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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).

https://en.wikipedia.org/wiki/Jean-Baptiste_Joseph_Fourier

https://en.wikipedia.org/wiki/Trigonometric_series

https://en.wikipedia.org/wiki/Trigonometric_series

https://en.wikipedia.org/wiki/Leonhard_Euler

https://en.wikipedia.org/wiki/Jean_le_Rond_d%27Alembert

https://en.wikipedia.org/wiki/Daniel_Bernoulli

https://en.wikipedia.org/wiki/Heat_equation

https://en.wikipedia.org/wiki/Acad%C3%A9mie_fran%C3%A7aise

https://en.wikipedia.org/wiki/Acad%C3%A9mie_fran%C3%A7aise

https://en.wikipedia.org/wiki/Deferent_and_epicycle

http://mathworld.wolfram.com/PeriodicFunction.html

http://mathworld.wolfram.com/Sine.html

http://mathworld.wolfram.com/Cosine.html

http://mathworld.wolfram.com/OrthogonalFunctions.html

http://mathworld.wolfram.com/Sine.html

http://mathworld.wolfram.com/Cosine.html

http://mathworld.wolfram.com/HarmonicAnalysis.html

https://en.wikipedia.org/wiki/Mathematics

https://en.wikipedia.org/wiki/Series_%28mathematics%29

https://en.wikipedia.org/wiki/Periodic_function

https://en.wikipedia.org/wiki/Sine_wave

https://en.wikipedia.org/wiki/Sine_wave

https://en.wikipedia.org/wiki/Complex_exponential



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 ( ).



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



(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



| 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



To find :

∫ ( )

∫ ( )

, -

*∫

∫ ( − )

+ 0 ∫( ) ( )

( )

( )1

[(

)

(( − )

− )

]

[(

− 0) (0 − (

− )+]

*

+

.

To find :

∫ ( ) .

/

∫ ( )

, -

*∫

∫ ( − )

+

∫ − .

−

−

0 −

0

,[ (

* − .

−

/]

[( − ) (

* − (− ) .

−

/]

-

{0

1

− 0

1

}



,( − 0) − ( − )- , -

,((− ) − ) − ( − (− ) )- , (− ) -

,((− ) − ) ((− ) − )- , 0 -

,(− ) − -

{

−

;

To find :

∫ ( ) .

/

∫ ( )

, -

*∫

∫ ( − )

+

∫ − .

−

−

−

0 −

0

,[ .

−

/ − (

−

*]

[( − ) .−

/ − (− ) (

−

*]

-

{0 .

−

/1

0( − ) .−

/1

}

,−( − 0) − (0 − )-

.

The required Fourier series is

( )

∑ (

−

*

.



Problem: 5

Find the Fourier series expansion ( ) 2 f 0 < < f < <

.

Answer:


( )

∑ .

/

Here 0;

0


( )

∑( )

( )

To find :

∫ ( )

(∫

∫

) , -

,( )

( ) -

,( − 0) ( − )-

, -

.

To find :

∫ ( ) .

/

(∫

∫

) , -

([

]

[

]

)

0 𝜋 𝜋

𝑓(𝑥) 2 ; 0 < 𝑥 < 𝜋 ; 𝜋 < 𝑥 < 𝜋



To find :

∫ ( ) .

/

(∫

∫

) , -

(0

−

1

0−

1

*

(− [

− 0

] − [

−

]*

(− *

(− ) −

+ − *

− (− )

+)

(− *

(− ) −

+ *

(− ) −

+)

,(− ) − -

{

−


( )

∑ (

−

*

.

Problem: 6

Obtain the Fourier series for ( ) ( − ) 0 < .

Answer:


( )

∑ .

/

Here 0;

0


( )

∑( )

( )

(𝝅− 𝒙)

𝜋 𝟎 𝜋

𝑓(𝑥) (𝜋 − 𝑥) ; 0 < 𝑥 < 𝜋



To find :

∫ ( )

∫ ( − )

*( − )

−3+

0 ∫( ) ( )

( )

( )1

−3 ,( − ) − -

−3 ,(− ) − -

−3 ,− -

.

To find :

∫ ( ) .

/

∫ ( − )

u u ∫ − .

