Thank you for the positive answer. May I have one more question?
Let b=fft(a) and P=plan_fft(a). Are the following two expressions equally good? a = inv(P) * b a = P \ b On Sunday, September 13, 2015 at 6:15:05 PM UTC+2, Yichao Yu wrote: > > On Sun, Sep 13, 2015 at 11:53 AM, <g...@szfki.hu <javascript:>> wrote: > > I am learning the new FFTW syntax of version 0.4.0. > > > > One thing that surprised me is the possibility of using > > inv(plan1) for ifft, where plan1 was created for fft by plan_fft. > > > > Could you confirm please, that this method is just as efficient > > as creating a separate plan2 for ifft by a separate plan_ifft call? > > When I measured it right around when it was merged, the difference > between `ifft_plan * ary` and `fft_plan \ ary` are smaller than what I > can measure. (and both of them are much faster compare to the old one) >
