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Fourier Series and Fourier Transforms

MingsianMingsian R. R. BaiBai22

0 1 2 cos particular solution of the form cos sin

How about an arbitrary periodic input? Fourier series

* ( ) is periodic ( 2 ) ( ) , : integer, 2 is called the e op riod

a y a y a y k tY A t B t

f t f t Pn f tn P

+ + = = +

+ =f ( )

* : periodic discrete freq.

f t


0

1

1

1

1( ) ( cos sin ) , 2

even periodic odd per

2 : ;22 , 1, 2, : superharmonic

fundamental frequency

History of Fouri

2

ierodic

Fourier Series

n nn

n

n t n tf t a a bP P

Pn n n

P

=

=

= = =

= + +

K s

Oscillates increasingly faster as .' & ' are called the Euler coefficients.n n

na s b s


2 2

2

2

Orthogonality ( ) cf. linear al

1. cos 0 , sin

gebra

0 , 0

2.

cos cos , , 0

sin sin , , 0

s

co

d P d P

d d

d P

mnd

d P

mnd

n t n tdt dt nP P

m t n t dt P m nP P

m t n t dt P m nP P

m t

+ +

+

+

= =

=

=

Note :

2sin 0

d P

d

n t dtP P

+=

( ) ( )2

cos cos 1 cos cos2

cos 2 1cos2

= + +

+=

(cos,sin are orthogonal functions, see Sturm-Liuille)


0

2 20

To find ,

c( ) os2

d P d Pnd d

aa n tdtf d

Pt t a

+ += + sinn

n tdt bP+

1

20

2 2

00 = 22

(DC bias, me1 ( an val )) ued

d P d

P

P

d dn

d

dt

a P

a f t dtP

Pa

+ +

=

+

=

=

DC0

2a

( )f t


2

0

To find ,

1co) c( s os2

n

d P

d

a

n t n tdt aP P

f t +

=2 2

1

cos cos

sin cos

d P d P

md dm

m

m t n tdt a dtP P

m t n tbP P

+ +

=

+ +

2 2

0 0

2

2cos

1 ( & can b1 ( )cos e combined here )2

d P

n d

d P d P

n nd d

nn ta

n tdt a dt a PP

a a af t dtP P

+ +

+

= =

=

Q

2

0

To find ,

1 sin sin2

( )

n

d P

d

b

n t n tdt aP

fP

t +

=2

cos sind P

md

m t n tdt aP P +

+2

1

2 2 2

2

sin sin sin

1 ) ( sin

d P

dm

d P d P

m n nd d

d P

n d

m t n t n tb dt b dt b P

n tb f t dtP P

P P P

+

+

+

=

+

+ = =

=


Thm 1 If ( ) is a bounded periodic function which, in any one period, has at most a finite number of local maxima and minima and a finite n

Fourier conve

umber of po

rg

in

ence

ts of

the

orem

discont

f t

01

inuity, then

Pf: see p.81 ,

(

)

Greenberg, Foundations of

( ) 1 cos sin2 2

Applied Mathematics.

n nn

f t f t n t n ta a bP P +

=

+ = + +

interchangibility of

Dirichlet cond. (sufficient cond.)

Q

converges to the mean


( )1

0

1

0

0

1 ( ) cos sin (synthesi

2 ( ) cos , (analysis equations)

* Alternative f

2 ( )

orm:

sin

s equation)

2 , 2

2T

n

n n n

n

n

n

n

n

T

n

a f t t dtT

b f t t

nn

dt

f t a a t b t

T

n TT P

=

= = = =

=

= + +

=

, 0,1, 2,

*Plot Time-domain: ( ) Frequency-domain: ,

*Parameters : sin sin , (rad/sec) 2 (cycle/sec, CPS, Hz)

