Uraian Fourier

8/4/2019 Uraian Fourier

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The Fourier Series


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Joseph Fourier 1768 - 1830

Joseph Fourier

Fourier wasobsessed with thephysics of heat anddeveloped theFourier series and

transform to modelheat-flow problems.


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Anharmonic waves are sums of sinusoids.

Consider the sum of two sine waves (i.e., harmonic waves)of different frequencies:

The resulting wave is periodic, but not harmonic.Essentially all waves are anharmonic.


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Fourierdecomposing

functions

Here, we write asquare wave asa sum of sinewaves.

sin(wt)

sin(3wt)

sin(5wt)


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A Fourier series is an expansion of aperiodicfunctionf(t) in terms of an infinite sum

of cosines and sines

Introduction

1

0 )sincos(2

)(

n

nn tnbtnaa

tf ww


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In other words, any periodicfunction can beresolved as a summation ofconstant value and cosine and sine functions:

1

0

)sincos(2)( nnn tnbtna

a

tf ww

)sincos( 11 tbta ww 2

0a

)2sin2cos( 22 tbta ww

)3sin3cos( 33 tbta ww


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The computation and study of Fourierseries is known as harmonic analysisandis extremely useful as a way to break upan arbitrary periodic function into a set ofsimple terms that can be plugged in,solved individually, and then recombinedto obtain the solution to the original

problem or an approximation to it towhatever accuracy is desired or practical.


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=

+ +

+ + +

Periodic Function

2

0a

ta wcos1

ta w2cos2

tb wsin1

tb w2sin2

f(t)

t


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1

0

)sincos(2)( nnn tnbtna

a

tf ww

where

T

dttfT

a0

0 )(2

frequencylFundementa2

T

w

T

n tdtntfTa0

cos)(2

w

T

n tdtntfTb0

sin)(2

w

*we can also use the integrals limit .

T

dttfT

a0

0 )(2

2/

2/

T

T


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Example 1

Determine the Fourier series representation of thefollowing waveform.


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Solution

First, determine the period & describe the one periodof the function:

T= 2

21,0

10,1)(

t

ttf )()2( tftf


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Then, obtain the coefficients a0, an and bn:

10101)(2

2)(

22

1

1

0

2

00

0 dtdtdttfdttfTa

T

Or, sincey =f(t) over the interval [a,b], hence

b

a

dttf )( is the total area below graph

1)11(22

],0[overgraphbelowArea2)(2

0

0

TT

dttfT

a

T


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w

n

n

n

tndttdtn

tdtntfT

an

sinsin0cos1

cos)(2

1

0

2

1

1

0

2

0

Notice that n is integer which leads ,since

0sin n03sin2sinsin

Therefore, .0na


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w

n

n

n

tndttdtn

tdtntfTbn

cos1cos0sin1

sin)(

2

1

0

2

1

1

0

2

0

15cos3coscos 16cos4cos2cos

Notice that

Therefore,

even,0

odd,/2)1(1

n

nn

nb

n

n

orn

n )1(cos


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ttt

tnn

tnbtnaa

tf

n

n

n

nn

ww

5sin5

23sin

3

2sin

2

2

1

sin)1(1

2

1

)sincos(2

)(

1

1

0

Finally,


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Some helpful identities

For n integers,n

n )1(cos 0sin n

02sin

n 12cos

n

xx sin)sin( xx cos)cos(


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[Supplementary]

The sum of the Fourier series terms canevolve (progress) into the original

waveform

From Example 1, we obtain

ttttf

5sin5

2

3sin3

2

sin

2

2

1

)(

It can be demonstrated that the sum willlead to the square wave:


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tttt

7sin7

25sin

5

23sin

3

2sin

2ttt

5sin5

23sin

3

2sin

2

tt

3sin3

2sin

2t

sin2

(a) (b)

(c) (d)


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ttttt

9sin9

27sin

7

25sin

5

23sin

3

2sin

2

ttt

23sin23

23sin

3

2sin

2

2

1

(e)

(f)


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Example 2

Given ,)( ttf 11 t

)()2( tftf

Sketch the graph off(t) such that .33 t

Then compute the Fourier series expansion off(t).


