th
> > Subject: Re: [Numpy-discussion] second 2d fft gives the same result as
> fft+ifft
> > To: "Discussion of Numerical Python"
> > Date: Tuesday, June 9, 2009, 11:01 AM
> >
> > --- On Tue, 6/9/09, Matthieu Brucher
> > wrote:
> >
> > &
Sorry, I meant:
Im(iFT(FT(f))) = Im(FT^2(f)), Re(iFT(FT(f))) != Re(FT^2(f))
DG
--- On Tue, 6/9/09, David Goldsmith wrote:
> From: David Goldsmith
> Subject: Re: [Numpy-discussion] second 2d fft gives the same result as
> fft+ifft
> To: "Discussion of Numerica
--- On Tue, 6/9/09, Matthieu Brucher wrote:
> Hi,
>
> Is it really ?
> You only show the imaginary part of the FFT, so you can't
> be sure of
> what you are saying.
Indeed, is there not a "label" for a function f which satisfies
Im(iFFT(f)) = Im(FFT^2(f)), Re(iFFT(f)) != Re(FFT^2(f))?
Hi,
Is it really ?
You only show the imaginary part of the FFT, so you can't be sure of
what you are saying.
Don't forget that the only difference between FFT and iFFT is (besides
of teh scaling factor) a minus sign in the exponent.
Matthieu
2009/6/9 bela :
>
> I tried to calculate the second fo
I tried to calculate the second fourier transformation of an image with the
following code below:
---
import pylab
import numpy
### Create a simple image
fx = numpy.zeros( 128**2 ).reshape(128,128).astype( numpy.float )
for i in xrang
