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*(m1)
, 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. plan: object, 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 zeropadded 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 >>> 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.])