16384 bins, and the second half of them can be thrown away, but the first half can later be used as FIR coefficients.

But if I do the same in Ooura using Real DFT rdft() function, I get coefficients which create a mirror effect in the resulting AFR, all the frequencies on the AFR plot are divided by 2.

So an idea came to me: in my C++ code I made my H twice as big (16384 points) and filled them all with frequency data without mirroring. And voila! it worked, now I got 16384 points, throw away everything after 8192 point and now the resulting AFR matches the Matlab's.

I was convinced that all standard FFT implementations need this mirroring. Is it just a quirk of Ooura that it DOES NOT need mirroring data in input, or maybe there is something else going on?

hotpaw2 hotpaw2 · Accepted Answer · 2011-10-20T19:40:15

A "real" FFT often does the mirroring internally by automatically using the complex conjugates of the input for the mirrored values. A standard FFT/IFFT doesn't do this because it has twice the degrees of freedom (e.g. to produce a complex output or, say, a complex domain filter from an IFFT) of an FFT/IFFT implementation constrained to real-only output.

The frequency bin step size of an FFT is controlled by its length. Half the length produces half the frequency step size, as in your original trial.

Ooura inverse FFT differences from Matlab ifft

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