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Solution

The function is described by the following graph:

T= 2

w

T

2We find that


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Then we compute the coefficients:

02

11

22

2

)(2

1

1

21

1

1

1

0

ttdt

dttfT

a


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0coscos

)cos(cos0

cos)]sin([sin

sinsin

coscos)(2

22

22

1

1

22

1

1

1

1

1

1

1

1

w

n

nnn

nn

n

tn

n

nn

dtn

tn

n

tnt

tdtnttdtntf

T

an

since xx cos)cos(


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w

nnn

n

n

nn

n

n

n

tn

n

nn

dtn

tn

n

tnt

tdtnttdtntfT

b

nn

n

1

22

1

1

22

1

1

1

1

1

1

1

1

)1(2)1(2cos2

)sin(sincos2

sin)]cos([cos

coscos

sinsin)(2


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ttt

tnn

tnbtnaa

tf

n

n

n

nn

ww

3sin

3

22sin

2

2sin

2

sin)1(2

)sincos(2

)(

1

1

1

0

Finally,


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Example 3

Given

42,0

20,2)(

t

tttv

)()4( tvtv

Sketch the graph of v (t) such that .120 t

Then compute the Fourier series expansion of v (t).


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Solution


T= 4

2

2 w

T

We find that

0 2 4 6 8 10 12t

v (t)

2


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Then we compute the coefficients:

1

2

2

2

1)2(

2

1

0)2(42

)(2

2

0

22

0

4

2

2

0

4

0

0

ttdtt

dtdtt

dttvT

a


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222222

2

0

22

2

0

2

0

4

2

2

0

4

0

])1(1[2)cos1(2

2

2cos1

cos

2

10

sin

2

1sin)2(

2

1

0cos)2(21cos)(2

w

w

w

w

w

w

w

w

ww

nn

n

n

n

n

tn

dt

n

tn

n

tnt

tdtnttdtntvT

a

n

n


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ww

w

w

w

w

w

w

w

w

w

ww

nnnn

n

n

tn

n

dtn

tn

n

tnt

tdtnttdtntvTbn

212

2sin1

sin

2

11

cos

2

1cos)2(

2

1

0sin)2(2

1sin)(

2

22

2

0

22

2

0

2

0

4

2

2

0

4

0

since 0sin2sin w nn


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122

1

0

2sin

2

2cos

])1(1[2

2

1

)sincos(2

)(

n

n

n

nn

tn

n

tn

n

tnbtnaa

tv

ww

Finally,


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Symmetry Considerations

Symmetry functions:

(i) even symmetry

(ii) odd symmetry


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Even symmetry

Any functionf(t) is even if its plot issymmetrical about the vertical axis, i.e.

)()( tftf


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Even symmetry (cont.)

The examples of even functions are:

2)( ttf

t t

t

||)( ttf

ttf cos)(


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Even symmetry (cont.)

The integral of an even function from A to+A is twice the integral from 0 to +A

t

AA

Adttfdttf

0

eveneven )(2)(A +A

)(even tf


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Odd symmetry

Any functionf(t) is odd if its plot isantisymmetrical about the vertical axis, i.e.

)()( tftf


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Odd symmetry (cont.)

The examples of odd functions are:

3)( ttf

t t

t

ttf )(

ttf sin)(


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Odd symmetry (cont.)

The integral of an odd function from A to+A is zero

t 0)(odd

A

A

dttfA

+A

)(odd tf


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Even and odd functions

(even)(even) = (even)

(odd)(odd) = (even)

(even)

(odd) = (odd) (odd)(even) = (odd)

The product properties of even and oddfunctions are:


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Symmetry consideration

From the properties of even and oddfunctions, we can show that:

for even periodic function;

2/

0

cos)(4

T

n tdtntfT

a w 0nb

for odd periodic function;

2/

0

sin)(4

T

n tdtntfT

b w00 naa


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How?? [Even function]

2

T

2

T

2/

0

2/

2/

cos)(4

cos)(2

TT

T

n tdtntfT

tdtntfT

a ww

(even) (even)

| |

(even)

0sin)(2

2/

2/

T

T

n tdtntfT

b w

(even) (odd)

| |

(odd)

)(tf

t


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How?? [Odd function]

2

T

2

T

2/

0

2/

2/

sin)(4

sin)(2

TT

T

n tdtntfT

tdtntfT

b ww

(odd)(odd)

| |

(even)

