python - How to calculate a Fourier series in Numpy - Stack Overflow

16/12/2020
python - How to calculate a Fourier series in Numpy? - Stack Overflow
How to calculate a Fourier series in Numpy?
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I have a periodic function of period T and would like to know how to obtain the list of the Fourier
coefficients. I tried using fft module from numpy but it seems more dedicated to Fourier
transforms than series. Maybe it a lack of mathematical knowledge, but I can't see how to
calculate the Fourier coefficients from fft.
Help and/or examples appreciated.
13
python
numpy
fft
edited May 12 '13 at 21:43
1
Saullo G. P. Castro
Mermoz
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numpy does not provide options for Fourier series, whereas Sympy does (sympy fouries documentation).
Here are some examples where I used sympy to work out some problems for Krezig Advanced
engineering mathematics! – Pradeep Padmanaban C May 20 '18 at 6:41
5 Answers
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asked Nov 23 '10 at 16:10
Active Oldest Votes
In the end, the most simple thing (calculating the coefficient with a riemann sum) was the most
portable/efficient/robust way to solve my problem:
import numpy as np
def cn(n):
c = y*np.exp(-1j*2*n*np.pi*time/period)
return c.sum()/c.size
def f(x, Nh):
f = np.array([2*cn(i)*np.exp(1j*2*i*np.pi*x/period) for i in range(1,Nh+1)])
return f.sum()
y2 = np.array([f(t,50).real for t in time])
plot(time, y)
plot(time, y2)
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gives me:
edited Jul 22 at 22:34
answered Nov 24 '10 at 11:59
zabop
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Thanks for posting this solution. It saved me some time :) – zega Jun 28 '13 at 15:44
Thank you. Exactly what I wanted – Thoth19 Dec 29 '14 at 18:15
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Any chance you could include a commented minimum working example? – SeSodesa Jun 22 '18 at 15:23
This is an old question, but since I had to code this, I am posting here the solution that uses the
numpy.fft module, that is likely faster than other hand-crafted solutions.
The DFT is the right tool for the job of calculating up to numerical precision the coefficients of the
Fourier series of a function, defined as an analytic expression of the argument or as a numerical
interpolating function over some discrete points.
This is the implementation, which allows to calculate the real-valued coefficients of the Fourier
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def fourier_series_coeff_numpy(f, T, N, return_complex=False):
"""Calculates the first 2*N+1 Fourier series coeff. of a periodic function.
Given a periodic, function f(t) with period T, this function returns the
coefficients a0, {a1,a2,...},{b1,b2,...} such that:
f(t) ~= a0/2+ sum_{k=1}^{N} ( a_k*cos(2*pi*k*t/T) + b_k*sin(2*pi*k*t/T) )
If return_complex is set to True, it returns instead the coefficients
{c0,c1,c2,...}
such that:
f(t) ~= sum_{k=-N}^{N} c_k * exp(i*2*pi*k*t/T)
where we define c_{-n} = complex_conjugate(c_{n})
Refer to wikipedia for the relation between the real-valued and complex
valued coeffs at http://en.wikipedia.org/wiki/Fourier_series.
Parameters
---------f : the periodic function, a callable like f(t)
T : the period of the function f, so that f(0)==f(T)
N_max : the function will return the first N_max + 1 Fourier coeff.
Returns
------if return_complex == False, the function returns:
a0 : float
a,b : numpy float arrays describing respectively the cosine and sine coeff.
if return_complex == True, the function returns:
c : numpy 1-dimensional complex-valued array of size N+1
"""
# From Shanon theoreom we must use a sampling freq. larger than the maximum
# frequency you want to catch in the signal.
f_sample = 2 * N
# we also need to use an integer sampling frequency, or the
# points will not be equispaced between 0 and 1. We then add +2 to f_sample
t, dt = np.linspace(0, T, f_sample + 2, endpoint=False, retstep=True)
y = np.fft.rfft(f(t)) / t.size
if return_complex:
return y
else:
y *= 2
return y[0].real, y[1:-1].real, -y[1:-1].imag
This is an example of usage:
from numpy import ones_like, cos, pi, sin, allclose
T = 1.5 # any real number
def f(t):
"""example of periodic function in [0,T]"""
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return a0 / 2 * ones_like(t) + a1 * cos(2 * pi * n1 * t / T) + b4 * sin(
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2 * pi * n2 * t / T) + a7 * cos(2 * pi * n3 * t / T)
N_chosen = 10
a0, a, b = fourier_series_coeff_numpy(f, T, N_chosen)
# we have as expected that
assert allclose(a0, 4)
assert allclose(a, [2, 0, 0, 0, 0, 0, -3, 0, 0, 0])
assert allclose(b, [0, 0, 0, -1, 0, 0, 0, 0, 0, 0])
And the plot of the resulting a0,a1,...,a10,b1,b2,...,b10 coefficients:
This is an optional test for the function, for both modes of operation. You should run this after the
example, or define a periodic function f and a period T before running the code.
