Transcript Chapter 13 Integral transforms The

Chapter 13 Integral transforms
13.1 Fourier transforms:

f (t ) 

c r exp(
2  irt

)
T
r  
as T   frequency

c r exp( i  r t ) period  T and ω r 
r  
 
quantum
2
T
cr 
1
T

T /2
T / 2

f (t ) 

[
r  
f ( t ) exp(
 2  irt
) dt 

2
T

2

T /2
T / 2

 0   r continuous
T /2
T / 2
f ( t ) exp(  i  r t ) dt
f ( u ) exp(  i  r u ) du ] exp( i  r t )
 r  2  r / T is a discrete function
of r
( 2  / T ) c r exp( i  r t ) : the area of the rth broken - line rectangle
as T       2  / T  0



r  

2
 f (t ) 
g (  r ) exp( i  r t ) 
1
2



exp( i  t ) d  
1
2





g (  ) exp( i  t ) d 
f ( u ) exp(  i  u ) du
2 πr
T
, 

Chapter 13 Integral transforms
f (t ) 
1
2



exp( i  t ) d  [
1
2



f ( u ) exp(  i  u ) du ]  Fourier
inversion
theorem
The Fourier transform of f(t)
~
f ( ) 
1

2

f ( t ) exp(  i  t ) dt

Inverse Fourier transform of f(t)
f (t ) 
1
2



~
f (  ) exp( i  t ) d 
Ex: Find the Fourier transform of the exponential decay function f ( t )  0 for t  0
and f ( t )  A exp(   t ) for t  0 (   0 )
Sol:
~
f ( ) 
1
2
 0

0

A
2
0 exp(  i  t ) dt 
[
A
2
 exp(  (   i  ) t )
  i



exp(   t ) exp(  i  t ) dt
0
]0 
A
2 (   i  )
Chapter 13 Integral transforms
Properties of distribution:
Mean : E [ X ] 

x i f ( x i ) discrete distributi on
i

 xf ( x ) dx
Variance : V [ X ] 
 (x
continuous
distribut ion
  ) f ( x i ) discrete distributi on
2
i
i

 ( x   ) f ( x ) dx
S tan dard devia tion :  
2
V [X ]
continuous
distributi on
Chapter 13 Integral transforms

The uncertainty principle:
Gaussian distribution: probability density function
f (x) 
1

2
exp[
1 x  2
(
) ]
2




f ( x ) dx  1
(1) f ( x ) is symmetric about the point x  
 : the standard deviation describes the width of a
curve
(2) at x     , f ( x ) falls to e  1 / 2  0 . 61 of the peak value,
2
these points are points of inflection d f |
0
x   
2
dx
  f ( x ) 
 
 (x  )
 xf ( x ) dx
2
f ( x ) dx
Chapter 13 Integral transforms
Ex: Find the Fourier transform of the normalized Gaussian distribution.
f (t ) 
1

2
exp(
t
2
2
2
  t  
)
Sol: the Gaussian distribution is centered on t=0, and has a root
mean square deviation  t  
~
f ( ) 



~
f ( )
1
2
1
2
1
2
1
2




1



2
1

exp(
2
1
exp(
exp{
  ){
2
2
exp(
2
2
2
) exp(  i  t ) dt
1
2
[ t  2 i  t  ( i  )  ( i  ) ]} dt
2
2
1
2
2
 
t

2
2

2
exp[
2
 ( t  i  )
2


2
2
2
2
2
=1
2
]dt }
2
)
2
is a Gaussian distribution centered on zero and with a root
mean square deviation    1 /      t  1 /    is a constant.
Chapter 13 Integral transforms
Applications of Fourier transforms:
(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
y
known as a Fraunhofer diffraction pattern.
at a position


r0  x 0 iˆ  y 0 ˆj  z 0 kˆ , for large | r0 |
the total light amplitude

is A ( r0 ) 

Y
Y
' 
f ( y ) exp[ i k  ( r0  y ˆj )]
dy

| r0  y ˆj |

k
f ( y ) : aperture function
' 

exp[ i k  ( r0  y ˆj )] : the phase change traveling from y ˆj to r0
'

k  k cos  iˆ  k sin  ˆj and | r0 | Y
 
exp( i k  r0 ) 

A ( r0 ) 
  f ( y ) exp(  iky sin  ) dy , for | y | Y  f ( y )  0
r0
the intensity
I ( )  | A | 
2
in the direction
2
2
r0
 is
~
2
| f ( q ) | , q  k sin 
'
k
Y

x
Y
Chapter 13 Integral transforms
Ex: EvaluateI ( ) 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  .
f ( y)
1
ab
a
ab
ab
a
ab
f ( y)  1
1
~
f (q ) 


