SRIT / UICM003 - TPDE / Fourier Series
SRIT / M & H / M. Vijaya Kumar 1
SRI RAMAKRISHNA INSTITUTE OF TECHNOLOGY
(AN AUTONOMOUS INSTITUTION)
COIMBATORE- 641010
UICM003 & TRANSFORMS AND PARTIAL DIFFERENTIAL EQUATIONS
Fourier Series
History:
The Fourier series is named in honour of Jean-Baptiste Joseph Fourier (1768–1830),
French mathematician who made important contributions to the study of trigonometric
series, after preliminary investigations by Leonhard Euler, Jean le Rond d'Alembert, and
Daniel Bernoulli. Fourier introduced the series for the purpose of solving the heat equation
in a metal plate, publishing his initial results in the year 1807. The Mémoire introduced
Fourier analysis, specifically Fourier series. Through Fourier's research the fact was
established that an arbitrary function can be represented by a trigonometric series. The
first announcement of this great discovery was made by Fourier in 1807, before the French
Academy. Early ideas of decomposing a periodic function into the sum of simple oscillating
functions date back to the 3rd century BC, when ancient astronomers proposed an empiric
model of planetary motions, based on deferents and epicycles. Fourier was a buddy of
Napoleon and worked as scientific adviser for Napoleon's army.
Introduction:
A Fourier series is an expansion of a periodic function ( ) in terms of an infinite
sum of sines and cosines. Fourier series make use of the orthoganality relationships of the
sine and cosine functions. The computation and study of Fourier series is known as
harmonic analysis and is extremely useful as a way to break up an arbitrary periodic
function into a set of simple terms that can be plugged in, solved individually, and then
recombined to obtain the solution to the original problem or an approximation to it to
whatever accuracy is desired or practical.
In mathematics, a Fourier series is a way to represent a (wave-like) function as the
sum of simple sine waves. More formally, it decomposes any periodic function or periodic
signal into the sum of a (possibly infinite) set of simple oscillating functions, namely sines
and cosines (or, equivalently, complex exponentials).
SRIT / UICM003 - TPDE / Fourier Series
SRIT / M & H / M. Vijaya Kumar 2
Periodic function
A function ( ) is said to be periodic with period , if and only if ( ) ( ), for
all values of , where is a positive constant.
Example:
are periodic functions with period .
Dirichlet’s Condition:
A function ( ) is defined in , it can be expanded as an infinite
trigonometric series of the form
∑ .
/
v
a. ( ) is defined and single valued except possibly at a finite number of points in ( ).
b. ( ) is periodic in ( ).
c. ( ) and ( ) are piece wise continuous in ( )
d. ( ) has no or finite number of maxima or minima in ( )
Fourier series:
The infinite trigonometric series
∑ .
/
is called Fourier series of ( ) which satisfies Dirichlet’s Conditions in ( ).
SRIT / UICM003 - TPDE / Fourier Series
SRIT / M & H / M. Vijaya Kumar 3
w
∫ ( )
∫ ( ) .
/
∫ ( ) .
/
The values are also called Euler’s co-efficient.
Convergence of Fourier Series
The Fourier Series ( ) in < < converges to
( ) ( ) if is a point of continuity.
( ) ( 0) ( − 0)
f a t f t u ty.
Problem: 1
Find the sum of Fourier Series ( ) ( − ) in 0 < < at
1. 0; 2. ; 3. .
Answer:
( ) 0, a point of discontinuity
u f . ( 0) ( − 0)
(0 0) (0 − 0)
( − 0 0) ( − 0 − 0)
) , a point of continuity.
u f . ( )
( )
( − )
0
0 𝑙 𝑙
0 𝑙 𝑙
SRIT / UICM003 - TPDE / Fourier Series
SRIT / M & H / M. Vijaya Kumar 4
(3) 2l, a point of discontinuity
u f . ( 0) ( − 0)
( 0) ( − 0)
( − 0) ( − − 0)
(− ) (− )
Problem: 2
Find the sum of Fourier Series ( ) | | in (− ) at 1. 0; 2. − ; 3. .
Answer:
( ) 0, a point of continuity
u f . ( )
(0)
|0|
0
( ) − , a point of discontinuity
u f . ( 0) ( − 0)
(− 0) (− − 0)
|− 0| |− − 0|
(3) , a point of discontinuity
u f . ( 0) ( − 0)
( 0) ( − 0)
0 𝑙 𝑙
𝜋 0
−𝜋
𝜋 0
−𝜋
𝜋 0
−𝜋
SRIT / UICM003 - TPDE / Fourier Series
SRIT / M & H / M. Vijaya Kumar 5
| 0| | − 0|
Problem: 3
Find the sum of Fourier Series of ( ) ,− ; − < < 0
; 0 < < at 0.
Answer:
0, a point of discontinuity
u f . ( 0) ( − 0)
(0 0) (0 − 0)
(− )
0 −
−
Problem: 4
Obtain the Fourier series for ( ) 2 f 0 < <
− f < < .
Answer:
The Fourier series of ( ) in < < is given by
( )
∑ .
/
Here 0;
0
The Fourier series of ( ) in (0 ) is
( )
∑( )
( )
−𝜋 0
𝜋
𝑥 𝜋 − 𝑥
0 𝜋 𝜋
𝑓(𝑥) 2𝑥; 0 < 𝑥 < 𝜋
𝜋 − 𝑥; 𝜋 < 𝑥 < 𝜋
SRIT / UICM003 - TPDE / Fourier Series
SRIT / M & H / M. Vijaya Kumar 6
To find :
∫ ( )
∫ ( )
, -
*∫
∫ ( − )
+ 0 ∫( ) ( )
( )
( )1
[(
)
(( − )
− )
]
[(
− 0) (0 − (
− )+]
*
+
.
To find :
∫ ( ) .
/
∫ ( )
, -
*∫
∫ ( − )
+
∫ − .
−
−
0 −
0
,[ (
* − .
−
/]
[( − ) (
* − (− ) .
−
/]
-
{0
1
− 0
1
}
SRIT / UICM003 - TPDE / Fourier Series
SRIT / M & H / M. Vijaya Kumar 7
,( − 0) − ( − )- , -
,((− ) − ) − ( − (− ) )- , (− ) -
,((− ) − ) ((− ) − )- , 0 -
,(− ) − -
{
−
;
To find :
∫ ( ) .
/
∫ ( )
, -
*∫
∫ ( − )
+
∫ − .
−
−
−
0 −
0
,[ .
−
/ − (
−
*]
[( − ) .−
/ − (− ) (
−
*]
-
{0 .
−
/1
0( − ) .−
/1
}
,−( − 0) − (0 − )-
.
The required Fourier series is
( )
∑ (
−
*
.
SRIT / UICM003 - TPDE / Fourier Series
SRIT / M & H / M. Vijaya Kumar 8
Problem: 5
Find the Fourier series expansion ( ) 2 f 0 < < f < <
.
Answer:
The Fourier series of ( ) in < < is given by
( )
∑ .
/
Here 0;
0
The Fourier series of ( ) in (0 ) is
( )
∑( )
( )
To find :
∫ ( )
(∫
∫
) , -
,( )
( ) -
,( − 0) ( − )-
, -
.
To find :
∫ ( ) .
/
(∫
∫
) , -
([
]
[
]
)
0 𝜋 𝜋
𝑓(𝑥) 2 ; 0 < 𝑥 < 𝜋 ; 𝜋 < 𝑥 < 𝜋
SRIT / UICM003 - TPDE / Fourier Series
SRIT / M & H / M. Vijaya Kumar 9
To find :
∫ ( ) .
/
(∫
∫
) , -
(0
−
1
0−
1
*
(− [
− 0
] − [
−
]*
(− *
(− ) −
+ − *
− (− )
+)
(− *
(− ) −
+ *
(− ) −
+)
,(− ) − -
{
−
The required Fourier series is
( )
∑ (
−
*
.
