Fourier Series - National Tsing Hua...

© 2012, Ching-Han Hsu, Ph.D.

Fourier Series

Ching-Han Hsu


Periodic Function

• A function f(x) is called periodic, if it is

defined for every x in the domain of f and if

there is some positive number p such that

f(x+p)= f(x)

• The number p is called a period of f(x).

• If a periodic function f has a smallest

period p, this is often called the

fundamental period of f(x).


Periodic Function

p

f(x)

x


Periodic Function

• For any integer n,

f(x+np) = f(x), for all x

• If f(x) and g(x) have period p, then so does

h(x) = af(x) + bg(x),

a and b are constants.


Trigonometric Functions

• Consider the following periodic functions

with period 2π: 1, sin x, cos x, sin2x,

cos2x, …, sin nx, cos nx, …


Trigonometric Series• The trigonometric series is of the form

a0, a1, a2, a3, …, b1, b2, b3, … are real

constants and are called the coefficients of

the series.

• If the series converges, its sum will be a

function of period 2π.

1

0

22110

sincos1

2sin2cossincos1

n

nn nxbnxaa

xbxaxbxaa


Fourier Series

• Assume f(x) is a periodic function of

period 2π that can be represented by the

trigonometric series:

That is, we assume that the series

converges and has f(x) as its sum.

• Question: how to compute the coefficients?

1

0 sincos1n

nn nxbnxaaxf


Properties of Trigonometric

Functions

• Recall some basic properties of

trigonometric functions

0sin1sin

0cos1cos

dxnxnxdx

dxnxnxdx

0sin2

1sin

2

1

sincos

xdxmnxdxmn

nxdxmx


Properties of Trigonometric

Functions

mn

mnxdxmnxdxmn

nxdxmx

0cos

2

1cos

2

1

sinsin

mn

mnxdxmnxdxmn

nxdxmx

0cos

2

1cos

2

1

coscos


Determination of the Constant

Term a0

• Integrating both sides from – to :

0

1

0

2

sincos

a

dxnxbnxaadxxfn

nn

dxxfa

2

10


Determination of the Coefficients an

of the Cosine Term

• Multiply by cosmx with fixed positive integer

and then integrate both sides from – to :

m

n

nn

a

mxdxnxbnxaa

mxdxxf

cossincos

cos

1

0

mxdxxfam cos

1


Determination of the Coefficients bn

of the Sine Term

• Multiply by sinmx with fixed positive integer

and then integrate both sides from – to :

m

n

nn

b

mxdxnxbnxaa

mxdxxf

sinsincos

sin

1

0

mxdxxfbm sin

1


Summary

1

0 sincosn

nn nxbnxaaxf

dxxfa

2

10

nxdxxfbn sin

1

nxdxxfan cos

1


Fourier Series: Square Wave

• Ex. Find the Fourier coefficients of the

periodic function f(x)

xfxfxk

xkxf

2,

0

0



02

1

2

1

2

1

0

0

0

kk

kdxkdxdxxfa

0sinsin1

coscos1

cos1

0

0

0

0

nxn

knx

n

k

nxdxknxdxk

nxdxxfan

Q: Is f(x) even or odd?!


nn

k

nn

kn

n

k

nxn

knx

n

k

nxdxknxdxk

nxdxxfbn

cos12

1coscos11

coscos1

sinsin1

sin1

0

0

0

0

nn coscos

2,4,0

,3,12cos1

n

nn



,5

4,0,

3

4,0,

454321

kbb

kbb

kb

xxxk

nxnn

k

nxbnxaaxf

n

n

nn

5sin5

13sin

3

1sin

4

sincos12

sincos)(

1

1

0



• Consider the series expansion of finite

sine terms:

xxxk

S

xxk

S

xk

S

5sin5

13sin

3

1sin

4

3sin3

1sin

4

sin4

3

2

1


Representation by Fourier Series

• Thm. If a periodic function f(x) with period

2 is peicewise continuous in the interval

-≦x≦ and has a left-hand derivative and

right-hand derivative at each point of that

interval, then Fourier series of f(x) is

convergent. Its sum is f(x), except at a

point of x0 at which f(x) is discontinuous

and the sum of the series is the average of

the left- and right-hand limits of f(x) at x0.


