hfft — SciPy v1.15.3 Manual (original) (raw)
scipy.fft.
scipy.fft.hfft(x, n=None, axis=-1, norm=None, overwrite_x=False, workers=None, *, plan=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{“backward”, “ortho”, “forward”}, optional
Normalization mode (see fft). Default is “backward”.
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.
planobject, optional
This argument is reserved for passing in a precomputed plan provided by downstream FFT vendors. It is currently not used in SciPy.
Added in version 1.5.0.
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
Compute the 1-D FFT for real input.
The inverse of hfft.
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 import numpy as np 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.])