Problem: 6
Obtain the Fourier series for ( ) ( − ) 0 < .
Answer:
The Fourier series of ( ) in < < is given by
( )
∑ .
/
Here 0;
0
The Fourier series of ( ) in (0 ) is
( )
∑( )
( )
(𝝅− 𝒙)
𝜋 𝟎 𝜋
𝑓(𝑥) (𝜋 − 𝑥) ; 0 < 𝑥 < 𝜋
SRIT / UICM003 - TPDE / Fourier Series
SRIT / M & H / M. Vijaya Kumar 10
To find :
∫ ( )
∫ ( − )
*( − )
−3+
0 ∫( ) ( )
( )
( )1
−3 ,( − ) − -
−3 ,(− ) − -
−3 ,− -
.
To find :
∫ ( ) .
/
∫ ( − )
u u ∫ − .
( − )
( − )(− )
− ( − )
− (0 − )
−
0 −
[( − ) (
* − ,− ( − )- .
−
/ , - (
−
*]
−
*, ( − )- .
/+
SRIT / UICM003 - TPDE / Fourier Series
SRIT / M & H / M. Vijaya Kumar 11
−
[ ( − ) (
* − ( − 0) (
0
*]
−
[ (− ) (
* − (
*]
−
[
−
]
.
To find :
∫ ( ) .
/
∫ ( − )
u u ∫ − .
( − )
( − )(− )
− ( − )
−
− (0 − )
−
0
[( − ) .
−
/ − ,− ( − )- (
−
* , - .
/]
0−( − ) .
/ .
/1
*(−( − ) (
* (
*) − (−( − 0) (
0
* (
0
*)+
*(− (
* (
*) − (− (
* (
*)+
.
The required Fourier series is
( ) . 3
/
∑
SRIT / UICM003 - TPDE / Fourier Series
SRIT / M & H / M. Vijaya Kumar 12
( )
∑
.
Problem: 7
Expand ( ) ( − ) as a Fourier series in (0 ) and deduce the sum of the
3
Answer:
The Fourier series of ( ) in < < is given by
( )
∑ .
/
Here 0;
0
The Fourier series of ( ) in (0 ) is
( )
∑( )
( )
To find :
∫ ( )
∫ ( − )
*
−
3+
*(
( )
−
( )
3) − (0)+
* −
3+
* −
3+
.
𝑥( 𝜋 − 𝑥)
𝜋 𝟎 𝜋
𝑓(𝑥) 𝑥( 𝜋 − 𝑥); 0 < 𝑥 < 𝜋
SRIT / UICM003 - TPDE / Fourier Series
SRIT / M & H / M. Vijaya Kumar 13
To find :
∫ ( ) .
/
∫ ( − )
u u ∫ − .
−
−
− −
0 −
[( − ) (
* − , − - .
−
/ ,− - (
−
*]
*, ( − )- .
/+
[( − ) (
* − ( − 0) (
0
*]
[(− ) (
* − (
*]
[−
]
−
.
To find :
∫ ( ) .
/
∫ ( − )
SRIT / UICM003 - TPDE / Fourier Series
SRIT / M & H / M. Vijaya Kumar 14
u u ∫ − .
−
− −
− −
0
[( − ) .
−
/ − ,( − )- (
−
* ,− - .
/]
0−( − ) .
/ − .
/1
−
*(( − ) (
* (
*) − ( (
0
*)+
*(0
* − ( (
*)+
.
The required Fourier series is
( ) . 3
/
∑
−
( − )
− ∑
. ( )
Deduction:
Put 0, a point of discontinuity
u f . ( 0) ( − 0)
(0 0) (0 − 0)
( 0 − 0) ( 0 − 0)
0
0 𝜋 𝜋
SRIT / UICM003 - TPDE / Fourier Series
SRIT / M & H / M. Vijaya Kumar 15
( ) 0
3− ∑
0
∑
3
3
.
Problem: 8
Find the Fourier series for the function ( ) defined in (− ).
Answer:
The Fourier series of ( ) in < < is given by
( )
∑ .
/
Here − ;
−
The Fourier series of ( ) in (− ) is
( )
∑( )
( )
To find :
∫ ( )
∫
, -
, − - *
−
+
0 −𝝅
𝑒𝑥
𝜋
𝑓(𝑥) 𝑒𝑥; −𝜋 < 𝑥 < 𝜋
SRIT / UICM003 - TPDE / Fourier Series
SRIT / M & H / M. Vijaya Kumar 16
To find :
∫ ( ) .
/
∫
[
( )]
, f. u a ( )-
[
( ) −
( (− ))]
[
( ) −
( )]
( ), − -
(− )
( )
To find :
∫ ( ) .
/
∫
[
( − )]
, f. u a ( )-
[
( − ) −
( (− ) − )]
[
(− ) −
(− )]
−
( ), − -
− (− )
( )
(− )
( ) .
SRIT / UICM003 - TPDE / Fourier Series
SRIT / M & H / M. Vijaya Kumar 17
The required Fourier series is
( )
∑ (*
(− )
( ) + − *
(− )
( ) + )
(− )
∑ (
( )−
( )*
.
CHANGE OF INTERVAL
Problem: 9
Find the Fourier series for ( ) ( − ) in 0 < < , hence
u t at ∑
.
Answer:
The Fourier series for the function ( ) in
< < is given by
( )
∑ .
/
To find :
∫ ( )
∫ ( − )
*( − )
3(− )+
*(− ) − ( )
−3+
*−
−3+
.
𝑙 𝟎 𝑙
𝑓(𝑥) (𝑙 − 𝑥) ; 0 < 𝑥 < 𝑙
SRIT / UICM003 - TPDE / Fourier Series
SRIT / M & H / M. Vijaya Kumar 18
To find :
∫ ( ) .
/
∫ ( − ) .
/
u u ∫ − .
( − ) .
/
( − )(− )
− ( − )
.
/
− (0 − )
− .
/
0 − .
/
[( − ) .
/ ( − )(−
, (−
,]
−
*
( − )
+
−
,( − )( ) − ( − 0)( 0)-
−
,− − -
.
To find :
∫ ( ) .
/
∫ ( − ) .
/
SRIT / UICM003 - TPDE / Fourier Series
SRIT / M & H / M. Vijaya Kumar 19
u u ∫ − .
( − ) .
/
( − )(− )
− ( − )
− .
/
− (0 − )
− .
/
0 .
/
[( − ) .
−
/ ( − )(−
, (
,]
[ ,( )(− ) (
* ( )(
)-
−,( )(− 0) (
* ( )( 0) (
)-
]
*,
−
- − ,
−
-+
.
The required Fourier series is
( )
3
∑
.
/
( )
3
∑
.
/
. ( )
Deduction:
Put 0, a point of discontinuity, we have
SRIT / UICM003 - TPDE / Fourier Series
SRIT / M & H / M. Vijaya Kumar 20
u f . ( 0) ( − 0)
(0 0) (0 − 0)
( − 0 0) ( − 0 − 0)
( ) becomes
3
∑ (
0*
−
3
∑ (
*
3
∑ (
*
∑ (
*
.
Problem: 10
Obtain the Fourier series for ( ) of period and defined as follows
( ) { − 0 <
0 < <
Answer:
The Fourier series for the function ( ) in
< < is given by
( )
∑ .
/
To find :
∫ ( )
0 𝑙 𝑙
𝑙 𝟎
𝟎
𝑙
𝑓(𝑥) {𝑙 − 𝑥; 0 < 𝑥 < 𝑙0; 𝑙 < 𝑥 < 𝑙
𝑙 − 𝑥
SRIT / UICM003 - TPDE / Fourier Series
SRIT / M & H / M. Vijaya Kumar 21
∫ ( − )
*( − )
(− )+
*0 −
− +
*−
− +
.
To find :
∫ ( ) .
/
∫ ( − ) .
/
u u ∫ − .
− .
