Periodic Functions and
Fourier Series
Periodic Functions
A function
f  
is periodic
if it is defined for all real

and if there is some positive number,
T
such that
f   T   f  
.
f  
0

T
f  
0

T
f  
0

T
Fourier Series
f  
be a periodic function with period
2
The function can be represented by
a trigonometric series as:


n 1
n 1
f    a0   a n cos n   bn sin n


n 1
n 1
f    a0   a n cos n   bn sin n
What kind of trigonometric (series) functions
are we talking about?
cos  , cos 2 , cos 3  and
sin  , sin 2 , sin 3 
0

0
cos 
cos 2
2
cos 3
0

0
sin 
sin 2
2
sin 3
We want to determine the coefficients,
an
and
bn
.
Let us first remember some useful
integrations.

  cos n cos m d

1 
1 
  cosn  m  d   cosn  m  d
2 
2 


cos n cos m d  0
nm



cos n cos m d  

nm

  sin n cos m d

1 
1 
  sin n  m  d   sin n  m  d
2 
2 


sin n cos m d  0

for all values of m.

  sin n sin m d

1 
1 
  cosn  m  d   cosn  m  d
2 
2 


sin n sin m d  0



sin n sin m d  

nm
nm
Determine
a0
Integrate both sides of (1) from

to


  f   d





   a0   an cos n   bn sin n  d

n 1
n 1




  f   d



  a0d     an cos n  d


 n 1






    bn sin n  d

 n 1





  f  d    a d  0  0


0


f   d  2 a0  0  0

1
a0 
2
a0


f  d

is the average (dc) value of the
function,
f  
.
You may integrate both sides of (1) from
0

to
2
instead.
f   d
0

2
2
0


 a0   an cos n   bn sin n  d
n 1
n 1




It is alright as long as the integration is
performed over one period.

2
f   d
0
2
2
0
0
  a0d  

2
0

2
0


  an cos n  d
 n 1




  bn sin n  d
 n 1


f   d  

2
0
a0 d   0  0

2
0
f   d   2 a0  0  0
1
a0 
2

2
0
f   d
Determine
Multiply (1) by
an
cos m 
and then Integrate both sides from

to


  f   cos m d





   a0   an cos n   bn sin n  cos m d

n 1
n 1



Let us do the integration on the right-hand-side
one term at a time.
First term,


a0 cos m  d  0

Second term,



  n 1
a n cos n cos m  d
Second term,


  a

n
cos n cos m  d  am
n 1
Third term,


b



n 1
n
sin n  cos m  d  0
Therefore,



f   cos m  d  am
1
am 




f   cos m  d
m  1, 2, 
Determine
Multiply (1) by
bn
sin m 
and then Integrate both sides from

to


  f   sin m  d





   a 0   a n cos n   bn sin n  sin m  d

n 1
n 1



Let us do the integration on the right-hand-side
one term at a time.
First term,



a 0 sin m  d  0
Second term,


 a

n 1
n
cos n  sin m  d
Second term,


 a

n 1
n
cos n  sin m  d  0
Third term,


b



n
n 1
sin n sin m  d  bm
Therefore,



f   sin m d  bm 
1
bm 




f   sin m d
m  1 , 2 ,
The coefficients are:
1
a0 
2
1
am 

1
bm 



f  d


  f   cos m  d




f   sin m  d
m  1, 2, 
m  1, 2, 
We can write n in place of m:
1
a0 
2
1
an 

1
bn 




f  d


f   cos n  d
n  1 , 2 ,

f   sin n  d
n  1 , 2 ,



The integrations can be performed from
0
2
to
2
1
a0 
2

1
an 


2

2
1
bn 

0
0
0
instead.
f   d
f   cos n  d
f   sin n  d
n  1 , 2 ,
n  1 , 2 ,
Example 2. Find the Fourier series of
the following periodic function.
f  
A
0
-A


2
3
4
5
f    A when 0    
  A when     2 
f   2   f  
1 2


a0 
f

d


2 0
2
1  






f

d


f

d





0


2
2
1  


A
d



A
d





0


2
0
an 
1


2
0
f   cos n d
2
1 




A
cos
n

d



A
cos
n

d





0




2
1  sin n 
1
sin n 
 A


A

0

n  0  
n  
bn 
1

2
0
f   sin n d
2
1 




A
sin
n

d



A
sin
n

d



  0

2
1
cos n 
1  cos n 
  A
 A



n 0  
n 
A
 cos n  cos 0  cos 2n  cos n 

n
A
 cos n  cos 0  cos 2n  cos n 
bn 
n
A
1  1  1  1

n
4A

when n is odd
n
A
 cos n  cos 0  cos 2n  cos n 
bn 
n
A
 1  1  1  1

n
 0 when n is even
Therefore, the corresponding Fourier series is
4A 
1
1
1

 sin   sin 3  sin 5  sin 7   
 
3
5
7

In writing the Fourier series we may not be
able to consider infinite number of terms for
practical reasons. The question therefore, is
– how many terms to consider?
When we consider 4 terms as shown in the
previous slide, the function looks like the
following.
1.5
1
0.5
f()
0
0.5
1
1.5

When we consider 6 terms, the function looks
like the following.
1.5
1
0.5
f()
0
0.5
1
1.5

When we consider 8 terms, the function looks
like the following.
1.5
1
0.5
f ( )
0
0.5
1
1.5

When we consider 12 terms, the function looks
like the following.
1.5
1
0.5
f()
0
0.5
1
1.5

