Thanks! Now all is clear.
On Sunday, September 13, 2015 at 6:46:25 PM UTC+2, Kristoffer Carlsson wrote: > > 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) >>> >>