/
0 −
−
.
/
0 − .
/
[( − ).
/ − (− )(−
,]
*(0 (− )
) − (0 (− 0)
)+
, − (− ) -
{
SRIT / UICM003 - TPDE / Fourier Series
SRIT / M & H / M. Vijaya Kumar 22
To find :
∫ ( ) .
/
∫ ( − ) .
/
u u ∫ − .
− .
/
0 −
−
− .
/
0 − .
/
[( − ).
−
/ − (− )(−
,]
[*0+ − {( − 0)(− 0) (
*}]
*(
)+
.
The required Fourier series is
( )
∑ (
.
/*
.
∑
.
/
( )
∑
.
/
∑
.
/
.
SRIT / UICM003 - TPDE / Fourier Series
SRIT / M & H / M. Vijaya Kumar 23
Odd and Even Function
A function ( ) is said to be even if (− ) ( )
E.g. ( − ) ..
A function ( ) is said to be odd if (− ) − ( )
E.g. ..
Note:
u f ( ) 2 −
t ta ( ) { ( )
( )
f (− ) ( ) t ( ) v
f (− ) − ( ) t ( )
Remarks:
A function ( ) is said to be even, then 0.
A function ( ) is said to be odd then 0; 0.
Problem: 11
Obtain the Fourier series for ( ) in (− ).
Answer:
The Fourier series of ( ) in < < is given by
( )
∑ .
/
Here − ;
−
The Fourier series of ( ) in (− ) is
( )
∑( )
( )
Since the given interval is symmetric. So we can verify odd and even function. Here
is even function, but is odd function. So we split the function in to two parts as
show below.
𝑥 𝑥
0 −𝜋 𝜋
SRIT / UICM003 - TPDE / Fourier Series
SRIT / M & H / M. Vijaya Kumar 24
To find :
∫ ( )
∫ ( )
∫
∫ ( )
0 ( fu t )
(
3)
*(
3) − 0+
(
).
To find :
∫ ( ) .
/
∫ ( )
∫
∫ ( )
0 ( fu t )
u u ∫ − .
−
0 −
[( ) (
* − .
−
/ ( ) (
−
*]
0 .
/1
SRIT / UICM003 - TPDE / Fourier Series
SRIT / M & H / M. Vijaya Kumar 25
, − 0-
(− ) .
To find :
∫ ( ) .
/
∫ ( )
∫
0 ( v fu t )
∫
u u ∫ − .
−
0 −
[ .
−
/ − (
−
*]
0 .
−
/1
,− − 0-
−
(− ) .
u u
( ) (
3 *
∑ (
(− ) −
(− ) *
( )
∑ (
(− )
(− )
)
.
SRIT / UICM003 - TPDE / Fourier Series
SRIT / M & H / M. Vijaya Kumar 26
Problem: 12
Obtain the Fourier series for ( ) 2 − < 0 − 0 <
.
u t at
3
.
Answer:
Since the given interval is symmetric. So we
can verify odd and even function.
Take ( ) 2 −
{ ( )
( )
(− ) (− ) − ( )
( ) is even function, then 0.
The Fourier series of ( ) in (− ) is
( )
∑ .
/
The Fourier series of ( ) in (− ) is
( )
∑
. . ( )
To find :
∫ ( )
∫ ( − )
∫ ( − )
( −
)
(( − ) − 0)
.
𝜋 𝑥
−𝜋 0 𝜋
𝜋 − 𝑥
SRIT / UICM003 - TPDE / Fourier Series
SRIT / M & H / M. Vijaya Kumar 27
To find :
∫ ( ) .
/
∫ ( − )
u u ∫ − .
−
−
0 −
[( − ) (
* − (− ) .
−
/]
0− .
/1
−
( − 0)
−
,(− ) − -
{
u u
( ) ∑
. . ( )
Deduction:
Put 0, a point of continuity
u f . ( )
(0)
− (0)
( ) becomes
−𝜋 0
𝜋
SRIT / UICM003 - TPDE / Fourier Series
SRIT / M & H / M. Vijaya Kumar 28
∑
0
∑
(
3
*
3
Problem: 13
Obtain the Fourier series of period for the function ( ) in (− ).
u t u f t
3 a −
3 − .
Answer:
Since the given interval is symmetric. So we
can verify odd and even function.
(− ) (− ) ( )
( ) is even function.
0.
The Fourier series of ( ) (− ) is
( )
∑ .
/
( )
∑
To find :
∫ ( )
∫
(
3)
𝑥
−𝜋 0 𝜋
𝑓(𝑥) 𝑥 in (−𝜋 𝜋)
SRIT / UICM003 - TPDE / Fourier Series
SRIT / M & H / M. Vijaya Kumar 29
(
3)
.
To find :
∫ ( ) .
/
∫
u u ∫ − .
−
0 −
[ (
* − .
−
/ (
−
*]
0 .
/1
, − 0-
(− ) .
u u
( )
3
∑ (
(− ) *
( )
∑
(− )
. ( )
SRIT / UICM003 - TPDE / Fourier Series
SRIT / M & H / M. Vijaya Kumar 30
Deduction: 1
Put , a point of discontinuity
u f . ( 0) ( − 0)
( 0) ( − 0)
( 0) ( − 0)
( ) becomes
3 [−
−
3 3 ]
−
3 [− (− )
( ) −
3 (− ) ]
3 [
3 ]
3
Deduction: 2
Put 0 , a point of continuity
u f . ( )
(0)
0
( ) becomes
0
3 ∑ (
(− )
0)
3 ∑ (
(− )
)
0
∑ ((− )
)
−
3
−𝜋 0
𝜋
−𝜋 0
𝜋
SRIT / UICM003 - TPDE / Fourier Series
SRIT / M & H / M. Vijaya Kumar 31
[−
−
3 ] −
3
−
3 −
Problem: 14
Obtain the Fourier series of the function ( ) | | – < < . Deduce the
u f t
3
.
Answer:
Since the given interval is symmetric. So
we can verify odd and even function.
(− ) |− | | | ( )
( ) is even function.
0.
The Fourier series of ( ) (− ) is
( )
∑ .
/
( )
∑
. ( )
To find :
∫ ( )
∫
(
)
(
)
.
−𝜋 0 𝜋
𝑓(𝑥) |𝑥| −𝜋 < 𝑥 < 𝜋
|𝑥|
SRIT / UICM003 - TPDE / Fourier Series
SRIT / M & H / M. Vijaya Kumar 32
To find :
∫ ( ) .
/
∫
u u ∫ −
0 −
[ (
* − .
−
/]
0
1
, − 0-
,(− ) − -
{ −
u u
( )
∑ (
−
*
| |
−
∑
. . ( )
Deduction: Put 0 a point of continuity
u f . ( )
(0)
0
−𝜋 0
𝜋
SRIT / UICM003 - TPDE / Fourier Series
SRIT / M & H / M. Vijaya Kumar 33
0
−
∑ (
0*
∑
3
3
Problem: 15
ta t u x a f ( ) {
− 0
−
0
u t at
3
. .
Answer:
Since the given interval is symmetric. So
we can verify odd and even function.
a ( ) {
−
{ ( )
( )
(− ) −
( )
( ) is even function, then 0.
The Fourier series of ( ) in (− ) is
( )
∑ .
/
( )
∑
To find :
∫ ( )
𝑥
𝜋
−𝜋 0 𝜋
− 𝑥
𝜋
SRIT / UICM003 - TPDE / Fourier Series
SRIT / M & H / M. Vijaya Kumar 34
∫ ( −
*
* −
+
*( −
) − (0)+
,0-
.
To find :
∫ ( ) .
/
∫ ( −
*
x
u u ∫ − .
−
−
0 −
[( −
* (
* − (−
* .
−
/]
−
[(
* .
/]
−
, − 0-
−
,(− ) − -
, − (− ) -
{
SRIT / UICM003 - TPDE / Fourier Series
SRIT / M & H / M. Vijaya Kumar 35
u u
( ) ∑ (
*
..
