Fourier series are essential in electronics engineering for analyzing and synthesizing periodic signals, which underpin communication systems, signal processing, and control systems. In Kenya’s growing electronics industry, understanding how to decompose complex waveforms into simpler sinusoidal components enables engineers to design efficient filters, modulators, and amplifiers. This chapter covers the mathematical foundation of Fourier series, focusing on determining the series for functions with period \(2\pi\) or \(T\), and extends to non-periodic functions to address real-world signal analysis challenges.

5.1 Definition of Fourier Series for a Function

Fourier series allow representation of a periodic function as a sum of sine and cosine terms. This decomposition is fundamental in electronics engineering for frequency domain analysis of signals. We begin with the classical Fourier series for functions of period \(2\pi\), then generalize to any period \(T\).

5.1.1 Fourier Series for Functions of Period \(2\pi\)

A periodic function \(f(x)\) with period \(2\pi\) can be expressed as a sum of sines and cosines. The Fourier series is given by:

$$ f(x) = \frac{a_0}{2} + \sum_{n=1}^\infty ( a_n \cos nx + b_n \sin nx ) $$

where the coefficients are:

$$ a_0 = \frac{1}{\pi} \int_{-\pi}^\pi f(x) dx $$

$$ a_n = \frac{1}{\pi} \int_{-\pi}^\pi f(x) \cos nx \, dx $$

$$ b_n = \frac{1}{\pi} \int_{-\pi}^\pi f(x) \sin nx \, dx $$

Worked Examples

Example 1: Find the Fourier series of \(f(x) = x\) on \([- \pi, \pi]\).

Given: \(f(x) = x\), period \(2\pi\)

Calculate \(a_0\):

$$ a_0 = \frac{1}{\pi} \int_{-\pi}^\pi x \, dx $$

$$ = \frac{1}{\pi} [ \frac{x^2}{2} ]_{-\pi}^\pi = \frac{1}{\pi} ( \frac{\pi^2}{2} - \frac{\pi^2}{2} ) = 0 $$

Calculate \(a_n\):

$$ a_n = \frac{1}{\pi} \int_{-\pi}^\pi x \cos nx \, dx $$

Since \(x \cos nx\) is an odd function over symmetric limits, integral is zero:

$$ a_n = 0 $$

Calculate \(b_n\):

Use integration by parts:

Let \(u = x\), \(dv = \sin nx \, dx\)

Then:

$$ du = dx, \quad v = -\frac{\cos nx}{n} $$

So,

$$ b_n = \frac{1}{\pi} [ -\frac{x \cos nx}{n} \Big|_{-\pi}^\pi + \frac{1}{n} \int_{-\pi}^\pi \cos nx \, dx ] $$

Since \(\int_{-\pi}^\pi \cos nx \, dx = 0\),

$$ b_n = \frac{1}{\pi} ( -\frac{\pi \cos n \pi}{n} + \frac{\pi \cos n \pi}{n} ) = \frac{2}{\pi n} (-1)^{n+1} $$

$$ b_n = \frac{2}{n} (-1)^{n+1} $$

Answer:

$$ f(x) = \sum_{n=1}^\infty \frac{2}{n} (-1)^{n+1} \sin nx $$

Example 2: Determine the Fourier series for the square wave defined by

\[ f(x) = \begin{cases} 1, & 0 < x < \pi \\ -1, & -\pi < x < 0 \end{cases} \]

with period \(2\pi\).

Calculate \(a_0\):

$$ a_0 = \frac{1}{\pi} \int_{-\pi}^\pi f(x) dx = \frac{1}{\pi} ( \int_{-\pi}^0 -1 dx + \int_0^\pi 1 dx ) $$

$$ = \frac{1}{\pi} (-\pi + \pi) = 0 $$

Calculate \(a_n\):

$$ a_n = \frac{1}{\pi} ( \int_{-\pi}^0 -1 \cos nx \, dx + \int_0^\pi 1 \cos nx \, dx ) $$

$$ = \frac{1}{\pi} ( - \int_{-\pi}^0 \cos nx \, dx + \int_0^\pi \cos nx \, dx ) $$

$$ = \frac{1}{\pi} ( - \frac{\sin nx}{n} \Big|_{-\pi}^0 + \frac{\sin nx}{n} \Big|_0^\pi ) = 0 $$

