By the end of this chapter, you will be able to:
Mastering these skills will help you solve complex engineering problems with confidence and precision, making you a valuable asset in your trade.
Analysis of periodic signals is essential for industrial control and installation professionals. Many physical signals in automation, such as alternating currents, sensor outputs, and rotating machine vibrations, are periodic and must be analyzed to optimize control systems. Fourier series provides a powerful mathematical tool for breaking down these periodic signals into sums of simple sine and cosine components, making complex waveforms easier to understand and manipulate.
Fourier series expresses any periodic function \(f(x)\) with period \(2\pi\) as a sum of sines and cosines. This is crucial for analyzing signals in PLC (Programmable Logic Controller) systems, AC motor drives, and sensor signal processing.
A function \(f(x)\) is periodic with period \(T\) if \(f(x + T) = f(x)\) for all \(x\). In engineering, signals like voltages and currents are often periodic. For analysis, we commonly use \(T = 2\pi\).
The basic Fourier series for a function of period \(2\pi\) is:
$$ f(x) = a_0 + \sum_{n=1}^{\infty} [a_n \cos(nx) + b_n \sin(nx)] $$
where the coefficients are:
$$ a_0 = \frac{1}{2\pi} \int_{-\pi}^{\pi} f(x)\,dx $$
$$ a_n = \frac{1}{\pi} \int_{-\pi}^{\pi} f(x) \cos(nx)\,dx $$
$$ b_n = \frac{1}{\pi} \int_{-\pi}^{\pi} f(x) \sin(nx)\,dx $$
Example 1:
Find the period of \( f(x) = \sin(2x) \).
Given: \( f(x) = \sin(2x) \)
To find the period, set \( f(x + T) = f(x) \):
\( \sin(2(x + T)) = \sin(2x) \)
So, \( 2(x + T) = 2x + 2T \). The sine function repeats every \(2\pi\):
Set \( 2T = 2\pi \Rightarrow T = \pi \)
Answer: \( T = \pi \)
Example 2:
Check if the function \( f(x) = 3\cos(x) + 2\sin(3x) \) is periodic with period \(2\pi\).
Given: \( f(x) = 3\cos(x) + 2\sin(3x) \)
\( \cos(x) \) has period \(2\pi\), \( \sin(3x) \) has period \( \frac{2\pi}{3} \).
The overall period is the least common multiple (LCM):
LCM of \(2\pi\) and \( \frac{2\pi}{3} \) is \( 2\pi \).
Answer: The function has period \(2\pi\).
Example 3:
Given \( f(x) \) is periodic with period \(2\pi\), show that \( f(x + 2\pi) = f(x) \).
Given: \( f(x) \) periodic, period \(2\pi\).
By definition: \( f(x + 2\pi) = f(x) \).
Answer: Verified by definition.
Example 4:
If \( f(x) = x^2 \) for \( -\pi < x < \pi \), does \( f(x) \) repeat every \(2\pi\)?
Given: \( f(x) = x^2 \)
Check: \( f(x + 2\pi) = (x + 2\pi)^2 = x^2 + 4\pi x + 4\pi^2 \neq x^2 \)
Answer: Not periodic with period \(2\pi\).
Example 5:
A PLC output waveform repeats every \(0.02\) seconds. What is its period in terms of \(\pi\) if the waveform is represented as a function of angle \(x\) where one cycle corresponds to \(2\pi\)?
Given: \( T = 0.02 \) s, one cycle \(= 2\pi \).
Answer: The period is \(2\pi\) in angular terms.
To find the Fourier series, compute the coefficients:
$$ a_0 = \frac{1}{2\pi} \int_{-\pi}^{\pi} f(x)\,dx $$
$$ a_n = \frac{1}{\pi} \int_{-\pi}^{\pi} f(x) \cos(nx)\,dx $$
$$ b_n = \frac{1}{\pi} \int_{-\pi}^{\pi} f(x) \sin(nx)\,dx $$
Example 1:
Find the Fourier coefficients for \( f(x) = x \) on \( -\pi < x < \pi \).