( )
Deduction:
Put 0 , a point of continuity
u f . ( )
(0)
− (0)
( ) becomes
∑ (
0*
..
[
3
]
3
.
Problem: 16
Find the Fourier series for ( ) (− ).
Answer:
Since the given interval is symmetric. So
we can verify odd and even function.
(− ) (− ) ( )
( ) is even function.
0.
The Fourier series of ( ) (− ) is
( )
∑ .
/
−𝜋 0
𝜋
𝑥
0 −𝒍 𝑙
𝑓(𝑥) 𝑥 ; −𝑙 < 𝑥 < 𝑙
SRIT / UICM003 - TPDE / Fourier Series
SRIT / M & H / M. Vijaya Kumar 36
To find :
∫ ( )
∫
(
3)
(
3)
.
To find :
∫ ( ) .
/
∫ .
/
u u ∫ − .
.
/
.
/
. /
− .
/
(
*
0 − .
/
(
*
[ .
/ − (−
, ( )(−
,]
SRIT / UICM003 - TPDE / Fourier Series
SRIT / M & H / M. Vijaya Kumar 37
*
+
, − 0-
(− ) .
u u
( )
3
∑ (
(− ) .
/)
( )
∑
(− )
.
/
.
Problem: 17
Find the Fourier series expansion for the function ( ) define as follows
( ) 2− − < < 0 0 < <
.
Answer:
Since the given interval is symmetric. So we
can verify odd and even function.
a ( ) 2 −
{ ( )
( )
(− ) − (− ) ( )
( ) is even function, then 0.
The Fourier series of ( ) in (− ) is
( )
∑ .
/
( )
∑
. ( )
To find :
∫ ( )
𝑥
−𝜋 0 𝜋
𝑓(𝑥) 2−𝑥 ; −𝜋 < 𝑥 < 0𝑥 ; 0 < 𝑥 < 𝜋
− 𝑥
SRIT / UICM003 - TPDE / Fourier Series
SRIT / M & H / M. Vijaya Kumar 38
∫( )
*
+
(
)
.
To find :
∫ ( ) .
/
∫ ( )
u u ∫ − .
0 −
[( ) (
* − ( ) .
−
/]
0
1
, − 0-
,(− ) − -
{
−
u u
( )
∑
−
..
.
SRIT / UICM003 - TPDE / Fourier Series
SRIT / M & H / M. Vijaya Kumar 39
Problem: 18
Find the Fourier series expansion for the function ( ) define as follows
( ) { − < < 0 − 0 < <
a u t at
3
. . .
.
Answer:
Since the given interval is symmetric. So we
can verify odd and even function.
a ( ) 2 −
{ ( )
( )
(− ) − ( )
( ) is even function, then 0.
The Fourier series of ( ) in (− ) is
( )
∑ .
/
. ( )
To find :
∫ ( )
∫ ( − )
*( − )
(− )+
*0 −
− +
*−
− +
.
To find :
∫ ( ) .
/
∫ ( − ) .
/
𝑙 𝑥
−𝑙 0 𝑙
𝑓(𝑥) {𝑙 𝑥; −𝑙 < 𝑥 < 0𝑙 − 𝑥; 0 < 𝑥 < 𝑙
𝑙 − 𝑥
SRIT / UICM003 - TPDE / Fourier Series
SRIT / M & H / M. Vijaya Kumar 40
u u ∫ − .
− .
/
− .
/
. /
0 − .
/
(
*
[( − ).
/ − (− )(−
,]
−
*
+
−
, − 0-
, − (− ) -
{
u u
( )
∑
.
/
..
( )
∑
.
/
.
( )
Deduction:
Put 0 a point of discontinuity
u f . ( 0) ( − 0)
(0 0) (0 − 0)
−𝑙 0
𝑙
SRIT / UICM003 - TPDE / Fourier Series
SRIT / M & H / M. Vijaya Kumar 41
0 − 0
( ) becomes
∑
0
.
−
∑
.
∑
.
. . .
.
Problem: 19
Expand the function ( ) as a Fourier series in the interval (− ).
Answer:
Since the given interval is symmetric. So we
can verify odd and even function.
(− ) − (− )
( )
( ) is even function, then 0.
The Fourier series of ( ) in (− ) is
( )
∑
. ( )
To find :
∫ ( )
−𝜋 0 𝜋
𝑓(𝑥) 𝑥 𝑥 ; (−𝜋 𝜋)
𝑥 𝑥
SRIT / UICM003 - TPDE / Fourier Series
SRIT / M & H / M. Vijaya Kumar 42
∫
, (− ) − (− )-
,− − 0-
,− (− )-
.
To find :
∫ ( )
∫
∫
∫ (
( ) − ( − )
)
∫ , ( ) − ( − ) -
[( (
− ( )
) – (
− ( )
( ) )+ − ( (
− ( − )
− ) – (
− ( − )
( − ) )+]
* (
− ( )
) − (
− ( − )
− )+
* (
− ( )
) (
( − )
− )+
* −(− )
(− )
− + , ( ) ( − ) (− ) -
(− ) [ −
− ]
(− ) [ –
− ]
(− ) [
− ]
SRIT / UICM003 - TPDE / Fourier Series
SRIT / M & H / M. Vijaya Kumar 43
(− ) [
− ]
When , we have
∫
∫
, -
∫
[ (
−
* − (
−
*]
[ (
−
* − 0]
−
.
u u
( )
∑
( )
−
∑
(− )
−
.
Problem: 20
Expand the function ( ) as a Fourier series in the interval (− ).
Answer:
Since the given interval is symmetric. So we
can verify odd and even function.
(− ) − (− )
−
− ( )
( ) is odd function, then
0.
The Fourier series of ( ) in (− ) is
−𝜋 0 𝜋
𝑓(𝑥) 𝑥 𝑥 ; (−𝜋 𝜋)
𝑥 𝑥
SRIT / UICM003 - TPDE / Fourier Series
SRIT / M & H / M. Vijaya Kumar 44
( ) ∑
. ( )
To find :
∫ ( )
∫
∫
∫ (
( ) ( − )
)
∫ , ( ) ( − ) -
[( (
− ( )
) − (
− ( )
( ) )+ ( (
− ( − )
− ) − (
− ( − )
( − ) )+]
* (
− ( )
) (
− ( − )
− )+
−
* (
( )
) (
( − )
− )+
−
* (− )
(− )
− + , ⟨ ( ) ( − ) (− ) ⟩-
−(− ) [
− ]
−(− ) [ −
− ]
−(− ) (− ) [
− ]
(− ) [
− ]
When , we have
∫
∫
, -
SRIT / UICM003 - TPDE / Fourier Series
SRIT / M & H / M. Vijaya Kumar 45
∫
[ (
−
* − (
−
*]
[ (
−
* − 0]
−
.
u u
( ) ∑
. ( )
( ) −
∑
(− )
−
.
Problem: 21
Obtain the Fourier series expansion of ( ) | | t t va (− ).
Answer:
Since the given interval is symmetric. So we
can verify odd and even function.
(− ) | |
( )
( ) is even function, then 0.
The Fourier series of ( ) in (− ) is
( )
∑
. ( )
To find :
∫ ( )
∫
,− -
−𝜋 0 𝜋
𝑓(𝑥) | 𝑥|; (−𝜋 𝜋)
SRIT / UICM003 - TPDE / Fourier Series
SRIT / M & H / M. Vijaya Kumar 46
−
, − 0-
−
,− − -
.