The red curve was drawn with 12 terms and
the blue curve was drawn with 4 terms.
1.5
1
0.5
0
0.5
1
1.5

The red curve was drawn with 12 terms and
the blue curve was drawn with 4 terms.
1.5
1
0.5
0
0.5
1
1.5
0
2
4
6

8
10
The red curve was drawn with 20 terms and
the blue curve was drawn with 4 terms.
1.5
1
0.5
0
0.5
1
1.5
0
2
4
6

8
10
Even and Odd Functions
(We are not talking about even or
odd numbers.)
Even Functions
f(
The value of the
function would
be the same
when we walk
equal distances
along the X-axis

in opposite
directions.
Mathematically speaking -
f      f  
Odd Functions
f(

Mathematically speaking -
f       f  
The value of the
function would
change its sign
but with the
same magnitude
when we walk
equal distances
along the X-axis
in opposite
directions.
Even functions can solely be represented
by cosine waves because, cosine waves
are even functions. A sum of even
functions is another even function.
5
0
5
10
0

10
Odd functions can solely be represented by
sine waves because, sine waves are odd
functions. A sum of odd functions is another
odd function.
5
0
5
10
0

10
The Fourier series of an even function f  
is expressed in terms of a cosine series.

f    a0   a n cos n
n 1
The Fourier series of an odd function f  
is expressed in terms of a sine series.

f     bn sin n
n 1
Example 3. Find the Fourier series of
the following periodic function.
f(x)

0

f x   x
3
2
5
7
9
when    x  
f   2   f  
x
1
a0 
2
1

2

1
f  x  dx 
2


x 
x 


3
3
  x  
3
2



2
x dx
an 
1


 

f  x  cos nx dx
1  2


x
cos
nxdx



  
Use integration by parts. Details are shown
in your class note.
4
an  2 cos n
n
4
an   2
n
4
an  2
n
when n is odd
when n is even
This is an even function.
Therefore,
bn  0
The corresponding Fourier series is

cos 2 x cos 3 x cos 4 x


 4 cos x 


 
2
2
2
3
2
3
4


2
Functions Having Arbitrary Period
Assume that a function f t  has
period, T . We can relate angle
(  ) with time ( t ) in the following
manner.
  t
 is the angular velocity in radians per
second.
  2 f
f is the frequency of the periodic function,
f t 
  2 f t
Therefore,
where
2

t
T
1
f 
T
2

t
T
2
d 
dt
T
Now change the limits of integration.
  
2
 
t
T
T
t
2
 
2

t
T
T
t
2
1
a0 
2
1
a0 
T


f  d


T
2
f t dt
T

2
1
an 

2
an 
T



T
2
f   cos n  d
 2 n 
 Tf t  cos T t dt

2
n  1 , 2 ,
n  1, 2, 
1
bn 

2
bn 
T

  f   sin n d

T
2
 2 n 
 Tf t  sin T t dt

2
n  1 , 2 ,
n  1, 2, 
Example 4. Find the Fourier series of
the following periodic function.
f(t)
3T/4
0
-T/2
t
T/4
T/2
T
2T
T
T
f t   t when   t 
4
4
T
T
3T
 t 
when
t
2
4
4
f t  T   f t 
This is an odd function. Therefore,
2
bn 
T
4

T

0

0
 2n 
f t  sin 
t  dt
 T 
T
T
2
 2n 
f t  sin 
t  dt
 T 
an  0
4
bn 
T
4

T
T
4
 2n 
t
sin
t
dt


 T 
0
T
2
T   2n 


t

sin
t
dt




T 
2  T 
4
Use integration by parts.
  T 
 n
 sin
 2.
 2
  2n 
2T
 n 
 2 2 sin

n
 2 
4
bn 
T
bn  0
2
when n is even.


 
Therefore, the Fourier series is
2T
2
  2  1

 6  1
 10  
 sin T t   2 sin T t   2 sin T t   
 3

 5


 

The Complex Form of Fourier Series


n 1
n 1
f    a0   an cos n   bn sin n
Let us utilize the Euler formulae.
j
 j
e
cos  
2
j
 j
e e
sin  
2j
e
n
th harmonic component of (1) can be
The
expressed as:
an cos n  bn sin n
 an
 an
e
e
jn
e
2
 jn
jn
e
2
 jn
 bn
e
 jbn
jn
e
e
2j
jn
 jn
e
2
 jn
an cos n  bn sin n
 an  jbn  jn  an  jbn   jn

e  
e
2 
2 


Denoting
 a n  jbn
c n  
2

and
c0  a0



,
cn
 a n  jbn 


2


a n cos n  bn sin n
 cne
jn 
 c n e
 jn 
The Fourier series for
can be expressed as:


f  
f    c0   cne
n 1


c e
n
n  
jn
jn
 c ne
 jn

The coefficients can be evaluated in
the following manner.
 an  jbn 
cn  

2 

1 
j

f   cos nd 

2 
2

1

2

1

2





  f   sin n d

f  cos n  j sin n d
f   e
 jn 
d
 an  jbn 
c n  

2 

1 
j

f   cos nd 

2 
2
1

2
1

2


  f   sin n d

  f  cos n  j sin n d


  f   e

jn 
d
 an  jbn 
cn  

2 

Note that
cn .
cn
c n
 an  jbn 


2 

is the complex conjugate of
Hence we may write that
1
cn 
2


f   e
 jn 
d

.
n  0 ,  1,  2 ,
The complex form of the Fourier series of
f   with period
f   
2

c e
n  
n
is:
jn 