[Numpy-discussion] numpy.fft, yet again
Tue Jul 20 19:02:58 CDT 2010
On Thu, Jul 15, 2010 at 9:41 AM, David Goldsmith <email@example.com>wrote:
> On Thu, Jul 15, 2010 at 3:20 AM, Martin Raspaud <firstname.lastname@example.org>wrote:
>> -----BEGIN PGP SIGNED MESSAGE-----
>> Hash: SHA1
>> 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
>> > Having seen no post of a "good reason," I'm going to go ahead and file
>> > enhancement tickets.
>> I have worked on fourier transforms and I think normalization is generally
>> 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
>> 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
>> for the different components of the sound (sinusoids).
>> My guess is that normalization requirements are different for every user:
>> why I like the no normalization approach of fftw, such that anyone does
>> he/she/it wants.
> I get the picture: in the docstring, refer people to fftw.
I can't find this fftw function in either numpy or scipy - where is it?
-------------- next part --------------
An HTML attachment was scrubbed...
More information about the NumPy-Discussion