Fjdkdkdkdd

1. FOURIER SERIES
PERIODIC FUNCTIONS
A function f(x) is said to be
periodic with period T if
f(x+T)=f(x) x , where T is a
positive constant . The least
value of T>0 is called the period
of f(x).

2. f(x+2T) =f ((x+T)+T)
=f (x+T)=f(x)
f(x+nT)=f(x) for all x
Ex.1 f(x)=sin x has periods 2,
4, 6, …. and 2 is the period
of f(x).
Ex.2 The period of sin nx and
cos nx is 2/n.

4. FOURIER SERIES
Let be defined in the
interval and outside the
interval by i.e
assume that has the
period .The Fourierseries
corresponding to is
given by











1
0
sin
cos
2 n
n
n
l
x
n
b
l
x
n
a
a 

)
(x
f
)
(
)
2
( x
f
l
x
f 

)
(x
f
l
2
)
,
( l
l

)
(x
f

5. ,....
3
,
2
,
1
sin
)
(
1
cos
)
(
1
)
(
1
0












n
dx
l
x
n
x
f
l
b
dx
l
x
n
x
f
l
a
dx
x
f
l
a
l
l
n
l
l
n
l
l


where the Fourier coeffecients are

6. dx
l
x
n
x
f
l
b
dx
l
x
n
x
f
l
a
dx
x
f
l
a
l
c
c
n
l
c
c
n
l
c
c









2
2
2
0
sin
)
(
1
cos
)
(
1
)
(
1


If is defined in the interval
(c,c+2 ), the coefficients can be
determined equivalently from
)
(x
f
l

7. DIRICHLET CONDITIONS
Suppose that
1. f(x) is defined and single valued
except possibly at finite number of
points in (-l,+l)
2. f(x) is periodic outside (-l,+l) with
period 2l
3. f(x) and f’(x) are piecewise
continuous in(-l,+l)

8. T h e n t h e F o u r i e r s e r i e s o f
f ( x ) c o n v e r g e s t o
a ) f ( x ) i f x i s a p o i n t o f
c o n t i n u i t y
b ) [ f ( x + 0 ) + f ( x - 0 ) ] / 2 i f x i s a
p o i n t o f d i s c o n t i n u i t y

9. METHOD OF OBTAINING FOURIER
SERIES OF

)
(x
f
1.
,....
3
,
2
,
1
sin
)
(
1
cos
)
(
1
)
(
1
0












n
dx
l
x
n
x
f
l
b
dx
l
x
n
x
f
l
a
dx
x
f
l
a
l
l
n
l
l
n
l
l


2.
4.
3.
)
(x
f











1
0
sin
cos
2 n
n
n
l
x
n
b
l
x
n
a
a 


10. SOLVED PROBLEMS
1. Expand f(x)=x2,0<x<2 in Fourier series
if the period is 2 . Prove that
6
...
3
1
2
1
1
1 2
2
2
2






12. Period = 2 = 2  thus =  and choosing
c=0
0
4
sin
2
cos
2
sin
1
cos
1
cos
)
(
1
cos
)
(
1
2
2
0
3
2
2
2
0
2
2
0
2













 






 









n
n
n
nx
n
nx
x
n
nx
x
dx
nx
x
dx
nx
x
f
dx
l
x
n
x
f
l
a
c
c
n








l
l
SOLUTION

13. n
n
nx
n
nx
x
n
nx
x
nxdx
x
nxdx
x
f
dx
l
x
n
x
f
l
b
dx
x
a
c
c
n












4
cos
2
sin
2
cos
1
sin
1
sin
)
(
1
sin
)
(
1
3
8
1
2
0
3
2
2
2
0
2
2
0
2
2
0
2
2
0














 




















14.  



















1
2
2
2
1
0
sin
4
cos
4
3
4
)
(
sin
cos
2
)
(
n
n
n
n
nx
n
nx
n
x
x
f
nx
b
nx
a
a
x
f


At x=0 the above Fourier series
reduces to




1
2
2
4
3
4
n n

X=0 is the point of discontinuity

15. By Dirichlet conditions, the series converges
at x=0 to (0+4 2)/2 = 2 2











1
2
2
1
2
2
2
6
1
4
3
4
2
n
n
n
n




16. 2. Find the Fourier series expansion for the
following periodic function of period 4.











