[SciPy-user] FFT speed ?

Stefan van der Walt stefan@sun.ac...
Wed Feb 7 08:39:21 CST 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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