Fourier transform in scipy and Parseval's identity

Question

While working on fft in Python, I'm running into problems with the normalization of the function and calculating the squared sum of the coefficients.

My code is as follows:

import pandas as pd
import numpy as np
import scipy.fftpack 

# sample points
N = 1024

# sample spacing
T = 1.0/512.0

# example function
x = np.linspace(0.0, N*T, N)
y = 1.5 * np.sin(5*2*np.pi*x) + 3 * np.sin(6*2*np.pi*x)

df = pd.DataFrame({'X': x, 'Y': y})

sumy = sum(abs(df['Y'])**2)

df['FFT (Y)'] = scipy.fftpack.fft(df['Y'])/(N/2)
df['FFT (freq)'] = scipy.fftpack.fftfreq(N, T)

sumffty = sum(abs(df['FFT (Y)'])**2)

print sumy, sumffty

According to Parseval's identity, the sums should be equal; however, they're off by a factor of 256 (sumffty being the smaller one). What am I missing?

Answer 1

Parseval's theorem states that the total power in the two domains (space/time and frequency) are equal. Its statement, for DFT is this. Notice, that in spite of the 1/N on the right, this is not mean power, but total power. The reason for the 1/N is the normalization convention for the DFT.

Put in Python, this means that for a time/space signal sig, Parseval equivalence may be stated as:

np.sum(np.abs(sig)**2) == np.sum(np.abs(np.fft.fft(sig))**2)/sig.size

Below is a complete example, starting with some toy cases (one and two dimensional arrays filled one 1s). Note that I used the .size property of numpy.ndarray, which returns the total number of elements in the array. Hope this helps!

import numpy as np

# 1-d, 4 elements:
ones_1d = np.array([1.,1.,1.,1.])
ones_1d_f = np.fft.fft(ones_1d)

# compute total power in space and frequency domains:
space_power_1d = np.sum(np.abs(ones_1d)**2)
freq_power_1d = np.sum(np.abs(ones_1d_f)**2)/ones_1d.size
print 'space_power_1d = %f'%space_power_1d
print 'freq_power_1d = %f'%freq_power_1d

I'm sorry I couldn't use your example directly--I'm not a Pandas user and couldn't understand it perfectly.