To find :
∫ ( )
∫
∫
∫ (
( ) − ( − )
)
∫ , ( ) − ( − ) -
*− ( )
−
− ( − )
− +
*(
− ( )
( − )
− ) − (
− 0
0
− *+
*(
−(− )
(− )
− ) − (
−
− *+
*−(− )
(− )
−
−
− +
[
−
((− ) − )
− ((− ) − )]
((− ) − )
[
− −
]
((− ) − )
[ −
− ]
SRIT / UICM003 - TPDE / Fourier Series
SRIT / M & H / M. Vijaya Kumar 47
(−(− ) − )
[
− ]
−
( − ) ,(− ) -
−
− ,(− ) - f
2
f −
( − ) f v
When , we have
∫
∫
, -
∫
[(
−
*]
−
[ (
* − (
0
*]
[
−
]
.
u u
( )
∑
( )
− ∑
( − )
..
.
Problem: 22
Obtain the Fourier series expansion of ( ) | | t t va (− ).
Answer:
SRIT / UICM003 - TPDE / Fourier Series
SRIT / M & H / M. Vijaya Kumar 48
Since the given interval is symmetric. So
we can verify odd and even function.
(− ) | |
( )
( ) is even function, then
0.
The Fourier series of ( ) in (− ) is
( )
∑
. ( )
( ) | | { 0
−
The Fourier series of ( ) (− ) is
( )
∑
− − − −( )
To find :
∫ ( )
∫ | |
[∫
∫ −
]
*( )
− ( )
+
,( − 0) − (0 − )-
.
−𝜋 0 𝜋
𝑓(𝑥) | 𝑥|; (−𝜋 𝜋)
𝜋
−𝜋
SRIT / UICM003 - TPDE / Fourier Series
SRIT / M & H / M. Vijaya Kumar 49
To find :
∫ ( )
∫ | |
[∫
∫ −
]
[∫
− ∫
]
[∫ (
( ) ( − )
)
− ∫ ( ( ) ( − )
)
]
[∫ ( ( ) ( − ) )
− ∫ ( ( ) ( − ) )
]
0(
( )
( − )
− )
− ( ( )
( − )
− )
1
0.
( )
( − )
− − 0/ − .0 −
( )
−
( − )
− /1
0 ( )
( − )
−
( )
( − )
− 1
0 ( )
( − )
− 1
0
−
− 1
0
−
− 1
[
−
− ]
SRIT / UICM003 - TPDE / Fourier Series
SRIT / M & H / M. Vijaya Kumar 50
[−
− ]
−
( − )
When , we have
∫ | |
[∫
− ∫
]
[∫
− ∫
]
[∫ (
*
− ∫ (
*
]
[∫ ( )
− ∫ ( )
]
[
]
−
[
]
0
− 01 −
0
− .
/1
0
1 −
0 − .
/1
−
.
u u
( )
∑
( )
∑ .−
( − ) /
( )
−
∑ .
( − ) /
SRIT / UICM003 - TPDE / Fourier Series
SRIT / M & H / M. Vijaya Kumar 51
Half range expansion
Cosine series:
The half range cosine series for the function ( ) in the interval (0 ) is defined as
( )
∑ .
/
w
∫ ( )
a
∫ ( ) .
/
Sine series:
The half range sine series for the function ( ) in the interval (0 ) is defined as
( ) ∑ .
/
w
∫ ( ) .
/
Problem: 23
Find the half range sine series for the function ( ) ( − ) (0 ).
Answer:
The half range sine series for the function ( ) in
the interval (0 ) is given by
( ) ∑ .
/
w
∫ ( ) .
/
Here
∫ ( − )
∫ ( − )
𝑥(𝜋 − 𝑥)
𝟎 𝜋
𝑓(𝑥) 𝑥(𝜋 − 𝑥); 0 < 𝑥 < 𝜋
SRIT / UICM003 - TPDE / Fourier Series
SRIT / M & H / M. Vijaya Kumar 52
∫ − .
−
− −
0 − −
0
[( − ) .
–
/ − ( − ) (
–
* (− ) .
/]
−
0( − ) .
/ .
/1
−
[.0
/ − (0
0
*]
−
, − 0-
−
,(− ) − -
{
f
0 f v
u u
( ) ∑ (
*
.
Problem: 24
t a f a f t fu t ( ) { 0
( − )
.
Answer:
SRIT / UICM003 - TPDE / Fourier Series
SRIT / M & H / M. Vijaya Kumar 53
The half range sine series for the function
( ) in the interval (0 ) is given by
( ) ∑ .
/
w
∫ ( ) .
/
Here
∫ ( )
[∫
∫ ( − )
]
∫ − .
−
−
−
0 −
0
([ .
–
/ − (
–
*]
[( − ) .
–
/ − (− ) (
–
*]
+
(0.
−
/ − (0)1 [(0) − (. −
/−
−
,],
.−
/
.
/
.
𝑓(𝑥) {𝑘𝑥; 0 < 𝑥 <
𝜋
𝑘(𝜋 − 𝑥); 𝜋
< 𝑥 < 𝜋
𝑥 𝜋 − 𝑥
0 𝜋
𝜋
SRIT / UICM003 - TPDE / Fourier Series
SRIT / M & H / M. Vijaya Kumar 54
u u
( ) ∑
.
Problem: 25
Find the half range sine series for the function ( ) ( − ) in 0 .
Answer:
The half range sine series for the function ( )
in the interval (0 ) is given by
( ) ∑ .
/
w
∫ ( ) .
/
∫ ( − ) .
/
∫ ( − ) .
/
u u ∫ − .
− .
/
− − .
/
. /
− .
/
(
*
0 .
/
(
*
[( − ).
−
/ − ( − )(−
, ( )(
,]
𝑘𝑥(𝑥 − 𝑙)
𝟎 𝑙
𝑓(𝑥) 𝑘𝑥(𝑥 − 𝑙); 0 < 𝑥 < 𝑙
SRIT / UICM003 - TPDE / Fourier Series
SRIT / M & H / M. Vijaya Kumar 55
[
(
0 − 0 (
,
)
−
(
0 − 0 ( 0
,
)
]
*
−
0+
,(− ) − -
,(− ) − -
{ −
u u
( ) ∑ (−
) .
/
.
Root mean square value
What is the meaning of RMS value?
The RMS value is the effective value of a varying voltage or current. It is the
equivalent steady DC (constant) value which gives the same effect. For example, a lamp
connected to a 6V RMS AC supply will shine with the same brightness when connected to a
steady 6V DC supply.
Definition: RMS Value
The Root mean square value or effective value over the interval ( ) is defined by
. . √∫ , ( )-
−
− ∫ , ( )-
SRIT / UICM003 - TPDE / Fourier Series
SRIT / M & H / M. Vijaya Kumar 56
Parseval’s Identity
Full range ( ):
Let ( ) be a periodic function with period defined in the interval ( ), we
define the Parseval’s Identity as follows
∫ , ( )-
∑(
)
Half range Cosine series in ( ):
∫ , ( )-
∑
Half range Sine series in ( ):
∫ , ( )-
∑
Problem: 26
Find the Fourier series of ( ) in − < < .
u t va u f
3 .
Answer:
Since the given interval is symmetric. So we can verify odd and even function.
(− ) (− ) ( )
( ) is even
.
The Fourier series of ( ) (− ) is
( )
∑ .
/
( )
Here − ; ; − ; ;
SRIT / UICM003 - TPDE / Fourier Series
SRIT / M & H / M. Vijaya Kumar 57
To find :
∫ ( )
, -
∫
(
3)
(
3− 0)
.
To find :
∫ ( ) .
/
∫
u u ∫ − .
−
− −
0 −
[ (
* − .
−
/ (− ) (
−
*]
0 .
/1
, − 0-
(− )
SRIT / UICM003 - TPDE / Fourier Series
SRIT / M & H / M. Vijaya Kumar 58
u u
( ) (
3 *
∑ [
(− ) ]
.
( )
∑
(− )
.
Deduction:
Pa va ’ t ty f Fourier series even function in (− ) is given by
∫ , ( )-
∑
∫ ( )
(
3 *
∑ [
(− ) ]
∑
(− )
∫
∑
*
+
∑
*
+ −
∑
−
∑
(
*
3
0.