1
2
2
[

ab
ab
 exp(  iqx )
iq
4 cos qa sin qb
q 2
1
exp(  iqx ) dx 
ab
] a  b 
for q 
1
2
2
[
2  sin 


ab
ab
exp(  iqx ) dx
 exp(  iqx )
iq
ab
]a  b
 k sin   I ( ) 
2
2
r0
2
2
16 cos qa sin qb
~
2
| f (q ) | 
2 2
q r0
Chapter 13 Integral transforms
 The Diracδ-function:
 ( t )  0 for t  0

x2
x1
f ( t ) ( t  a ) dt  f ( a ) if a  [ x 1 , x 2 ]
 0 if a  [ x 1 , x 2 ]
other useful
(1)
(2)

b
:
 ( t ) dt  1 for all a , b  0
a

properties
x2
x1
 ( t  a ) dt  1 a  [ x 1 , x 2 ]
 0 a  [ x1 , x 2 ]
(3)
δ(t)   (  t )
(4)
 ( at )   ( t ) / | a |
(5)
t ( t )  0
Chapter 13 Integral transforms
Ex: Prove that  ( bt )   ( t ) / | b |
(1) for b  0


set t  bt
'
f ( t ) ( bt ) dt 
-


1
-
f ( t /  c ) ( t )( dt /  c ) 
1
'

b
  ( bt )   ( t ) / b   ( t ) / | b |
(2)
dt
'
f (0) 
f ( t / b ) ( t )
'
1
b

b


f ( t ) ( t ) dt
for b  0
for b   c  0 set t  bt   ct
'



f ( t ) ( bt ) dt 




1
c
'
'
'
c
f (0) 
1
|b|
f (0) 
1

|b|





f ( t /  c ) ( t ) dt
f ( t ) ( t ) dt
'
'
'
Chapter 13 Integral transforms

consider an integral  f ( t ) ( h ( t )) dt to obtain
  ( h ( t )) 
  (t  t
'
i
) / | h (ti ) |
i
for
h ( t i )  0 t i is the zeros of h ( t ), i  1 , 2 , 3 .... N , and h ( t i )  0
'
Proof:
set z  h ( t )  dz  h ( t ) dt  dt  dz / h ( t )
'

f ( t ) ( h ( t )) dt 


i
f (ti )
'

h (ti )
  ( h ( t )) 





dz
f ( z ( t ))  ( z )
'
h (t )
f ( t ) ( t  t i )
'
dt 
h (ti )
i
i


'

for t  t i  h ( t i )  z  0
f ( t )
 (t  ti )
'
h (ti )
i
 (t  ti )
'
| h (ti ) |
Define the derivative of   function
f ( t ) ( t ) dt  [ f ( t ) ( t )]
'






f ( t ) ( t ) dt   f ( 0 )
'
'
dt
Chapter 13 Integral transforms

Physical examples for δ-function:
(1) an impulse of magnitude J  j ( t )  J  ( t  t 0 ) applied at time t 0

(2) a point charge q at a point r0



 ( r )  q  ( r  r0 )  q  ( x  x 0 ) ( y  y 0 ) ( z  z 0 )
(3) total charge in volume V


 ( r ) dV 
V

V
 

q  ( r  r0 ) dV  q if r0 lies in V
 0 otherwise

unit step (Heviside) function H(t)
H ( t )  1 for
t 0
 0 for
t 0
at t  0 , it is continuous
and take H ( 0 )  1 / 2
Chapter 13 Integral transforms
H (t )   (t )
'
Proof:


f ( t ) H ( t ) dt  [ f ( t ) H ( t )]
'

 f ( ) 









0
'
f ( t ) H ( t ) dt
f ( t ) dt  f (  )  f ( t ) |  f ( 0 ) 
'
0



f ( t ) ( t ) dt
 H (t )   (t )
'

Relation of the δ-function to Fourier transforms
f (t ) 


1
2


i t




d e
duf ( u ){
  (t  u) 
1
2
1
2


e



e



duf ( u ) e
i ( t  u )
 i u
d } 
i ( t  u )
d



f ( u ) ( t  u ) du
Chapter 13 Integral transforms
1
f (t ) 


1
2
2



1 e



i t
~
i t
f  (  )e dt
d 
2  sin  t
2
t
for large  , f  ( t ) becomes very
large at t=0 and also very narrow
about t=0

as     lim f  ( t ) 

  ( t )  lim

sin  t
t
1
2


e

i t
dt 
2  ( t )
Chapter 13 Integral transforms
 (t ) 
*
~
 ( ) 

1
2

1
2

e
 i t




dt   (  t ) 
 ( t )e
 i t
dt 
1
2


e

1
2
i t
(  dt )   ( t )  - function
Fourier
transform
of  - function
Properties of Fourier transforms:
denote the Fourier transform of
~
'
F [ f ( t )]  i  f (  ) for
(i) Differenti ation :
F [ f ( t )] 
'


1
2
1
2
1
2



[e
e
f ( t ) by ~f (  )
'
f ( t )e
 i t
 i t
 i t
f (t ) |



dt
 i 


f ( t )e
 i t
dt ]
~
~

f ( t ) |   i  f ( )  i  f ( )
''
'
2 ~
F [ f ( t )]  i  F [ f ( t )]    f (  )
or F [ f ( t )]


is real
| f ( t ) | dt is finite
Chapter 13 Integral transforms
1 ~
f ( )  2 c  ( )
t
F [  f ( s ) ds ] 
(ii) Integratio n :
1
t
F [  f ( s ) ds ] 
2
1