0cos)(22/

2/

T

T

n tdtntfT

a w

(odd) (even)

| |(odd)

)(tf

t

0)(2

2/

2/

0

T

T

dttfT

a

(odd)


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L 0 0

-L 0 0

0

0 0

0 0

- 0

If f(x) is an even function

f(x)dx ( ) ( ) ( ) ( ) ( )

( ) ( ) 2 ( )

If f(x) is an odd function, then

( ) ( ) ( ) ( ) ( ) ( )

L L

L L

L L

L

L L

L L L

f x dx f x dx f x d x f x dx

f x dx f x dx f x dx

f x dx f x dx f x dx f x d x f x dx

00

0( ) ( ) 0

If f(x) is even and g(x) is odd, then

h(x)=f(x)g(x) is an odd function

h(x)=f(x)g(x)=f(-x)[-g(-x)]=-[f(-x)g(-x)]=-h(-x)

L

L

Lf x dx f x dx


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Example 4

Given

21,1

11,

12,1

)(

t

tt

t

tf

)()4( tftf

Sketch the graph off(t) such that .66 t

Then compute the Fourier series expansion off(t).


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Solution


T= 4

2

2 w

T

We find that

046 2 4 6t

f(t)

2

1

1


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Then we compute the coefficients. Sincef(t) is

an odd function, then

0)(2

2

2

0

dttfT

a

0cos)(2

2

2

tdtntf

T

an w

and


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w

w

w

w

w

ww

w

w

w

w

w

w

w

w

w

w

ww

ww

n

n

n

n

n

nn

nn

n

tn

n

n

ntndt

ntn

ntnt

tdtntdtnt

tdtntfT

tdtntfT

bn

cos2sin2cos

cos2cossincos

coscoscos

sin1sin4

4

sin)(4

sin)(2

22

1

0

22

2

1

1

0

1

0

2

1

1

0

2

0

2

2

since 0sin2sin w

nn


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1

1

1

1

0

2sin)1(2

2sin

cos2

)sincos(

2

)(

n

n

n

n

nn

tnn

tn

n

n

tnbtnaa

tf

ww

Finally,


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Example 5

Compute the Fourier series expansion off(t).


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Solution

The function is described by

T= 3

3

22 w

Tand

32,1

21,2

10,1

)(

t

t

t

tf

)()3( tftf

T= 3


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Then we compute the coefficients.

3

81

2

32)01(

3

421

3

4)(

4)(

22/3

1

1

0

2/3

0

3

0

0

dtdtdttfT

dttfT

a

3

8)23()12(2)01(

3

2121

3

2)(

2 3

2

2

1

1

0

3

0

0

dtdtdtdttfTa

Or, sincef(t) is an even function, then

Or, simply

3

84

3

2

periodain

graphbelowareaTotal2)(

23

0

0

TdttfTa


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3

2sin

2

3

2sinsin2

2

sin2

3sin2

3

4

sin2

3sin2sin

3

4

sin2

3

4sin

3

4

cos2cos13

4

cos)(4

cos)(2

2/3

1

1

0

2/3

1

1

0

2/3

0

3

0

w

w

w

w

w

w

w

w

w

w

w

ww

ww

n

n

nn

n

nn

n

nn

nn

n

tn

n

tn

tdtntdtn

tdtntfT

tdtntfT

an

;3

2w


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1

1

1

0

3

2cos

3

2sin

12

3

4

3

2cos

3

2sin

2

3

4

)sincos(2

)(

n

n

n

nn

tnn

n

tnn

n

tnbtnaa

tf

ww

Finally,

and 0nb sincef(t) is an even function.


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Function defines over a finite interval

Fourier series only support periodic functions

In real application, many functions are non-

periodic The non-periodic functions are often can be

defined over finite intervals, e.g.

y = 1 y = 1

y = 2


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Therefore, any non-periodic function must be

extended to a periodic function first, beforecomputing its Fourier series representation

Normally, we prefer symmetry (even or odd)periodic extension instead of normal periodic

extension, since symmetry function will providezero coefficient of either an or bn

