[Numpy-discussion] numpy.fft, yet again

[David Goldsmith]

On Thu, Jul 15, 2010 at 9:41 AM, David Goldsmith <d.l.goldsmith at gmail.com>wrote:

> On Thu, Jul 15, 2010 at 3:20 AM, Martin Raspaud <martin.raspaud at 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
>

I can't find this fftw function in either numpy or scipy - where is it?

DG
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