Given: \( f(x) = x \)
Calculate \( a_0 \):
$$ a_0 = \frac{1}{2\pi} \int_{-\pi}^{\pi} x\,dx $$$$ = \frac{1}{2\pi} [ \frac{1}{2} x^2 ]_{-\pi}^{\pi} $$$$ = \frac{1}{2\pi} ( \frac{1}{2} \pi^2 - \frac{1}{2} \pi^2 ) = 0 $$
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, the integral is zero:
\( a_n = 0 \)
Calculate \( b_n \):
$$ b_n = \frac{1}{\pi} \int_{-\pi}^{\pi} x \sin(nx)\,dx $$
Let’s solve for general \( n \):
$$ b_n = \frac{1}{\pi} [ -\frac{x \cos(nx)}{n} \Big|_{-\pi}^{\pi} + \int_{-\pi}^{\pi} \frac{\cos(nx)}{n} dx ] $$
But \( \int_{-\pi}^{\pi} \cos(nx) dx = 0 \), and \( x \cos(nx) \) at bounds evaluates to zero (since \( \cos(n\pi) = (-1)^n \)), so:
$$ b_n = \frac{2}{n} (-1)^{n+1} $$
Answer: \( a_0 = 0 \), \( a_n = 0 \), \( b_n = \frac{2}{n} (-1)^{n+1} \)
Example 2:
Find the Fourier coefficients for the square wave:
\( f(x) = 1 \) for \( 0 < x < \pi \), \( f(x) = -1 \) for \( -\pi < x < 0 \).
Given: \( f(x) \) as above.
Calculate \( a_0 \):
$$ a_0 = \frac{1}{2\pi} ( \int_{-\pi}^{0} (-1) dx + \int_{0}^{\pi} 1 dx ) $$$$ = \frac{1}{2\pi} ( [-x]_{-\pi}^{0} + [x]_{0}^{\pi} ) $$$$ = \frac{1}{2\pi} (\pi + \pi) = \frac{1}{2\pi} (2\pi) = 1 $$
Calculate \( a_n \):
$$ a_n = \frac{1}{\pi} ( \int_{-\pi}^{0} (-1) \cos(nx) dx + \int_{0}^{\pi} 1 \cos(nx) dx ) $$$$ = \frac{1}{\pi} ( -\int_{-\pi}^{0} \cos(nx) dx + \int_{0}^{\pi} \cos(nx) dx ) $$
But \( \int_{-a}^{a} \cos(nx) dx = 0 \), so \( a_n = 0 \).
Calculate \( b_n \):
$$ b_n = \frac{1}{\pi} ( \int_{-\pi}^{0} (-1) \sin(nx) dx + \int_{0}^{\pi} \sin(nx) dx ) $$$$ = \frac{1}{\pi} ( -\int_{-\pi}^{0} \sin(nx) dx + \int_{0}^{\pi} \sin(nx) dx ) $$
\( \int \sin(nx) dx = -\frac{1}{n} \cos(nx) \)
Evaluate:
For \( 0 < x < \pi \):
$$ \int_{0}^{\pi} \sin(nx) dx = [-\frac{1}{n} \cos(nx)]_0^{\pi} = -\frac{1}{n} [\cos(n\pi) - 1] $$
For \( -\pi < x < 0 \):
$$ \int_{-\pi}^{0} \sin(nx) dx = [-\frac{1}{n} \cos(nx)]_{-\pi}^{0} = -\frac{1}{n} [1 - \cos(-n\pi)] $$ But \( \cos(-n\pi) = \cos(n\pi) \).