2
0
2
0
2
2
)
(
x
x
x
x
x
f
   
2
2
2
2
2
2
1
2
2
2
1
)
(
1
)
(
)
4
(
2
0
2
0
2
2
2
0
0
2
0
















































x
x
x
x
dx
x
dx
x
dx
x
f
l
a
x
f
x
f
l
l
Solution

17.    
 
 
 







































































odd
n
for
n
even
n
for
n
n
x
n
n
x
n
x
n
x
n
n
x
n
x
dx
x
n
x
dx
x
n
x
dx
l
x
n
x
f
l
a
n
l
l
n
2
2
2
2
2
0
2
0
2
2
2
0
0
2
8
0
)
1
(
1
4
4
)
(
2
cos
)
1
(
2
2
sin
2
4
)
(
2
cos
)
1
(
2
2
sin
2
2
1
2
cos
2
2
cos
2
2
1
cos
)
(
1














18.    
 
 
 








































































































1
2
2
2
0
2
2
0
2
2
2
2
0
0
2
2
1
2
cos
1
8
1
)
(
0
4
2
sin
)
1
(
2
2
cos
2
4
2
sin
)
1
(
2
2
cos
2
2
1
sin
2
sin
2
2
1
sin
)
(
1
n
l
l
n
x
n
n
x
f
n
x
n
n
x
n
x
n
x
n
n
x
n
x
dx
l
x
n
x
dx
l
x
n
x
dx
l
x
n
x
f
l
b














19. EVEN AND ODD FUNCTIONS
A f u n c t i o n f ( x ) i s c a l l e d o d d i f
f ( - x ) = - f ( x )
E x : x 3 , s i n x , t a n x , x 5 + 2 x + 3
A f u n c t i o n f ( x ) i s c a l l e d e v e n i f
f ( - x ) = f ( x )
E x : x 4 , c o s x , e x + e - x , 2 x 6 + x 2 + 2

20. EXPANSIONS OF EVEN AND ODD
PERIODIC FUNCTIONS
If is a periodic function defined in the
interval , it can be represented by the
Fourier series
Case1. If is an even function
dx
x
f
l
dx
x
f
l
a
l
l
l





0
0
)
(
2
)
(
1
)
(x
f
)
,
( l
l

)
(x
f

21. 

















function
odd
is
l
x
n
x
f
dx
l
x
n
x
f
l
b
function
even
also
is
l
x
n
x
f
dx
l
x
n
x
f
l
a
l
l
n
l
n




sin
)
(
0
sin
)
(
1
cos
)
(
cos
)
(
2
0


If a periodic function is even in
, its Fourier series expansion
contains only cosine terms
)
(x
f
)
,
( l
l


22. 








l
n
l
n
n
dx
l
x
n
x
f
l
a
dx
x
f
l
a
l
x
n
a
a
x
f
0
0
0
1
0
cos
)
(
2
)
(
2
cos
2
)
(


Case 2. When is an odd function




l
l
dx
x
f
l
a 0
)
(
1
0
)
(x
f

23. 

















even
is
l
x
n
x
f
dx
l
x
n
x
f
l
b
odd
is
l
x
n
x
f
dx
l
x
n
x
f
l
a
l
n
l
l
n




sin
)
(
sin
)
(
2
cos
)
(
0
cos
)
(
1
0


If a periodic function is odd in
,its Fourier expansion
contains only sine terms
dx
l
x
n
x
f
l
b
l
x
n
b
x
f
l
n
n
n


sin
)
(
2
sin
)
(
0
1






)
(x
f
)
,
( l
l


24. 1.For a function defined by
obtain a Fourier series. Deduce that

 


 x
x
x
f ,
)
(
SOLVED PROBLEMS
8
....
5
1
3
1
1
1 2
2
2
2





Solution
x
x
f 
)
( is an even function





1
0
cos
2
)
(
n
n
l
x
n
a
a
x
f


25.  
 
1
1
2
cos
sin
2
cos
2
cos
2
2
2
2
2
2
0
2
0
0
0
2
0
0
0














 


























n
n
n
n
nx
n
nx
x
nxdx
x
nxdx
x
a
x
xdx
dx
x
a














SOLUTION

26.  
 