Representation by Fourier Series

01012

1 ff


Functions of Any Period p=2L

• A function f(x) of period p=2L has a Fourier

series

1

0

1

0

sincos

2

2sin

2

2cos

n

nn

n

nn

xL

nbx

L

naa

nxL

bnxL

aaxf

L

Ldxxf

La

2

10

L

Ln dx

L

xnxf

La

cos

1

L

Ln dx

L

xnxf

Lb

sin

1

,3,2,1n


Fourier Series: Periodic Square Wave

• Ex. Find the Fourier series of the function

2,42,

210

11

120

LLp

x

xk

x

xf



24

1

4

1 1

1

2

20

kkdxdxxfa

2sin

2

2sin

2cos

2

1

2cos

2

1

1

1

1

1

2

2

n

n

kx

n

n

k

dxxn

kdxxn

xfan

,11,7,3,2

,9,5,1,2

nn

kan

n

ka nn



xxx

kkxf

2

5cos

5

1

2

3cos

3

1

2cos

2

2)(

02

cos2

cos

2cos

2cos

2cos

2sin

2

1

2sin

2

1

1

1

1

1

2

2

n

n

kn

n

k

n

n

kn

n

kx

n

n

k

dxxn

kdxxn

xfbn



k=5


Fourier Series: Half-wave Rectifier

• Ex. A sinusoidal voltage Esinwt, is passed

through a half-wave rectifier that clips the

negative portion of the wave. Find the

Fourier series of the resulting periodic

function:

w

w

w

LLp

LttE

tLtu ,

22,

0sin

00



,4,3,20

12

1

1sin

1

1sin

2

1

1sin

1

1sin

2

1cos1cos2

sinsinsin

0

0

0

n

nE

n

n

n

nE

n

tn

n

tnE

dttntnE

tdtntEtdtntubn

w

w

w

w

w

ww

w

ww

ww

w

w

w

w

w

w

nndn

d

n

n

n

nn

1cos1sin

1

1coslim

1

1sinlim

11



w

w

w

w

w

w

w

w

EEt

E

tdtEdttua

0coscos2

cos2

sin22

0

00

w

w

w

w

w

w

w

w

w

w

ww

w

ww

ww

w

0

0

0

1

1cos

1

1cos

2

1sin1sin2

cossincos

n

tn

n

tnE

dttntnE

tdtntEtdtntuan



,6,4,211

2,5,3,10

,6,4,21

2

1

2

2

,5,3,10

1

1cos1

1

1cos1

2

nnn

En

nnn

E

n

n

n

n

nEan

tt

Et

EEtu ww

w

4cos

53

12cos

31

12sin

2



k=5 =1


Even and Odd Functions

• A function e(x) is even, if e(-x)=e(x).

• A function o(x) is odd, if o(-x)=-o(x).

xe xo


Even and Odd Functions: Facts

• If e(x) is an even function, then

• If o(x) is an odd function, then

• The product of an even and an odd

function is odd

LL

Ldxxedxxe

02

0L

Ldxxo

xoxexoxe


Fourier Cosine Series

• Thm. The Fourier series of an even

function of period 2L is a Fourier cosine

series

with coefficients

1

0 cosn

n xL

naaxf

LL

Ldxxf

Ldxxf

La

00

1

2

1

LL

Ln dx

L

xnxf

Ldx

L

xnxf

La

0cos

2cos

1


Fourier Sine Series

• Thm. The Fourier series of an odd

function of period 2L is a Fourier sine

series

with coefficients

1

sinn

n xL

nbxf

LL

Ln dx

L

xnxf

Ldx

L

xnxf

Lb

0sin

2sin

1


Fourier Series: Sum of Functions

• Thm. The Fourier coefficients of a sum

f1+f2 are the sums of the corresponding

Fourier coefficients of f1 and f2.


xxxk

kxf

xxxk

xf

5sin5

13sin

3

1sin

4)(

5sin5

13sin

3

1sin

4)(

*


Sawtooth Wave


xfxfxxxf 2,,

xfxxffff 2121 ,


Sawtooth Wave

• Since f1 is odd, an=0 for n=1,2,3,… and

nn

nxn

nn

dxnxn

nxxn

nxxdn

nxdxx

dxL

xnxf

Lb

L

n

cos2

sin2

cos2

cos2

cos2

cos2

sin2

sin2

0

2

00

00

01

11

1 sincos2

sinnn

n nxnn

nxbxf


Sawtooth Wave

xfxf

xxxxf

21

3sin3

12sin

2

1sin2


Half-Range Expansion• In some applications, we often need to

employ a Fourier series for a function f(x)

that is given only on some interval, say

0≦x≦L.

Suppose that we have a choice!! We can

use even or odd periodic expansion.