So total is:
$$ b_n = \frac{1}{\pi} ( -(-\frac{1}{n} [1 - \cos(n\pi)]) + (-\frac{1}{n} [\cos(n\pi) - 1]) ) $$$$ = \frac{1}{\pi} ( \frac{1}{n} [1 - \cos(n\pi)] - \frac{1}{n} [\cos(n\pi) - 1] ) $$$$ = \frac{1}{\pi} \frac{2}{n} [1 - \cos(n\pi)] $$
But \( 1 - \cos(n\pi) = 2 \) for odd \( n \), \( 0 \) for even \( n \), so:
For odd \( n \):
$$ b_n = \frac{4}{n\pi} $$
For even \( n \):
$$ b_n = 0 $$
Answer: \( a_0 = 0 \), \( a_n = 0 \), \( b_n = \frac{4}{n\pi} \) for odd \( n \)
Example 3:
Find \( a_0 \) for \( f(x) = \cos(x) \) on \( -\pi < x < \pi \).
Given: \( f(x) = \cos(x) \)
$$ a_0 = \frac{1}{2\pi} \int_{-\pi}^{\pi} \cos(x) dx $$
But \( \int_{-\pi}^{\pi} \cos(x) dx = 0 \):
Answer: \( a_0 = 0 \)
Example 4:
Find \( a_1 \) for \( f(x) = 2\cos(x) \) on \( -\pi < x < \pi \).
Given: \( f(x) = 2\cos(x) \)
$$ a_1 = \frac{1}{\pi} \int_{-\pi}^{\pi} 2\cos(x) \cos(x) dx $$$$ = \frac{2}{\pi} \int_{-\pi}^{\pi} \cos^2(x) dx $$
Recall \( \cos^2(x) = \frac{1 + \cos(2x)}{2} \):
$$ = \frac{2}{\pi} \int_{-\pi}^{\pi} \frac{1 + \cos(2x)}{2} dx $$$$ = \frac{1}{\pi} ( \int_{-\pi}^{\pi} dx + \int_{-\pi}^{\pi} \cos(2x) dx ) $$$$ = \frac{1}{\pi} (2\pi + 0) = 2 $$
Answer: \( a_1 = 2 \)
Example 5:
Find \( b_2 \) for \( f(x) = x \) on \( -\pi < x < \pi \).
Given: \( f(x) = x \)
From earlier, \( b_n = \frac{2}{n} (-1)^{n+1} \):
For \( n = 2 \):
$$ b_2 = \frac{2}{2} (-1)^{3} = 1 \times (-1) = -1 $$
Answer: \( b_2 = -1 \)
Once the coefficients \( a_0, a_n, b_n \) are known, substitute into the Fourier series:
$$ f(x) = a_0 + \sum_{n=1}^{\infty} [a_n \cos(nx) + b_n \sin(nx)] $$
Example 1:
Construct the Fourier series for \( f(x) = x \) on \( -\pi < x < \pi \).
From earlier:
So,
$$ f(x) = \sum_{n=1}^{\infty} \frac{2}{n} (-1)^{n+1} \sin(nx) $$
Write first three terms:
For \( n=1 \): \( b_1 = 2 \)
For \( n=2 \): \( b_2 = -1 \)
For \( n=3 \): \( b_3 = \frac{2}{3} \)
So,
$$ f(x) \approx 2\sin(x) - \sin(2x) + \frac{2}{3} \sin(3x) $$
Answer: Fourier series is \( f(x) = 2\sin(x) - \sin(2x) + \frac{2}{3} \sin(3x) + \cdots \)
Example 2:
Construct the Fourier series for the square wave:
\( f(x) = 1 \) for \( 0 < x < \pi \), \( f(x) = -1 \) for \( -\pi < x < 0 \).
From earlier: Only \( b_n \) nonzero for odd \( n \):
So,
$$ f(x) = \frac{4}{\pi} ( \sin(x) + \frac{1}{3} \sin(3x) + \frac{1}{5} \sin(5x) + \cdots ) $$
Answer: Fourier series is \( f(x) = \frac{4}{\pi} ( \sin(x) + \frac{1}{3} \sin(3x) + \frac{1}{5} \sin(5x) + \cdots ) \)
Example 3:
Construct the Fourier series for \( f(x) = \cos(x) \) on \( -\pi < x < \pi \).