1
2
cos
1
1
2
2
)
(
n
n
nx
n
x
f


At x=0 the above series reduces to
 
 






1
2
1
1
2
2 n
n
n


x=0 is a point of continuity, by
Dirichlet condition the Fourier
series converges to f(0) and
f(0)=0

27.  
8
....
5
1
3
1
1
1
....
5
2
3
2
1
2
2
2
0
1
1
2
2
0
2
2
2
2
2
2
2
1
2

























 


 


n
n
n










3
0
0
3
)
(
x
when
k
x
when
k
x
f
Is the function even or odd. Find
the Fourier series of f(x)
PROBLEM 2

28. SOLUTION
is odd function
dx
x
n
k
dx
l
x
n
x
f
l
b
a
a
l
n
n
3
sin
3
2
sin
)
(
2
0
0
3
0
0
0








)
(x
f

29. 





1 3
sin
]
)
1
(
1
[
2
)
(
n
n
x
n
n
k
x
f


]
)
1
(
1
[
2
3
3
cos
3
2
3
0
n
n
k
n
x
n
k



















30. 





1 3
sin
]
)
1
(
1
[
2
)
(
n
n
x
n
n
k
x
f


]
)
1
(
1
[
2
3
3
cos
3
2
3
0
n
n
k
n
x
n
k



















31. 





1 3
sin
]
)
1
(
1
[
2
)
(
n
n
x
n
n
k
x
f


]
)
1
(
1
[
2
3
3
cos
3
2
3
0
n
n
k
n
x
n
k



















32. 0
,
cos
)
(
2
cos
2
)
(
0
1
0








n
dx
l
x
n
x
f
l
a
where
l
x
n
a
a
x
f
l
n
n
n


HALF RANGE SERIES
COSINE SERIES
A function defined in can be expanded
as a Fourier series of period containing only
cosine terms by extending suitably in .
(As an even function)
)
(x
f
)
(x
f
)
,
0
( l
l
2
)
0
,
( l


33. SINE SERIES
A function defined in can be expanded
as a Fourier series of period containing
only sine terms by extending suitably in
[As an odd function]
1
,
sin
)
(
2
sin
)
(
0
1







n
dx
l
x
n
x
f
l
b
where
l
x
n
b
x
f
l
n
u
n


)
(x
f
)
(x
f
)
,
0
( l
).
0
,
( l

l
2

34. SOLVED PROBLEMS
Obtain the Fourier expansion of (x sinx )as a
cosine series in
.Hence find the value of
SOLUTION
Given function represents an even function in
)
,
0
( 





7
.
5
2
5
.
3
2
3
.
1
2
1
nx
n
n
a
a
x
f cos
1
2
0
)
( 




)
,
( 



35.  


 0
)
sin
(
1
)
cos
(
2
0
2
0
sin x
x
x
a xdx
x 













0
)
(
2
0
0
cos
)
(
2
dx
x
f
a
nxdx
x
f
n
a
2




 0
cos
sin
2
nxdx
x
x
an

36.  










 0
)
1
sin(
)
1
sin(
2
1
2
dx
x
n
x
n
x













































































0
2
0
2
)
1
(
)
1
sin(
.
1
1
)
1
cos(
)
1
(
)
1
sin(
.
1
1
)
1
cos(
1
n
x
n
n
x
n
x
n
x
n
n
x
n
x

37.  
1
1
2
1
1
1
1
)
1
( 2
1
1
















n
n
n
n
n
n
n
n
n








 1
1
)
1
(
1
1
1
)
1
(
1
if
1

n
1

n

38. dx
x
x
xdx
x
x
a 
 




 0
0
1 2
sin
1
cos
sin
2

 0
2
2
2
sin
.
1
2
2
cos
1











 






 

x
x
x
2
1
2
1 



















2
2
1
cos
1
)
1
(
2
cos
2
1
1
sin
n
n
nx
n
x
x
x
in )
,
0
( 

39. At
2


x the above series reduces to
 
2
cos
1
1
2
1
2
2
1

n
n
n
n







is a point of continuity
2


x
 The given series converges to
2
)
2
(



f

40.  
2
cos
1
1
2
1
2 2
2
1

 n
n
n
n








4
2
.
..........
7
.
5
1
5
.
3
1
3
.
1
1 








41. 2) Expand in half range
(a) sine Series (b) Cosine series.
SOLUTION
(a)
Extend the definition of given function to that of an
odd function of period 4
i.e












