13.1 Fourier transforms: Chapter 13 Integral transforms.

13.1 Fourier transforms:

Chapter 13 Integral transforms

)exp(])exp()(2

[)(

)exp()(2

)2

exp()(1

,continuous 02

quantum frequency as

2

and period )exp()2

exp()(

2/

2/

2/

2/

2/

2/

tiduuiuftf

dttitfdtT

irttf

Tc

TT

T

πr ωTtic

T

irtctf

r

T

T rr

T

T r

T

Tr

r

rrr

rr

r

duuiufdtitf

dtigtig

TT

ticT

rTr

rrr

rr

r

)exp()()exp(2

1)(

)exp()(2

1)exp()(

2

0/2 as

rectangle line-broken rth the of area the :)exp()/2(

of function discrete a is /2


theorem inversionFourier ])exp()(2

1[)exp(

2

1)(

duuiufdtitf

The Fourier transform of f(t)

dttitff )exp()(

2

1)(

~

Inverse Fourier transform of f(t)

dtiftf )exp()(

~

2

1)(

Ex: Find the Fourier transform of the exponential decay function

and

0for 0)( ttf

)0(0for )exp()( ttAtf

)(2]

))(exp([

20

)exp()exp(2

)exp(02

1)(

~

0

0

0

i

A

i

tiA

dttitA

dttif

Sol:


][ :tan

)()(

)()(][:

)(

)(][:

2

2

XVtiondard deviaS

ondistributicontinuousdxxfx

ondistributidiscretexfxXVVariance

ion distributcontinuousdxxxf

on distributidiscretexfxXEMean

ii

i

iii

Properties of distribution:

The uncertainty principle:


Gaussian distribution: probability density function

1)(])(

2

1exp[

2

1)( 2 dxxf

xxf

(1) is symmetric about the point

the standard deviation describes the width of a curve

(2) at falls to of the peak value,

these points are points of inflection

)(xf x

)( , xfx 61.02/1 e

0|2

2

xdx

fd

:

dxxfx

dxxxfxf

)()(

)()(

2

Ex: Find the Fourier transform of the normalized Gaussian distribution.

tt

tf )2

exp(2

1)( 2

2

Sol: the Gaussian distribution is centered on t=0, and has a root

mean square deviation

)2

exp(2

1

}]2

)(exp[

2

1){

2

1exp(

2

1

]})()(2[2

1exp{

2

1

2

1

)exp()2

exp(2

1

2

1)(

~

22

2

22222

2222222

2

2

dtit

dtiitit

dttit

f


t

=1

is a Gaussian distribution centered on zero and with a root

mean square deviation is a constant.

)(~ f

/1/1 t

(1) Fraunhofer diffraction:When the cross-section of the object is small compared with the distance at which the light is observed the pattern is known as a Fraunhofer diffraction pattern.

Applications of Fourier transforms:


sin ,|)(~

|2

||)(

is direction the in intensity the

0)(||for ,)sinexp()()exp(

)(

|| and ˆsinˆcos

to ˆ from traveling change phase the :)]ˆexp[

function aperture :)(

|ˆ|

)]ˆ(exp[)()( is amplitudelight total the

|| largefor ,ˆˆˆ position aat

22

0

2

0

00

0'

00

0

0'

0

00000

kqqfr

AI

yfYydyikyyfr

rkirA

Yrjkikk

rjyjyr(ki

yf

dyjyr

jyrkiyfrA

rkzjyixr

'

Y

Y

k

'k

x

y

Y

Y

Ex: Evaluate for an aperture consisting of two long slits each of width 2b whose centers are separated by a distance 2a, a>b; the slits illuminated by light of wavelength .