( − )

( − )(− )

− ( − )

− (0 − )

−

0 −

[( − ) (

* − ,− ( − )- .

−

/ , - (

−

*]

−

*, ( − )- .

/+



−

[ ( − ) (

* − ( − 0) (

0

*]

−

[ (− ) (

* − (

*]

−

[

−

]

.

To find :

∫ ( ) .

/

∫ ( − )

u u ∫ − .

( − )

( − )(− )

− ( − )

−

− (0 − )

−

0

[( − ) .

−

/ − ,− ( − )- (

−

* , - .

/]

0−( − ) .

/ .

/1

*(−( − ) (

* (

*) − (−( − 0) (

0

* (

0

*)+

*(− (

* (

*) − (− (

* (

*)+

.


( ) . 3

/

∑



( )

∑

.

Problem: 7

Expand ( ) ( − ) as a Fourier series in (0 ) and deduce the sum of the

3

Answer:


( )

∑ .

/

Here 0;

0


( )

∑( )

( )

To find :

∫ ( )

∫ ( − )

*

−

3+

*(

( )

−

( )

3) − (0)+

* −

3+

* −

3+

.

𝑥( 𝜋 − 𝑥)

𝜋 𝟎 𝜋

𝑓(𝑥) 𝑥( 𝜋 − 𝑥); 0 < 𝑥 < 𝜋



To find :

∫ ( ) .

/

∫ ( − )

u u ∫ − .

−

−

− −

0 −

[( − ) (

* − , − - .

−

/ ,− - (

−

*]

*, ( − )- .

/+

[( − ) (

* − ( − 0) (

0

*]

[(− ) (

* − (

*]

[−

]

−

.

To find :

∫ ( ) .

/

∫ ( − )



u u ∫ − .

−

− −

− −

0

[( − ) .

−

/ − ,( − )- (

−

* ,− - .

/]

0−( − ) .

/ − .

/1

−

*(( − ) (

* (

*) − ( (

0

*)+

*(0

* − ( (

*)+

.


( ) . 3

/

∑

−

( − )

− ∑

. ( )

Deduction:

Put 0, a point of discontinuity

u f . ( 0) ( − 0)

(0 0) (0 − 0)

( 0 − 0) ( 0 − 0)

0

0 𝜋 𝜋



( ) 0

3− ∑

0

∑

3

3

.

Problem: 8

Find the Fourier series for the function ( ) defined in (− ).

Answer:


( )

∑ .

/

Here − ;

−

The Fourier series of ( ) in (− ) is

( )

∑( )

( )

To find :

∫ ( )

∫

, -

, − - *

−

+

0 −𝝅

𝑒𝑥

𝜋

𝑓(𝑥) 𝑒𝑥; −𝜋 < 𝑥 < 𝜋



To find :

∫ ( ) .

/

∫

[

( )]

, f. u a ( )-

[

( ) −

( (− ))]

[

( ) −

( )]

( ), − -

(− )

( )

To find :

∫ ( ) .

/

∫

[

( − )]

, f. u a ( )-

[

( − ) −

( (− ) − )]

[

(− ) −

(− )]

−

( ), − -

− (− )

( )

(− )

( ) .




( )

∑ (*

(− )

( ) + − *

(− )

( ) + )

(− )

∑ (

( )−

( )*

.

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



To find :

∫ ( ) .

/

∫ ( − ) .

/

u u ∫ − .

( − ) .

/

( − )(− )

− ( − )

.

/

− (0 − )

− .

/

0 − .

/

[( − ) .

/ ( − )(−

, (−

,]

−

*

( − )

+

−

,( − )( ) − ( − 0)( 0)-

−

,− − -

.

To find :

∫ ( ) .

/

∫ ( − ) .

/



u u ∫ − .

( − ) .

/

( − )(− )

− ( − )

− .

/

− (0 − )

− .

/

0 .

/

[( − ) .

−

/ ( − )(−

, (

,]

[ ,( )(− ) (

* ( )(

)-

−,( )(− 0) (

* ( )( 0) (

)-

]

*,

−

- − ,

−

-+

.


( )

3

∑

.

/

( )

3

∑

.

