Fourier series play a critical role in electrical engineering by enabling the decomposition of complex periodic signals into simpler sinusoidal components. This technique is essential for analyzing electrical circuits, signal processing, and communications systems widely used in Kenya’s energy and telecommunications sectors. Understanding how to determine the Fourier series of functions, including non-periodic ones, equips engineers to design efficient filters, analyze harmonic content, and optimize system performance.

1.1 Definition of Fourier series for a function

Fourier series represent a periodic function as a sum of sine and cosine terms, each multiplied by coefficients that capture the function’s frequency components. This mathematical tool is particularly useful in electrical engineering for analyzing alternating current (AC) waveforms and signal harmonics in power systems and communication signals.

1.1.1 Fourier series representation of a periodic function with period \(2\pi\)

A function \(f(x)\) with period \(2\pi\) can be expressed as a Fourier series:

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

where the coefficients are given by

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

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

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

Worked Examples

Example 1: Find the Fourier series of the function \(f(x) = x\) defined on \([- \pi, \pi]\) with period \(2\pi\).

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,

$$ a_n = 0 $$

Calculate \(b_n\):

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

Integration by parts where:

• \(u = x \Rightarrow du = dx\)
• \(dv = \sin nx \, dx \Rightarrow v = -\frac{\cos nx}{n}\)

So,

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

Evaluate boundary term:

$$ -\frac{x \cos nx}{n} \Big|_{-\pi}^{\pi} = -\frac{\pi \cos n\pi}{n} + \frac{\pi \cos(-n\pi)}{n} = 0 $$

Since \(\cos n\pi = \cos(-n\pi)\).

Evaluate integral term:

$$ \int_{-\pi}^{\pi} \cos nx \, dx = 0 $$

$$ b_n = \frac{1}{\pi} \times 0 = 0 $$

But this contradicts the known result; re-examining, the integral of \(x \sin nx\) is even, so calculate directly:

$$ b_n = \frac{2}{\pi} \int_{0}^{\pi} x \sin nx \, dx $$

Integration by parts:

$$ u = x, \quad dv = \sin nx \, dx $$

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

So,

$$ b_n = \frac{2}{\pi} [ -\frac{x \cos nx}{n} \Big|_0^{\pi} + \frac{1}{n} \int_0^{\pi} \cos nx \, dx ] $$

Evaluate boundary term:

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

Integral term:

$$ \int_0^{\pi} \cos nx \, dx = \frac{\sin n\pi}{n} = 0 $$

So,

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

Answer:

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

Example 2: Find the Fourier series of \(f(x) = x^2\) over \([- \pi, \pi]\).

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

Calculate \(a_0\):

$$ a_0 = \frac{1}{\pi} \int_{-\pi}^{\pi} x^2 \, dx = \frac{2}{\pi} \int_0^{\pi} x^2 \, dx = \frac{2}{\pi} [ \frac{x^3}{3} ]_0^{\pi} = \frac{2}{\pi} \times \frac{\pi^3}{3} = \frac{2 \pi^2}{3} $$

Calculate \(a_n\):

$$ a_n = \frac{1}{\pi} \int_{-\pi}^{\pi} x^2 \cos nx \, dx = \frac{2}{\pi} \int_0^{\pi} x^2 \cos nx \, dx $$

Use integration by parts twice:

First integration:

• \(u = x^2\), \(du = 2x dx\)
• \(dv = \cos nx dx\), \(v = \frac{\sin nx}{n}\)

$$ \int x^2 \cos nx \, dx = \frac{x^2 \sin nx}{n} - \frac{2}{n} \int x \sin nx \, dx $$

Second integration for \(\int x \sin nx \, dx\):

• \(u = x\), \(du = dx\)
• \(dv = \sin nx dx\), \(v = -\frac{\cos nx}{n}\)

$$ \int x \sin nx \, dx = - \frac{x \cos nx}{n} + \frac{1}{n} \int \cos nx \, dx = - \frac{x \cos nx}{n} + \frac{1}{n^2} \sin nx + C $$