2
0
;
0
2
;
)
(
x
x
x
x
x
f
2
0
,
)
( 

 x
x
x
f

43. Here
dx
l
x
n
x
f
l
b
a
l
n
n



0
sin
)
(
2
0







n
n
x
n
n
x
n
x
dx
x
n
x
f
b
n
n
)
1
(
4
)
2
2
sin
(
1
)
2
2
cos
(
2
sin
)
(
2
2
2
0
2
2
2
2
0



















 

44. 2
sin
)
1
(
4
)
(
1
x
n
n
x
f
n
n








(b)
Extend the definition of given function to that of
an even function of period 4













2
0
;
0
2
;
)
(
x
x
x
x
x
f

46. dx
l
x
n
x
f
l
a
b
l
n
n



0
cos
)
(
2
0

 
0
;
1
)
1
(
4
)
2
2
cos
(
1
)
2
2
sin
(
2
cos
)
(
2
2
2
2
2
0
2
2
2
2
0



















 
n
n
n
x
n
n
x
n
x
dx
x
n
x
f
a
n
n







47.  

2
0
0 2
xdx
a
 
2
cos
1
)
1
(
4
1
)
(
1
2
2
x
n
n
x
f
n
n










48. Exercise problems
1.














x
x
o
x
x
f
0
;
sin
;
0
)
(
Find Fourier series of )
(x
f
2. )
(x
f x
e
 in )
2
,
0
( 
Find Fourier series of )
(x
f

49. 3.Find the Fourier series of
3
)
( x
x
f 
in )
,
( 


4.Find the Fourier series of
2
4
)
( x
x
f 

in (-2 ,2)

50. 5.Represent function
L
x
x
f

sin
)
( 
In (0,L) by a Fourier cosine series
6.Determine the half range sine series for





x
x
x
f
8
)
(
4
0 
 x
8
4 
 x

51. PARSEVAL’S IDENTITY
• To prove that
  







 



 1
2
2
2
0
2
)
(
2
1
)
(
n
n
n
l
l
b
a
a
l
dx
x
f
Provided the Fourier series for f(x) converges uniformly in (-l, I).
The Fourier Series for f(x) in (-l,l) is
)
1
......(
..........
sin
cos
2
)
(
1
1
0
l
x
n
b
l
x
n
a
a
x
f
n
n
n
n











Multiplying both sides of (1) by f(x)and integrating term from – l to l
( which is justified because f(x) is uniformly convergent)

52. )
2
.(
..........
..........
)
(
2
)
(
)
(
)
(
2
sin
)
(
cos
)
(
2
)
(
1
2
2
2
0
2
1
1
0
0
1
1
1 1
0
2



































 

















n
n
n
l
l
n
n
n
n
n
n
n
n
n
n n
n
l
l
b
a
a
l
dx
x
f
lb
b
la
a
la
a
dx
l
x
n
x
f
b
dx
l
x
n
x
f
a
a
dx
x
f



53. CASE-I
  )
3
.....(
..........
)
(
2
1
)
(
1
2
2
2
0
2
2
0








 



n
n
n
l
b
a
a
l
dx
x
f
If f(x) is defined in (0,2l) then Parseval’s
Identity is given by

54. CASE-II
Ifhalf range cosine series in (o,l) for f(x) is
l
x
n
a
a
x
f
n
n

cos
2
)
(
1
0





Then Parseval’s Identity is given by
  )
4
....(
..........
2
1
2
)
(
1
2
2
0
2
0







 



n
n
l
a
a
l
dx
x
f
.
.

55.   )
5
....(
..........
2
)
(
1
2
2
0






 



n
n
l
b
l
dx
x
f
CASE-III
If the half range Sine sereies in (0,l) for f(x) is
l
x
n
b
x
f
n
n

sin
)
(
1




Then Parseval,s Identity is given by

56. RMS VALUE OF FUNCTION
If a function y=f(x) is defined in ( c , c+2l ),then
dx
y
l
l
c
c

 2
2
2
1
is called the root mean square value (RMS value) of y in
( c , c+2l ).It is denoted by
y
dx
x
f
l
y
l
c
c



2
2
2
)
(
2
1
.