)(I

20

2

222

20

sincos16|)(

~|

2)(sin

sin2for

2

sincos4

])exp(

[2

1]

)exp([

2

1

)exp(2

1)exp(

2

1)(

~

1)(

rq

qbqaqf

rIkq

q

qbqa

iq

iqx

iq

iqx

dxiqxdxiqxqf

yf

baba

baba

ba

ba

ba

ba

)( yf

1

baaba baaba

The Diracδ-function:


0)( (5)

||/)()( (4)

)( (3)

],[ 0

],[ 1)( )2(

0, allfor 1)( (1)

: properties usefulother

],[ if 0

],[ if )()()(

0for 0)(

21

21

21

21

2

1

2

1

tt

atat

tδ(t)

xxa

xxadtat

badtt

xxa

xxaafdtattf

tt

x

x

b

a

x

x

Ex: Prove that


||/)()( btbt

dtttfb

fb

fc

dttctfc

cdttctfdtbttf

ctbttcb

bbtbtbt

dtttfb

fbb

dttbtfdtbttf

bttb

--

)()(||

1)0(

||

1)0(

1

)()/(1

)/)(()/()()(

set 0for (2)

0for ||/)(/)()(

)()(1

)0(1

)()/()()(

set 0for (1)

''''''

'

'''

'

consider an integral to obtain


dtthtf ))(()(

0)( and ,....3,2,1 ),( of zeros the is 0)(for

|)(|/)())((

'

'

iii

ii

i

thNithtth

thttth

Proof:

i i

i

i i

i

i i

i

i i

i

ii

th

ttth

dtth

tttfdt

th

tttf

th

tf

zthttth

dzztzfdtthtf

thdzdtdtthdzthz

|)(|

)())((

)(

)()(

)(

)()(

)(

)(

0)(for )(

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

)(/)()(set

'

'''

'

''

Define the derivative of function

)0()()()]()([)()( ''' fdtttfttfdtttf

Physical examples for δ-function:


(1) an impulse of magnitude applied at time

(2) a point charge at a point

(3) total charge in volume V

)()( 0ttJtjJ 0t

q

)()()()()( 0000 zzyyxxqrrqr 0r

otherwise 0

V in lies if )()( 00

rqdVrrqdVrV V

unit step (Heviside) function H(t)

2/1)0( take and continuous isit ,0at

0for 0

0for 1)(

Ht

t

ttH


)()(' ttH

Proof:

)()(

)()()0(|)()()()(

)()()]()([)()(

'

00

'

''

ttH

dtttfftffdttff

dttHtftHtfdttHtf

Relation of the δ-function to Fourier transforms

deut

duutufdeuduf

eudufedtf

uti

uti

uiti

)(

)(

2

1)(

)()(}2

1){(

)(2

1)(


t

tde

dteftf

ti

ti

sin

2

21

2

1

)(~

2

1)(

for large becomes very

large at t=0 and also very narrow

about t=0

as

)( , tf

t

tt

tdtetf ti

sin

lim)(

)(22

1)(lim

Properties of Fourier transforms:

denote the Fourier transform of by or

function- of transformFourier 2

1)(

2

1)(

~

real is function- )()(2

1)(

2

1)(*

dtet

tdtetdtet

ti

titi

)(~

)]([)]([

)(~

)(~

|)(2

1

])(|)([2

1

)(2

1)]([

finite is |)(|for )(~

)]([ :ationDifferenti (i)

2'''

''

'

ftfFitfF

fifitfe

dtetfitfe

dtetftfF

dttffitfF

ti

titi

ti


)(tf )(~ f )]([ tfF


)(~1

1)(

2

1

set )(2

1)]([

)(~1

)]([ :Scaling (iii)

)(2)(~1

2

1)(

2

11

}1

)(1

])(1

{[2

1

])([2

1])([

)(2)(~1

])([ :nIntegratio (ii)

/

'

'

af

a

da

ef π

atdteatfatfF

af

aatfF

cfi

dtecdtetfi

dteci

dtetfi

dsesfi

dtedssfdssfF

cfi

dssfF

-

ai

ti

titi

titit ti

t tit

t


)(~

)(2

1

)(2

1)]([

)(~

)]([ :tionmultiplica lExponentia (V)