Apply definite integration:

$$ a_n = \frac{2}{\pi} [ \frac{x^2 \sin nx}{n} + \frac{2x \cos nx}{n^2} - \frac{2 \sin nx}{n^3} ]_0^{\pi} $$

Evaluate terms at \(x = \pi\) and \(x=0\):

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

$$ a_n = \frac{2}{\pi} ( \frac{2 \pi (-1)^n}{n^2} - 0 ) = \frac{4 (-1)^n}{n^2} $$

Calculate \(b_n\):

Since \(x^2\) is even and \(\sin nx\) is odd,

$$ b_n = 0 $$

Answer:

$$ f(x) = \frac{\pi^2}{3} + \sum_{n=1}^\infty \frac{4 (-1)^n}{n^2} \cos nx $$

Example 3: Find 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 \(2\pi\).

Given: \(f(x)\) as above.

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 \(a_n\):

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

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

Calculate each integral:

$$ \int_{-\pi}^0 \cos nx \, dx = [ \frac{\sin nx}{n} ]_{-\pi}^0 = \frac{\sin 0}{n} - \frac{\sin(-n\pi)}{n} = 0 - 0 = 0 $$

$$ \int_0^{\pi} \cos nx \, dx = [ \frac{\sin nx}{n} ]_0^{\pi} = \frac{\sin n\pi}{n} - 0 = 0 $$

$$ a_n = 0 $$

Calculate \(b_n\):

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

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

Calculate each integral:

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

$$ \int_0^{\pi} \sin nx \, dx = [ -\frac{\cos nx}{n} ]_0^{\pi} = -\frac{\cos n\pi}{n} + \frac{\cos 0}{n} = -\frac{(-1)^n}{n} + \frac{1}{n} = \frac{1 - (-1)^n}{n} $$

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

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

Answer:

$$ f(x) = \sum_{\substack{n=1 \\ n \text{ odd}}}^\infty \frac{4}{\pi n} \sin nx $$

1.1.2 Fourier series for a function with arbitrary period \(T\)

For a function \(f(x)\) with period \(T\), the Fourier series is expressed as:

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

where coefficients are

$$ a_0 = \frac{2}{T} \int_{c}^{c+T} f(x) \, dx $$

$$ a_n = \frac{2}{T} \int_{c}^{c+T} f(x) \cos \frac{2 n \pi x}{T} \, dx $$

$$ b_n = \frac{2}{T} \int_{c}^{c+T} f(x) \sin \frac{2 n \pi x}{T} \, dx $$

for any real number \(c\).

Worked Examples

Example 1: Find the Fourier series of \(f(x) = x\) with period \(T=4\), defined on \([-2,2]\).

Given: \(f(x) = x\), \(T=4\), interval \([-2,2]\)

Calculate \(a_0\):

$$ a_0 = \frac{2}{4} \int_{-2}^{2} x \, dx = \frac{1}{2} [ \frac{x^2}{2} ]_{-2}^2 = \frac{1}{2} ( 2 - 2 ) = 0 $$

Calculate \(a_n\):

$$ a_n = \frac{2}{4} \int_{-2}^{2} x \cos \frac{n \pi x}{2} \, dx = \frac{1}{2} \int_{-2}^{2} x \cos \frac{n \pi x}{2} \, dx $$

Since \(x \cos \alpha x\) is an odd function over symmetric limits,

$$ a_n = 0 $$

Calculate \(b_n\):

$$ b_n = \frac{1}{2} \int_{-2}^{2} x \sin \frac{n \pi x}{2} \, dx = \frac{1}{2} \times 2 \int_0^{2} x \sin \frac{n \pi x}{2} \, dx = \int_0^{2} x \sin \frac{n \pi x}{2} \, dx $$

Use integration by parts:

• \(u = x\), \(du = dx\)
• \(dv = \sin \frac{n \pi x}{2} dx\), \(v = - \frac{2}{n \pi} \cos \frac{n \pi x}{2}\)

So,

$$ b_n = [ -\frac{2 x}{n \pi} \cos \frac{n \pi x}{2} ]_0^{2} + \frac{2}{n \pi} \int_0^{2} \cos \frac{n \pi x}{2} \, dx $$

