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 + 1input 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- mis 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 - fftfor more details.
- workersint, optional
- Maximum number of workers to use for parallel computation. If negative, the value wraps around from - os.cpu_count(). See- fftfor 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- mis 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 - 1in the typical case,
 
- Raises:
- IndexError
- If axis is larger than the last axis of a. 
 
 - See also - Notes - hfft/- ihfftare 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.])