07 periodic functions and fourier series

Periodic Functions and Fourier Series

Periodic Functions

A function f is periodic

if it is defined for all real and if there is some positive number,

T such that fTf .

0

T

f

0

T

f

0

T

f

Fourier Series

f be a periodic function with period 2

The function can be represented by a trigonometric series as:

11

0 sincosn

nn

n nbnaaf

11

0 sincosn

nn

n nbnaaf

What kind of trigonometric (series) functions are we talking about?

32

and32

sin,sin,sin

cos,cos,cos

0

0

2

cos cos 2 cos 3

0

0 2

sin sin 2 sin 3

We want to determine the coefficients,

na and nb .

Let us first remember some useful integrations.

dmndmn

dmn

cos2

1cos

2

1

coscos

0

dmn coscos mn

dmn coscos mn

dmndmn

dmn

sin2

1sin

2

1

cossin

0

dmn cossin

for all values of m.

dmndmn

dmn

cos2

1cos

2

1

sinsin

0

dmn sinsin mn

dmn sinsin mn

Determine 0aIntegrate both sides of (1) from

to

dnbnaa

df

nn

nn

110 sincos

dnb

dnada

df

nn

nn

1

10

sin

cos

000

dadf

002 0

adf

dfa

2

10

0a is the average (dc) value of the

function, f .

dnbnaa

df

nn

nn

2

011

0

2

0

sincos

It is alright as long as the integration is performed over one period.

You may integrate both sides of (1) from

0 to 2 instead.

dnb

dnada

df

nn

nn

2

01

2

01

2

0 0

2

0

sin

cos

002

0 0

2

0

dadf

002 0

2

0 adf

dfa

2

00 2

1

Determine naMultiply (1) by mcosand then Integrate both sides from

to

dmnbnaa

dmf

nn

nn

cossincos

cos

110

Let us do the integration on the right-hand-side one term at a time.

First term,

00 dma cos

Second term,

dmnan

n coscos1

m

nn admna

coscos1

Second term,

Third term,

01

dmcosnsinb

nn

Therefore,

madmf cos

,2,1cos1

mdmfam

Determine nbMultiply (1) by msinand then Integrate both sides from

to

dmsinnsinbncosaa

dmsinf

nn

nn

110

Let us do the integration on the right-hand-side one term at a time.

First term,

00 dmsina

Second term,

dmsinncosa

nn

1

01

dmsinncosa

nn

Second term,

Third term,

m

nn bdmnb

sinsin1

Therefore,

mbdmsinf

,,mdmsinfbm 211

dfa

2

10

The coefficients are:

,2,1cos1

mdmfam

,2,1sin1

mdmfbm

dfa

2

10

We can write n in place of m:

,,ndncosfan 211

,,ndnsinfbn 211

The integrations can be performed from

0 to 2 instead.

dfa

2

00 2

1

,,ndncosfan 211 2

0

,,ndnsinfbn 211 2

0

Example 1. Find the Fourier series of the following periodic function.

0

f

2 3 4 5

A

-A

2

0

whenA

whenAf

ff 2

02

12

12

1

2

0

2

0

2

00

dAdA

dfdf

dfa

011

1

1

2

0

2

0

2

0

n

nsinA

n

nsinA

dncosAdncosA

dncosfan

ncosncoscosncosn

A

n

ncosA

n

ncosA

dnsinAdnsinA

dnsinfbn

20

11

1

1

2

0

2

0

2

0

oddisnwhen4

1111

20

n

An

A

ncosncoscosncosn

Abn

evenisnwhen0

1111

20

n

A

ncosncoscosncosn

Abn

Therefore, the corresponding Fourier series is

7sin

7

15sin

5

13sin

3

1sin

4A

In writing the Fourier series we may not be able to consider infinite number of terms for practical reasons. The question therefore, is – how many terms to consider?

When we consider 4 terms as shown in the previous slide, the function looks like the following.

1.5

1

0.5

0

0.5

1

1.5

f ( )

When we consider 6 terms, the function looks like the following.

1.5

1

0.5

0

0.5

1

1.5

f ( )

When we consider 8 terms, the function looks like the following.

1.5

1

0.5

0

0.5

1

1.5

f ( )

When we consider 12 terms, the function looks like the following.

1.5

1

0.5

0

0.5

1

1.5

f ( )

The red curve was drawn with 12 terms and the blue curve was drawn with 4 terms.

1.5

1

0.5

0

0.5

1

1.5

The red curve was drawn with 12 terms and the blue curve was drawn with 4 terms.

0 2 4 6 8 101.5

1

0.5

0

0.5

1

1.5

The red curve was drawn with 20 terms and the blue curve was drawn with 4 terms.

0 2 4 6 8 101.5

1

0.5

0

0.5

1

1.5

Even and Odd Functions

(We are not talking about even or odd numbers.)

Even Functions

f(

The value of the function would be the same when we walk equal distances along the X-axis in opposite directions.

ff Mathematically speaking -

Odd Functions The value of the function would change its sign but with the same magnitude when we walk equal distances along the X-axis in opposite directions.

ff Mathematically speaking -

f(

Even functions can solely be represented by cosine waves because, cosine waves are even functions. A sum of even functions is another even function.

10 0 10

5

0

5

Odd functions can solely be represented by sine waves because, sine waves are odd functions. A sum of odd functions is another odd function.

10 0 10

5

0

5

The Fourier series of an even function fis expressed in terms of a cosine series.

1

0 cosn

n naaf

The Fourier series of an odd function fis expressed in terms of a sine series.

1

sinn

n nbf

Example 2. Find the Fourier series of the following periodic function.

ff 2

x

f(x)

3 5 7 90

xwhenxxf 2

332

1

2

1

2

1

23

20

x

x

x

dxxdxxfa

nxdxx

dxnxxfan

cos1

cos1

2

Use integration by parts. Details are shown in your class note.

nn

an cos4

2

odd is n when4

2nan

even is n when4

2nan

This is an even function.

Therefore, 0nbThe corresponding Fourier series is

222

2

4

4cos

3

3cos

2

2coscos4

3

xxxx

Functions Having Arbitrary Period

Assume that a function tf has period, T . We can relate angle ( ) with time ( t ) in the following manner.

t

is the angular velocity in radians per second.

f 2f is the frequency of the periodic function,

tf

tf 2 Tf

1where

tT

2Therefore,

tT

2 dt

Td

2

Now change the limits of integration.

2

Tt t

T

2

2

Tt t

T

2

dfa

2

10

dttfT

a

T

T

2

2

0

1

,,ndncosfan 211

,2,12

cos2 2

2

ndttT

ntf

Ta

T

Tn

,,ndnsinfbn 211

,2,12

sin2 2

2

ndttT

ntf

Tb

T

Tn

Example 4. Find the Fourier series of the following periodic function.

0 t

T/2

f(t)

T/4

3T/4

T-T/2 2T

4

3

42

44T

tT

whenT

t

Tt

Twhenttf

tfTtf

This is an odd function. Therefore, 0na

dttT

ntf

T

dttT

ntf

Tb

T

T

n

2sin

4

2sin

2

2

0

0

dttT

nTt

T

dttT

nt

Tb

T

T

T

n

2sin

2

4

2sin

4

2

4

4

0

Use integration by parts.

2sin

2

2sin

2.2

4

22

2

n

n

T

n

n

T

Tbn

0nb when n is even.

Therefore, the Fourier series is

t

Tt

Tt

T

T 10

5

16

3

122222sinsinsin

The Complex Form of Fourier Series

11

0 sincosn

nn

n nbnaaf

Let us utilize the Euler formulae.

2

jj ee

cos

i

eesin

jj

2

The nth harmonic component of (1) can beexpressed as:

i

eeb

eea

nbnajnjn

n

jnjn

n

nn

22

sincos

22

jnjn

n

jnjn

n

eeib

eea

jnnnjnnn

nn

ejba

ejba

nbna

22

sincos

Denoting

2nn

n

jbac

2

nnn

jbac,

and 00 ac

jn

njn

n

nn

ecec

nsinbncosa

The Fourier series for f can be expressed as:

n

jnn

n

jnn

jnn

ec

ececcf

1

0

The coefficients can be evaluated in the following manner.

dnf

jdnf

jbac nnn

sin2

cos2

1

2

dnjnf sincos

2

1

def jn

2

1

dnf

jdnf

jbac nnn

sin2

cos2

1

2

dnjnf sincos

2

1

def jn

2

1

.

2

nnn

jbac

2nn

n

jbac

nc nc

Note that is the complex conjugate of

. Hence we may write that

defc jnn 2

1

,,,n 210

The complex form of the Fourier series of

f 2 with period is:

n

jnnecf

Example 1. Find the Fourier series of the following periodic function.

0

f

2 3 4 5

A

-A

2

0

whenA

whenAf

ff 2

A 5

f x( ) A 0 x if

A x 2 if

0 otherwise

A01

2 0

2

xf x( )

d

A0 0

n 1 8

An1

0

2

xf x( ) cos n x( )

d

A1 0 A2 0 A3 0 A4 0

A5 0 A6 0 A7 0 A8 0

Bn1

0

2

xf x( ) sin n x( )

d

B1 6.366 B2 0 B3 2.122 B4 0

B5 1.273 B6 0 B7 0.909 B8 0

Complex Form

n

jnnecf

defc jnn 2

1

,,, 210 n

C n( )1

2 0

2

xf x( ) e1i n x

d

C 0( ) 0 C 1( ) 3.183i C 2( ) 0 C 3( ) 1.061i

C 4( ) 0 C 5( ) 0.637i C 6( ) 0 C 7( ) 0.455i

C 1( ) 3.183i C 2( ) 0 C 3( ) 1.061i

C 4( ) 0 C 5( ) 0.637i C 6( ) 0 C 7( ) 0.455i

C n( )1

2 0

2

xf x( ) e1i n x

d