Only \( a_1 \) is nonzero, \( a_1 = 2 \):
So,
$$ f(x) = 2\cos(x) $$
Answer: Fourier series is \( f(x) = 2\cos(x) \)
Example 4:
A temperature sensor in a bakery outputs a periodic sawtooth signal \( f(x) = x \) for \( -\pi < x < \pi \). Write the Fourier series up to \( n = 3 \).
From earlier: \( f(x) \approx 2\sin(x) - \sin(2x) + \frac{2}{3} \sin(3x) \)
Answer: \( f(x) \approx 2\sin(x) - \sin(2x) + \frac{2}{3} \sin(3x) \)
Example 5:
Construct the Fourier series for \( f(x) = |\sin(x)| \) on \( -\pi < x < \pi \).
First, \( f(x) \) is even, only cosine terms appear. This is more advanced; for brevity, state that the solution involves only cosine terms.
Answer: Fourier series contains only cosine terms with coefficients determined by integration.
If a function has period \(T\) (not \(2\pi\)), change variables to fit the standard form. Let \( x = \frac{2\pi}{T} t \).
For a function \( f(t) \) of period \(T\):
$$ f(t) = a_0 + \sum_{n=1}^{\infty} [a_n \cos(\frac{2\pi n t}{T}) + b_n \sin(\frac{2\pi n t}{T})] $$
Where:
$$ a_0 = \frac{1}{T} \int_{t_0}^{t_0 + T} f(t) dt $$
$$ a_n = \frac{2}{T} \int_{t_0}^{t_0 + T} f(t) \cos(\frac{2\pi n t}{T}) dt $$
$$ b_n = \frac{2}{T} \int_{t_0}^{t_0 + T} f(t) \sin(\frac{2\pi n t}{T}) dt $$
Example 1:
A sensor voltage repeats every \( T = 4 \) seconds and is given by \( f(t) = t \) for \( 0 < t < 4 \). Find \( a_0 \).
Given: \( f(t) = t \), \( T = 4 \)
$$ a_0 = \frac{1}{4} \int_{0}^{4} t dt $$$$ = \frac{1}{4} [ \frac{1}{2} t^2 ]_0^4 $$$$ = \frac{1}{4} ( \frac{1}{2} \times 16 ) = \frac{1}{4} \times 8 = 2 $$
Answer: \( a_0 = 2 \)
Example 2:
For the same function, find \( a_1 \):
$$ a_1 = \frac{2}{4} \int_{0}^{4} t \cos(\frac{2\pi t}{4}) dt $$$$ = \frac{1}{2} \int_{0}^{4} t \cos(\frac{\pi t}{2}) dt $$
This integral can be solved by integration by parts:
Let \( u = t \), \( dv = \cos(\frac{\pi t}{2}) dt \)
\( du = dt \), \( v = \frac{2}{\pi} \sin(\frac{\pi t}{2}) \)
So,
$$ \int t \cos(\frac{\pi t}{2}) dt = t \frac{2}{\pi} \sin(\frac{\pi t}{2}) - \int \frac{2}{\pi} \sin(\frac{\pi t}{2}) dt $$
\( \int \sin(\frac{\pi t}{2}) dt = -\frac{2}{\pi} \cos(\frac{\pi t}{2}) \)
So,
$$ = t \frac{2}{\pi} \sin(\frac{\pi t}{2}) + \frac{4}{\pi^2} \cos(\frac{\pi t}{2}) $$
Evaluate from \( t = 0 \) to \( t = 4 \):
At \( t = 4 \):
\( \sin(2\pi) = 0 \), \( \cos(2\pi) = 1 \)
At \( t = 0 \):
\( \sin(0) = 0 \), \( \cos(0) = 1 \)
So,
$$ \int_{0}^{4} t \cos(\frac{\pi t}{2}) dt = [0 + \frac{4}{\pi^2} \times 1] - [0 + \frac{4}{\pi^2} \times 1] = 0 $$
So \( a_1 = \frac{1}{2} \times 0 = 0 \)
Answer: \( a_1 = 0 \)
Example 3:
Given \( f(t) = 3 \) for \( 0 < t < 4 \), period \( T = 4 \), find \( a_0 \).
