Determining Fourier Coefficients from Time Series Data

My solution to this problem is to apply a discrete Fourier transform to the data using NumPy’s implementation of the Fast Fourier Transform, numpy.fft.fft(); this is why it’s critical that the data is evenly spaced in time, as FFT requires this. While the FFT is typically used to perform analysis of the frequency spectrum, the desired Fourier coefficients are directly related to the output of this function.

Specifically, this function outputs a series of i complex-valued coefficients c. The Fourier series coefficients are found using the relations:

Therefore the FFT allows the Fourier coefficients to be directly computed. Here is the MWE of my solution to this problem, expanding the example given above:

import numpy as np
import matplotlib.pyplot as plt

# Set the number of equal-time bins to create.
n_bins = 101

# Set the number of Fourier coefficients to use.
n_coeff = 51

# Define a function to generate a Fourier series based on the coefficients determined by the Fast Fourier Transform.
# This also includes a series of phases x to pass through the function.
def create_fourier_series(x, coefficients):

    # Begin the series with the zeroeth-order Fourier coefficient.
    fourier_series = coefficients[0][0] / 2

    # Now generate the first through n_coeff'th terms.  The period is defined to be 1 since we're operating in phase
    # space.
    for n in range(1, n_coeff):
        fourier_series += (fourier_coeff[n][0] * np.cos(2 * np.pi * n * x) + fourier_coeff[n][1] * 
                           np.sin(2 * np.pi * n * x))

    return fourier_series

# Create a dependent test variable to define the x-axis of the test data.
test_array = np.linspace(0, 1, n_bins) - 0.5

# Define some test data to try to apply a Fourier series to.
test_data = [0.9783883464566918, 0.979599093567252, 0.9821424606299206, 0.9857575507812502, 0.9899278899999995,
             0.9941848228346452, 0.9978438300395263, 1.0003009205426352, 1.0012208923679058, 1.0017130521235522,
             1.0021799664031628, 1.0027475606936413, 1.0034168260869563, 1.0040914266144825, 1.0047781181102355,
             1.005520348837209, 1.0061899214145387, 1.006846206627681, 1.0074483048543692, 1.0078691461988312,
             1.008318736328125, 1.008446947572815, 1.00862051262136, 1.0085134881422921, 1.008337095516569,
             1.0079539881889774, 1.0074857334630352, 1.006747783037474, 1.005962048923679, 1.0049115434782612,
             1.003812267822736, 1.0026427549407106, 1.001251963531669, 0.999898555335968, 0.9984976286266923,
             0.996995982142858, 0.9955652088974847, 0.9941647321428578, 0.9927727076023389, 0.9914750532544377,
             0.990212467710371, 0.9891098035363466, 0.9875998927875242, 0.9828093773946361, 0.9722532524271845,
             0.9574084365384614, 0.9411012303149601, 0.9251820309477757, 0.9121488392156851, 0.9033119748549322,
             0.9002445803921568, 0.9032760564202343, 0.91192435882353, 0.9249696964980555, 0.94071381372549,
             0.957139088974855, 0.9721083392156871, 0.982955287937743, 0.9880613320235758, 0.9897455322896282,
             0.9909590626223097, 0.9922601592233015, 0.9936513112840472, 0.9951442427184468, 0.9967071285988475,
             0.9982921493123781, 0.9998775465116277, 1.001389230174081, 1.0029109110251453, 1.0044033691406251,
             1.0057110841487276, 1.0069551867704276, 1.008118776264591, 1.0089884470588228, 1.0098663972602735,
             1.0104514566473979, 1.0109849223300964, 1.0112043902912626, 1.0114717968750002, 1.0113343036750482,
             1.0112205972495087, 1.0108811786407768, 1.010500276264591, 1.0099054552529192, 1.009353759223301,
             1.008592596116505, 1.007887223091976, 1.0070715634615386, 1.0063525891472884, 1.0055587861271678,
             1.0048733732809436, 1.0041832862669238, 1.0035913326848247, 1.0025318871595328, 1.000088536345776,
             0.9963596140350871, 0.9918380684931506, 0.9873937281553398, 0.9833394624277463, 0.9803621496062999,
             0.9786476100386117]

# Determine the fast Fourier transform for this test data.
fast_fourier_transform = np.fft.fft(test_data[n_bins / 2:] + test_data[:n_bins / 2])

# Create an empty list to hold the values of the Fourier coefficients.
fourier_coeff = []

# Loop through the FFT and pick out the a and b coefficients, which are the real and imaginary parts of the
# coefficients calculated by the FFT.
for n in range(0, n_coeff):
    a = 2 * fast_fourier_transform[n].real / n_bins
    b = -2 * fast_fourier_transform[n].imag / n_bins
    fourier_coeff.append([a, b])

# Create the Fourier series approximating this data.
fourier_series = create_fourier_series(test_array, fourier_coeff)

# Create a figure to view the data.
fig, ax = plt.subplots(1, 1, figsize=(6, 6))

# Plot the data.
ax.scatter(test_array, test_data, color="k", s=1)

# Plot the Fourier series approximation.
ax.plot(test_array, fourier_series, color="b", lw=0.5)

This outputs the following:

Note that how I defined the FFT (importing the second half of the data followed by the first half) is a consequence of how this data was generated. Specifically, the data runs from -0.5 to 0.5, but the FFT assumes it runs from 0.0 to 1.0, necessitating this shift.

I’ve found that this works quite well for data that doesn’t include very sharp and narrow discontinuities. I would be interested to hear if anyone has another suggested solution to this problem, and I hope people find this explanation clear and helpful.