/

. ( )

Deduction:

Put 0, a point of discontinuity, we have



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



∫ ( − )

*( − )

(− )+

*0 −

− +

*−

− +

.

To find :

∫ ( ) .

/

∫ ( − ) .

/

u u ∫ − .

− .

/

0 −

−

.

/

0 − .

/

[( − ).

/ − (− )(−

,]

*(0 (− )

) − (0 (− 0)

)+

, − (− ) -

{



To find :

∫ ( ) .

/

∫ ( − ) .

/

u u ∫ − .

− .

/

0 −

−

− .

/

0 − .

/

[( − ).

−

/ − (− )(−

,]

[*0+ − {( − 0)(− 0) (

*}]

*(

)+

.


( )

∑ (

.

/*

.

∑

.

/

( )

∑

.

/

∑

.

/

.



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:


( )

∑ .

/

Here − ;

−


( )

∑( )

( )

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



To find :

∫ ( )

∫ ( )

∫

∫ ( )

0 ( fu t )

(

3)

*(

3) − 0+

(

).

To find :

∫ ( ) .

/

∫ ( )

∫

∫ ( )

0 ( fu t )

u u ∫ − .

−

0 −

[( ) (

* − .

−

/ ( ) (

−

*]

0 .

/1



, − 0-

(− ) .

To find :

∫ ( ) .

/

∫ ( )

∫

0 ( v fu t )

∫

u u ∫ − .

−

0 −

[ .

−

/ − (

−

*]

0 .

−

/1

,− − 0-

−

(− ) .

u u

( ) (

3 *

∑ (

(− ) −

(− ) *

( )

∑ (

(− )

(− )

)

.



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.


( )

∑ .

/


( )

∑

. . ( )

To find :

∫ ( )

∫ ( − )

∫ ( − )

( −

)

(( − ) − 0)

.

𝜋 𝑥

−𝜋 0 𝜋

𝜋 − 𝑥



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



∑

0

∑

(

3

*

3

Problem: 13

Obtain the Fourier series of period for the function ( ) in (− ).

u t u f t

3 a −

3 − .

Answer:



(− ) (− ) ( )

( ) is even function.

0.

The Fourier series of ( ) (− ) is

( )

∑ .

/

( )

∑

To find :

∫ ( )

∫

(

3)

𝑥

−𝜋 0 𝜋

𝑓(𝑥) 𝑥 in (−𝜋 𝜋)



(

3)

.

To find :

∫ ( ) .

/

∫

u u ∫ − .

−

0 −

[ (

* − .

−

/ (

−

*]

0 .

/1

, − 0-

(− ) .

u u

( )

3

∑ (

(− ) *

( )

∑

(− )

. ( )



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



[−

−

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.

(− ) |− | | | ( )


0.


( )

∑ .

/

( )

∑

. ( )

To find :

∫ ( )

∫

(

)

(

)

.

−𝜋 0 𝜋

𝑓(𝑥) |𝑥| −𝜋 < 𝑥 < 𝜋

|𝑥|



To find :

∫ ( ) .

/

∫

u u ∫ −

0 −

[ (

* − .

−

/]

0

1

, − 0-

,(− ) − -

{ −

u u

( )

∑ (

−

*

| |

−

∑

. . ( )

Deduction: Put 0 a point of continuity

u f . ( )

(0)

0

−𝜋 0

𝜋



0

−

∑ (

0*

∑

3

3

Problem: 15

ta t u x a f ( ) {

− 0

−

0

u t at

3

. .

Answer:



a ( ) {

−

{ ( )

( )

(− ) −

( )



( )

∑ .

/

( )

∑

To find :

∫ ( )

𝑥

𝜋

−𝜋 0 𝜋

− 𝑥

𝜋



∫ ( −

*

* −

+

*( −

) − (0)+

,0-

.

To find :

∫ ( ) .

/

∫ ( −

*

x

u u ∫ − .

−

−

0 −

[( −

* (

* − (−

* .

−

/]

−

[(

* .

/]

−

, − 0-

−

,(− ) − -

, − (− ) -

{



u u

( ) ∑ (

*

..

( )

Deduction:


u f . ( )

(0)

− (0)

( ) becomes

∑ (

0*

..