Problem: 27
Obtain the Fourier cosine series expansion for ( ) ( − ) in 0 < < .
u t va u f
3 .
SRIT / UICM003 - TPDE / Fourier Series
SRIT / M & H / M. Vijaya Kumar 59
Answer:
The Fourier cosine series of ( ) (0 ) is
( )
∑ .
/
. ( )
To find :
∫ ( )
, -
∫ ( − )
*
−
3+
*(
−
3) − 0+
*
+
.
To find :
∫ ( ) .
/
∫ ( − )
u u ∫ − .
−
−
− −
0 −
[( − ) (
* − ( − ) .
−
/ (− ) (
−
*]
SRIT / UICM003 - TPDE / Fourier Series
SRIT / M & H / M. Vijaya Kumar 60
0( − ) .
/1
,(− ) − 0-
−
,(− ) -
{ −
u u
( ) (
3 *
∑ [
−
]
.
( )
− ∑
.
Deduction:
Pa va ’ t ty f a f a (0 ) is given by
∫ , ( )-
∑
∫ ( − )
(
3 *
∑ (
−
*
∑
∫ ( − )
∑
*
3
−
+
∑
*
3
−
− 0+ −
∑
*
30+ −
[
]
−
SRIT / UICM003 - TPDE / Fourier Series
SRIT / M & H / M. Vijaya Kumar 61
[
( )
( 3) ]
0
[
3 ]
0
[
3 ]
0
3
0.
Problem: 28
Obtain the Fourier cosine series expansion for ( ) in 0 < < .
u t va u f
3
.
Answer:
The Fourier cosine series of ( ) (0 )
is given by
( )
∑ .
/
To find :
∫ ( )
∫
, -
(
)
(
− 0*
.
To find :
∫ ( ) .
/
∫ .
/
, -
𝑥
0
𝑓(𝑥) 𝑥; 0 < 𝑥 <
SRIT / UICM003 - TPDE / Fourier Series
SRIT / M & H / M. Vijaya Kumar 62
u u ∫ − .
.
/
.
/
0 − .
/
[ .
. /
/ − .− .
/
/]
[ .
/
]
*
( − 0)+
,(− ) − -
,(− ) − -
{ −
The required Fourier series is
( )
∑
−
.
Deduction:
By Parseval’s Identity, we have
∫ , ( )-
∑
∫ ( )
∑ (
−
*
SRIT / UICM003 - TPDE / Fourier Series
SRIT / M & H / M. Vijaya Kumar 63
(
3)
∑
(
3* −
∑
3
∑
3
.
Harmonic Analysis
The process of finding the Fourier series for the function given by numerical
value is known as harmonic analysis. In harmonic analysis the Fourier co-efficient
and of the function ( ) in < < are defined by
(∑
) ; (
∑
) ; (
∑
)
( )
(
* (
* (
* (
*
First Harmonic
(
* (
*
Amplitude of First Harmonic
√
Second Harmonic
(
* (
*
Amplitude of Second Harmonic
√
SRIT / UICM003 - TPDE / Fourier Series
SRIT / M & H / M. Vijaya Kumar 64
Third Harmonic
(3
* (
3
*
Amplitude of Third Harmonic
√
Type I Given data in Angle Form ( )
Problem: 29
Determine the first two harmonics of the Fourier series for the following data:
0
3
3
3
3
. .30 .0 .30 −0. −0.
Answer:
The Fourier series expansion upto second harmonic is given by
( )
(
* (
* (
* (
* ( )
Since first value and last value must be same.
0
3
3
3
3
( ) . .30 .0 .30 −0. −0. .
Here 0; ; 0 ;
Equation ( ) becomes
( )
( )
Angle 0
3
3
3
3
degree 0
0
3
0
0
3
0
0
0
3
0
0
3
300
0
3 0
Here (No. of intervals )
Since first value and last value must be same. We omit last column .
We form the table as follows
SRIT / UICM003 - TPDE / Fourier Series
SRIT / M & H / M. Vijaya Kumar 65
degree
. . 0 . 0
. 0. . 3 −0. . 3
. −0. 3 0. −0. 3 −0.
. − .3 0.00 .30 0.00
− . 0. 0. 0. −0.
− . −0. 3 0. 0. 3 0.
∑ . . 3.0 . −0.3
From table we see that
∑ . ; ∑ . ; ∑ 3.0 ;
∑ . ; ∑ −0.3
(∑
) (
.
* .
(∑
) (
.
* 0.3
(∑
) (
.
* .0
(∑
) (
3.0
* 0.
(∑
) (
−0.3
* −0.
Using the value a in equation ( ) we get
( )
3 3
.
0.3 .0 0. (−0. )
0. 0.3 .0 0. − 0. .
SRIT / UICM003 - TPDE / Fourier Series
SRIT / M & H / M. Vijaya Kumar 66
Problem: 30
Find the Fourier series expansion of period for the function ( ) which is
defined in (0 ) by means of the table of values given below. Find the series upto the third
harmonic.
0
3
3
3
3
. . . . .
Answer:
The Fourier series expansion upto third harmonic is given by
( )
(
* (
* (
* (
*
(3
* (
3
* ( )
Here 0; ; 0 ;
Equation ( ) becomes
( )
3 3 ( )
Angle 0
3
3
3
3
degree 0
0
3
0
0
3
0
0
0
3
0
0
3
300
0
3 0
Here (No. of intervals)
Since first value and last value must be same. We omit last column .
We form the table as follows
SRIT / UICM003 - TPDE / Fourier Series
SRIT / M & H / M. Vijaya Kumar 67
degree
. 0 0 0
. 0. . −0. . − . 0
. −0. . −0. − . . 0
. − . 0 . 0 − . 0
. −0. − .3 −0. .3 . 0
. 0. − .0 −0. − .0 − . 0
∑ . − . . − . − . .
From table we see that
∑ . ; ∑ − . ; ∑ 0. ; ∑ −0.3;
∑ −0. ; ∑ 3 0. ; ∑ 3 0
(∑
) (
.
* .
(∑
) (
− .
* −0.3
(∑
) (−
0.3
* −0.
(∑ 3
)
(0. ) 0.03
(∑
) (
0.
* 0.
(∑
) (
−0.
* −0.0
(∑ 3
) (
0
* 0
Using the value a in equation ( ) we get
( )
3 3
.
(−0.3 ) (−0. ) (0.03) 3 (0. ) (−0.0 ) (0) 3
. − 0.3 − 0. 0.03 3 0. − 0.0
SRIT / UICM003 - TPDE / Fourier Series
SRIT / M & H / M. Vijaya Kumar 68
Type II Given data in form
Problem: 31
The following table gives the variations of a periodic function over a period T:
0
3
3
( ) . .3 .0 .3 −0. −0. .
w t at ( ) 0. 0.3 .00 w
.
Answer:
0; ; 0 ;
The Fourier series expansion upto first harmonic is given by
( )
.
/ .
/ ( )
.
/ .
/
(
* (
* [ a
]
( )
( )
in 0
3
3
degree 0
0
(
*
0
0
(
3*
0
0
(
*
0
0
(
3*
0
0
(
*
300
0
( )
3 0
Here (No. of intervals)
Since first value and last value must be same. We omit last column .
We form the table as follows
SRIT / UICM003 - TPDE / Fourier Series
SRIT / M & H / M. Vijaya Kumar 69
. . 0
. 0. .
. −0. 0. 0 3
. − .3 0
− . 0. 0.
− . −0. 0.
Total 4.5 1.12 3.013
From table we see that
∑ . ; ∑ . ; ∑ 3.0 3
(∑
) (
.
* .
(∑
) (
.
* 0.3
(∑
) (
3.0 3
* .00
Using the value a in equation ( ) we get
( )
.
0.3 .00
0. 0.3 (
* .00 (
*. [ au
]
Type II Given data in Length Form ( )
Problem: 32
Find the Fourier series as far as the second harmonic to represent the function given
in the following data.