57. l
y 2
2

.
)
(
2
1
4
1
1
2
2
2
0 







 


n
n
n b
a
a
.
)
(
2
1
1
2
2
2
0 







 


n
n
n b
a
a
l
2
y
Equation(2) becomes

58. Equation(3) becomes
.
)
(
2
1
1
2
2
2
0 







 


n
n
n b
a
a
l
l
y 2
2

2
y .
)
(
2
1
4
1
1
2
2
2
0 







 


n
n
n b
a
a

59. Equation(4) becomes
l
y 
2
.
2
1
2 1
2
2
0 






 


n
n
a
a
l
2
y .
2
1
4
1
1
2
2
0 






 


n
n
a
a
Equation(5)becomes
2
y .
2
1
1
2






 


n
n
b

60. SOLVED PROBLEMS
1) Find the Fourier series of periodic function
x
x
x
f 
 2
)
(




 .......
4
1
3
1
2
1
1
1
4
4
4
4
in )
,
( 


Hence deduce the sum of series
Assuming that 6
1 2
1
2





n n

61. SOLUTION
.
sin
cos
2
)
(
1
1
0
l
x
n
b
l
x
n
a
a
x
f
n
n
n
n











in )
,
( 


3
2
1
)
(
1 2
2
0








 



 dx
x
dx
x
f
a

62. n
n
n
n
nx
x
n
nx
x
nxdx
x
x
nxdx
x
f
a
)
1
(
4
cos
2
)
sin
(
1
cos
)
(
1
cos
)
(
1
2
2
2
2







 
















 


if 0

n
n
n
n
n
nx
n
nx
x
nxdx
x
nxdx
x
x
nxdx
x
f
b
)
1
(
2
)
sin
(
1
)
cos
(
1
sin
1
0
sin
)
(
1
sin
)
(
1
2
2







 




























nx
xcos
 is odd function
nx
x sin
2
 is odd function

63. in
nx
n
nx
n
x
f
n
n
n
n
.
sin
)
1
(
2
cos
)
1
(
4
3
)
(
1
1
2
2












)
,
( 


Using the Parseval’s Identity
.
)
(
2
1
4
1
1
2
2
2
0 







 


n
n
n b
a
a
2
y
3
5
)
(
2
1 2
4
2
2
2 







 

dx
x
x
y

64. 















1
2
1
4
4
2
4
2
2
0
4
2
1
16
2
1
9
4
4
1
3
5
)
1
(
2
;
)
1
(
4
;
3
2
n
n
n
n
n
n
n
n
n
b
n
a
a




90
1 4
1
4


 


n n
6
1 2
1
2





n n


65. 2)By using sine series for 1
)
( 
x
f
i


 x
0
in
Show that 




 .......
7
1
5
1
3
1
1
8 2
2
2
2

SOLUTION

 

 l
x
n
b
x
f
n
n

sin
)
(
1
 
n
n
n
n
nx
nxdx
nxdx
x
f
b
)
1
(
1
2
)
cos
(
2
sin
2
sin
)
(
2
0
0
0








 


 








for 0

n
nx
x
f
x
f
n
sin
)
(
)
(
1





66. nx
n
x
f
n
n
sin
)
1
(
1
2
1
)
(
1








.
2
1
1
2






 


n
n
b
2
y
 2
1
2
2
0
2
2
)
1
(
1
4
2
1
1
1
1 n
n n
dx
y 



 


 


