The frequencies produced by the two recipies are not the same. But the DFT is periodic in both frequency and time. So whether you think that the number in bin in n/2 should correspond to frequency n/2 or -n/2, it's the same number. w On Mon, 5 Feb 2007, Hanno Klemm wrote:
Hi there,
I have a question regarding the definitions surrounding FFTs. The help to numpy.fft.fft says:
help(N.fft.fft) Help on function fft in module numpy.fft.fftpack:
fft(a, n=None, axis=-1) fft(a, n=None, axis=-1)
Will return the n point discrete Fourier transform of a. n defaults to the length of a. If n is larger than a, then a will be zero-padded to make up the difference. If n is smaller than a, the first n items in a will be used.
The packing of the result is "standard": If A = fft(a, n), then A[0] contains the zero-frequency term, A[1:n/2+1] contains the positive-frequency terms, and A[n/2+1:] contains the negative-frequency terms, in order of decreasingly negative frequency. So for an 8-point transform, the frequencies of the result are [ 0, 1, 2, 3, 4, -3, -2, -1].
This is most efficient for n a power of two. This also stores a cache of working memory for different sizes of fft's, so you could theoretically run into memory problems if you call this too many times with too many different n's.
However, the help to numpy.fft.helper.fftfreq says:
help(N.fft.helper.fftfreq) Help on function fftfreq in module numpy.fft.helper:
fftfreq(n, d=1.0) fftfreq(n, d=1.0) -> f
DFT sample frequencies
The returned float array contains the frequency bins in cycles/unit (with zero at the start) given a window length n and a sample spacing d:
f = [0,1,...,n/2-1,-n/2,...,-1]/(d*n) if n is even f = [0,1,...,(n-1)/2,-(n-1)/2,...,-1]/(d*n) if n is odd
So one claims, that the packing goes from [0,1,...,n/2,-n/2+1,..,-1] (fft) and the other one claims the frequencies go from [0,1,...,n/2-1,-n/2,...-1]
Is this inconsistent or am I missing something here?
Hanno
-- Hanno Klemm klemm@phys.ethz.ch
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