FSn n

P n

f t a b

t t f

=

=

K


Ex. Half-wave rectifierFind the Fourier series of a 2 -periodic function

0 0 ( )

sin 0t

f tt t

=

t

( )tf

0 2 3

2

0

2 20

1 00

Sol: 2 / 2 Let

1 1 ( ) cos sin cos

1 1 cos(1 ) cos(1 ) 1 cos 1 ( 1) , 12 1 1 (1 ) (1 )

1 cos 2 1: sin cos 04

n

n

Pd

n ta f t dt t nt dt

n t n t n nn n n n

tn a t t dt

= ==

= =

+ + + = + = = +

= = = =


( ) ( )

0

0

21 0

0

1 1( )sin sin sin

1 1 sin(1 ) sin(1 ) = 0, 12 1 1

1 1 sin 2 11: sin (or By l'

sin

Hos2

s

4 2

in 1 cos cos2

nn tb f t dt t nt dt

n t n t nn n

t tn b t dt

= =

+ = +

= = = =

= +

Q

pital rule)


1 sin 2 cos 2 cos 4 cos6 cos8Thus, ( )2 3 15 35 63

* The "smoother' the function, the fewer terms the F.S. takes to converge. Low-frequency components prevail.

Thm 4 term-by-term integra ti

t t t t tf t

= + + + + +

L

The integral of any periodic function ( ) which satisfies the Dirichlet condition can be found by term-by-term integration of the F.S. of the fu

on

term-by-term differentia

n

t

ction.

Thm 5 If ( ) i

ion

f t

f t s a periodic function which satisfies the Dirichlet condition, and is everywhere , and if ( ) also satisfies the Dirichlet condition, then ( ) can be found by term-by-te

continrm dif

uoferentia

ust

f tf t

ion of the F.S. of ( ).

Ex. Square wave is a counter-example.f t


Half-Range Expansions0

1

1( ) cos sin2

== + + n n

n

n t n tf t a a bP P

even odd

0

0

( ) is an function, i.e., ( ) ( ) Let

1 1 1 ( )cos ( )cos ( )cos

ev

( )

e

n

P Pn P P

f t f t f td P

n t n t n ta f t dt f t dt f t dtP P P P P P

t t

==

= = +

0

0

0 0 0

1 ( ) 1 ( )cos ( ) ( )cos

1 1 2 ( )cos ( )cos ( )cos

P

P

P P P

n t n tf t d t f t dtP P P P

n t n t n tf t dt f t dt f t dtP P P P P P

= +

+= + =


0

01

1( ) is an function, ( ) ( ) ( ) cos2

,

1 ( )sin 0 (Fourier cosine series)

If ( ) is an f

even

even odd

odd

2 ( )c

unction, )

os

(

=

= = +

=

=

=

nn

n

Pn

P

P

n ta

n tf t f t f t f t a aP

n tb f t dtP P

f t dtP P

f t f t1

0

( ) ( ) sin

1 ( )cos 0,

2

odd even

(Fourier( ) sine ser esi is )n

=

=

=

=

n

Pn

n

Pn P

n tb f t dtP

n tf t f t bP

n ta f t dtP P

P


: consider ( ) only in (0, p] for solving PDEs

( ), 0 : arbitrary func( ), 0

tion( )

( 2 ) ( ), periodic extension

Half-range expansion

<


Alternative forms of Fourier Series

2 2 2 2

0

1

2 20

1

cos sin

cos sin

( )2

Single trigonometri

c form:

2

n n

n n

n n n n

n

n nn

n t n ta bp p

a bn t n tp pa b a b

af t

a a b

=

=

+

+ + +

= +

= + +

n

nnb

na

2 2= +n n nA a b

( )2 2 100

0 01 1

0 01 1

cos cos sin sin

sin cos cos sin

Let , , tan /2 2

( ) cos

( ) sin

n n n

n n n

n n n n n n n

n nn n

n nn n

n t n t n tp p p

n t n t n tp p p

aA A a b b a

f t A A A A

f t A A A A

= =

= =

+

+ +

= = + = =

= + = +

= + = +


0

( ) (synthesis equation)