)(~

)(2

1

set )(2

1)]([

)(~

)]([ :nTranslatio (iV)

)(

if

dtetf π

dtetfetfeF

iftfeF

fe

deef

tadteatfatfF

featfF

-

ii

titt

t

ai

aii

ti

ia


spwctrum whole the ofshift -frequency a )(~

)(2

)(2

)(~

at only oncontributi has which , term the ignoring

)]([

)(

c

-

ti

titi

cti

ti

fA

dtetfA

dteetfA

g

Aae

etfaAg(t)

c

c

c

c

Consider an amplitude-modulated radio wave initial, a message is represent

by , then add a constant signal atfa )()(tf


Convolution and deconvolution

ondistributi observed the :)(

apparatus measuring of function resolution :

ondistributi true :)(

zh

g(y)

xf

Note: x, y, z are the same physical variable (length or angle), but each of them appears three different roles in the analysis.

)()()()( if (5)

function. resolution alexperiment and ondistributi

true the of nconvolutio the is ondistributi observed The (4)

* written often is and and

function the of nconvolutio the called )()()(

is ondistributi observed the , to region the in (3)

)( is which of interval small

a into amount an by resolution alinstrument the by moved (2)

)( yprobabilit the has and between reading true the (1)

zfzhyyg

gfgf

dxxzgxfzh

dzzz

dzxzgdz

xz

dxxfdxxx


Ex: Find the convolution of the function with the function in the above figure.

)()()( axaxxf )( yg

)()(

)()]()([

)()()(

azgazg

dxxzgaxax

dxxzgxfzh

Sol:


The Fourier transform of the convolution

)(~)(~

2

)(~2)(~

22

1

)()(2

1

})(){(2

1

set })(){(2

1

})()({2

1)(

~

)(

kgkf

kgkf

dueugexdxf

dueugxdxf

xzudzexzgxdxf

dxxzgxfdzekh

ikuikx

xuik

ikz

ikz

The Fourier transform of the product is given by)()( xgxf

)(~*)(~

2

1)]()([ kgkfxgxfF

)( and )( of nconvolutio the )(~*)(~

2

1

)(~)(~

2

1

)()(~)(~

2

1

2

1)(~)(

~

2

1

})(~{2

1})(

~{

2

1

2

1

})(~2

1{)(

2

1

)()(2

1)]()([

'''

'''''''''

)(''''''

''''''

''''

'''

'''

''

kgkfkgkf

kkgkfdk

kkkkgkfdkdk

dxekgkfdkdk

dxdkekgedkekf

dxdkekgexf

dxexgxfxgxfF

xkkki

xikikxxik

xikikx

ikx



Ex: Find the Fourier transform of the function representing two wide slits by considering the Fourier transforms of (i) two δ-functions, at , (ii) a rectangular function of height 1 and width 2b centered on x=0

ax

2

sincos4

)(~)(~

2]*[

])()([)]([

2

sin2)(

2

1

][2

1

2

1)(~ (ii)

2

cos2)(

2

1

)(2

1)(

2

1)(

~ (i)

q

qbqa

qgqfgfF

dxxzgxfFzhF

q

qbee

iq

iq

edxeqg

qaee

dxeaxdxeaxqf

iqbiqb

bb

iqxb

b

iqx

iqaiqa

iqxiqx


Deconvolution is the inverse of convolution, allows us to find a true

distribution f(x) given an observed distribution h(z) and a resolution

unction g(y).

Ex: An experimental quantity f(x) is measured using apparatus with a known resolution function g(y) to give an observed distribution h(z). How may f(x) be extracted from the measured distribution.