# #### test that it works with real coefficients:
from numpy import linspace, allclose, cos, sin, ones_like, exp, pi, \
complex64, zeros
def series_real_coeff(a0, a, b, t, T):
"""calculates the Fourier series with period T at times t,
from the real coeff. a0,a,b"""
tmp = ones_like(t) * a0 / 2.
for k, (ak, bk) in enumerate(zip(a, b)):
tmp += ak * cos(2 * pi * (k + 1) * t / T) + bk * sin(
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2 * pi * (k + 1) * t / T)
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. tmp
return
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t = linspace(0, T, 100)
f_values = f(t)
a0, a, b = fourier_series_coeff_numpy(f, T, 52)
# construct the series:
f_series_values = series_real_coeff(a0, a, b, t, T)
# check that the series and the original function match to numerical precision:
assert allclose(f_series_values, f_values, atol=1e-6)
# #### test similarly that it works with complex coefficients:
def series_complex_coeff(c, t, T):
"""calculates the Fourier series with period T at times t,
from the complex coeff. c"""
tmp = zeros((t.size), dtype=complex64)
for k, ck in enumerate(c):
# sum from 0 to +N
tmp += ck * exp(2j * pi * k * t / T)
# sum from -N to -1
if k != 0:
tmp += ck.conjugate() * exp(-2j * pi * k * t / T)
return tmp.real
f_values = f(t)
c = fourier_series_coeff_numpy(f, T, 7, return_complex=True)
f_series_values = series_complex_coeff(c, t, T)
assert allclose(f_series_values, f_values, atol=1e-6)
edited May 12 '16 at 18:00
answered Dec 31 '14 at 11:12
gg349
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Numpy isn't the right tool really to calculate fourier series components, as your data has to be
discretely sampled. You really want to use something like Mathematica or should be using fourier
transforms.
To roughly do it, let's look at something simple a triangle wave of period 2pi, where we can easily
calculate the Fourier coefficients (c_n = -i ((-1)^(n+1))/n for n>0; e.g., c_n = { -i, i/2, -i/3, i/4, -i/5,
i/6, ... } for n=1,2,3,4,5,6 (using Sum( c_n exp(i 2 pi n x) ) as Fourier series).
import numpy
x = numpy.arange(0,2*numpy.pi, numpy.pi/1000)
y = (x+numpy.pi/2) % numpy.pi - numpy.pi/2
fourier_trans = numpy.fft.rfft(y)/1000
If you look at the first several Fourier components:
array([ -3.14159265e-03 +0.00000000e+00j,
2.54994550e-16 -1.49956612e-16j,
3.14159265e-03 -9.99996710e-01j,
1.28143395e-16 +2.05163971e-16j,
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-3.14159265e-03
+4.99993420e-01j
5.28320925e-17
-2.74568926e-17j
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3.14159265e-03 -3.33323464e-01j,
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7.73558750e-17
-3.14159265e-03
1.73758496e-16
3.14159265e-03
-1.74044469e-16
-3.14159265e-03
-1.02291982e-16
3.14159265e-03
1.96729377e-17
-3.14159265e-03
-7.50516717e-17
3.14159265e-03
-1.27900121e-16
-3.14159265e-03
-3.41761974e-16j,
+2.49986840e-01j,
+1.55882418e-17j,
-1.99983550e-01j,
-1.22437710e-17j,
+1.66646927e-01j,
-2.05092972e-16j,
-1.42834113e-01j,
+5.35550532e-17j,
+1.24973680e-01j,
+3.33475329e-17j,
-1.11081501e-01j,
-3.32193126e-17j,
+9.99670992e-02j,
First neglect the components that are near 0 due to floating point accuracy (~1e-16, as being
zero). The more difficult part is to see that the 3.14159 numbers (that arose before we divide by
the period of a 1000) should also be recognized as zero, as the function is periodic). So if we
neglect those two factors we get:
fourier_trans = [0,0,-i,0,i/2,0,-i/3,0,i/4,0,-i/5,0,-i/6, ...
and you can see the fourier series numbers come up as every other number (I haven't
investigated; but I believe the components correspond to [c0, c-1, c1, c-2, c2, ... ]). I'm using
conventions according to wiki: http://en.wikipedia.org/wiki/Fourier_series.
Again, I'd suggest using mathematica or a computer algebra system capable of integrating and
dealing with continuous functions.
edited Nov 23 '10 at 17:59
answered Nov 23 '10 at 17:12
dr jimbob
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Excellent, excellent point about having to put some effort into understanding the result. +1. – mtrw Nov 23
'10 at 17:29
As other answers have mentioned, it seems that what you are looking for is a symbolic computing
package, so numpy isn't suitable. If you wish to use a free python-based solution, then either
sympy or sage should meet your needs.
answered Nov 24 '10 at 9:40
DaveP
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here is the reference for fourier series using sympy: docs.sympy.org/dev/modules/mpmath/calculus/…. It
requires mpmath which is not even in my sympy distribution. Although a good hint, i won't choose this
the sake of portability
of theread
code.
– understand
Mermoz Nov
'10 at 10:35
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python - How to calculate a Fourier series in Numpy? - Stack Overflow
Do you have a list of discrete samples of your function, or is your function itself discrete? If so, the
Discrete Fourier Transform, calculated using an FFT algorithm, provides the Fourier coefficients
directly (see here).
On the other hand, if you have an analytic expression for the function, you probably need a
symbolic math solver of some kind.
answered Nov 23 '10 at 16:49
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