[Numpy-discussion] numpy.fft, yet again
David Goldsmith
d.l.goldsmith@gmail....
Thu Jul 15 11:41:47 CDT 2010
On Thu, Jul 15, 2010 at 3:20 AM, Martin Raspaud <martin.raspaud@smhi.se>wrote:
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> David Goldsmith skrev:
> >
> >
> > Interesting comment: it made me run down the fftpack tutorial
> > <http://docs.scipy.org/scipy/docs/scipy-docs/tutorial/fftpack.rst/>
> > josef has alluded to in the past to see if the suggested pointer
> > could point there without having to write a lot of new content.
> > What I found was that although the scipy basic fft functions don't
> > support it (presumably because they're basically just wrappers for
> > the numpy fft functions), scipy's discrete cosine transforms support
> > an "norm=ortho" keyword argument/value pair that enables the
> > function to return the unitary versions that you describe above.
> > There isn't much narrative explanation of the issue yet, but it got
> > me wondering: why don't the fft functions support this? If there
> > isn't a "good" reason, I'll go ahead and submit an enhancement
> ticket.
> >
> >
> > Having seen no post of a "good reason," I'm going to go ahead and file
> > enhancement tickets.
>
> Hi,
>
> I have worked on fourier transforms and I think normalization is generally
> seen
> as a whole : fft + ifft should be the identity function, thus the necessity
> of a
> normalization, which often done on the ifft.
>
> As one of the previous poster mentioned, sqrt(len(x)) is often seen as a
> good
> compromise to split the normalization equally between fft and ifft.
>
> In the sound community though, the whole normalization often done after the
> fft,
> such that looking at the amplitude spectrum gives the correct amplitude
> values
> for the different components of the sound (sinusoids).
>
> My guess is that normalization requirements are different for every user:
> that's
> why I like the no normalization approach of fftw, such that anyone does
> whatever
> he/she/it wants.
>
I get the picture: in the docstring, refer people to fftw.
DG
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