1. Periodic Functions and
Fourier Series

2. 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   .

3. f
0
T

4. f
0
T

5. f
0
T

6. 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   an cos n   bn sin n

7. 

n 1
n 1
f    a0   an 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 

8. 0
0
cos
cos
cos

9. 0
0
sin
sin
sin

10. We want to determine the coefficients,
an
and
bn
.
Let us first remember some useful
integrations.

11. 
  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

12. 
  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.

13. 
  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

14. Determine
a0
Integrate both sides of (1) from

to


  f  d





  a0   an cos n   bn sin n  d

n1
n 1




15. 
  f   d



  a0d     an cos n  d


 
 n 1






    bn sin n  d

 
 n 1





  f  d    a d  0  0


0

16. 

f d  2a0  0  0

1
a0 
2
a0


f d

is the average (dc) value of the
function,
f   .

17. 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.

18. 
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


2
f   d   a0d  0  0
0

19. 
2
0
f   d  2a0  0  0
1
a0 
2

2
0
f   d

20. 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




21. 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
an cos n cos m d

22. Second term,


  a

n
cos n cos m d  am
n1
Third term,


b


n 1
n
sin n cos m d  0

23. Therefore,



f   cos m d  am
am 
1




f   cos m d
m  1, 2, 

24. Determine
Multiply (1) by
bn
sin m
and then Integrate both sides from

to


  f   sin m d





  a0   a n cos n   bn sin n  sin m d

n 1
n 1




25. Let us do the integration on the right-hand-side
one term at a time.
First term,



a0 sin m d  0
Second term,


  a

n 1
n
cos n sin m d

26. Second term,


  a

n 1
n
cos n sin m d  0
Third term,


b


n
n1
sin n sin m d  bm

27. Therefore,



f   sin md  bm
bm 
1




f   sin md
m  1, 2 ,

28. The coefficients are:
1
a0 
2


f d


am 
  f  cos m d

bm 
1
1





f   sin m d
m  1, 2, 
m  1, 2, 

29. We can write n in place of m:
1
a0 
2


f d


an 
  f  cos n d

bn 
1
1





f   sin n d
n  1, 2 ,
n  1, 2 ,

30. The integrations can be performed from
0
to
2
1
a0 
2

2
0
an 

2
bn 
1
2
1
0


0
instead.
f   d
f   cos n d
f   sin n d
n  1, 2 ,
n  1, 2 ,

31. Example 1. Find the Fourier series of
the following periodic function.
f
A
0
-A
f   A when 0    
  A when     2
f   2   f  

32. 1 2
a0 
f   d
0
2
2
1  

f   d   f   d 
 0



2 
2
1  

A d    A d 
 0



2 
0

33. 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 n   0

n 0



34. 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

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

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

37. 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?

38. 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


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

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

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

42. 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


43. 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

44. 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

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

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

47. 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.

48. 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

49. 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

50. The Fourier series of an even function f  
is expressed in terms of a cosine series.

f    a0   an 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

51. Example 2. Find the Fourier series of
the following periodic function.
f(x)
x
0
f x   x
2
when    x  
f   2   f  

52. 1
a0 
2
1

2

1
f  x  dx 
2


x 
x 


3
3
  x  
3
2



2
x dx

53. an 
1




f  x  cos nx dx
1  2


x cos nxdx
 



Use integration by parts.

54. 4
an  2 cos n
n
4
an   2
n
4
an  2
n
when n is odd
when n is even

55. 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

56. 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 frequency in radians per
second.

57.   2 f
f is the frequency of the periodic function,
f t 
  2 f t
Therefore,
where
2

t
T
1
f 
T

58. 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

59. 1
a0 
2
1
a0 
T




f d

T
2
f t dt
T
2

60. an 
1



2
an 
T

T
2
f   cos n d
 2 n 
 Tf t cos T t dt



2
n  1, 2 ,
n  1, 2, 

61. bn 
1

f   sin n d
n  1, 2 ,
 2 n 
 Tf t sin T t dt



n  1, 2, 


2
bn 
T

T
2
2

62. 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

63. f t  T   f t 
This is an odd (periodic) function. Therefore,
an  0
2
 2n 
bn   f t  sin
t dt
T 0
 T 
T
4

T

0
T
2
 2n 
f t  sin
t dt
 T 

64. 4
bn 
T
4

T
T
4
 2n 
t sin
t  dt
 T 
0
T
2
T   2n 

t dt
  t   sin
T 
2  T 
4
Use integration by parts.

65.   T 
 n
 sin
 2.
 2
  2n 

2T
 n 
 2 2 sin

n
 2 
4
bn 
T
bn  0
2
when n is even.





66. 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


 


67. The Complex Form of Fourier Series


n 1
n 1
f    a0   an cos n   bn sin n
Let us utilize the Euler formulae.
cos  
sin 
e
e
j
j
e
2
 j
e
2i
 j

68. n
th harmonic component of (1) can be
The
expressed as:
an cos n  bn sin n
 an
 an
e
e
jn
jn
e
2
 jn
e
2
 jn
 bn
e
 ibn
jn
e
e
2i
jn
 jn
e
2
 jn

69. an cos n  bn sin n
 an  jbn  jn  an  jbn   jn

e  
e
2 
2 


Denoting
 a n  jb n
cn  

2

and
c0  a0




,
cn
 a n  jb n 


2



70. an cos n  bn sin n
 cn e
jn
 c n e
 jn

71. The Fourier series for
can be expressed as:


f  
f    c0   cne
n 1


c e
n
n 
jn
jn
 c ne
 jn


72. 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

73.  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

74.  an  jbn 
cn  

2 

Note that
cn .
cn
 an  jbn 
c n  

 2 
is the complex conjugate of
Hence we may write that
1
cn 
2


f e
 jn
d

.
n  0 ,  1,  2 , 

75. The complex form of the Fourier series of
f   with period
f   
2

c e
n  
n
is:
jn

76. Example 1. Find the Fourier series of
the following periodic function.
f
A
0
-A
f   A when 0    
  A when     2
f   2   f  

77. A  5
f ( x) 
A if 0  x  
A if   x  2 
0 otherwise
2
1 
A0 
 f ( x) dx
2 0
A0  0

78. n  1  8
2
1 
An   f ( x) cos( nx) dx
 0
A1  0
A2  0
A3  0
A4  0
A5  0
A6  0
A7  0
A8  0

79. 2
1 
Bn  
 0
f ( x) sin( nx) dx
B1  6.366
B2  0
B3  2.122
B4  0
B5  1.273
B6  0
B7  0.909
B8  0

80. Comple x Form
f   

c e
1
cn 
2
jn
n
n  


f e  jnd

n  0,  1,  2,
2
1 
C ( n) 

2  0
 1i n  x
f ( x) e
dx

81. 2
1 
C ( n) 

2  0
 1i n  x
f ( x) e
dx
C (0)  0
C (1)  3.183i
C (2)  0
C (3)  1.061i
C (4)  0
C (5)  0.637i
C (6)  0
C (7)  0.455i
C (1)  3.183i
C (2)  0
C (3)  1.061i
C (5)  0.637i
C (6)  0
C (7)  0.455i
C (4)  0