[SciPy-user] FFT speed ?
Stefan van der Walt
stefan at sun.ac.za
Wed Feb 7 09:39:21 EST 2007
On Wed, Feb 07, 2007 at 03:18:19PM +0100, Samuel GARCIA wrote:
> yes I was thinking of doing something like that but
> fft(a,pow(2,18)).shape is of course 262144 (2**18)
> and when I use ifft after that The length of my signal have changed I have an
> interpolated signal ... new problem for me ...
An interpolated signal?
In [19]: N.real(N.fft.ifft(N.fft.fft(N.arange(11),16)))
Out[19]:
array([ 0., 1., 2., 3., 4., 5., 6., 7., 8., 9., 10.,
-0., 0., 0., 0., 0.])
Padding in the Fourier-domain, on the other hand:
In [20]: N.real(N.fft.ifft(N.fft.fft(N.arange(11)),16))
Out[20]:
array([ 0. , 2.10035464, 3.04455839, 1.35780382, 3.23688237,
3.12291155, 2.35790278, 4.12133152, 3.4375 , 3.29213721,
5.00323313, 3.69035768, 4.32561763, 5.98165513, 3.3443057 ,
6.58344845])
Cheers
Stéfan
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