1 CHAPTER 5 : FOURIER SERIES Introduction Periodic Functions Fourier Series of a function Dirichlet Conditions Odd and Even Functions Relationship.

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CHAPTER 5 : FOURIER SERIES

Introduction Periodic Functions Fourier Series of a function Dirichlet Conditions Odd and Even Functions Relationship Between Even and Odd

functions to Fourier series

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5.1 : Introduction

Fourier Series ?

A Fourier series is a representation of a function as a series of constants times sine and/or cosine functions of different frequencies.

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5.2 : Periodic Functions

A function f(x) is said to be periodic if its function values repeat at regular intervals of the independent variables.

For the following example, a function f(x) has the period p.

p

x

y

x1 x1 + p x1 + 2p x1 + 3p

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In general, a function f(x) is called periodic if there is some positive number p such that ;

f(x) = f(x + np)

for any integer n.This number p is called a period of f(x).

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5.3 : Fourier Series of a Function

If f(x) is defined within the interval c < x < c+2L. The Fourier Series corresponding to f(x) is given by

where

1

0 sincos2 n

nn xL

nbx

L

na

axf

Lc

c

n

Lc

c

n

Lc

c

xdxL

nxf

Lb

xdxL

nxf

La

dxxfL

a

2

2

2

0

sin1

cos1

1

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5.4 : Dirichlet Conditions

If a function f(x) defined within the interval c < x < c+2L, the following conditions must be satisfied;

1. f(x) is defined and single-valued.

2. f(x) is continuous or finite discontinuity in the corresponding periodic interval.

3. f(x) and f ’(x) are piecewise continuous .

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Example 5.1:

Determine whether the Dirichlet conditions are satisfied in the following cases :

i.Yes.

ii.No, because there is infinite discontinuity

at

xx

xf ; 1

xxxf ; 2

.0x

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iii.

No, the function is not defined for x < 0.

iv.

Yes.

xxxf ; ln

11 ; 5

1

x

xxf

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Example 5.2:

1. Find the Fourier series for the function defined within the interval

Answer :

2

22

2

,0

,4

,0

x

x

x

xf

xxxxxxf 9cos

9

17cos

7

15cos

5

13cos

3

1cos

82

. x

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Solution

The Fourier series is

where

1

0 sincos2 n

nn nxbnxaa

xf

.sin)(1

cos)(1

)(1

0

dxnxxfb

dxnxxfa

dxxfa

n

n

. Therefore, .2 and get We LLcc

11

.4

)4(1

]4[1

0401

)(1

2

2

2

2

2

2

0

x

dxdxdx

dxxfa

12

.2

sin8

sin14

cos41

,cos)(1

2

2

2

2

n

n

nxn

dxnx

dxnxxfan

.0

cos14

sin41

sin)(1

2

2

2

2

nxn

dxnx

dxnxxfbn

13

.9cos9

17cos

7

15cos

5

13cos

3

1cos

82

9cos9

87cos

7

85cos

5

83cos

3

8cos

82

cos2

sin8

2

4

)sincos(2

is seriesFourier theTherefore,

1

1

0

xxxxx

xxxxx

nxn

n

nxbnxaa

xf

n

nnn

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Graph of a Function

2

22

2

,0

,4

,0

x

x

x

xf

What will we obtain as more terms are included in the series ?

2

y

4

x- 0

2

15

Graph of a Function

xxf cos8

2

4

O

x

16

Graph of a Function

xxxf 3cos

3

1cos

82

4

x O

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Graph of a Function

xxxxf 5cos

5

13cos

3

1cos

82

4

x O

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Graph of a Function

xxxxxf 7cos

7

15cos

5

13cos

3

1cos

82

4

x

O

The above figure show that the graph is merely to the shape …..

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The above function is defined within the interval

As the number of terms increases, the graph of Fourier series gradually approaches the shape of the original square waveform.