])(~)(

~[

2

1)(

)(~)(

~

2

1)(

~

)(~)(~

2)(~

1

kg

khFxf

kg

khkf

kgkfkh

the Fourier transform of the measured distribution

extract the true distribution


Correlation functions and energy spectra

The cross-correlation of two functions and is defined by f g

gfcdxzxgxfzc

)()()( *

It provides a quantitative measurement of the similarity of two functions and as one is displaced through a distances relative to the other.

g f z

ecommutativ-non )(][)]([ * zfgzgf

theorem Kinchin-r Wiene)(~)](~

[2)(~ * kgkfkc

})(){(2

1

set })(){(2

1

})()({2

1)(~

)(*

*

*

dueugxdxf

zxudzezxgxfdx

dxzxgxfdzekc

xuik

ikz

ikz


)(~)](~

[2

)(~2)](~

[22

1

)()(2

1)(~

*

*

*

kgkf

kgkf

dueugdxexfkc ikuikx

)(~)](~

[2)]()([ * kgkfxgxfF

)( of function ncorrelatio-auto )()()( * xfdxzxfxfza

of spectrum energy the : |)(~

|2

|)(~

|22

1

)(~

)](~

[22

1

)(~2

1)(

2

2

*

fkf

dkekf

dkekfkf

dkekaza

ikz

ikz

ikz


Parseval’s theorem:

dkkfdxxffg

dkkgkfdxxgxfz

dkekgkfdxzxgxfzc ikz

22

**

**

|)(~

||)(|let

)(~)](~

[)()(0set

)(~)](~

[)()()(

Ex: The displacement of a damped harmonic oscillator as a

function of time is given by

Find the Fourier transform of this function and so give a physical interpretation of Parseval’s theorem.

0for sin

0for 0)(

0

tte

ttft/τ

]/

1

/

1[

2

1sin)(

~

0000

/

iidtteef tit

oscillator the of V)(K energy total |)(|

spectrum energy the :|)(~

|2

2

tf

f

Sol:


Fourier transforms in higher dimensions:

kdekfrf

dkdkdkeeekkkfzyxf

rderfkf

dxdydzeeezyxfkkkf

rki

zyxzikyikxik

zyx

rki

zikyikxikzyx

zyx

zyx

32/3

2/3

32/3

2/3

)(~

)2(

1)(

),,(~

)2(

1),,(

)()2(

1)(

~

),,()2(

1),,(

~

three dimensional δ-function:

kder rki

33)2(

1)(


Ex: In three-dimensional space a function possesses spherical symmetry, so that . Find the Fourier transform of as a one-dimensional integral.

)(rf

)()( rfrf )(rf

Sol:

drk

krrfr

ikr

errfdr

errfdddr

rderfkf

krrkddrdrrd

ikr

ikr

rki

sin)(4

)2(

1

][)(2)2(

1

sin)()2(

1

)()2(

1)(

~

cos and sin

02/3

00

cos2

2/3

cos22

0002/3

32/3

23


13.2 Laplace transforms:

existnot does transformFourier

diverge ~

as 0 if :)( function a

ftftf

Laplace transform of a function f(t) is defined by

dtetfsftfL st

0)()()]([

define a linear transformation of L

)()()]([)]([)]()([ 212121 sfbsfatfbLtfaLtbftafL

Ex: Find the Laplace transforms of the functions:

..2,1,0for )( (iii) )( (ii) 1)( (i) nttfetftf nat


0for

!)( .......

3)(

!2)(

2)(

1)(

0for 1

)(1for

0for 0

|)(][ (iii)

for 1

|][ (ii)

0for 1

|1

]1[ (i)

1133121221

00

11

0

100

0

)(

0

)(

0

00

ss

nsf

s

!sf

ssf

ssf

ssf

ss

sft

s(s)fs

n]L[t

s

n

dtets

n

s

etdtetsftL

asassa

edtedteeeL

ss

es

dteL

nn

nn

stnstn

stnn

n

tsatsastatat

stst


Standard Laplace transforms

220

)(

0

22

220

)(

0

)(

0

22

1

00

1ReRecos)(

)(cos)( (4)