This can provide a simpler Fourier seriesexpansion


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)(ty

t

)(tf

t

)(even tf

)(odd tf

t

t

lttytf 0,)()(

)()( tfltf lT

0,)(

0,)()(tlty

lttytf

)()2( tfltf

lT 2

0,)(

0,)()(

tlty

lttytf

)()2( tfltf

lT 2

l0

l0 l2ll2

l0 l2ll2

l3l3

l3l3

l0 l2ll2 l3l3

T

T

T

Periodic extension

Even periodic extension

Odd periodic extension

Non-periodicfunction


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Half-range Fourier series expansion

The Fourier series of the even or oddperiodic extension of a non-periodic

function is called as the half-range Fourierseries

This is due to the non-periodic function is

considered as the half-range before it isextended as an even or an odd function


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If the function is extended as an even

function, then the coefficient bn= 0, hence

1

0 cos2

)(n

n tnaa

tf w

which only contains the cosine harmonics. Therefore, this approach is called as the

half-range Fourier cosine series


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If the function is extended as an odd

function, then the coefficient an= 0, hence

1

sin)(n

n tnbtf w

which only contains the sine harmonics. Therefore, this approach is called as the

half-range Fourier sine series


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Example 6

ttf 0,1)(

Compute the half-range Fourier sine series expansionoff(t), where


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Solution

Since we want to seek the half-range sine series,the function to is extended to be an odd function:

T= 2

12

T

w

0 t

f(t)

1

1

220 t

f(t)

1


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Hence, the coefficients are

00 naa

w

0

2/

0sin12

4

sin)(

4

ntdttdtntfTb

T

n

and

even,0

odd,/4)cos1(

2cos2

0 n

nnn

nn

nt

Therefore,

ntn

ntnn

tf

nnn

sin4

sin)cos1(2

)(

odd11


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Example 7

10,12)(

tttf

Determine the half-range cosine series expansionof the function

Sketch the graphs of bothf(t) and the periodicfunction represented by the series expansion for3 < t< 3.


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Solution

Since we want to seek the half-range cosine series,the function to is extended to be an even function:

T= 2

w T

2

t

f(t)

t

f(t)

1

2233 1

1

11

11

2

1


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Hence, the coefficients are

02)12(24

)(

4 1

0

2

1

0

2/

00

ttdttdttfTa

T

1

0

22

1

0

1

0

1

0

2/

0

cos4sin2

2sin

2sin)12(

2

cos)12(2

4cos)(

4

w

ntn

nn

dtn

tn

n

tnt

tdtnttdtntfT

a

T

n

even,0

odd,/8)1(cos422

22

n

nn

n

n


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0nb

Therefore,

odd1

22

odd1

22

1

0

cos18

cos8

0

cos)(

nn

nn

n

n

tnn

tnn

tnaatf

w


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Exponential Fourier series

Recall that, from the Eulers identity,

xjxejx sincos

yields

2cos

jxjx eex

2

sin

j

eex

jxjx and


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Then the Fourier series representation becomes

11

0

1

0

1

0

1

0

1

0

222

222

222

222

)sincos(2

)(

n

tjnnn

n

tjnnn

n

tjnnntjnnn

n

tjntjn

n

tjntjn

n

n

tjntjn

n

tjntjn

n

n

nn

ejba

ejbaa

ejbaejbaa

eejb

eea

a

j

eeb

eea

a

tnbtnaa

tf

ww

ww

wwww

wwww

ww


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Here, let we name

11

0

222)(

n

tjnnn

n

tjnnn ejba

ejbaa

tf ww

2

nnn

jbac

,

2

nnn

jbac

Hence,

n

tjn

n

n

tjn

n

n

tjn

n

n

tjn

n

n

tjn

n

n

tjn

n

n

tjn

n

ececcec

ececc

ececc

www

ww

ww

1

0

1

11

0

11

0

and .20

0

ac

c0 cncn


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Then, the coefficient cn can be derived from

T

tjn

T

TT

TT

nnn

dtetfT

dttnjtntfT

tdtntfjtdtntfT

tdtntfT

jtdtntf

T

jbac

0

0

00

00

)(1

]sin)[cos(1

sin)(cos)(1

sin)(2

2cos)(

2

2

1

2

w

ww

ww

ww


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In fact, in many cases, the complexFourier series is easier to obtain rather

than the trigonometrical Fourier series

In summary, the relationship between thecomplex and trigonometrical Fourier series

are:

