This document discusses Fourier series and periodic functions. It defines a periodic function as one where f(x+T)=f(x) for some positive number T. The Fourier series of a periodic function with period T can be expressed as the sum of sines and cosines with frequencies that are integer multiples of 1/T. It also discusses even and odd functions, noting that the Fourier series of an even function contains only cosine terms, while the Fourier series of an odd function contains only sine terms. Examples are provided of finding the Fourier series for specific periodic functions.

1. Fourier Series

2. Periodic Functions
 
   
x
f
T
x
f
that
such
T
number
positive
some
is
there
if
and
x
real
for
defined
is
it
if
periodic
is
x
f
function
A


,

3. 0

T
f 
 
x
 
x
f

4. 0

T
f 
 
x
 
x
f

5. 0

T
f 
 
x
 
x
f

8. Example

12. Fourier Series-(-l<x<l)
  




















1
1
0 sin
cos
2
1
n
n
n
n
l
x
n
b
l
x
n
a
a
x
f


 
:
,
2
as
series
ric
trigonomet
by
d
represente
be
can
function
The
l
period
with
function
periodic
a
be
x
f

13.  dx
x
f
l
a
l
l


1
0
  
,
2
,
1
cos
1







 
m
dx
l
x
n
x
f
l
a
l
l
n

  
,
2
,
1
sin
1







 
n
dx
l
x
n
x
f
l
b
l
l
n

We can determine the coefficients of FS by using,
Fourier Series-Fourier Coefficient

14. Fourier Series-(- π< x <π)
 
:
,
2
as
series
ric
trigonomet
by
d
represente
be
can
function
The
period
with
function
periodic
a
be
x
f 
     









1
1
0 sin
cos
2
1
n
n
n
n nx
b
nx
a
a
x
f

15.  dx
x
f
a 




1
0
    
,
2
,
1
cos
1

 
m
dx
nx
x
f
an



    
,
2
,
1
sin
1

 
n
dx
nx
x
f
bn



We can determine the coefficients of FS by using,
Fourier Series-Fourier Coefficient

16. Even and Odd Functions
(We are not talking about even or
odd numbers.)
Fourier Series

25. The value of the
function would
be the same
when we walk
equal distances
along the X-axis
in opposite
directions.
   
x
f
x
f 

Mathematically speaking -

f(
x
 
x
f
Fourier Series-Even Function

26. Even Functions-example 1
 
  





























1
0
2
1
1
2
1
1
3
1
1
2
2
3
2
3
1
3
1
3
1
1
dx
x
dx
x
x
f
OR
x
x
dx
x
x
f
1

1
1
x
 
x
f

27. 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.
   
x
f
x
f 


Mathematically speaking -

f(
x
 
x
f
Fourier Series-Odd Function

28. Odd Functions- example 2
 
0
2
1
2
1
2
1
1
1
1
2
1
1

















x
x
xdx
x
f
1

1

1
1
x
 
x
f

29. Even functions can solely be represented
by cosine waves because, cosine waves
are even functions. A sum of even
functions is another even function.
10 0 10
5
0
5


30. Odd functions can solely be represented by
sine waves because, sine waves are odd
functions. A sum of odd functions is another
odd function.
10 0 10
5
0
5


32. The Fourier series of an even function  
x
f
is expressed in terms of a cosine series.
  )
0
(
,
cos
2
1
1
0 

 


n
n
n b
l
x
n
a
a
x
f

The Fourier series of an odd function  
x
f
is expressed in terms of a sine series.
  )
0
,
0
(
,
sin 0
1


 


n
n
n a
a
l
x
n
b
x
f

Fourier Series-Odd & Even Function

39. Find the Fourier series of the following
periodic function.
   
x
f
x
f 
 
2
x
f(x)
  3 5 7 9
0
  
 


 x
when
x
x
f 2
Fourier Series

40.  
3
2
3
1
1
1
2
3
2
0
























 

x
x
x
dx
x
dx
x
f
a
 




















nxdx
x
nxdx
x
dx
nx
x
f
an
cos
2
cos
1
cos
1
0
2
2








3
2
2
sin
0
cos
2
sin
2
cos
n
nx
n
nx
n
nx
x
nx
x






method
tabular
By

41.  
 