1Im|ImImsin)(

)(sin)( (3)

!)()( (2)

/|)(

)()( (1)

bs

s

sibdtedtbtesf

bs

ssfbttf

bs

b

sibsib

edtedtbtesf

bs

bsfbttf

s

cnsfcttf

sces

cdtcesf

s

csfctf

tsibst

tsibtsibst

nn

stst


])[()(cos)( (10)

])[()(sin)( (9)

)(cosh)( (8)

0for

][2

1][

2

1sinh)(

)(sinh)( (7)

)(

!)()( (6)

for 1

|)(

1)()( (5)

22

22

22

22

0

)()()(

0

)(

0

22

1

0

)(

0

bas

assfbtetf

bas

bsfbtetf

as

ssfbttf

asas

asa

e

sa

edteedtatesf

as

asfattf

as

nsfettf

asassa

edteesf

assfetf

at

at

tsatsatsatsast

natn

tsastat

at


2/12/10

2/1

2/12/1

2/133

000

0 0

)(

00

2

0

00

12

12

2/13

2/1

)()()2

1()1

2

1()!

2

1-( ,

)!2/1()(

)()()( (12)

)(2

1)!2/1()(!

2

1

2)

2

3(

|)2

1(2

set

2

1

2

1)

2

1(

2

1)

2

3(

)2

3()1

2

1(!

2

1)(!)1( :function gamma using

)(2

1)()( (11)

22

2222

2

22

ssf

sdtetsf

ssfttf

sssfI

eded

dudvedvedueI

dyeI

dyedyey

nnnn

ssfttf

st

vuvu

y

yy


)()(

for 0

)(for 1)()( (14)

)()(

)()()( (13)

0

0

0

0

0

0 0

0

00

0 0

0

s

edtedtettHsf

tts

esfttttHtf

edtettsf

esftttf

st

t

stst

st

stst

st

The inverse Laplace transform is unique and linear

)()()]()([ 21211 tbftafsfbsfaL

0for 23]1

2[]

3[)(

1

23)(

?)()1(

3)(

11

ses

Ls

Ltfss

sf

tfss

ssf

t


Laplace transforms of derivatives and integrals

][1

)( 1

)(1

])(1

[)(])([ (2)

0for )0(.............)0()0()(] [

0for )0()0()())()0(()0(

][)(][

0for )()0(

)(])([][ (1)

0

00000 0

1

121

2

0002

2

000

fLs

dttfes

dttfes

duufes

duufdteduufL

sdt

df

dt

dfsfssfs

dt

fdL

sdt

dfsfsfssfsfs

dt

df

dtedt

dfse

dt

dfdte

dt

df

dt

d

dt

fdL

ssfsf

dtetfsetfdtedt

df

dt

df L

st

sttsttt st

n

nnnn

n

n

ststst

ststst


Other properties of Laplace transforms:

exists ]/)([limfor )(})({

}){()(

])(

[ (5)

....3,2,1for )(

)1()]([ (4)

)(1

)]([ (3)

for )(

0for 0)(

by defined )( of transform Laplace the is )(

for )()()(0for (2)

)()()()]([ (1)

0t0

00

00

)(

0

)(

0

ttfduufdudtetf

dtduetfdtet

tf

t

tfL

nds

sfdtftL

a

sf

aatfL

btbtf

bttg

tgsfe

zbtdzbzfedttfesfeb

asfdtetfdteetftfeL

ss

ut

s

utst

n

nnn

bs

szbtsbs

tasstatat


Ex: Find the expression for the Laplace transform of )/( 22 dtfdt

)0()(2)(

)]0()0()([

][

2

'2

2

2

02

2

02

2

fsfssfds

ds

fsfsfsds

ddt

fde

ds

ddt

dt

fdte

dt

fdtL stst

Sol:

13.1 Fourier transforms: Chapter 13 Integral transforms.

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Transcript of 13.1 Fourier transforms: Chapter 13 Integral transforms.