By Parseval’s Identity






 .......
7
1
5
1
3
1
1
8 2
2
2
2


67. 3)Prove that in l
x 

0
)
..........
5
cos
5
1
3
cos
3
1
(cos
4
2 2
2
2





l
x
l
x
l
x
n
l
l
x



and deduce that 96
.......
7
1
5
1
3
1
1
1 4
4
4
4
4







SOLUTION
In Half range cosine series

 


l
l
l
xdx
l
dx
x
f
l
a
0
0
0
2
)
(
2

68.  
1
)
1
(
2
)
cos
.(
1
)
sin
(
2
cos
2
cos
)
(
2
2
2
0
2
2
2
0
0



















 

n
l
l
l
n
n
l
l
n
l
x
n
l
n
l
x
n
x
l
dx
l
x
n
x
l
dx
l
x
n
x
f
l
a







l
x
n
a
a
x
f
n
n

cos
2
)
(
1
0





l
x
n
n
l
x
n


cos
4
1
2 ...
5
,
3
,
1
2
2 






69. .
2
1
4
1
1
2
2
0 






 


n
n
a
a
2
y
By Parseval’s Identity
 2
1
4
4
2
2
2
2
0
2
0
2
2
1
)
1
(
4
2
1
4
3
3
1
)
(
1













n
n
l
l
n
l
l
l
l
dx
x
l
dx
x
f
l
y

96
.......
7
1
5
1
3
1
1
1 4
4
4
4
4









70. COMPLEX FORM OF FOURIER
SERIES
The Fourier series of a periodic
function of period 2l is






































1
0
1
0
2
2
2
)
(
2
sin
2
cos
sin
cos
2
)
(
n
l
x
n
i
l
x
n
i
n
l
x
n
i
l
x
n
i
n
i
i
i
i
n
n
n
i
e
e
b
e
e
a
a
x
f
i
e
e
e
e
l
x
n
b
l
x
n
a
a
x
f













71.    
,....
2
,
1
,
0
)
(
2
1
2
1
,
2
1
,
2
1
)
(
0
0
1
0





























n
for
dx
e
x
f
l
c
ib
a
c
ib
a
c
a
c
e
c
e
c
c
x
f
l
l
l
x
in
n
n
n
n
n
n
n
n
l
x
n
i
n
l
x
n
i
n



The Fourier series can be represented in the
following way

72. 









l
l
l
x
in
n
n
n
l
x
in
n
dx
e
x
f
l
c
where
dx
e
c
x
f


)
(
2
1
)
(
SOLVED PROBLEM
1.Find the complex form of the Fourier series
of the periodic function








l
x
l
when
l
x
when
x
f
2
0
0
)
(


73.  












ni
in
l
l
l
x
in
l
l
l
x
in
l
l
x
in
n
l
l
x
in
n
n
n
e
e
i
n
e
in
l
l
dx
e
dx
e
l
c
dx
e
x
f
l
c
l
x
in
e
c
x
f














































2
2
2
0
2
0
2
2
0
2
1
)
(
2
1
)
(
SOLUTION

74. 2.Find the complex form of Fourier seriesof
f(x)=sinx in (0,)
 
 
 
 
l
x
in
e
n
i
x
f
n
i
i
n
n
n
n
n





















)
)
1
(
1
(
2
)
(
1
1
2
1
1
2

75.  









0
2
2
0
2
0
2
2
2
cos
sin
2
4
1
1
sin
1
)
(
1
2
)
(


































x
x
in
n
e
dx
xe
dx
e
x
f
c
e
c
l
e
c
x
f
inx
inx
inx
n
n
inx
n
n
x
in
n 
SOLUTION

76.   
 
  )
,
0
(
1
4
1
2
)
(
1
4
2
1
1
4
1
2
2
2
2
2





in
e
n
x
f
n
e
n
c
n
inx
in
n
















77. HARMONIC ANALYSIS
 


















2
0
2
0
2
0
0
1
0
sin
)
(
1
cos
)
(
1
)
(
1
sin
cos
2
)
(
nxdx
x
f
b
nxdx
x
f
a
dx
x
f
a
nx
b
nx
a
a
x
f
n
n
n
n
n

78.    
   
   









2
,
0
sin
)
(
2
sin
)
(
2
1
2
2
,
0
cos
)
(
2
cos
)
(
2
1
2
2
,
0
)
(
2
)
(
2
1
2
2
0
2
0
2
0
0
in
nx
x
f
of
mean
nxdx
x
f
b
in
nx
x
f
of
mean
nxdx
x
f
a
in
x
f
of
mean
dx
x
f
a
n
n


































79. The term a1cosx+b1sinx is called the
fundamental or first harmonic,
the term a2cosx+b2sinx is called the
second harmonic and so on.
Solved Problem
1.Find first two harmonics of Fourier
Series from the following table