. x

x-5 -3 - 0 3 5

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Example 5.3:

Find the Fourier series for defined within the interval

Answer :

2xxf . x

12

2

cos1

43 n

n

nxn

xf

O

f(x)

x

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Solution

The Fourier series is

where

1

0 sincos2 n

nn nxbnxaa

xf

.sin)(1

cos)(1

)(1

0

dxnxxfb

dxnxxfa

dxxfa

n

n

. Therefore, .2 and get We LLcc

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.3

2

3

21

3

11

1

)(1

2

3

3

2

0

x

dxx

dxxfa

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.4

)1(

.cos4

cos1

,cos)(1

2

2

2

na

nn

dxnxx

dxnxxfa

nn

n

.0

.0

sin1

sin)(1

2

n

n

b

dxnxx

dxnxxfb

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.cos)1(

43

cos4

)1(3

2

2

1

)sincos(2

)(

is seriesFourier The

12

2

12

2

1

0

n

n

n

n

n

nn

nxn

nxn

nxbnxaa

xf

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Find the Fourier series for defined within the interval

Answer :

xxf .11 x

1

1

sin12

n

n

xnn

xf

x

f (x)

5 3 1 1 3 5

O

Example 5.4:

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Solution

The Fourier series is

.sincos2

sincos2

1

0

1

0

nnn

nnn

xnbxnaa

xL

nbx

L

na

axf

.1 Therefore, .12 and 1get We LLcc

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.0

2

1

1

1

)(1

1

1

2

1

1

2

0

x

dxx

dxxfL

aLc

c

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.0

.0

)cos(11

)cos(1

1

,cos)(1

1

1

1

1

2

n

Lc

c

n

a

xnnn

dxxnx

dxxL

nxf

La

.2

)1(

.cos2

)sin(1

1

sin)(1

1

1

1

2

nb

nn

dxxnx

dxxL

nxf

Lb

nn

Lc

c

n

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.sin)1(2

sin2

)1(

)sincos(2

is seriesFourier theTherefore,

1

1

1

1

1

0

n

n

n

n

n

nn

xnn

xnn

xnbxnaa

xf

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Exercise

Find the Fourier series for defined by

Answer :

xxf . x

1

1

sin1

2n

n

nxn

xf

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5.5 : Odd and Even Functions

A function f(x) is said to be even if :

f(-x) = f(x)

i.e the function value for a particular negative value of x is the same as that for the corresponding positive value of x.

A function f(x) is said to be odd if :

f(-x) = - f(x)

i.e the function value for a particular negative value of x is numerically equal to that for the corresponding positive value of x but opposite in sign.

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Example 5.5 :

Even function :

y = f(x) = x2 is an even function

because

f(-2) = 4 = f(2)

f(-6) = 36 = f(6)

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Odd function :

y = f(x) = x3 is an odd function

because

f(-2) = -8 = -f(2)

f(-3) = -27 = -f(3)

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Properties of An Even function

The graph of an even function is symmetrical about the y-axis.

Hence, the areas under curves is twice the area from 0 to a :

aa

a

dxxfdxxf0

2

y

x a 0 a

y = x2

.0

0

a

a

dxxfdxxf

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Properties of An Odd function

The graph of an odd function is symmetrical about the origin.

Thus, the integral is 0 because the areas cancel.

0

a

a

dxxf

y

x 0 a

y = x3

a

.0

0

a

a

dxxfdxxf

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Moreover :If n is positive integer, thus

x2n is an even function,and x2n+1 is an odd function.

Function is an even function.

Function is an odd function.

x

L

cos

x

L

sin

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The product of two even functions is even.Example :

The product of two odd functions is even Example :

The product of an even function and an odd function is odd. Example :

.642 xxx

.43 xxx

.532 xxx

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5.6 : Relationship between Even and Odd functions to Fourier series

We could simply obtain the Fourier series for a function, f(x) defined within the interval –a<x<a. In this case, c = -a and c+2L = a, thus L = a, if we could identify whether f(x) is an even or odd function and use the properties of these functions in order to find the coefficients of Fourier series, a0, an and bn

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i. If f(x) is an even function,

and is an even function, thus

an even function.

and is an odd function, thus

an odd function.

xa

ncos

xa

nxf

cos)(

xa

nxf

sin)(

xa

nsin

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If f(x) is an even function

aa

a

n dxxa

nxf

adxx

a

nxf

aa

0

cos2

cos1

aa

a

dxxfa

dxxfa

a0

0

21

.0sin1

a

a

n dxxa

nxf

ab

series. CosinesFourier called is

cos2

series,Fourier 1

0

nn x

a

na

axf

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ii. If f(x) is an odd function, thus :

and an even function, thus

is an odd function.

and an odd function, thus

is an even function

xa

ncos

xa

nxf

cos)(

xa

nxf

sin)(

xa

nsin

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If f(x) is an odd function

01

0

a

a

dxxfa

a

0cos1

a

a

n dxxa

nxf

aa

.sin2

sin1

0

aa

a

n dxxa

nxf

adxx

a

nxf

ab

series. SineFourier called is

sin series,Fourier 1

n

n xa

nbxf