.....)
6
,
4
,
2
(
4
.....)
5
,
3
,
1
(
4
)
1
(
4
cos
4
0
cos
2
0
2
0
2
sin
2
cos
2
sin
2
sin
2
cos
2
sin
2
2
2
2
2
2
3
2
2
0
3
2
2
even
is
n
when
n
a
and
odd
is
n
when
n
a
n
n
n
n
n
n
n
n
n
n
n
n
nx
n
nx
x
n
nx
x
a
n
n
n
n






























































42. This is an even function.
Therefore, 0

n
b
The corresponding Fourier series is
     
 





































2
2
2
2
1
2
2
1
1
2
2
4
4
cos
3
3
cos
2
2
cos
cos
4
3
cos
)
1
(
4
3
sin
)
0
(
cos
)
1
(
4
3
2
2
1
x
x
x
x
nx
n
nx
nx
n
x
f
n
n
n
n
n



     









1
1
0 sin
cos
2
1
n
n
n
n nx
b
nx
a
a
x
f

43. Find the Fourier series of the following
periodic function.
 



2
0







x
when
A
x
when
A
x
f
   
x
f
x
f 
 
2
0

f 
 
 2 3 4 5
A
-A
x
 
x
f

44.  dx
x
f
a 



2
0
0
1
    




 
 
 dx
x
f
dx
x
f




2
0
1
0
1 2
0






 

 
 dx
A
dx
A




  dx
nx
x
f
an cos
1 2
0




0
sin
1
sin
1
2
0




















 n
nx
A
n
nx
A
  




 

 
 dx
nx
A
dx
nx
A cos
cos
1 2
0





45.   dx
nx
x
f
bn sin
1 2
0




  




 

 
 dx
nx
A
dx
nx
A sin
sin
1 2
0









2
0
cos
1
cos
1















n
nx
A
n
nx
A
 




n
n
n
n
A
cos
2
cos
0
cos
cos 





46.  
 
odd
is
n
when
4
1
1
1
1
cos
2
cos
0
cos
cos






n
A
n
A
n
n
n
n
A
bn










 
 
even
is
n
when
0
1
1
1
1
cos
2
cos
0
cos
cos
















n
A
n
n
n
n
A
bn

47. Therefore, the corresponding Fourier series is
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?
     









1
1
0 sin
cos
2
1
n
n
n
n nx
b
nx
a
a
x
f
   




1
sin
4
n
nx
n
A
x
f

  









 





7
sin
7
1
5
sin
5
1
3
sin
3
1
sin
4A
x
f

48. When we consider 4 terms as shown in the
previous slide, the function looks like the
following.
1.5
1
0.5
0
0.5
1
1.5
f 
( )

x
 
x
f

49. When we consider 6 terms, the function looks
like the following.
1.5
1
0.5
0
0.5
1
1.5
f 
( )

x
 
x
f

50. When we consider 8 terms, the function looks
like the following.
1.5
1
0.5
0
0.5
1
1.5
f 
( )

x
 
x
f

51. When we consider 12 terms, the function looks
like the following.
1.5
1
0.5
0
0.5
1
1.5
f 
( )

x
 
x
f

52. 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

x
 
x
f

53. The red curve was drawn with 12 terms and
the blue curve was drawn with 4 terms.
0 2 4 6 8 10
1.5
1
0.5
0
0.5
1
1.5

x
 
x
f

54. The red curve was drawn with 20 terms and
the blue curve was drawn with 4 terms.
0 2 4 6 8 10
1.5
1
0.5
0
0.5
1
1.5

x
 
x
f