Fourier
Series
Part 1
Intended Learning Outcomes
1. Define the concepts and properties Fourier
series.
2. Apply the formula of Fourier series to express
elementary function to periodic functions
Introduction
Fourier series are used in the analysis of periodic
functions.
Many of the phenomena studied in engineering and
science are periodic in nature e.g., the current and
voltage in an alternating current circuit. These
periodic functions can be analyzed into their
constituent components (fundamentals and
harmonics) by a process called Fourier analysis.
Introduction
We are aiming to find an approximation using
trigonometric functions for various square, saw
tooth, etc. waveforms that occur in electrical
engineering problems. We do this by adding more
and more trigonometric functions together. The sum
of these special trigonometric functions is called the
Fourier Series.
Jean Fourier
Fourier was a French mathematician, who
was taught by Lagrange and Laplace.
He almost died on the guillotine in the
French Revolution. Fourier was a buddy of
Napoleon and worked as scientific adviser
for Napoleon's army.
He worked on theories of heat and expansions of
functions as trigonometric series... but these were
controversial at the time. Like many scientists, he had to
battle to get his ideas accepted.
What are Periodic Functions?
A function 𝑓 is said periodic if there exists a
smallest positive number 𝑇 such that:
𝑓 𝑥 + 𝑇 = 𝑓(𝑥)
for all x in the domain.
Fourier Series
The Fourier Series is an infinite series expansion
involving Trigonometric Functions.
A periodic waveform 𝑓(𝑡) of period 𝑝 = 2𝐿 has
a Fourier Series given by:
∞
𝑓 𝑡 = 𝑎0 +
𝑛=1
𝑛𝜋𝑡
𝑎𝑛 cos
+
𝐿
∞
𝑛=1
𝑛𝜋𝑡
𝑏𝑛 sin
𝐿
Where 𝑎𝑛 and 𝑏𝑛 are the Fourier coefficients and
𝑎0
is the mean value.
2
Fourier Series
The Fourier Series is an infinite series expansion involving
Trigonometric Functions.
It can be proved that 𝑓(𝑡) can be expressed as the sum of an
infinite number of sine and/or cosine functions.
This periodic sum is known as a Fourier Series.
A periodic waveform 𝑓(𝑡) of period 𝑇 = 2𝐿 has a Fourier
Series given by:
∞
𝑓 𝑡 = 𝑎0 +
𝑛=1
𝑛𝜋𝑡
𝑎𝑛 cos
+
𝐿
∞
𝑛=1
𝑛𝜋𝑡
𝑏𝑛 sin
𝐿
𝑎0
Where 𝑎0, 𝑎𝑛 and 𝑏𝑛 are the Fourier coefficients and
is the
2
mean value.
Fourier Series
∞
𝑓 𝑡 = 𝑎0 +
𝑛=1
𝑛𝜋𝑡
𝑎𝑛 cos
+
𝐿
∞
𝑛=1
𝑛𝜋𝑡
𝑏𝑛 sin
𝐿
𝑎0
Where 𝑎𝑛 and 𝑏𝑛 are the Fourier coefficients and
is the
2
mean value. They can be obtained by:
1
𝑎0 =
2𝐿
1
𝑎n =
𝐿
1
𝑏n =
𝐿
for 𝑛 = 1, 2, 3, …
𝐿
−𝐿
𝐿
−𝐿
𝐿
𝑓 𝑡 𝑑𝑡
−𝐿
𝑛𝜋𝑡
𝑓 𝑡 cos
𝑑𝑡
𝐿
𝑛𝜋𝑡
𝑓 𝑡 sin
𝑑𝑡
𝐿
Fourier Series
∞
𝑓 𝑡 = 𝑎0 +
𝑛=1
𝑛𝜋𝑡
𝑎𝑛 cos
+
𝐿
∞
𝑛=1
𝑛𝜋𝑡
𝑏𝑛 sin
𝐿
Dirichlet Conditions
Any periodic waveform of period 𝑝 = 2𝐿, can be expressed in
a Fourier series provided that
a. it has a finite number of discontinuities within the period
2𝐿;
b. it has a finite average value in the period 2𝐿;
c. it has a finite number of positive and negative maxima and
minima.
