numpy.fft.ihfft()

numpy.fft.ihfft(a, n=None, axis=-1, norm=None) [source]

Compute the inverse FFT of a signal which has Hermitian symmetry.

Parameters:

Parameters:	a : array_like Input array. n : int, optional Length of the inverse FFT. Number of points along transformation axis in the input to use. If `n` is smaller than the length of the input, the input is cropped. If it is larger, the input is padded with zeros. If `n` is not given, the length of the input along the axis specified by `axis` is used. axis : int, optional Axis over which to compute the inverse FFT. If not given, the last axis is used. norm : {None, ?ortho?}, optional New in version 1.10.0. Normalization mode (see `numpy.fft`). Default is None.
Returns:	out : complex ndarray The truncated or zero-padded input, transformed along the axis indicated by `axis`, or the last one if `axis` is not specified. If `n` is even, the length of the transformed axis is `(n/2)+1`. If `n` is odd, the length is `(n+1)/2`.

a : array_like

Input array.

n : int, optional

Length of the inverse FFT. Number of points along transformation axis in the input to use. If n is smaller than the length of the input, the input is cropped. If it is larger, the input is padded with zeros. If n is not given, the length of the input along the axis specified by axis is used.

axis : int, optional

Axis over which to compute the inverse FFT. If not given, the last axis is used.

norm : {None, ?ortho?}, optional

New in version 1.10.0.

Normalization mode (see numpy.fft). Default is None.

Returns:

out : complex ndarray

The truncated or zero-padded input, transformed along the axis indicated by axis, or the last one if axis is not specified. If n is even, the length of the transformed axis is (n/2)+1. If n is odd, the length is (n+1)/2.

Notes

hfft/ihfft are a pair analogous to rfft/irfft, but for the opposite case: here the signal has Hermitian symmetry in the time domain and is real in the frequency domain. So here it?s hfft for which you must supply the length of the result if it is to be odd: ihfft(hfft(a), len(a)) == a, within numerical accuracy.

Examples

>>> spectrum = np.array([ 15, -4, 0, -1, 0, -4])
>>> np.fft.ifft(spectrum)
array([ 1.+0.j,  2.-0.j,  3.+0.j,  4.+0.j,  3.+0.j,  2.-0.j])
>>> np.fft.ihfft(spectrum)
array([ 1.-0.j,  2.-0.j,  3.-0.j,  4.-0.j])

Links:

https://docs.scipy.org/doc/numpy-1.11.0/reference/generated/numpy.fft.ihfft.html

doc_NumPy

2017-01-10 18:13:56

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