[

3

]

3

.

Problem: 16

Find the Fourier series for ( ) (− ).

Answer:



(− ) (− ) ( )


0.


( )

∑ .

/

−𝜋 0

𝜋

𝑥

0 −𝒍 𝑙

𝑓(𝑥) 𝑥 ; −𝑙 < 𝑥 < 𝑙



To find :

∫ ( )

∫

(

3)

(

3)

.

To find :

∫ ( ) .

/

∫ .

/

u u ∫ − .

.

/

.

/

. /

− .

/

(

*

0 − .

/

(

*

[ .

/ − (−

, ( )(−

,]



*

+

, − 0-

(− ) .

u u

( )

3

∑ (

(− ) .

/)

( )

∑

(− )

.

/

.

Problem: 17

Find the Fourier series expansion for the function ( ) define as follows

( ) 2− − < < 0 0 < <

.

Answer:



a ( ) 2 −

{ ( )

( )

(− ) − (− ) ( )



( )

∑ .

/

( )

∑

. ( )

To find :

∫ ( )

𝑥

−𝜋 0 𝜋

𝑓(𝑥) 2−𝑥 ; −𝜋 < 𝑥 < 0𝑥 ; 0 < 𝑥 < 𝜋

− 𝑥



∫( )

*

+

(

)

.

To find :

∫ ( ) .

/

∫ ( )

u u ∫ − .

0 −

[( ) (

* − ( ) .

−

/]

0

1

, − 0-

,(− ) − -

{

−

u u

( )

∑

−

..

.



Problem: 18

Find the Fourier series expansion for the function ( ) define as follows

( ) { − < < 0 − 0 < <

a u t at

3

. . .

.

Answer:



a ( ) 2 −

{ ( )

( )

(− ) − ( )



( )

∑ .

/

. ( )

To find :

∫ ( )

∫ ( − )

*( − )

(− )+

*0 −

− +

*−

− +

.

To find :

∫ ( ) .

/

∫ ( − ) .

/

𝑙 𝑥

−𝑙 0 𝑙

𝑓(𝑥) {𝑙 𝑥; −𝑙 < 𝑥 < 0𝑙 − 𝑥; 0 < 𝑥 < 𝑙

𝑙 − 𝑥



u u ∫ − .

− .

/

− .

/

. /

0 − .

/

(

*

[( − ).

/ − (− )(−

,]

−

*

+

−

, − 0-

, − (− ) -

{

u u

( )

∑

.

/

..

( )

∑

.

/

.

( )

Deduction:

Put 0 a point of discontinuity

u f . ( 0) ( − 0)

(0 0) (0 − 0)

−𝑙 0

𝑙



0 − 0

( ) becomes

∑

0

.

−

∑

.

∑

.

. . .

.

Problem: 19

Expand the function ( ) as a Fourier series in the interval (− ).

Answer:



(− ) − (− )

( )



( )

∑

. ( )

To find :

∫ ( )

−𝜋 0 𝜋

𝑓(𝑥) 𝑥 𝑥 ; (−𝜋 𝜋)

𝑥 𝑥



∫

, (− ) − (− )-

,− − 0-

,− (− )-

.

To find :

∫ ( )

∫

∫

∫ (

( ) − ( − )

)

∫ , ( ) − ( − ) -

[( (

− ( )

) – (

− ( )

( ) )+ − ( (

− ( − )

− ) – (

− ( − )

( − ) )+]

* (

− ( )

) − (

− ( − )

− )+

* (

− ( )

) (

( − )

− )+

* −(− )

(− )

− + , ( ) ( − ) (− ) -

(− ) [ −

− ]

(− ) [ –

− ]

(− ) [

− ]



(− ) [

− ]

When , we have

∫

∫

, -

∫

[ (

−

* − (

−

*]

[ (

−

* − 0]

−

.

u u

( )

∑

( )

−

∑

(− )

−

.

Problem: 20

Expand the function ( ) as a Fourier series in the interval (− ).

Answer:



(− ) − (− )

−

− ( )

( ) is odd function, then

0.