0 1 2 3 4 5
( ) 4 8 15 7 6 2
Answer:
SRIT / UICM003 - TPDE / Fourier Series
SRIT / M & H / M. Vijaya Kumar 70
Since first value and last value must be same.
0 1 2 3 4 5 6
( ) 4 8 15 7 6 2 4
Here 0; 6; 0 ;
The Fourier series expansion upto second harmonic is given by
( )
.
/ .
/ .
/ .
/
.
3/ .
3/ .
3/ .
3/ ; 0
31
( )
In 0 1 2 3 4 5 6
3
degree 0
0
3( )
0
0
3( )
0
0
3(3)
0
0
3( )
0
0
3( )
300
0
3( )
3 0
Here (No. of intervals
The Fourier series takes the form
Since first value and last value must be same. We omit last column .
We form the table as follows
4 4 0 4 0
8 . 3 − . 3
15 − . . − . − .
7 − 0 0
6 −3 − . −3 .
2 − . 3 − − . 3
Total − . . − . − .
From table we see that
SRIT / UICM003 - TPDE / Fourier Series
SRIT / M & H / M. Vijaya Kumar 71
∑ ; ∑ − . ; ∑ . ; ∑ − . ;
∑ − .
(∑
) (
*
(∑
) (
− .
* − . 3
(∑
) (
− .
* − .
(∑
) (
.
* .33
(∑
) (
− .
* −0.
Using the value and in equation ( ) we get
( )
− . 3 .33 − . − 0.
− . .
/ . .
/ − . .
/ − . .
/ .
Problem: 33
Find the constant of the first and the co-efficient of sine and cosine term in the
Fourier expansion of y as given in the following table.
0 1 2 3 4 5
( ) 9 18 24 28 26 20
Answer:
Since first value and last value must be same.
0 1 2 3 4 5 6
( ) 9 18 24 28 26 20 9
Here 0; 6; 0 ;
The Fourier series expansion upto second harmonic is given by
SRIT / UICM003 - TPDE / Fourier Series
SRIT / M & H / M. Vijaya Kumar 72
( )
.
/ .
/ .
/ .
/
.
3/ .
3/ .
3/ .
3/ ; 0
31
( )
In 0 1 2 3 4 5 6
3
degree 0
0
3( )
0
0
3( )
0
0
3(3)
0
0
3( )
0
0
3( )
300
0
3( )
3 0
Here (No. of intervals)
The Fourier series takes the form
Since first value and last value must be same. We omit last column .
We form the table as follows
9 0 0
18 . − .
24 − 0. − − 0.
28 − 0 0
26 − 3 − . − 3 .
20 0 − .3 − 0 − .3
Total − −3. − 0.0
From table we see that
∑ ; ∑ − ; ∑ −3. ;
∑ − ; ∑ 0.0
(∑
) (
* .
SRIT / UICM003 - TPDE / Fourier Series
SRIT / M & H / M. Vijaya Kumar 73
(∑
) (
−
* − .33
(∑
) (
−
* − .
(∑
) (
−3.
* − .33
(∑
) (
0.0
* 0.003
Using the value and in equation ( ) we get
( )
.
− .33 − .33 − . 0.003
0. 3 − .33 − .33 − . 0.003 .
Short Answer Questions
1. Define periodic function with examples.
Answer:
A function ( ) is said to be periodic with period , if and only if ( ) ( ),
for all values of , where is a positive constant.
Eg:
is a periodic function with period .
. . ( )
2. State the Dirichlet’s conditions for Fourier series.
(OR)
3. Write the conditions for a function ( ) to satisfy for the existence of a Fourier series.
Answer:
A function ( ) is defined in , it can be expanded as an infinite
trigonometric series of the form
∑ .
/
v
SRIT / UICM003 - TPDE / Fourier Series
SRIT / M & H / M. Vijaya Kumar 74
i. ( ) is defined and single valued except possibly at a finite number of points in ( ).
ii. ( ) is periodic in ( ).
iii. ( ) and ( ) are piece wise continuous in ( )
iv. ( ) has no or finite number of maxima or minima in ( )
4. Write the Fourier coefficients and .
Answer:
∫ ( )
∫ ( ) .
/
∫ ( ) .
/
.
5. Find the constant term in the expansion of as a Fourier series in the
interval (− ).
Answer:
ta t t
( ) a v fu t
∫
∫
( )
[
]
, 0-
ta t t
.
SRIT / UICM003 - TPDE / Fourier Series
SRIT / M & H / M. Vijaya Kumar 75
6. Find the constant term in the Fourier expansion of expansion of ( ) in (0 ).
Answer:
ta t t
∫
, -
−
−
ta t t
−
.
7. Find the value of from ( ) in (0 ).
Answer:
. . t
∫ ( )
∫ ( )
,( )( ) − ( )(− ) ( )(− )-
[( ( )) − ( 0) 0]
,( ) − -
.
.
8. Can ta be expressed as a Fourier series.
Answer:
ta cannot be expressed as a Fourier series, because it has infinite number of
discontinuities
SRIT / UICM003 - TPDE / Fourier Series
SRIT / M & H / M. Vijaya Kumar 76
9. Define Root mean square value of the function ( ) over the interval ( ).
Answer:
Root Mean Square value of ( ) over the interval ( ) is
√
− ∫, ( )-
10. Find the root mean square value of ( ) in (– ).
Answer:
. . √
− ∫, ( )-
√
− (− )∫( )
√
*
+
√
( − (− )
) √
(
) √
t a ua va u
√ .
11. Find the root mean square value of ( ) in (0 ).
Answer:
. . √
− ∫( ( ))
√
− 0∫( )
√
*
+
√
(
) √
t a ua va u
√ .
12. Find the root mean square value of the function ( ) in (– ) .
Answer:
SRIT / UICM003 - TPDE / Fourier Series
SRIT / M & H / M. Vijaya Kumar 77
. . √
− ∫( ( ))
√
− (− )∫( )
√
*
3+
√
( − (− )
3) √
(
3) √
3
t a ua va u
√3.
13. Find the sum of the Fourier series for ( ) 2 ; 0 < ; < <
at .
Answer:
, a point of discontinuity
u f . ( 0) ( − 0)
( 0) ( − 0)
3
14. Find the sum of the Fourier series for ( ) in the interval (– ) at .
Answer:
, a point of discontinuity
u f . ( 0) ( − 0)
( 0) ( − 0)
,( 0) ( 0)- ,( − 0) ( − 0)-
0
𝑥
−
SRIT / UICM003 - TPDE / Fourier Series
SRIT / M & H / M. Vijaya Kumar 78
15. Find the sum of the Fourier series for ( ) − from – at 0.
Answer:
Put 0 , a point of continuity
u f . ( )
(0)
0 − 0
0
. f ( − )
3 ∑
0 t u t va u f ∑
.
Answer:
Put 0 , a point of continuity
u f . ( )
(0)
( − 0)
3 ∑
−
3 ∑
3 ∑
∑
.
17. Find the half range sine series of the function ( ) in the interval (0 ).
Answer:
a f a ( ) (0 ) ( ) ∑ .
/ .
−𝜋 0
𝜋
0 0
𝜋
SRIT / UICM003 - TPDE / Fourier Series
SRIT / M & H / M. Vijaya Kumar 79
∫ ( ) .
/
∫ .
/
, -
0− .
/
1
−
( − 0)
−
,(− ) − -
{
; f
0; f v
( ) ∑
.
/ .
18. Expand ( ) as a half range sine series in the interval (0 ).
Answer:
a f a ( ) (0 ) ( ) ∑ .
∫ ( ) .
/
∫
, -
0−
1
−
( − 0)
−
,(− ) − -
{
; f
0; f v
( ) ∑
.
/ .