Calculate \(b_n\):

$$ b_n = \frac{1}{\pi} ( \int_{-\pi}^0 -1 \sin nx \, dx + \int_0^\pi 1 \sin nx \, dx ) $$

$$ = \frac{1}{\pi} ( - \int_{-\pi}^0 \sin nx \, dx + \int_0^\pi \sin nx \, dx ) $$

$$ = \frac{1}{\pi} ( - [ -\frac{\cos nx}{n} ]_{-\pi}^0 + [ -\frac{\cos nx}{n} ]_0^\pi ) $$

$$ = \frac{1}{\pi n} ( \cos 0 - \cos (-n \pi) - \cos \pi + \cos 0 ) $$

Since \(\cos(-n \pi) = \cos n \pi = (-1)^n\),

$$ b_n = \frac{1}{\pi n} ( 1 - (-1)^n - (-1) + 1 ) = \frac{2}{\pi n} (1 - (-1)^n) $$

For even \(n\), \(b_n=0\); for odd \(n\), \(b_n = \frac{4}{\pi n}\).

Answer:

$$ f(x) = \sum_{n=1,3,5...}^\infty \frac{4}{\pi n} \sin nx $$

Example 3: Find the Fourier series of \(f(x) = \cos x\) on \([- \pi, \pi]\).

Calculate \(a_0\):

$$ a_0 = \frac{1}{\pi} \int_{-\pi}^\pi \cos x \, dx = \frac{1}{\pi} [\sin x]_{-\pi}^\pi = 0 $$

Calculate \(a_n\):

$$ a_n = \frac{1}{\pi} \int_{-\pi}^\pi \cos x \cos nx \, dx $$

Using orthogonality:

• For \(n=1\),

$$ a_1 = \frac{1}{\pi} \int_{-\pi}^\pi \cos^2 x \, dx = \frac{1}{\pi} \cdot \pi = 1 $$

• For \(n eq 1\), \(a_n = 0\)

Calculate \(b_n\):

$$ b_n = \frac{1}{\pi} \int_{-\pi}^\pi \cos x \sin nx \, dx = 0 $$

Answer:

$$ f(x) = \cos x $$

5.1.2 Fourier Series for Functions of Period \(T\)

For functions periodic with period \(T\), the Fourier series generalizes as:

$$ f(t) = \frac{a_0}{2} + \sum_{n=1}^\infty ( a_n \cos n \omega t + b_n \sin n \omega t ) $$

where \(\omega = \frac{2\pi}{T}\) is the fundamental angular frequency, and coefficients are:

$$ a_0 = \frac{2}{T} \int_{t_0}^{t_0 + T} f(t) dt $$

$$ a_n = \frac{2}{T} \int_{t_0}^{t_0 + T} f(t) \cos n \omega t \, dt $$

$$ b_n = \frac{2}{T} \int_{t_0}^{t_0 + T} f(t) \sin n \omega t \, dt $$

with \(t_0\) any convenient interval start.

Worked Examples

Example 1: Find the Fourier series of \(f(t) = t\) over \([- \frac{T}{2}, \frac{T}{2}]\) with period \(T = 2\).

Given: \(f(t) = t\), \(T = 2\), \(\omega = \pi\)

Calculate \(a_0\):

$$ a_0 = \frac{2}{2} \int_{-1}^1 t \, dt = \int_{-1}^1 t \, dt = [ \frac{t^2}{2} ]_{-1}^1 = \frac{1}{2} - \frac{1}{2} = 0 $$

Calculate \(a_n\):

Because \(t \cos n \pi t\) is an odd function over symmetric limits,

$$ a_n = 0 $$

Calculate \(b_n\):

Use integration by parts:

$$ b_n = \frac{2}{2} \int_{-1}^1 t \sin n \pi t \, dt = \int_{-1}^1 t \sin n \pi t \, dt $$

Let \(u = t\), \(dv = \sin n \pi t dt\),

$$ du = dt, \quad v = -\frac{\cos n \pi t}{n \pi} $$

Then,

$$ b_n = [ -\frac{t \cos n \pi t}{n \pi} ]_{-1}^1 + \frac{1}{n \pi} \int_{-1}^1 \cos n \pi t \, dt $$