−𝜋 0 𝜋

𝑓(𝑥) 𝑥 𝑥 ; (−𝜋 𝜋)

𝑥 𝑥



( ) ∑

. ( )

To find :

∫ ( )

∫

∫

∫ (

( ) ( − )

)

∫ , ( ) ( − ) -

[( (

− ( )

) − (

− ( )

( ) )+ ( (

− ( − )

− ) − (

− ( − )

( − ) )+]

* (

− ( )

) (

− ( − )

− )+

−

* (

( )

) (

( − )

− )+

−

* (− )

(− )

− + , ⟨ ( ) ( − ) (− ) ⟩-

−(− ) [

− ]

−(− ) [ −

− ]

−(− ) (− ) [

− ]

(− ) [

− ]

When , we have

∫

∫

, -



∫

[ (

−

* − (

−

*]

[ (

−

* − 0]

−

.

u u

( ) ∑

. ( )

( ) −

∑

(− )

−

.

Problem: 21

Obtain the Fourier series expansion of ( ) | | t t va (− ).

Answer:



(− ) | |

( )



( )

∑

. ( )

To find :

∫ ( )

∫

,− -

−𝜋 0 𝜋

𝑓(𝑥) | 𝑥|; (−𝜋 𝜋)



−

, − 0-

−

,− − -

.

To find :

∫ ( )

∫

∫

∫ (

( ) − ( − )

)

∫ , ( ) − ( − ) -

*− ( )

−

− ( − )

− +

*(

− ( )

( − )

− ) − (

− 0

0

− *+

*(

−(− )

(− )

− ) − (

−

− *+

*−(− )

(− )

−

−

− +

[

−

((− ) − )

− ((− ) − )]

((− ) − )

[

− −

]

((− ) − )

[ −

− ]



(−(− ) − )

[

− ]

−

( − ) ,(− ) -

−

− ,(− ) - f

2

f −

( − ) f v

When , we have

∫

∫

, -

∫

[(

−

*]

−

[ (

* − (

0

*]

[

−

]

.

u u

( )

∑

( )

− ∑

( − )

..

.

Problem: 22

Obtain the Fourier series expansion of ( ) | | t t va (− ).

Answer:





(− ) | |

( )

( ) is even function, then

0.


( )

∑

. ( )

( ) | | { 0

−


( )

∑

− − − −( )

To find :

∫ ( )

∫ | |

[∫

∫ −

]

*( )

− ( )

+

,( − 0) − (0 − )-

.

−𝜋 0 𝜋

𝑓(𝑥) | 𝑥|; (−𝜋 𝜋)

𝜋

−𝜋



To find :

∫ ( )

∫ | |

[∫

∫ −

]

[∫

− ∫

]

[∫ (

( ) ( − )

)

− ∫ ( ( ) ( − )

)

]

[∫ ( ( ) ( − ) )

− ∫ ( ( ) ( − ) )

]

0(

( )

( − )

− )

− ( ( )

( − )

− )

1

0.

( )

( − )

− − 0/ − .0 −

( )

−

( − )

− /1

0 ( )

( − )

−

( )

( − )

− 1

0 ( )

( − )

− 1

0

−

− 1

0

−

− 1

[

−

− ]



[−

− ]

−

( − )

When , we have

∫ | |

[∫

− ∫

]

[∫

− ∫

]

[∫ (

*

− ∫ (

*

]

[∫ ( )

− ∫ ( )

]

[

]

−

[

]

0

− 01 −

0

− .

/1

0

1 −

0 − .

/1

−

.

u u

( )

∑

( )

∑ .−

( − ) /

( )

−

∑ .

( − ) /



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



∫ − .

−

− −

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:



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



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



[

(

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

. . √∫ , ( )-

−

− ∫ , ( )-



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

.


( )

∑ .

/

( )

Here − ; ; − ; ;



To find :

∫ ( )

, -

∫

(

3)

(

3− 0)

.

To find :

∫ ( ) .

/

∫

u u ∫ − .

−

− −

0 −

[ (

* − .

−

/ (− ) (

−

*]

0 .

/1

, − 0-

(− )



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 .



Answer:

The Fourier cosine series of ( ) (0 ) is

( )

∑ .

