SciPy

scipy.fft.hfft

scipy.fft.hfft(x, n=None, axis=-1, norm=None, overwrite_x=False, workers=None)[source]

Compute the FFT of a signal that has Hermitian symmetry, i.e., a real spectrum.

Parameters
xarray_like

The input array.

nint, optional

Length of the transformed axis of the output. For n output points, n//2 + 1 input points are necessary. If the input is longer than this, it is cropped. If it is shorter than this, it is padded with zeros. If n is not given, it is taken to be 2*(m-1), where m is the length of the input along the axis specified by axis.

axisint, optional

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

norm{None, “ortho”}, optional

Normalization mode (see fft). Default is None.

overwrite_xbool, optional

If True, the contents of x can be destroyed; the default is False. See fft for more details.

workersint, optional

Maximum number of workers to use for parallel computation. If negative, the value wraps around from os.cpu_count(). See fft for more details.

Returns
outndarray

The truncated or zero-padded input, transformed along the axis indicated by axis, or the last one if axis is not specified. The length of the transformed axis is n, or, if n is not given, 2*m - 2, where m is the length of the transformed axis of the input. To get an odd number of output points, n must be specified, for instance, as 2*m - 1 in the typical case,

Raises
IndexError

If axis is larger than the last axis of a.

See also

rfft

Compute the 1-D FFT for real input.

ihfft

The inverse of hfft.

hfftn

Compute the N-D FFT of a Hermitian signal.

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. * even: ihfft(hfft(a, 2*len(a) - 2) == a, within roundoff error, * odd: ihfft(hfft(a, 2*len(a) - 1) == a, within roundoff error.

Examples

>>> from scipy.fft import fft, hfft
>>> a = 2 * np.pi * np.arange(10) / 10
>>> signal = np.cos(a) + 3j * np.sin(3 * a)
>>> fft(signal).round(10)
array([ -0.+0.j,   5.+0.j,  -0.+0.j,  15.-0.j,   0.+0.j,   0.+0.j,
        -0.+0.j, -15.-0.j,   0.+0.j,   5.+0.j])
>>> hfft(signal[:6]).round(10) # Input first half of signal
array([  0.,   5.,   0.,  15.,  -0.,   0.,   0., -15.,  -0.,   5.])
>>> hfft(signal, 10)  # Input entire signal and truncate
array([  0.,   5.,   0.,  15.,  -0.,   0.,   0., -15.,  -0.,   5.])

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