2

nnn

jbac

2

nnn

jbac

T

dttfT

ac

0

00 )(

1

2

T

tjn

n dtetfT

c0

)(1

w

nn cc or


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Example 8

Obtain the complex Fourier series of the followingfunction

2 44 2 0

2e

1

)(tf

t


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Since , . Hence

Solution

2

1

2

1

2

1

)(1

22

0

2

0

0

0

ee

dte

dttf

T

c

t

t

T

1w2T


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)1(21

)1(21

)1(21

12

1

2

1

2

1

)(1

222)1(2

2

0

)1(

2

0

)1(

2

0

0

jn

e

jn

ee

jn

e

jn

e

dtedtee

dtetf

T

c

njjn

tjn

tjnjntt

T

tjn

n

w

since1012sin2cos

2 njne

nj


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jnt

nn

tjn

n ejn

eectf

)1(2

1)(

2

w

Therefore, the complex Fourier series off(t) is

0

2

0

2

0 2

1

)1(2

1

c

e

jn

e

cn

nn

*Notes: Even though c0 can be found by substitutingcn with n = 0, sometimes it doesnt works (as shownin the next example). Therefore, it is always better to

calculate c0 alone.


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2

2

121n

ecn

cn is a complex term, and it depends on n.Therefore, we may plot a graph of |cn| vs n.

In other words, we have transformed the functionf(t)in the time domain (t), to the function cn in thefrequency domain (n).


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Example 9

Obtain the complex Fourier series of the function inExample 1.


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Solution

2

11

2

1)(

11

00

0 dtdttfTc

T

w

)1(22

1

012

1)(

1

1

0

2

1

1

00

w

jntjn

tjn

T

tjn

n

en

j

jn

e

dtedtetfT

c


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)1(2

jn

n en

jc

But njn nnjne )1(cossincos

Thus,

even,0

odd,/]1)1[(2 n

nnjnj n

Therefore,

odd0

2

1)(

nn

n

tjn

n

tjn

n en

jectf

w

*Here notice that .00

ccn

n


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even,0

odd,1

n

nncn

The plot of |cn| vs n is shown below

2

10 c

0.5


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Exercise 1

Obtain the complex Fourier series of thefunction bellow;

f(x) =1, - x < 0

0, 0 x <

S l ti


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dxedxedxec inxinxinxn

0

0

0

)1(2

1)0(

2

1)1(

2

1

0

02

1)1(

2

1

dxc

)cos1(2

12

nnie

nic

in

n

Solution

Kita tinjau untuk n = 0 dan n 0

a. Untuk n = 0

b. Untuk n 0

= n

i

n ganjil

0, n genap


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...5

1

3

1

3

1

5

1

...2

1

)(5335 ixixixixixix

eeeeee

i

xf

...

5

5sin

3

3sin

1

sin2

2

1)(

xxxxf


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Parsevals Theorem

Parservals theorem states that the

average power in a periodic signal is equal

to the sum of the average power in its DCcomponent and the average powers in itsharmonics


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=

+ +

+ + +

2

0a

ta wcos1

ta w2cos2

tb wsin1

tb w2sin2

f(t)

t

PavgPdc

Pa1 Pb1

Pa2 Pb2


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For sinusoidal (cosine or sine) signal,

R

V

R

V

R

VP

2

peak

2peak

2

rm s

2

12

For simplicity, we often assumeR= 1,

which yields2

peak21 VP


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For sinusoidal (cosine or sine) signal,

2

2

2

2

2

1

2

1

2

0

dcavg

2

1

2

1

2

1

2

1

2

2211

babaa

PPPPPP baba

1

222

0avg )(2

1

4

1

n

nn baaP


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)()2/(2

1)(

1 22

1

20

2

n

n

n

La

a

baaxfL

inpx

necxf

)(

22

)(1

n

n

La

a

cxfL

Parsevals Identity

For exponential form

The Parsevals identity


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Example 10

Use Parseval theorem to determine sum of number series in

the exercise 1

ii


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C0= )cos1(2

12

nn

ie

n

ic

in

n

22

)(1

n

n

La

a

cxfL

2

10

2

11

2

1)(

2

12

0

202

dxdxdxxf

Parseval identity

2

2

2

)cos1(2

12/1

2

1

nn

cn

n

.. .

7

1

5

1

3

11

2

2

12222

2

12

S

Uraian Fourier

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