Complex Fourier series example

The complex Fourier analysis of will be applied in a moment. However, it is convenient to first convert f into an exponential. We can write f over a single period t ∈ [ − T/2, T/2) as $$\begin{align*} |A \cos(\omega t)| &= |A| |\cos(\omega t)| \tag{absolute value property} \\ &= |A| \cos(\omega t) \tag{already positive} \\ &= |A| \frac{1}{2} \left(e^{j \omega t} + e^{-j \omega t}\right) \tag{Euler}. \end{align*}$$

Applying Fourier analysis à la with harmonic frequency ω_n = 2πn/T, $$\begin{align*} c_{n} &= \frac{1}{T} \int_{-T/2}^{T/2} f(t) e^{-j \omega_n t} \diff t \\ &= \frac{1}{T} \int_{-T/2}^{T/2} |A| \frac{1}{2} \left(e^{j \omega t} + e^{-j \omega t}\right) e^{-j \omega_n t} \diff t \\ &= \frac{|A|}{2 T} \int_{-T/2}^{T/2} \left(e^{j \omega t} + e^{-j \omega t}\right) e^{-j \omega_n t} \diff t\\ &= \frac{|A|}{2 T} \int_{-T/2}^{T/2} \left(e^{j (\omega - \omega_n) t} + e^{-j (\omega + \omega_n) t}\right) \diff t \\ &= \frac{|A|}{2 T} \left.\left( \frac{1}{j (\omega - \omega_n)} e^{j (\omega - \omega_n) t} - \frac{1}{j (\omega + \omega_n)} e^{-j (\omega + \omega_n) t} \right)\right|_{-T/2}^{T/2} \\ &= \frac{|A|}{2 T} \left( \frac{1}{j (\omega - \omega_n)} e^{j (\omega - \omega_n) T/2} - \frac{1}{j (\omega + \omega_n)} e^{-j (\omega + \omega_n) T/2} + \right. \\ &\left.-\frac{1}{j (\omega - \omega_n)} e^{-j (\omega - \omega_n) T/2} + \frac{1}{j (\omega + \omega_n)} e^{j (\omega + \omega_n) T/2} \right) \\ &= \frac{|A|}{j 2 T (\omega - \omega_n)} \left( e^{j (\omega - \omega_n) T/2} - e^{-j (\omega - \omega_n) T/2} \right) + \\ &+\frac{|A|}{j 2 T (\omega + \omega_n)} \left( e^{j (\omega + \omega_n) T/2} - e^{-j (\omega + \omega_n) T/2} \right)\\ &= \frac{|A|}{T (\omega - \omega_n)} \sin((\omega - \omega_n) T/2) + \frac{|A|}{T (\omega + \omega_n)} \sin((\omega + \omega_n) T/2). \end{align*}$$

This can be simplified further if we substitute T = π/ω and ω_n = 2πn/T = 2nω, $$\begin{align*} c_{\pm n} &= \frac{|A|}{\pi (1 - 2 n)} \sin((1 - 2 n) \pi/2) + \frac{|A|}{\pi (1 + 2 n)} \sin((1 + 2 n) \pi/2). \end{align*}$$

Using a product-to-sum trigonometric identity (), this further simplifies to $$\begin{align} c_{n} &= \frac{- 2 |A|}{\pi (4 n^2 - 1)} \cos(\pi n), \end{align}$$ which, for n odd or even, $$\begin{align} c_{n} &= \begin{cases} \frac{2 |A|}{\pi (4 n^2 - 1)} & n\ \text{odd} \\ \frac{- 2 |A|}{\pi (4 n^2 - 1)} & n\ \text{even}. \end{cases} \end{align}$$

Let’s continue in Python. Load Python packages as follows:

import numpy as np
import matplotlib.pyplot as plt
import sympy as sp

Define the symbolic variables as follows:

A, T = sp.symbols("A, T", real=True, positive=True)
t, w, wn = sp.symbols("t, w, wn", real=True)
n = sp.symbols("n", integer=True)

Define the function f as a symbolic expression:

f = sp.Abs(A) * sp.cos(w * t)
f = f.rewrite(sp.exp)
props = {w: sp.pi / T, wn: 2 * sp.pi * n / T}  # Properties

Compute the Fourier coefficients c_n as follows:

c_n = sp.integrate(1/T * 
    f.subs(props) * sp.exp(-sp.I * wn * t), 
    (t, -T / 2, T / 2)
).subs(props).simplify()

$\displaystyle \begin{cases} \frac{A}{2} & \text{for}\: T = - \frac{T}{2 n} \vee T = \frac{T}{2 n} \\- \frac{2 \left(-1\right)^{n} A}{\pi \left(4 n^{2} - 1\right)} & \text{otherwise} \end{cases}$

The first piecewise condition is impossible because n is an integer. (SymPy should have caught this.) The second condition obtains for all n. We see that the sign changes with n even or odd. Manually extracting the second case, we have

c_n = c_n.args[1][0]

We will need the positive n and negative n coefficients separately. Let’s define them as follows:

c_n_pos = c_n.subs(n, sp.Abs(n))
c_n_neg = c_n.subs(n, -sp.Abs(n))

Let’s compute the harmonic amplitude |C_n| and phase ϕ_n as follows:

C_n = 2 * sp.sqrt(c_n_pos * c_n_neg)
phi_n = sp.atan2(sp.im(c_n), sp.re(c_n))

$\displaystyle \frac{4 A}{\pi \left|{4 n^{2} - 1}\right|}$

$\displaystyle \pi \left(1 - \theta\left(- \frac{2 \left(-1\right)^{n} A}{\pi \left(4 n^{2} - 1\right)}\right)\right)$

Here θ is the Heaviside step function (0 for negative arguments and 1 for positive arguments). The harmonic amplitude and phase can be plotted if we lambdify them as follows:

C_n_fn = sp.lambdify((A, T, n), C_n, "numpy")
phi_n_fn = sp.lambdify((A, T, n), phi_n, "numpy")

Letting A = 1 and T = 1, let’s plot the harmonic amplitude and phase for the first 10 harmonics as follows:

n_ = np.arange(0, 11)
A_, T_ = 1, 1
C_n_ = C_n_fn(A_, T_, n_)
phi_n_ = phi_n_fn(A_, T_, n_)
fig, ax = plt.subplots(2, 1, sharex=True)
ax[0].stem(n_, C_n_)
ax[0].set_ylabel(r"$|C_n|$")
ax[1].stem(n_, phi_n_)
ax[1].set_ylabel(r"$\angle C_n$")
ax[1].set_xlabel(r"$n$")
plt.show()

Now we convert the complex Fourier series coefficients c_n into real Fourier series coefficients a_n and b_n as follows:

a_0 = 2 * c_n.subs(n, 0)
a_n = (c_n_pos + c_n_neg).simplify()
b_n = (sp.I * (c_n_pos - c_n_neg)).simplify()

$\displaystyle \frac{4 A}{\pi}$

$\displaystyle - \frac{4 \left(-1\right)^{\left|{n}\right|} A}{\pi \left(4 n^{2} - 1\right)}$

0

The result that b_n = 0 is expected because the function f is even.