1 ( ) (analysis equation)

n

n

i tn

n

T i tn

f t C e

C f t e dtT

=

=

=

1

1

1

(rad/sec) 2 (Hz)

2 (period),2 ,

2Fundamental frequency:

Note that ( , ) Heisenberg uncertainty principle

n

f

T p

n n nT p

TT

=

=

= = =

=

Complex form of FS


0

1

0

1

0

1

00

cos sin

2 2

2 2

( )2

2

2

Defi

Complex FS de

ne

rivation:

, ,2 2

n n

in t p in t p in t p in t p

n n

in t p in t pn n n n

n

n

n

n nn

n t n ta bp p

e e e ea bi

a ib a ibe e

af t

a

a

a a ibC C

=

=

=

+

+

+

+ +

= +

= +

= +

= =

20 0

2 2

2cos

2

( ) ( ),

1 1 ( ) (1)2 2

1 1 1( )cos ( )sin2 2

1 1 ( ) sin2 2

n

n nn

in t p i tn n

d p

d

d p d pn nn d d

n

d p

d

n

n t i dtp

a ibC

f t C e C e

C a f t dtp

a ib n t n tC f t dt i f t dtp p p p

n tf tp p

=

+

+ +

=

+

+=

= =

= =

= =

= =

2( ) (2)

d p in t pd

f t e dtp

+


2

2 2

1Similarly, ( ) (3)21 1( ) ( )

2(1),(2),(3): ( ) , ,0,1, ,

n

d p in t pn d

d p d pin t p i tn d d

C f t e dtp

C f t e dt f t e dtp T

n

+

+ +

=

= =

=

K K

1(1 ) 11

11

1

2 2

Ex. ( ) 1 1 in one period ( 1)

1 1Sol: 2 2 (1 ) 2(1 )

( 1) ( 1) (1 )sinh(1) (1 2 1

t

in t in inin t pt

n

n nin

f t e t p

e e e e eC e e dtp in in

e e in ein n

+

= < =

= = = + +

= = = + +

( )1)

( )

n

in tn

nf t C e

==


0

( ) (synthesis equation)

1 ( ) (analysis equation)

n

n

i tn

n

T i tn

f t C e

C f t e dtT

=

=

=

( ) FS nf t C

t

( )f t

FS

nC

1n =0 1 21(time domain)continuous,periodic

(freq. domain) Spectrum( ) discrete, aperiodic

In Digital Signal Processing (DSP), the following Fourier Series pair is used to represent a continuous and periodic signal.


ApplicationEx. Steady-state response due to a periodic input

2 0 0.02 25 ( ) . ( )

2 0 ( 2 ) ( ),

t ty y y r t st r t

t tr t r t p

+ < < + + = =

+ <

=

=

+

1( ) { ( ) ( )}ih t FT E H =

t

1

( )h t


{ } { }{ }

{ }

1 1

* Three ways of descrbing a :: ( ) ( ) ( Impulse response function

Transfer function Frequency re

LTI)

: ( ) ( ): ( ) ( ) ( )sponse fun

sys

ction

tem

s j

h t LT H s FT HH s LT h t

H FT h t H s

=

= =

=

= =

FRF is only meaningful for stable LTI systems.

( )h t

( )H s ( )H

LT FT

s j=

system


Sampling theoremEx. fundamental theorem of DSPAssume fundamental theorem of DSP

( ) ( ) ( ), where ( ) ( )

being an

Nyquist-Shannon

implu , se train ( ) is the Dirac delta function a

s cn

x t x t s t s t t nT

t

=

= =

nd is the sampling period.Find the Fourier transform ( ) in terms of ( ).s c

TX X


1 2 ( ) 2 ( ) ( property)1 2( ) ( ) ( )

(impulse train to impluse tra

modulation

in)1Knowing ( ) ( ) ( ) ( ) ( )

