Inconsistency in ifft of FrequencySeries

Created by: MaxMelching

I am using gwpy 3.0.7 (Python 3.11.6) and noticed the following when using Fourier transformations: the f=0-component seems to be treated inconsistently. In TimeSeries.fft(), we multiply every Fourier coefficient by 2, except for the f=0 one (understandably so). In FrequencySeries.ifft(), however, every Fourier coefficient seems to be divided by 2, including the f=0 one (at least that is my understanding of the code).

Here is an example:

from gwpy.timeseries import TimeSeries
from gwpy.frequencyseries import FrequencySeries

timeseries1 = TimeSeries([1.0, 0.0, -1.0, 0.0], sample_rate=1.0)


print(f'first time series:')
print(timeseries1)
print(f'ifft of fft of first time series:')
print(timeseries1.fft().ifft())

print('----------------------------------------')

timeseries2 = TimeSeries([1.0, 2.0, 3.0, 2.0, 1.0, 0.0], sample_rate=1.0)


print(f'second time series:')
print(timeseries2)
print(f'ifft of fft of second time series:')
print(timeseries2.fft().ifft())

print('----------------------------------------')


frequencyseries1 = FrequencySeries([1.0, 0.0, -1.0, 0.0], df=1.0)


print(f'first frequency series:')
print(frequencyseries1)
print(f'fft of ifft of first frequency series')
print(frequencyseries1.ifft().fft())

print('----------------------------------------')

frequencyseries2 = FrequencySeries([1.0, 2.0, 3.0, 4.0], df=1.0)


print(f'second frequency series:')
print(frequencyseries2)
print(f'fft of ifft of second frequency series:')
print(frequencyseries2.ifft().fft())

which gives the following output:

first time series:
TimeSeries([ 1.,  0., -1.,  0.]
           unit: dimensionless,
           t0: 0.0 s,
           dt: 1.0 s,
           name: None,
           channel: None)
ifft of fft of first time series:
TimeSeries([ 1.,  0., -1.,  0.]
           unit: dimensionless,
           t0: 0.0 1 / Hz,
           dt: 1.0 1 / Hz,
           name: None,
           channel: None)
----------------------------------------
second time series:
TimeSeries([1., 2., 3., 2., 1., 0.]
           unit: dimensionless,
           t0: 0.0 s,
           dt: 1.0 s,
           name: None,
           channel: None)
ifft of fft of second time series:
TimeSeries([ 0.25,  1.25,  2.25,  1.25,  0.25, -0.75]
           unit: dimensionless,
           t0: 0.0 1 / Hz,
           dt: 1.0 1 / Hz,
           name: None,
           channel: None)
----------------------------------------
first frequency series:
FrequencySeries([ 1.,  0., -1.,  0.]
                unit: dimensionless,
                f0: 0.0 Hz,
                df: 1.0 Hz,
                epoch: None,
                name: None,
                channel: None)
fft of ifft of first frequency series
FrequencySeries([ 0.5+0.00000000e+00j,  0. -3.70074342e-17j,
                 -1. -3.70074342e-17j,  0. +0.00000000e+00j]
                unit: dimensionless,
                f0: 0.0 Hz,
                df: 1.0 Hz,
                epoch: 0.0,
                name: None,
                channel: None)
----------------------------------------
second frequency series:
FrequencySeries([1., 2., 3., 4.]
                unit: dimensionless,
                f0: 0.0 Hz,
                df: 1.0 Hz,
                epoch: None,
                name: None,
                channel: None)
fft of ifft of second frequency series:
FrequencySeries([0.5+0.00000000e+00j, 2. +3.70074342e-17j,
                 3. +3.70074342e-17j, 4. +0.00000000e+00j]
                unit: dimensionless,
                f0: 0.0 Hz,
                df: 1.0 Hz,
                epoch: 0.0,
                name: None,
                channel: None)

Replacing

dift = npfft.irfft(self.value * nout) / 2

in the ifft method with something like

coeffs = self.value.copy()
coeffs[1:] = coeffs[1:] / 2
dift = npfft.irfft(coeffs * nout)

seems to solve the issue, at least for this example code (although I am not sure how good this solution actually is with respect to "good" coding; perhaps, we could also divide the dift accordingly). This is also what my explanation/interpretation above is based on.