Bug with spectra when using real fft

I think there is a bug with the power spectrum estimates when a real dimension is provided.

For real fft, Hermitian symmetry is exploited to compute only the non-negative frequency terms. But we can’t ignore the energy in the negative frequency terms when computing spectra. To account for this, we need to multiply by 2 the energy of all but the first (DC) mode (and last mode if the signal length even - see https://numpy.org/doc/stable/reference/generated/numpy.fft.rfft.html).

I can demonstrate the issue and the fix through comparison to scipy.signal.periodogram:

import xrft
import numpy as np
import scipy.signal
import xarray as xr
import matplotlib.pyplot as plt 

x = xr.DataArray(np.random.normal(0, 1, size=100), 
                 dims=['t'], 
                 coords=[range(0,100)])


f_scipy, p_scipy = scipy.signal.periodogram(x.values,
                                            window='rectangular')
p_xrft = xrft.xrft.power_spectrum(x, 
                                  dim=['t'], 
                                  real='t',
                                  detrend='constant')

# For even length signals the last mode includes both the +ve and -ve contributions
correction = np.ones_like(f_scipy)
correction[1:-1] = 2

plt.plot(f_scipy, p_scipy,
         label='scipy psd')
plt.plot(p_xrft.freq_t, p_xrft, 
         color='C2', label='xrft psd')
plt.plot(p_xrft.freq_t, correction*p_xrft, 
         linestyle='--', color='C1', label='corrected xrft psd')
plt.xlabel('freq')
plt.ylabel('psd')
plt.legend()

[lanougue]

Hi @dougiesquire, I fully agree with you. We have to decide if we want to be compliant with scipy.signal.periodogram(). And populate the documentation for sure.

[rabernat]

Thanks for the explanations @lanougue! I think it’s super important we get all of this documented in the documentation.