2

s

s

jk tFT FTs

jk t FTs

k k

FTs c c

e k

s t e S kT T

x t x t s t X S

= =

= =

=

( theorem)convoluti

1 1( )

n

( )2

o

) (2s c

X X S

= = 2( )cX ( )s

kk

T

=

Sol: Since ( ) is a function, it has the following representation (Dirichlet conditions violated):

( ) ( ) , where (sampling frequency)

1 ( )

periodic FS

2sjk tk

n k

j

s

kk

T

s t

s t t nT a e

a t eT

= =

=

== = /2

/2

1 1( )s sT

t jk t

kTdt s t e

T T

=

= =


( ) ( ) ( ) [ ( )]( ) (freq-shifted impuse response)

1 1( ) ( ) ( ) ( )

( ) ( 1( ) ( ) )) (

c s c s

c s

s c s c sk k

FTs sc c s

k

X k X k dX k

X

x t s t X kT

X k X kT T

x t X

=

= =

=

=

= =

= =

Q

Oversampled, no aliasing: 2 > s N Undersampled, aliasing: 2 < s N

61 MingsianMingsian R. R. BaiBai

Demo of Aliasing: Alice in the Wonderland

20 0

0

Linear chirp/sweep sine:[ ] cos(0.5 ), * / /nsec

Instantaneous frequency

Time duration: 0 ~ sec (sec) Frequency sweeping range: 0 ~ (rad/sample)

sx n n r f

nn

nr

= =

= =

%MATLAB programfs=10000;nsec=4; %time durationr=2; %aliasing factorn=0:1:nsec*fs;w0=r*pi/fs/nsec;x=cos(0.5*w0.*n.*n);

No aliasing:

Aliased:

InterpretationFor the digital discrete-time signal, the sampling frequency must be greater than two times the maximum frequency of the analog continuous-time signal.

At least two samples within one cycle of a sinusoid must be sampled.


[ ]

Ex. Amplitude-modulated signals bandpass (narrowband) signals, Why modulation? 15 ( ) ( ) cos ( )cx t a t t t

= +

Q ( km at 20 kHz 15 mm at 20 GHz)

amplitude

carrier frequency

phase

t

( )x t( ): baseband info-bearing signal

(audio,speech,video )a t

K

carrier

( )a t

[ ]cos ( )ct t +

( )x t


*

( ) BasebandDefine complex amplitude (phasor)

( ) ( ) ( ) ( ) ( )cos ( ) Re ( ) ( )

( ) ( ) Re ( )

( ) ( )

RF ba

=

nd

2

c

c

c c

j t

j tc

j t j t j t

j t j t

u t a t ex t a t t t u t e

X j x t e dt u t e e dt

u t e u t e

=

= + =

= =

+

( )

( ) ( )

( )

-

( ) *

*

*

1 = ( ) ( )2

1 1 = ( ) ( )2 21 1 = ( )2 2

cc

c c

j t

j tj t

j t j t

c c

e dt

u t e u t e dt

u t e dt u t e dt

U j j U j j

+

+

+

+


( )X j12

c c

(RF band)

( )U j1

(Base band)


HW (Wylie)

Sec.9.1: 18, 30Sec.9.3: 4(a), (b), 11Sec.9.4: 18Sec.9.5: 13

Due date: 1/12/2009


Final Exam

: CD, 1/12/2009

:

Laplace transformFourier tranaform

Fourier Series and Fourier Transforms 2 3 4 5 6 7 8 9 10 11Half-Range Expansions 13 14Alternative forms of Fourier Series 16 17 18 19Application 21 22 23 24WylieFourier Integral and Fourier Transform 27 28 29Applications of Fourier Transform (FT) Vibration diagnostics 32 Time Versus Frequency Diagram 34 35 36 37 38 39 40 41 42 43Summary 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64HW (Wylie) Final Exam

ocw.nctu.edu.twocw.nctu.edu.tw/course/engineeringmathematicsi/ch04.pdf · mingsian r. bai 1 sin 2...

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