$$ a_0 = \frac{1}{4} \int_{0}^{4} 3 dt = \frac{1}{4} [3t]_{0}^{4} = \frac{1}{4} \times 12 = 3 $$
Answer: \( a_0 = 3 \)
Example 4:
A periodic current in a welding machine is \( f(t) = 2\sin(\frac{\pi t}{2}) \), \( T = 4 \) s. Find \( b_1 \).
$$ b_1 = \frac{2}{4} \int_{0}^{4} 2\sin(\frac{\pi t}{2}) \sin(\frac{\pi t}{2}) dt $$$$ = \frac{1}{2} \int_{0}^{4} 2\sin^2(\frac{\pi t}{2}) dt $$$$ = \int_{0}^{4} \sin^2(\frac{\pi t}{2}) dt $$
Recall \( \sin^2(\theta) = \frac{1 - \cos(2\theta)}{2} \):
$$ = \int_{0}^{4} \frac{1 - \cos(\pi t)}{2} dt $$$$ = \frac{1}{2} \int_{0}^{4} 1 dt - \frac{1}{2} \int_{0}^{4} \cos(\pi t) dt $$$$ = \frac{1}{2} [t]_{0}^{4} - \frac{1}{2} [ \frac{1}{\pi} \sin(\pi t) ]_{0}^{4} $$$$ = \frac{1}{2} (4 - 0) - \frac{1}{2} ( \frac{1}{\pi} [\sin(4\pi) - \sin(0)] ) $$ \( \sin(4\pi) = 0, \sin(0) = 0 \)
$$ = 2 - 0 = 2 $$
Answer: \( b_1 = 2 \)
Example 5:
A periodic square wave for a conveyor belt control signal has period \( T = 6 \) s, defined by \( f(t) = 1 \) for \( 0 < t < 3 \), \( f(t) = -1 \) for \( 3 < t < 6 \). Find \( a_0 \).
$$ a_0 = \frac{1}{6} \int_{0}^{6} f(t) dt = \frac{1}{6} ( \int_{0}^{3} 1 dt + \int_{3}^{6} -1 dt ) $$$$ = \frac{1}{6} (3 - 3) = 0 $$
Answer: \( a_0 = 0 \)
Create a free account to open more of this chapter.
Free: practical guides, quick cards, workplace scenarios and more.
Create a free accountThis chapter explored the application of Fourier series in various engineering mathematics contexts. Starting with the determination of Fourier series for periodic functions with period 2, the principles were extended to functions of arbitrary period \(T\), demonstrating how to express such functions as sums of sines and cosines. The chapter then addressed the calculation of Fourier series for non-periodic functions defined over specific intervals, highlighting the process of representing these functions within a given range. Techniques for finding Fourier series of even and odd functions were discussed, including the derivation of half-range series when a function is only defined over part of its domain. The final section focused on constructing Fourier series over any range, emphasizing adaptability in handling diverse boundary conditions and intervals. Throughout, practical examples illustrated each method, progressing from straightforward calculations to more complex applications relevant to civil engineering scenarios. By mastering these techniques, students are equipped to analyze and model a wide variety of engineering phenomena using Fourier series expansions.
Given: \( f(x) = x \), period \( 2 \), \( -1 \leq x \leq 1 \)
Step 1: Write the general Fourier series for period \( 2 \):
$$ f(x) = \frac{a_0}{2} + \sum_{n=1}^{\infty} [ a_n \cos(n\pi x) + b_n \sin(n\pi x) ] $$
Step 2: Calculate \( a_0 \):
$$ a_0 = \int_{-1}^{1} x \, dx = [ \frac{x^2}{2} ]_{-1}^{1} = \frac{1}{2} - \frac{1}{2} = 0 $$
Step 3: Calculate \( a_n \):
$$ a_n = \int_{-1}^{1} x \cos(n\pi x) \, dx $$
This is an odd function times an even function (odd), so integral over symmetric interval is 0.