SRIT / UICM003 - TPDE / Fourier Series
SRIT / M & H / M. Vijaya Kumar 80
19. Find the co-efficient of the Fourier series function ( ) in (− ).
(OR) 20. Give the expression for the Fourier Series co-efficient for the function
( ) defined in (− ).
Answer:
Since the given interval is symmetric. So we can verify odd and even function.
Given ( )
Now (− ) − (– )
( )
( ) v fu t
.
21. Obtain the first term of the Fourier series for the function ( ) in (− ).
Answer:
t t
( ) a v fu t
∫
*
3+
3
3
t t
(
3)
3.
22. Without finding the values of and , the Fourier coefficients of Fourier series,
for the function ( ) in the interval (0 ) find the value of
(
∑(
)
+ .
SRIT / UICM003 - TPDE / Fourier Series
SRIT / M & H / M. Vijaya Kumar 81
Answer:
Given that the interval is half range . . (0 )
By Parseval’s Identity,
∑(
)
∫, ( )-
∫
(
)
.
23. Write down Parseval’s formula on Fourier co-efficient.
Answer:
If the Fourier series corresponding to ( ) converges uniformly to ( ) in (– ), then
∫ , ( )-
∑(
)
24. What is Harmonic analysis?
Answer:
The process of finding the Fourier series for the function given by numerical value is
known as harmonic analysis. In harmonic analysis the Fourier co-efficient and of
the function ( ) in < < are defined as
(∑
) ; (
∑
) ; (
∑
)
( ) a
(
* (
* (
* (
*
SRIT / UICM003 - TPDE / Fourier Series
SRIT / M & H / M. Vijaya Kumar 82
. f
3 ∑
(− )
u t at
3
.
Answer:
v
3 ∑
(− )
a ut
3 ∑
(− )
−
3 ∑
(− )
(− )
3 ∑
∑
3
.
Application of Fourier series
Fourier Series in ECGs:
https://www.intmath.com/blog/mathematics/math-of-ecgs-fourier-series-4281
A person had a medical checkup that included an ECG (electro-cardiograph) shows like:
SRIT / UICM003 - TPDE / Fourier Series
SRIT / M & H / M. Vijaya Kumar 83
How it works
(Left) Electrodes used for an ECG and (Right) Nurse
performing an ECG. The electrodes are connected to
various parts of your anatomy (chest, legs, arms, and
feet) and voltage differences over time are measured
to give the ECG readout. The horizontal axis of the
ECG printout represents time and the vertical axis is
the amplitude of the voltage.
Amplitude units are millivolts (mV) and on the
graph, 1 mV = 10 mm high.
The time scale is 25 mm = 1 second (or 1 mm per
0.04 seconds on the graph).
So here's my readout for Lead II, representing the
voltage between the positive electrode on my left
leg and the electrode on my right arm. Each
thicker red vertical line represents a time of 1
second.
Apparently (according to the doctor), this indicates my heart is quite healthy.
In more detail, the features of the repeated pulse we are looking at are as follows.
The P wave is caused by contraction of the right atrium
followed by the left atrium (the chambers at the top of
the heart).
The QRS complex represents the point in time when
most of the heart muscles are in action, so has highest
amplitude.
The T wave represents the polarization of the ventricles
(the chambers at the bottom of the heart).
SRIT / UICM003 - TPDE / Fourier Series
SRIT / M & H / M. Vijaya Kumar 84
Human heart showing atria and ventricles.
[Image by UCSD, source page no longer available]
Modeling the Heartbeat Using Fourier Series
A heartbeat is roughly regular (if it isn't, it indicates something is wrong).
Mathematically, we say something that repeats regularly is periodic.
Such waves can be represented using a Fourier Series.
Assumptions
In my case, my heart rate was about 70 beats per minute. For the sake of simplicity,
I'll assume 60 beats per minute or 1 per second. So the period = 1 second = 1000
milliseconds.
Also for simplicity, I will only model the R wave for this article. To get a more
accurate model for the heartbeat, I would just need to do a similar process for the P, Q, S
and T waves and add them to my model.
I observed that my R wave was about 2.5 mV high and lasted for a total of 40 ms. The
shape of the R wave is almost triangular and so I could have used straight lines for my
model, but these don't give us a smooth curve (especially at the top - it must be
continuously differentiable).
A better approach is to use a polynomial (where the ascending and descending lines are
close enough to being straight), so my model is as follows (the time units are milliseconds):
( ) −0.0000 ( − 0) .
( ) ( 000)
SRIT / UICM003 - TPDE / Fourier Series
SRIT / M & H / M. Vijaya Kumar 85
Explanation of the Model
The model is based on a quartic (power 4) since this will give me close to the shape I
need (a parabola would be too broad). I'm using similar thinking to what I was doing in this
article, where I move a curve around to where I want it.
The ( − 0) term comes from deciding the curve should start at (0 0), (which makes
our lives easier), it will pass through ( 0 0) since the pulse is 40 ms long, and be centered
on 0.
The "+2.5" comes from the fact the amplitude of the pulse is 2.5 mV.
The −0.0000 comes from solving the following for when 0.
( − 0) . 0.
The ( ) ( 000) part means the function (pulse in this case) is repeated every
1000 ms.
Graph of the Model
This is the graph of part of one period
(the part above the t-axis from t = 0 to t = 40):
Of course. this is just one pulse. How do we
produce a graph that repeats this pulse at
regular intervals?
This is where we use Fourier Series.
I'll spare you all the details, but essentially the Fourier Series is an infinite series
involving trigonometric terms. When all the terms are added, you get a mathematical
model of the original periodic function.
To obtain the Fourier Series, wee need to find the mean value, a0, and 2 coefficient
expressions involving n, an and bn which are multiplied by trigonometric terms and
summed for n = 1 to infinity.
Mean Value Term
is obtained by integration as follows (L is half of the period):
∫ ( )
SRIT / UICM003 - TPDE / Fourier Series
SRIT / M & H / M. Vijaya Kumar 86
00∫ ( )
00∫ ,−0.0000 ( − 0) . -
0.
(The section of the curve we need for this part of the problem is from t = 0 to t = 40, so
that's why we chose those values for the limits of integration in the second last line.)
First Coefficient Term,
∫ ( ) (
*
00∫ ( ) (
00*
00∫ ,−0.0000 ( − 0) . - (
00*
Second Coefficient Term,
∫ ( ) (
*
00∫ ( ) (
00*
00∫ ,−0.0000 ( − 0) . - (
00*
Finally, we put it all together and obtain the Fourier Series for our simple model of a
heart beat:
( )
∑ [ (
* (
*]
0.
00∑
(
∫ ,−0.0000 ( − 0) . - (
00*
00∫ ,−0.0000 ( − 0) . - (
00*
)
When we graph this for just the first 5 terms (n = 1 to 5), we can see the beginnings
of a regular 1-second heart beat.
SRIT / UICM003 - TPDE / Fourier Series
SRIT / M & H / M. Vijaya Kumar 87
The above graph shows the "noise" you get in a Fourier Series expansion, especially
if you haven't taken enough terms.
Taking more terms (this time,
adding the first 100 terms) gives us
the following, and we see we get a
reasonable approximation for a
regular R wave with period 1 second.
The T wave for this next model (in blue).
Also we use a parabola for the
T wave because the shape of the T
wave is broader than the shape of the
R wave. We could keep going, adding
the P, Q and S waves to get an even
better model.
SRIT / UICM003 - TPDE / Fourier Series
SRIT / M & H / M. Vijaya Kumar 88
Some other applications:
1. Signal Processing. It may be the best application of Fourier analysis.
2. Approximation Theory. We use Fourier series to write a function as a trigonometric
polynomial.
3. Control Theory. The Fourier series of functions in the differential equation often gives
some prediction about the behavior of the solution of differential equation. They are useful
to find out the dynamics of the solution.
4. Partial Differential equation. We use it to solve higher order partial differential
equations by the method of separation of variables.
The past cannot be changed.
The future is yet in your power.