Half-Range Expansion

• If we apply even periodic extension, we

obtain a Fourier cosine series:

• If we apply odd periodic extension, we

obtain a Fourier sine series:


Triangle Function: Half-Range Expansion

• Ex. Find the two half-range expansion of

the triangle function

LxL

xLL

k

Lxx

L

k

xf

2

22

02


Triangle Function: Half-Range Expansion

Even Expansion

Odd Expansion


Triangle Function: Even Periodic Expansion

LxL

xLL

k

Lxx

L

k

xf

2

22

02

222

1

221

2

2

2

1

22

2/

2/

0

00

kL

L

kL

L

k

L

dxxLL

kxdx

L

k

L

dxxeL

dxxeL

a

L

L

L

L

f

L

Lf

L

L

L

L

f

L

Lfn

dxL

xnxL

L

kdx

L

xnx

L

k

L

dxL

xnxe

Ldx

L

xnxe

La

2/

2/

0

0

cos2

cos22

cos2

cos1



12

cos2

sin2

sinsincos

22

22

2/

0

2/

0

2/

0

n

n

Ln

n

L

dxL

xn

n

L

L

xnx

n

Ldx

L

xnx

LL

L

2coscos

2sin

2

sinsin

cos

22

22

2/2/

2/

nn

n

Ln

n

L

dxL

xn

n

L

L

xnxL

n

L

dxL

xnxL

L

L

L

L

L

L



,14,10,6,2,0

,10

16,

6

16

0,0,0,2

16,0

1cos2

cos24

2210226

5432221

22

na

ka

ka

aaak

aa

nn

n

ka

n

n

x

Lx

Lx

L

kkxf

10cos

10

16cos

6

12cos

2

116

2 2222


Triangle Function: Odd Periodic Expansion

x

Lx

Lx

L

kxf

5sin

5

13sin

3

1sin

1

182222

2sin

8

sin2

sin1

22

0

n

n

k

dxL

xnxo

Ldx

L

xnxo

Lb

L

f

L

Lfn


Triangle Function: Odd Periodic Expansion


Forced Oscillations

• Consider the forced oscillation of a body of mass m on spring of modulus are governed by the equation

where y(t) is the displacement from rest, c the damping constant, and r(t) the external force depending on time t.

trtkytyctym


Forced Oscillations

• The electric analog to the spring is RLC-

circuit: tEtI

CtIRtIL

1


Forced Oscillations• Ex. Let m=1(g), c=0.02(g/s), k=25(g/s2), so

that

where r(t) is measured in g‧㎝/s2. Let

Find the steady-state solution y(t).

trtytyty 2502.0

trtr

tt

tttr

2,

02

02

0

5sin5cos 01.001.0

tyt

tBetAety

h

tt

h


Forced OscillationsThe Fourier series of r(t) is

Now we the following linear differential

equation

ttttr 5cos

5

13cos

3

1cos

1

14222

tkk

tr

tytyty

k

12cos12

4

2502.0

12


Forced OscillationsThe steady-state (particular) solution yn(t)

corresponding to the sinusoidal input

cos(2k-1)t=cosnt can be computed using

the method of undetermined coefficients

tytytyty

nnD

DnB

Dn

nA

ntBntAty

p

nn

nnn

321

222

2

2

02.025

08.0,

254

sincos


Forced OscillationsLet us find the amplitude

We have

22

nnn BAC

0003.0

0011.0

5100.0

0088.0

0530.0

9

7

5

3

1

C

C

C

C

C


Parseval’s Theorem


Fourier Series Formulae

1

0 sincosn

nn xL

nbx

L

naaxf

L

Ldxxf

La

2

10

L

Ln dx

L

xnxf

La

cos

1

L

Ln dx

L

xnxf

Lb

sin

1

,


Parseval’s Theorem

• Show that:

