Hi Michael, > this documentation is saying that the difference between the equations > for the fft and ifft is a factor of 1/n (not the numpy implementations). > if you do > > output = numpy.ifft( numpy.fft( input ) ) > > and you get output = input, then the normalizations are appropriately > weighted.
what you say is of course correct, but I am wondering if there is a mistake in the user guide (p. 180 of http://numpy.scipy.org/numpybook.pdf): according to the expressions in the user guide, both fft and ifft are not normalized. The implementation if ifft, on the other hand, has the additional 1/n factor, consistent with the online documentation. Lutz _______________________________________________ Numpy-discussion mailing list Numpy-discussion@scipy.org http://mail.scipy.org/mailman/listinfo/numpy-discussion