Calculate each term:

$$ = -\frac{1 \cdot \cos n \pi}{n \pi} + \frac{(-1) \cdot \cos (-n \pi)}{n \pi} + \frac{1}{n \pi} [ \frac{\sin n \pi t}{n \pi} ]_{-1}^1 $$

Since \(\cos n \pi = (-1)^n\), \(\sin n \pi = 0\),

$$ b_n = -\frac{(-1)^n}{n \pi} + \frac{(-1)^{n+1}}{n \pi} + 0 = \frac{2}{n \pi} (-1)^{n+1} $$

Answer:

$$ f(t) = \sum_{n=1}^\infty \frac{2}{n \pi} (-1)^{n+1} \sin n \pi t $$

Example 2: Determine the Fourier series of the periodic pulse:

\[ f(t) = \begin{cases} 1, & 0 < t < \frac{T}{2} \\ 0, & \frac{T}{2} < t < T \end{cases} \]

with \(T=4\).

Calculate \(\omega = \frac{2\pi}{4} = \frac{\pi}{2}\).

Calculate \(a_0\):

$$ a_0 = \frac{2}{4} \int_0^4 f(t) dt = \frac{1}{2} \int_0^2 1 dt = \frac{1}{2} \times 2 = 1 $$

Calculate \(a_n\):

$$ a_n = \frac{2}{4} \int_0^2 \cos n \frac{\pi}{2} t \, dt = \frac{1}{2} \int_0^2 \cos ( \frac{n \pi t}{2} ) dt $$

$$ = \frac{1}{2} [ \frac{2}{n \pi} \sin ( \frac{n \pi t}{2} ) ]_0^2 = \frac{1}{n \pi} ( \sin n \pi - \sin 0 ) = 0 $$

Calculate \(b_n\):

$$ b_n = \frac{1}{2} \int_0^2 \sin \frac{n \pi t}{2} dt = \frac{1}{2} [ -\frac{2}{n \pi} \cos \frac{n \pi t}{2} ]_0^2 = -\frac{1}{n \pi} (\cos n \pi - 1) $$

Since \(\cos n \pi = (-1)^n\),

$$ b_n = \frac{1}{n \pi} (1 - (-1)^n) $$

For even \(n\), \(b_n=0\); for odd \(n\), \(b_n = \frac{2}{n \pi}\).

Answer:

$$ f(t) = \frac{1}{2} + \sum_{n=1,3,5...}^\infty \frac{2}{n \pi} \sin \frac{n \pi t}{2} $$

Example 3: Find the Fourier series of \(f(t) = t^2\) over \([-L, L]\) with period \(T=2L=2\).

Given: \(L=1\), \(T=2\), \(\omega = \pi\)

Calculate \(a_0\):

$$ a_0 = \frac{2}{2} \int_{-1}^1 t^2 dt = \int_{-1}^1 t^2 dt = [ \frac{t^3}{3} ]_{-1}^1 = \frac{1}{3} - ( -\frac{1}{3} ) = \frac{2}{3} $$

Calculate \(a_n\):

$$ a_n = \int_{-1}^1 t^2 \cos n \pi t \, dt $$

Integration by parts twice or using tabulated integrals yields:

$$ a_n = 2 ( \frac{(-1)^n}{n^2 \pi^2} ) $$

Calculate \(b_n\):

Since \(t^2 \sin n \pi t\) is odd, \(b_n=0\).

Answer:

$$ f(t) = \frac{1}{3} + \sum_{n=1}^\infty 2 \frac{(-1)^n}{n^2 \pi^2} \cos n \pi t $$

Chapter Summary

This chapter introduced the concept of the Fourier series as a method to represent a function as a sum of sine and cosine terms. It began by defining the Fourier series for a function with a standard period of \(2\pi\) and extended the determination process to functions with any general period \(T\). The chapter then explored how to determine the Fourier series for non-periodic functions by extending them periodically. Following this, the method was applied to find Fourier series representations of non-periodic functions over a specific finite range. Finally, the chapter examined the simplifications that occur when dealing with odd and even functions, including the derivation of half-range Fourier series, which use only sine or cosine terms depending on the function's symmetry. Throughout, the focus remained on the mathematical procedures needed to compute the coefficients and construct the series accurately. This foundation enables analysis and synthesis of complex periodic and non-periodic signals in engineering contexts.