Evaluate boundary term:

$$ -\frac{4}{n \pi} \cos n \pi + 0 = -\frac{4(-1)^n}{n \pi} $$

Calculate integral:

$$ \int_0^{2} \cos \frac{n \pi x}{2} \, dx = [ \frac{2}{n \pi} \sin \frac{n \pi x}{2} ]_0^{2} = \frac{2}{n \pi} ( \sin n \pi - 0 ) = 0 $$

$$ b_n = -\frac{4 (-1)^n}{n \pi} $$

Answer:

$$ f(x) = \sum_{n=1}^\infty -\frac{4 (-1)^n}{n \pi} \sin \frac{n \pi x}{2} $$

Example 2: Determine Fourier series of \(f(x) = x^2\), period \(T=6\), defined on \([-3,3]\).

Given: \(f(x) = x^2\), \(T=6\)

Calculate \(a_0\):

$$ a_0 = \frac{2}{6} \int_{-3}^{3} x^2 \, dx = \frac{1}{3} \times 2 \int_0^3 x^2 \, dx = \frac{2}{3} [ \frac{x^3}{3} ]_0^3 = \frac{2}{3} \times \frac{27}{3} = 6 $$

Calculate \(a_n\):

$$ a_n = \frac{2}{6} \int_{-3}^{3} x^2 \cos \frac{2 n \pi x}{6} \, dx = \frac{1}{3} \times 2 \int_0^3 x^2 \cos \frac{n \pi x}{3} \, dx = \frac{2}{3} \int_0^3 x^2 \cos \frac{n \pi x}{3} \, dx $$

Use integration by parts twice (similar to previous example):

Final result after evaluation is:

$$ a_n = \frac{12 (-1)^n}{n^2 \pi^2} $$

Calculate \(b_n\):

Since \(x^2\) is even and sine is odd,

$$ b_n = 0 $$

Answer:

$$ f(x) = 3 + \sum_{n=1}^\infty \frac{12 (-1)^n}{n^2 \pi^2} \cos \frac{n \pi x}{3} $$

Example 3: Fourier series for a sawtooth wave with period \(T=8\), defined on \([-4,4]\) by \(f(x) = x\).

Given: \(f(x) = x\), \(T=8\)

Calculate \(a_0\):

$$ a_0 = \frac{2}{8} \int_{-4}^{4} x \, dx = \frac{1}{4} \times 0 = 0 $$

Calculate \(a_n\):

$$ a_n = \frac{2}{8} \int_{-4}^{4} x \cos \frac{n \pi x}{4} \, dx = \frac{1}{4} \times 0 = 0 $$

Calculate \(b_n\):

$$ b_n = \frac{1}{4} \times 2 \int_0^{4} x \sin \frac{n \pi x}{4} \, dx = \frac{1}{2} \int_0^{4} x \sin \frac{n \pi x}{4} \, dx $$

Integration by parts yields:

$$ b_n = -\frac{8 (-1)^n}{n \pi} $$

Answer:

$$ f(x) = \sum_{n=1}^\infty -\frac{8 (-1)^n}{n \pi} \sin \frac{n \pi x}{4} $$

Chapter Summary

This chapter introduced the concept of the Fourier series as a method for representing functions as sums of sine and cosine terms. It began by defining the Fourier series for functions with period \(2x\) and generalized this to any period \(T\), providing the framework for expressing periodic functions in series form. The chapter then extended the discussion to non-periodic functions, explaining how their Fourier series can be determined by considering them over specific intervals. Techniques for finding Fourier series of non-periodic functions within a given range were presented, emphasizing the importance of interval selection in the analysis. The chapter concluded with methods for simplifying the Fourier series calculation for odd and even functions, including the development of half-range series that utilize only sine or cosine terms. These approaches allow for more efficient computation and better understanding of function behavior based on symmetry. Overall, the chapter equipped students with the tools to analyze and express a wide variety of functions using Fourier series in engineering contexts.