\( a_n = 0 \) for all \( n \).
Step 4: Calculate \( b_n \):
$$ b_n = \int_{-1}^{1} x \sin(n\pi x) \, dx $$
Here, \( x \sin(n\pi x) \) is even (odd × odd). So,
$$ b_n = 2 \int_{0}^{1} x \sin(n\pi x) \, dx $$
Integrate by parts:
Let \( u = x \), \( dv = \sin(n\pi x) dx \), \( du = dx \), \( v = -\frac{1}{n\pi} \cos(n\pi x) \)
So,
$$ b_n = 2 [ -\frac{x}{n\pi} \cos(n\pi x) \Big|_0^1 + \frac{1}{n\pi} \int_0^1 \cos(n\pi x) dx ] $$
Compute each term:
So,
$$ b_n = 2 ( -\frac{1}{n\pi} \cos(n\pi) ) = -\frac{2}{n\pi} \cos(n\pi) $$
But \( \cos(n\pi) = (-1)^n \), so
$$ b_n = -\frac{2}{n\pi} (-1)^n = \frac{2}{n\pi} (-1)^{n+1} $$
Final Answer:
$$ \boxed{f(x) = \sum_{n=1}^{\infty} \frac{2}{n\pi} (-1)^{n+1} \sin(n\pi x)} $$
Given: \( f(x) = x^2 \), period \( 2\pi \), \( -\pi < x < \pi \)
Fourier series:
$$ f(x) = \frac{a_0}{2} + \sum_{n=1}^2 a_n \cos(nx) $$
Since \( x^2 \) is even, all \( b_n = 0 \).
Step 1: Calculate \( a_0 \):
$$ a_0 = \frac{1}{\pi} \int_{-\pi}^{\pi} x^2 dx = \frac{1}{\pi} [ \frac{x^3}{3} ]_{-\pi}^{\pi} = \frac{1}{\pi} ( \frac{\pi^3}{3} - (-\frac{\pi^3}{3}) ) = \frac{2\pi^2}{3} $$
Step 2: Calculate \( a_n \):
$$ a_n = \frac{1}{\pi} \int_{-\pi}^{\pi} x^2 \cos(nx) dx $$
Since \( x^2 \cos(nx) \) is even,
$$ a_n = \frac{2}{\pi} \int_{0}^{\pi} x^2 \cos(nx) dx $$
Integrate by parts twice (standard result):
$$ \int_0^{\pi} x^2 \cos(nx) dx = \frac{2\pi}{n^2} \cos(n\pi) + \frac{2}{n^3} \sin(n\pi) $$
But \( \sin(n\pi) = 0 \) and \( \cos(n\pi) = (-1)^n \), so
$$ a_n = \frac{2}{\pi} \cdot \frac{2\pi}{n^2} (-1)^n = \frac{4}{n^2} (-1)^n $$
Step 3: Write the series up to \( n = 2 \):
Final Answer:
$$ \boxed{f(x) \approx \frac{\pi^2}{3} - 4\cos(x) + \cos(2x)} $$
A PLC control system at a Nairobi food processing plant produces a periodic voltage waveform \(f(t)\) over one cycle as follows: \(f(t) = 3\) for \(0 \le t < 1\), \(f(t) = -3\) for \(1 \le t < 2\), and the period is 2. Determine the Fourier series representation for \(f(t)\) up to the first two non-zero harmonics. (4 marks)
Find the Fourier series of the function \(f(x) = x\) defined on the interval \(-1 \le x \le 1\), and period 2. Show all steps. (4 marks)
At the start of this chapter we promised you would be able to:
Tick each one you can genuinely do.
So, are you there yet?
You're competent when you can confidently do 50% or more of what this chapter promised.
Sign in to record how you're doing.