When these conditions, called the Dirichlet conditions, are
satisfied, the Fourier series for the function 𝑓(𝑡) exists.
Example:
Find the function 𝑓(𝑡) define by the square waveform
shown below.
f(t)
1
t
-2π
−π
π
-1
2π
3π
Solution:
∞
𝑓 𝑡 = 𝑎0 +
𝑛=1
𝑛𝜋𝑡
𝑎𝑛 cos
+
𝐿
𝑎0 =
1
𝑎n =
𝐿
1
𝑏n =
𝐿
for period 𝑝
1
2𝐿
𝐿
𝑏𝑛 sin
𝑛=1
𝑓 𝑡 𝑑𝑡
−𝐿
𝐿
𝑓 𝑡 cos
𝑛𝜋𝑡
𝑑𝑡
𝐿
𝑓 𝑡 sin
𝑛𝜋𝑡
𝑑𝑡
𝐿
−𝐿
𝐿
−𝐿
∞
= 2𝐿
then
2𝐿 = 2𝜋
𝐿=𝜋
𝑛𝜋𝑡
𝐿
Solution:
Solving for 𝑎0
1
𝑎0 =
2𝐿
where
𝐿
𝑓 𝑡 𝑑𝑡
−𝐿
2𝐿 = 2𝜋
𝐿=𝜋
1
𝑎0 =
2𝜋
0
𝜋
−1 𝑑𝑡 +
−𝜋
1 𝑑𝑡
0
𝑎0 = 0
Solution:
∞
𝑓 𝑡 = 𝑎0 +
𝑛=1
𝑛𝜋𝑡
𝑎𝑛 cos
+
𝐿
𝑎0 =
𝑎n =
𝑏n =
1
𝐿
1
𝐿
1
2𝐿
𝐿
∞
𝑏𝑛 sin
𝑛=1
𝑓 𝑡 𝑑𝑡
−𝐿
𝐿
𝑓 𝑡 cos
𝑛𝜋𝑡
𝑑𝑡
𝐿
𝑓 𝑡 sin
𝑛𝜋𝑡
𝑑𝑡
𝐿
−𝐿
𝐿
−𝐿
Note:
 For modeling odd functions, use the sine terms
 For modeling even functions, use the cosine terms
𝑛𝜋𝑡
𝐿
Fourier Series of Even and Odd Functions
In some of the problems that we encounter, the Fourier
coefficients 𝑎0, 𝑎𝑛 and 𝑏𝑛 become zero after integration.
Finding zero coefficients in such problems is time
consuming and can be avoided. With knowledge of even
and odd functions, a zero coefficient may be predicted
without performing the integration.
∞
𝑓 𝑡 = 𝑎0 +
𝑛=1
𝑛𝜋𝑡
𝑎𝑛 cos
+
𝐿
∞
𝑛=1
𝑛𝜋𝑡
𝑏𝑛 sin
𝐿
Solution
∞
𝑓 𝑡 = 𝑎0 +
𝑛=1
𝑛𝜋𝑡
𝑎𝑛 cos
+
𝐿
𝑎0 =
1
𝑎n =
𝐿
𝑏n =
1
𝐿
𝐿
−𝐿
1
2𝐿
𝐿
∞
𝑏𝑛 sin
𝑛=1
𝑓 𝑡 𝑑𝑡
