13.1 Fourier transforms: Chapter 13 Integral transforms.
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Transcript of 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
Chapter 13 Integral transforms
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:
Chapter 13 Integral transforms
][ :tan
)()(
)()(][:
)(
)(][:
2
2
XVtiondard deviaS
ondistributicontinuousdxxfx
ondistributidiscretexfxXVVariance
ion distributcontinuousdxxxf
on distributidiscretexfxXEMean
ii
i
iii
Properties of distribution:
The uncertainty principle:
Chapter 13 Integral transforms
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
Chapter 13 Integral transforms
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:
Chapter 13 Integral 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 .
Chapter 13 Integral transforms
)(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:
Chapter 13 Integral transforms
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
Chapter 13 Integral transforms
||/)()( 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
Chapter 13 Integral transforms
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:
Chapter 13 Integral transforms
(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
Chapter 13 Integral transforms
)()(' 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)(
Chapter 13 Integral transforms
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
Chapter 13 Integral transforms
)(tf )(~ f )]([ tfF
Chapter 13 Integral transforms
)(~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
Chapter 13 Integral transforms
)(~
)(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
Chapter 13 Integral transforms
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
Chapter 13 Integral transforms
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
Chapter 13 Integral transforms
Ex: Find the convolution of the function with the function in the above figure.
)()()( axaxxf )( yg
)()(
)()]()([
)()()(
azgazg
dxxzgaxax
dxxzgxfzh
Sol:
Chapter 13 Integral transforms
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
Chapter 13 Integral transforms
Chapter 13 Integral transforms
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
Chapter 13 Integral transforms
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
Chapter 13 Integral transforms
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
Chapter 13 Integral transforms
)(~)](~
[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
Chapter 13 Integral transforms
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:
Chapter 13 Integral transforms
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)(
Chapter 13 Integral transforms
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
Chapter 13 Integral transforms
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
Chapter 13 Integral transforms
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
Chapter 13 Integral transforms
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
Chapter 13 Integral transforms
])[()(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
Chapter 13 Integral transforms
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
Chapter 13 Integral transforms
)()(
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
Chapter 13 Integral transforms
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
Chapter 13 Integral transforms
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
Chapter 13 Integral transforms
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: