Fourier Series Content

Fourier Series

Content

Periodic Functions

Fourier Series

Complex Form of the Fourier Series

Impulse Train

Analysis of Periodic Waveforms

Half-Range Expansion

Least Mean-Square Error Approximation

Fourier Series

Periodic Functions

The Mathematic Formulation

Any function that satisfies

( ) ( )f t f t T

where T is a constant and is called the period

of the function.

Example:

4cos

3cos)(

tttf Find its period.

)()( Ttftf )(4

1cos)(

3

1cos

4cos

3cos TtTt

tt

Fact: )2cos(cos m

mT

23

nT

24

mT 6

nT 8

24T smallest T

Example:

tttf 21 coscos)( Find its period.

)()( Ttftf )(cos)(coscoscos 2121 TtTttt

mT 21

nT 22

n

m

2

1

2

1

must be a

rational number

Example:

tttf )10cos(10cos)(

Is this function a periodic one?

10

10

2

1 not a rational

number

Fourier Series

Fourier Series

Introduction

Decompose a periodic input signal into

primitive periodic components.

A periodic sequence

T 2T 3T

t

f(t)

Synthesis

T

ntb

T

nta

atf

n

n

n

n

2sin

2cos

2)(

11

0

DC Part

Even Part

Odd Part

T is a period of all the above signals

)sin()cos(2

)( 0

1

0

1

0 tnbtnaa

tfn

n

n

n

Let 0=2/T.

Orthogonal Functions

Call a set of functions {k} orthogonal on an interval a < t < b if it satisfies

nmr

nmdttt

n

b

anm

0)()(

Orthogonal set of Sinusoidal Functions

Define 0=2/T.

0 ,0)cos(2/

2/0 mdttm

T

T0 ,0)sin(

2/

2/0 mdttm

T

T

nmT

nmdttntm

T

T 2/

0)cos()cos(

2/

2/00

nmT

nmdttntm

T

T 2/

0)sin()sin(

2/

2/00

nmdttntmT

T and allfor ,0)cos()sin(

2/

2/00

We now prove this one

Proof

dttntmT

T 2/

2/00 )cos()cos(

0

)]cos()[cos(2

1coscos

dttnmdttnmT

T

T

T

2/

2/0

2/

2/0 ])cos[(

2

1])cos[(

2

1

2/

2/0

0

2/

2/0

0

])sin[()(

1

2

1])sin[(

)(

1

2

1 T

T

T

Ttnm

nmtnm

nm

m n

])sin[(2)(

1

2

1])sin[(2

)(

1

2

1

00

nmnm

nmnm

0

0

Proof

dttntmT

T 2/

2/00 )cos()cos(

0

)]cos()[cos(2

1coscos

dttmT

T 2/

2/0

2 )(cos

2/

2/

0

0

2/

2/

]2sin4

1

2

1T

T

T

T

tmm

t

m = n

2

T

]2cos1[2

1cos2

dttmT

T 2/

2/0 ]2cos1[

2

1

nmT

nmdttntm

T

T 2/

0)cos()cos(

2/

2/00

Orthogonal set of Sinusoidal Functions

Define 0=2/T.

0 ,0)cos(2/

2/0 mdttm

T

T0 ,0)sin(

2/

2/0 mdttm

T

T

nmT

nmdttntm

T

T 2/

0)cos()cos(

2/

2/00

nmT

nmdttntm

T

T 2/

0)sin()sin(

2/

2/00

nmdttntmT


2/

2/00

,3sin,2sin,sin

,3cos,2cos,cos

,1

000

000

ttt

ttt

an orthogonal set.

Decomposition

dttfT

aTt

t

0

0

)(2

0

,2,1 cos)(2

0

0

0

ntdtntfT

aTt

tn

,2,1 sin)(2

0

0

0

ntdtntfT

bTt

tn

)sin()cos(2

)( 0

1

0

1

0 tnbtnaa

tfn

n

n

n

Proof

Use the following facts:

0 ,0)cos(2/

2/0 mdttm

T

T0 ,0)sin(

2/

2/0 mdttm

T

T

nmT

nmdttntm

T

T 2/

0)cos()cos(

2/

2/00

nmT

nmdttntm

T

T 2/

0)sin()sin(

2/

2/00

nmdttntmT


2/

2/00

Example (Square Wave)

112

2

00

dta

,2,1 0sin1

cos2

200

nntn

ntdtan

,6,4,20

,5,3,1/2)1cos(

1 cos

1sin

2

200

n

nnn

nnt

nntdtbn

2 3 4 5 - -2 -3 -4 -5 -6

f(t) 1

112

2

00

dta

,2,1 0sin1

cos2

200

nntn

ntdtan

,6,4,20

,5,3,1/2)1cos(

1 cos

1sin

2

100

n

nnn

nnt

nntdtbn

2 3 4 5 - -2 -3 -4 -5 -6

f(t) 1


ttttf 5sin

5

13sin

3

1sin

2

2

1)(

112

2

00

dta

,2,1 0sin1

cos2

200

nntn

ntdtan

,6,4,20

,5,3,1/2)1cos(

1 cos

1sin

2

100

n

nnn

nnt

nntdtbn

2 3 4 5 - -2 -3 -4 -5 -6

f(t) 1


-0.5

0

0.5

1

1.5

ttttf 5sin

5

13sin

3

1sin

2

2

1)(

Harmonics

T

ntb

T

nta

atf

n

n

n

n

2sin

2cos

2)(

11

0

DC Part

Even Part

Odd Part

T is a period of all the above signals

)sin()cos(2

)( 0

1

0

1

0 tnbtnaa

tfn

n

n

n

Harmonics

tnbtnaa

tfn

n

n

n 0

1

0

1

0 sincos2

)(

Tf

22 00

Define , called the fundamental angular frequency.

0 nnDefine , called the n-th harmonic of the periodic function.

tbtaa

tf n

n

nn

n

n

sincos2

)(11

0

Harmonics

tbtaa

tf n

n

nn

n

n

sincos2

)(11

0

)sincos(2 1

0 tbtaa

nnn

n

n

12222

220 sincos2 n

n

nn

nn

nn

nnn t

ba

bt

ba

aba

a

1

220 sinsincoscos2 n

nnnnnn ttbaa

)cos(1

0 n

n

nn tCC

Amplitudes and Phase Angles

)cos()(1

0 n

n

nn tCCtf

2

00

aC

22

nnn baC

n

nn

a

b1tan

harmonic amplitude phase angle

Fourier Series


Complex Exponentials

tnjtnetjn

00 sincos0

tjntjneetn 00

2

1cos 0

tnjtnetjn

00 sincos0

tjntjntjntjnee

jee

jtn 0000

22

1sin 0


tnbtnaa

tfn

n

n

n 0

1

0

1

0 sincos2

)(

tjntjn

n

n

tjntjn

n

n eebj

eeaa

0000

11

0

22

1

2

1

0 00 )(2

1)(

2

1

2 n

tjn

nn

tjn

nn ejbaejbaa

1

000

n

tjn

n

tjn

n ececc

)(2

1

)(2

1

2

00

nnn

nnn

jbac

jbac

ac


1

000)(

n

tjn

n

tjn

n ececctf

1

1

000

n

tjn

n

n

tjn

n ececc

n

tjn

nec 0

)(2

1

)(2

1

2

00

nnn

nnn

jbac

jbac

ac


2/

2/

00 )(

1

2

T

Tdttf

T

ac

)(2

1nnn jbac

2/

2/0

2/

2/0 sin)(cos)(

1 T

T

T

Ttdtntfjtdtntf

T

2/

2/00 )sin)(cos(

1 T

Tdttnjtntf

T

2/

2/

0)(1 T

T

tjndtetf

T

2/

2/

0)(1

)(2

1 T

T

tjn

nnn dtetfT

jbac )(2

1

)(2

1

2

00

nnn

nnn

jbac

jbac

ac


n

tjn

nectf 0)(

dtetfT

cT

T

tjn

n

2/

2/

0)(1

)(2

1

)(2

1

2

00

nnn

nnn

jbac

jbac

ac

If f(t) is real,

*

nn cc

nn j

nnn

j

nn ecccecc

|| ,|| *

22

2

1|||| nnnn bacc

n

nn

a

b1tan

,3,2,1 n

002

1ac

Complex Frequency Spectra

nn j

nnn

j

nn ecccecc

|| ,|| *

22

2

1|||| nnnn bacc

n

nn

a

b1tan ,3,2,1 n

002

1ac

|cn|

amplitude spectrum

n

phase

spectrum

Example

2

T

2

T TT

2

d

t

f(t)

A

2

d

dteT

Ac

d

d

tjn

n

2/

2/

0

2/

2/0

01

d

d

tjne

jnT

A

2/

0

2/

0

0011 djndjn

ejn

ejnT

A

)2/sin2(1

0

0

dnjjnT

A

2/sin1

0

021

dnnT

A

T

dn

T

dn

T

Adsin

T

dn

T

dn

T

Adcn

sin

82

5

1

T ,

4

1 ,

20

1

0

T

dTd

Example

40 80 120 -40 0 -120 -80

A/5

50 100 150 -50 -100 -150

T

dn

T

dn

T

Adcn

sin

42

5

1

T ,

2

1 ,

20

1

0

T

dTd

Example

40 80 120 -40 0 -120 -80

A/10

100 200 300 -100 -200 -300

Example

dteT

Ac

dtjn

n

0

0

d

tjne

jnT

A

00

01

00

110

jne

jnT

A djn

)1(1

0

0

djne

jnT

A

2/0

sindjn

e

T

dn

T

dn

T

Ad

TT d

t

f(t)

A

0

)(1 2/2/2/

0

000 djndjndjneee

jnT

A

Fourier Series

Impulse Train

Dirac Delta Function

0

00)(

t

tt and 1)(

dtt

0 t

Also called unit impulse function.

Property

)0()()(

dttt

)0()()0()0()()()(

dttdttdttt

(t): Test Function

Impulse Train

0 t

T 2T 3T T 2T 3T

n

T nTtt )()(

Fourier Series of the Impulse Train

n

T nTtt )()(T

dttT

aT

TT

2)(

2 2/

2/0

Tdttnt

Ta

T

TTn

2)cos()(

2 2/

2/0

0)sin()(2 2/

2/0 dttnt

Tb

T

TTn

n

T tnTT

t 0cos21

)(

Complex Form Fourier Series of the Impulse Train

Tdtt

T

ac

T

TT

1)(

1

2

2/

2/

00

Tdtet

Tc

T

T

tjn

Tn

1)(

1 2/

2/

0

n

tjn

T eT

t 01

)(

n

T nTtt )()(

Fourier Series

Analysis of Periodic Waveforms

Waveform Symmetry

Even Functions

Odd Functions

)()( tftf

)()( tftf

Decomposition

Any function f(t) can be expressed as the

sum of an even function fe(t) and an odd

function fo(t).

)()()( tftftf oe

)]()([)(21 tftftfe

)]()([)(21 tftftfo

Even Part

Odd Part

Example

00

0)(

t

tetf

t

Even Part

Odd Part

0

0)(

21

21

te

tetf

t

t

e

0

0)(

21

21

te

tetf

t

t

o

Half-Wave Symmetry

)()( Ttftf and 2/)( Ttftf

T T/2 T/2

Quarter-Wave Symmetry

Even Quarter-Wave Symmetry

T T/2 T/2

Odd Quarter-Wave Symmetry

T

T/2 T/2

Hidden Symmetry

The following is a asymmetry periodic function:

Adding a constant to get symmetry property.

A

T T

A/2

A/2

T T

Fourier Coefficients of Symmetrical Waveforms

The use of symmetry properties simplifies the

calculation of Fourier coefficients.

– Even Functions

– Odd Functions

– Half-Wave

– Even Quarter-Wave

– Odd Quarter-Wave

– Hidden

Fourier Coefficients of Even Functions

)()( tftf

tnaa

tfn

n 0

1

0 cos2

)(

2/

00 )cos()(

4 T

n dttntfT

a

Fourier Coefficients of Even Functions

)()( tftf

tnbtfn

n 0

1

sin)(

2/

00 )sin()(

4 T

n dttntfT

b

Fourier Coefficients for Half-Wave Symmetry


T T/2 T/2

The Fourier series contains only odd harmonics.

Fourier Coefficients for Half-Wave Symmetry


)sincos()(1

00

n

nn tnbtnatf

odd for )cos()(4

even for 02/

00 ndttntf

T

na T

n

odd for )sin()(4

even for 02/

00 ndttntf

T

nb T

n

Fourier Coefficients for Even Quarter-Wave Symmetry

T T/2 T/2

])12cos[()( 0

1

12 tnatfn

n

4/

0012 ])12cos[()(

8 T

n dttntfT

a

Fourier Coefficients for Odd Quarter-Wave Symmetry

])12sin[()( 0

1

12 tnbtfn

n

4/

0012 ])12sin[()(

8 T

n dttntfT

b

T

T/2 T/2

Example Even Quarter-Wave Symmetry

4/

0012 ])12cos[()(

8 T

n dttntfT

a 4/

00 ])12cos[(

8 T

dttnT

4/

0

0

0

])12sin[()12(

8T

tnTn

)12(

4)1( 1

n

n

T

T/2 T/2

1

1

T T/4 T/4

Example Even Quarter-Wave Symmetry

4/

0012 ])12cos[()(

8 T

n dttntfT

a 4/

00 ])12cos[(

8 T

dttnT

4/

0

0

0

])12sin[()12(

8T

tnTn

)12(

4)1( 1

n

n

T

T/2 T/2

1

1

T T/4 T/4

ttttf 000 5cos

5

13cos

3

1cos

4)(

Example

T

T/2 T/2

1

1

T T/4 T/4


4/

0012 ])12sin[()(

8 T

n dttntfT

b 4/

00 ])12sin[(

8 T

dttnT

4/

0

0

0

])12cos[()12(

8T

tnTn

)12(

4

n

Example

T

T/2 T/2

1

1

T T/4 T/4


4/

0012 ])12sin[()(

8 T

n dttntfT

b 4/

00 ])12sin[(

8 T

dttnT

4/

0

0

0

])12cos[()12(

8T

tnTn

)12(

4

n

ttttf 000 5sin

5

13sin

3

1sin

4)(

Fourier Series

Half-Range Expansions

Non-Periodic Function Representation

A non-periodic function f(t) defined over (0, )

can be expanded into a Fourier series which is

defined only in the interval (0, ).

Without Considering Symmetry




T

Expansion Into Even Symmetry




T=2

Expansion Into Odd Symmetry




T=2

Expansion Into Half-Wave Symmetry




T=2

Expansion Into Even Quarter-Wave Symmetry




T/2=2

T=4

Expansion Into Odd Quarter-Wave Symmetry




T/2=2 T=4

Fourier Series

Least Mean-Square

Error Approximation

Approximation a function

Use

k

n

nnk tnbtnaa

tS1

000 sincos

2)(

to represent f(t) on interval T/2 < t < T/2.

Define )()()( tStft kk

2/

2/

2)]([1 T

Tkk dtt

TE

Mean-Square

Error


Show that using Sk(t) to represent f(t) has

least mean-square property.

2/

2/

2)]([1 T

Tkk dtt

TE

2/

2/

2

1

000 sincos

2)(

1 T

T

k

n

nn dttnbtnaa

tfT

Proven by setting Ek/ai = 0 and Ek/bi = 0.


2/

2/

2)]([1 T

Tkk dtt

TE

2/

2/

2

1

000 sincos

2)(

1 T

T

k

n

nn dttnbtnaa

tfT

0)(1

2

2/

2/

0

0

T

T

k dttfT

a

a

E0cos)(

2 2/

2/0

T

Tn

n

k tdtntfT

aa

E

0sin)(2 2/

2/0

T

Tn

n

k tdtntfT

bb

E

Mean-Square Error

2/

2/

2)]([1 T

Tkk dtt

TE

2/

2/

2

1

000 sincos

2)(

1 T

T

k

n

nn dttnbtnaa

tfT

2/

2/1

222

02 )(2

1

4)]([

1 T

T

k

n

nnk baa

dttfT

E

Mean-Square Error

2/

2/

2)]([1 T

Tkk dtt

TE

2/

2/

2

1

000 sincos

2)(

1 T

T

k

n

nn dttnbtnaa

tfT

2/

2/1

222

02 )(2

1

4)]([

1 T

T

k

n

nn baa

dttfT

Mean-Square Error

2/

2/

2)]([1 T

Tkk dtt

TE

2/

2/

2

1

000 sincos

2)(

1 T

T

k

n

nn dttnbtnaa

tfT

2/

2/1

222

02 )(2

1

4)]([

1 T

Tn

nn baa

dttfT

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