/

. ( )

To find :

∫ ( )

, -

∫ ( − )

*

−

3+

*(

−

3) − 0+

*

+

.

To find :

∫ ( ) .

/

∫ ( − )

u u ∫ − .

−

−

− −

0 −

[( − ) (

* − ( − ) .

−

/ (− ) (

−

*]



0( − ) .

/1

,(− ) − 0-

−

,(− ) -

{ −

u u

( ) (

3 *

∑ [

−

]

.

( )

− ∑

.

Deduction:

Pa va ’ t ty f a f a (0 ) is given by

∫ , ( )-

∑

∫ ( − )

(

3 *

∑ (

−

*

∑

∫ ( − )

∑

*

3

−

+

∑

*

3

−

− 0+ −

∑

*

30+ −

[

]

−



[

( )

( 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 <



u u ∫ − .

.

/

.

/

0 − .

/

[ .

. /

/ − .− .

/

/]

[ .

/

]

*

( − 0)+

,(− ) − -

,(− ) − -

{ −


( )

∑

−

.

Deduction:

By Parseval’s Identity, we have

∫ , ( )-

∑

∫ ( )

∑ (

−

*



(

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

√



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



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. .



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)





degree

. 0 0 0

. 0. . −0. . − . 0

. −0. . −0. − . . 0

. − . 0 . 0 − . 0

. −0. − .3 −0. .3 . 0

. 0. − .0 −0. − .0 − . 0

∑ . − . . − . − . .


∑ . ; ∑ − . ; ∑ 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


( )

3 3

.

(−0.3 ) (−0. ) (0.03) 3 (0. ) (−0.0 ) (0) 3

. − 0.3 − 0. 0.03 3 0. − 0.0



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






. . 0

. 0. .

. −0. 0. 0 3

. − .3 0

− . 0. 0.

− . −0. 0.

Total 4.5 1.12 3.013


∑ . ; ∑ . ; ∑ 3.0 3

(∑

) (

.

* .

(∑

) (

.

* 0.3

(∑

) (

3.0 3

* .00


( )

.

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:




0 1 2 3 4 5 6

( ) 4 8 15 7 6 2 4

Here 0; 6; 0 ;


( )

.

/ .

/ .

/ .

/

.

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



4 4 0 4 0

8 . 3 − . 3

15 − . . − . − .

7 − 0 0

6 −3 − . −3 .

2 − . 3 − − . 3

Total − . . − . − .




∑ ; ∑ − . ; ∑ . ; ∑ − . ;

∑ − .

(∑

) (

*

(∑

) (

− .

* − . 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:


0 1 2 3 4 5 6

( ) 9 18 24 28 26 20 9

Here 0; 6; 0 ;




( )

.

/ .

/ .

/ .

/

.

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


The Fourier series takes the form



9 0 0

18 . − .

24 − 0. − − 0.

28 − 0 0

26 − 3 − . − 3 .

20 0 − .3 − 0 − .3

Total − −3. − 0.0


∑ ; ∑ − ; ∑ −3. ;

∑ − ; ∑ 0.0

(∑

) (

* .



(∑

) (

−

* − .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



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

.



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



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:



. . √

− ∫( ( ))

√

− (− )∫( )

√

*

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

−



15. Find the sum of the Fourier series for ( ) − from – at 0.

Answer:


u f . ( )

(0)

0 − 0

0

. f ( − )

3 ∑

0 t u t va u f ∑

.

Answer:


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



∫ ( ) .

/

∫ .

/

, -

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

( ) ∑

.

/ .



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

(

∑(

)

+ .



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

(

* (

* (

* (

*



. 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:

https://www.intmath.com/blog/mathematics/math-of-ecgs-fourier-series-4281

https://www.intmath.com/blog/wp-content/images/2010/03/ECG-full.jpg



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).



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)

https://www.intmath.com/fourier-series/fourier-intro.php



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):

∫ ( )

https://www.intmath.com/blog/mathematics/how-to-draw-y2-x-2-2301

https://www.intmath.com/blog/mathematics/how-to-draw-y2-x-2-2301



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.



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.



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.

ourier series - author...srit / uicm003 - tpde / fourier series srit / m & h / m. vijaya kumar 1...

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