1

222

0

22

1

n

nn

L

Lbaadxxf

L

1

0

1

0

2

sincos

sincos

n

nn

m

mm

xL

nbx

L

naa

xL

mbx

L

maaxf


nm

dxxL

mnx

L

mnba

dxxL

nx

L

mba

L

Lnm

L

Lnm

,,0

sinsin2

1

sincos

2

000 2LadxaaL

L


nmLa

dxxL

nadxx

L

na

dxxL

nx

L

naa

nmdxxL

nx

L

maa

n

L

Ln

L

Ln

L

Lnn

L

Lnm

,

2cos1

2

1cos

coscos

,0coscos

2

222

1,1 nm


nmLb

dxxL

nbdxx

L

nb

dxxL

nx

L

nbb

nmdxxL

nx

L

mbb

n

L

Ln

L

Ln

L

Lnn

L

Lnm

,

2cos1

2

1sin

sinsin

,0sinsin

2

222

1,1 nm


1,1,

1,1,

1

22222

0

1

0

1

0

2

sinsin

coscos

sincos2

sincos

sincos

sincos

nmnm

nm

nmnm

nm

nm

nm

n

nn

n

nn

n

mm

xL

nbx

L

mb

xL

nax

L

ma

xL

nbx

L

ma

xL

nbx

L

naa

xL

nbx

L

naa

xL

mbx

L

maaxf


1

222

0

2

1

222

0

1

22222

0

2

21

2

sincos

n

nn

L

L

n

nn

L

Ln

nn

L

L

baadxxfL

baaL

dxxL

nbx

L

naadxxf

1

222

0

2

2

1

2

1

n

nn

L

Lbaadxxf

L


Complex Fourier Series




series

1

0 sincosn

nn xL

nbx

L

naaxf

L

Ldxxf

La

2

10

L

Ln dx

L

xnxf

La

cos

1

L

Ln dx

L

xnxf

Lb

sin

1

,3,2,1n


Euler Formulae

iiii

i

eei

ee

ie

2

1sin,

2

1cos

sincos

L

xni

nnL

xni

nn

L

xni

L

xni

nL

xni

L

xni

n

nn

eibaeiba

eebi

eea

L

xnb

L

xna

2

1

2

1

)(2

1)(

2

1

sincos



0

)0(

0

)(

2

1

2

1

sincos2

1

2

1

2

1

sincos2

1

2

1

cdxexfL

a

cdxexfL

dxL

xni

L

xnxf

Liba

cdxexfL

dxL

xni

L

xnxf

Liba

L

L

L

xi

n

L

L

L

xni

L

Lnn

n

L

L

L

xni

L

Lnn



n

L

xni

n

n

L

xni

n

n

L

xni

n

n

L

xni

n

n

L

xni

n

n

L

xni

n

n

L

xni

n

n

L

xni

nn

n

L

xni

nn

n

nn

ec

ececc

ececcececc

eibaeibaa

L

xnb

L

xnaaxf

11

0

1

)(

1

0

11

0

11

0

1

0

2

1

2

1

sincos

L

L

L

xni

n dxexfL

c

2

1


Let Us Play Fourier Series!!




series

1

0 sincosn

nn xL

nbx

L

naaxf

L

Ldxxf

La

2

10

L

Ln dx

L

xnxf

La

cos

1

L

Ln dx

L

xnxf

Lb

sin

1

,3,2,1n


Periodic Square Wave (I)


Since the function f(x) is odd, there are no

constant and cosine terms in the Fourier's

series, i.e., a0=a1=a2=…=0!!

01

01

xL

Lxxf


Periodic Square Wave (I)

nn

L

xn

Ln

Ldx

L

xn

L

dxL

xn

Ldx

L

xn

L

dxL

xnxf

Lb

LL

L

L

L

Ln

cos12

cos2

sin12

sin11

sin11

sin1

00

0

0

1

sincos12

n

xL

n

n

nxf


Periodic Square Wave (II)


00

01

xL

Lxxg

1

sincos1

2

11

2

1

n

xL

n

n

nxfxg


Periodic Square Wave (III)


2/0

2/2/1

2/0

LxL

LxL

LxL

xh


Periodic Square Wave (III)

1

1

1

cos2

sincos1

2

1

2sincos

2cossin

cos1

2

1

2sin

cos1

2

1

2

n

n

n

xL

nn

n

n

L

Lnx

L

n

L

Lnx

L

n

n

n

Lx

L

n

n

n

Lxgxh


Triangle Wave (I)




series, i.e., a0=a1=a2=…=0!!

LxLxxf ,


nn

L

L

xn

n

L

nn

n

L

dxL

xn

nL

xnx

n

L

xnxd

n

L

Ldx

L

xnx

L

dxL

xnx

Ldx

L

xnxf

Lb

L

LL

LL

L

L

L

Ln

cos2

sin2

cos2

cos2

cos2

cos2

sin2

sin1

sin1

0

00

00

1

sincos2

n

xL

nn

n

Lxf


Triangle Wave (I)


Triangle Wave (II)




series, i.e., a0=a1=a2=…=0!!

LxLxxfxg ,

1

sincos2

n

xL

nn

n

Lxg


Triangle Wave (II)


Triangle Wave (III)


LxLLxLxfxh ,

1

sincos2

n

xL

nn

n

LLxh


Triangle Wave (III)


Triangle Wave (IV)


LxLLxL

xLxhxp

20

,


1

sincos2

n

LxL

nn

n

LL

Lxhxp

L

xnn

L

xn

L

Ln

L

Ln

L

xn

LxL

n

sincos

cossincossin

sin

1

sin2

n

xL

n

n

LLxp


Triangle Wave (IV)

Fourier Series - National Tsing Hua...

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