−𝐿
𝑛𝜋𝑡
𝑓 𝑡 cos
𝑑𝑡
𝐿
𝐿
𝑓 𝑡 sin
−𝐿
for period 𝑝 = 2𝐿
then
2𝐿 = 2𝜋
𝐿=𝜋
𝑛𝜋𝑡
𝑑𝑡
𝐿
𝑛𝜋𝑡
𝐿
Solution
1
𝑎n =
𝐿
𝐿
𝑛𝜋𝑡
𝑑𝑡
𝐿
𝑓 𝑡 cos
−𝐿
1
𝑎2 =
𝜋
0
𝜋
−1 cos 2𝑡 𝑑𝑡 +
−𝜋
0
for 𝐿 = 𝜋
1
𝑎1 =
𝜋
1
𝑎1 =
𝜋
𝑎2 =
𝜋
𝑓 𝑡 cos
−𝜋
(1)𝜋𝑡
𝑑𝑡
𝜋
0
𝜋
−1 cos 𝑡 𝑑𝑡 +
−𝜋
1 cos 𝑡 𝑑𝑡
0
𝑎1 =
1
0+0 =0
𝜋
1 cos 2𝑡 𝑑𝑡
1
0+0 =0
𝜋
𝑎1 = 𝑎2 = 𝑎3 = 𝑎n = 0
Solution
1
𝑏n =
𝐿
𝐿
−𝐿
𝑛𝜋𝑡
𝑓 𝑡 sin
𝑑𝑡
𝐿
1
𝑏2 =
𝜋
0
−1 sin 2𝑡 𝑑𝑡 +
−𝜋
for 𝐿 = 𝜋
𝜋
𝑓 𝑡 sin
−𝜋
(1)𝜋𝑡
𝑑𝑡
𝜋
0
𝑏1 =
1 sin 𝑡 𝑑𝑡
0
1
4
(−1)(−2) + (1)(2) =
𝜋
𝜋
1
0+0 =0
𝜋
0
𝜋
−1 sin 3𝑡 𝑑𝑡 +
−𝜋
𝜋
−1 sin 𝑡 𝑑𝑡 +
−𝜋
1
𝑏3 =
𝜋
1 sin 2𝑡 𝑑𝑡
0
𝑏2 =
1
𝑏1 =
𝜋
1
𝑏1 =
𝜋
𝜋
𝑏3 =
1
−1
𝜋
1 sin 3𝑡 𝑑𝑡
0
−
2
+ 1
3
2
3
=
4
3𝜋
Solution
𝑏2 = 𝑏4 = 𝑏6 = 𝑏even = 0
4
𝑏1 =
𝜋
𝑏3 =
4
3𝜋
1
𝑏5 =
𝜋
0
−1 sin 5𝑡 𝑑𝑡 +
−𝜋
𝑏5 =
1
𝑏7 =
𝜋
𝜋
1 sin 5𝑡 𝑑𝑡
0
1
−1
𝜋
−
2
+ 1
5
0
2
5
4
5𝜋
𝜋
−1 sin 7𝑡 𝑑𝑡 +
−𝜋
𝑏7 =
=
0
1
−1
𝜋
𝑏n =
1 sin 7𝑡 𝑑𝑡
4
𝑛𝜋
−
2
+ 1
7
2
7
=
4𝜋
7𝜋
Solution
∞
𝑓 𝑡 = 𝑎0 +
For 𝐿 = 𝜋
𝑛=1
𝑛𝜋𝑡
𝑎𝑛 cos
+
𝐿
∞
𝑏𝑛 sin
𝑛=1
∞
𝑓 𝑡 =
𝑏𝑛 sin 𝑛𝑡
𝑛=1
𝑓 𝑡 = 𝑏1 sin 1𝑡 + 𝑏3 sin 3𝑡 + 𝑏5 sin 5𝑡 + 𝑏7 sin 7𝑡 + 𝑏9 sin 9𝑡 + ⋯ + 𝑏𝑛 sin 𝑛𝑡
𝑓 𝑡 =
4
4
4
4
4
sin 𝑡 +
sin 3𝑡 +
sin 5𝑡 +
sin 7𝑡 +
sin 9𝑡 + ⋯
𝜋
3𝜋
5𝜋
7𝜋
9𝜋
𝑛𝜋𝑡
𝐿
Solution
∞
𝑓 𝑡 = 𝑎0 +
For 𝐿 = 𝜋
𝑓 𝑡 =
𝑛=1
𝑛𝜋𝑡
𝑎𝑛 cos
+
𝐿
4
4
4
4
4
sin 𝑡 +
sin 3𝑡 +
sin 5𝑡 +
sin 7𝑡 +
sin 9𝑡 + ⋯
𝜋
3𝜋
5𝜋
7𝜋
9𝜋
∞
𝑏𝑛 sin
𝑛=1
𝑛𝜋𝑡
𝐿
Odd Function
Even Function