>From dft.jl \(p::Plan, x::AbstractArray) = inv(p) * x
On Sunday, September 13, 2015 at 6:38:48 PM UTC+2, Gabor wrote: > > 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, <[email protected]> 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) >> >
