scipy.fft.hfft#
- 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 be2*(m-1)
, wherem
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()
. Seefft
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.
New 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
, wherem
is the length of the transformed axis of the input. To get an odd number of output points, n must be specified, for instance, as2*m - 1
in the typical case,
- Raises:
- IndexError
If axis is larger than the last axis of a.
See also
Notes
hfft
/ihfft
are a pair analogous torfft
/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’shfft
, 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.])