Hello, I am using FFTW v3.2 in C++ to compute the 3d DFT of a real image. I use the FFT as part of an iterative algorithm in which high accuracy is a must, so I checked the outputs of the FFTW and found some very small difference depending on whether I used the r2c or not. By small difference I mean about a 17% increase in numerical error versus by the traditional error. In pseudo-code, if I compute: b = fft(a); c = fft_r2c(a); a2 = ifft(b); a3 = ifft_c2r(c); e1 = sum( abs( a - a2)); e2 = sum( abs( a - a3)); e3 = sum( abs( a2 - a3)); Then e2 is greater than e1 by about 17%, and e3 is not equal to zero. E.g., e1 is around 10^(-5) when divided by the number of elements. I would expect the output of the FFTW code to be numerically identical. And indeed I find that it is if I do not use r2c. That is, again in pseudo-code, if I compute: b = fft(a); c = fft(a); a2 = ifft(b); a3 = ifft(c); e1 = sum( abs( a - a2)); e2 = sum( abs( a - a3)); e3 = sum( abs(a2 - a3)); Then of course e1 = e2 and e3 = 0. So my question is: has anyone heard of this before? Is the output of FFTW slightly different for r2c? Many thanks in advance, Kevin

# FFTW small difference for r2c ?

Started by ●January 20, 2009