2


t

{[
[  f ( s ) ds ]e
1

 i
1
1
i
2

t
f ( s ) dse

f ( t )e

 i t
 i t
dt
 i t
dt 
1

]  
1
2
i
c
1 ~
f ( )  2 c  ( )
i
F [ f ( at )] 
(iii) Scaling :
F [ f ( at )] 


i
1
2
1
2π





-
1 ~ 
f( )
a
a
f ( at ) e
f ( ) e
1 ~ 
f( )
a
a
 i t
set at  
dt
 i  / a
1
a
d
'





e

f ( t )e
 i t
dt
 i t
dt 
1
i


'
ce

 i t
dt }
Chapter 13 Integral transforms
(iV) Translatio
n:
F [ f ( t  a )] 
1
F [ f ( t  a )]  e

2


1

f ( t  a )e


2  
i a ~
 e
f ( )
(V) Exponentia
F [e
t

l multiplica
1
f ( t )] 
f (  )e
2
1
2π




e
t
-
f ( t )e
~
 f (  i  )
 i 
 i t
 i (   i )
dt
dt
~
f ( )
set a  t  
dt
e
tion :
f ( t )e

 i t
ia 
i a
d
F [e
t
~
f ( t )]  f (   i  )
Chapter 13 Integral transforms
Consider an amplitude-modulated radio wave initial, a message is represent
by f ( t ) , then add a constant signal a  f ( t )  a
g(t)  A [ a  f ( t )] e
ignoring
the term
g~ (  ) 
A

A
2

i c t
Aae



f ( t )e

f ( t )e
2
~
 A f (   c )
-
i c t
i c t
e
, which
 i t
 i (  c ) t
has contributi
on only at    c
dt
dt
a frequency
- shift of the whole
spwctrum
Chapter 13 Integral transforms

Convolution and deconvolution
f ( x ) : true distributi
g(y) : resolution
on
function
h ( z ) : the observed
of measuring
distributi
apparatus
on
Note: x, y, z are the same physical variable (length or angle), but each of
them appears three different roles in the analysis.
(1) the true reading
(2) moved
small
by the instrument
interval
(3) in the region
h( z ) 
f and
between



of which
x and
al resolution
z to z  dz , the observed
g and is often written
distributi
by an amount
distributi
y f ( x ) dx
z  x into a
dz is g ( z  x ) dz
f ( x ) g ( z  x ) dx called
(4) The observed
x  dx has the probabilit
distributi
the convolutio
(5) if g ( y )   ( y )  h ( z )  f ( z )
n of the function
f *g
on is the convolutio
on and experiment
on is
al resolution
n of the true
function.
Chapter 13 Integral transforms
Ex: Find the convolution of the function f ( x )   ( x  a )   ( x  a ) with
the function g ( y ) in the above figure.
Sol:
h( z ) 







f ( x ) g ( z  x ) dx
[ ( x  a )   ( x  a )] g ( z  x ) dx
 g(z  a)  g(z  a)
Chapter 13 Integral transforms
The Fourier transform of the convolution
~
h (k ) 





1
2
1
2
1
2
1
2
1
2





dze
 ikz





{


dxf ( x ){ 
dxf ( x ){ 



dxf ( x ) e


g ( z  x )e


 ikx
~
2 f ( k ) 
~
2  f ( k ) g~ ( k )
f ( x ) g ( z  x ) dx }
g ( u )e

 ik ( u  x )


 ikz
g ( u )e
2  g~ ( k )
 iku
dz } set u  z  x
du }
du
Chapter 13 Integral transforms
The Fourier transform of the product f ( x ) g ( x ) is given by
1
F [ f ( x ) g ( x )] 
F [ f ( x ) g ( x )] 






1
2
1
2
1
2
1
2
1
2
1
2
1
2






~
f ( k ) * g~ ( k )
2


f ( x ) g ( x )e


f ( x )e

1

2

dk
'
dk
'




 ikx
{


{
 ikx
dx
1
2



''
''
ik x
''
~
g ( k )e
dk } dx
~ ' ik ' x '  ikx
f ( k ) e dk } e


1
'' ~
' ~
''
dk
f
(
k
)
g
(
k
)

 
2




1
2
''
e
{


'
i(k  k k ) x

''
''
ik x
''
g~ ( k ) e
dk } dx
dx
'' ~
'
''
''
'
dk f ( k ) g~ ( k ) ( k  k  k )

' ~
'
'
dk f ( k ) g~ ( k  k )

~
f ( k ) * g~ ( k )
the convolutio
n of f ( k ) and
g(k )
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 x   a, (ii)
a rectangular function of height 1 and width 2b centered on x=0
~
(i) f ( q ) 

(ii) g~ ( q ) 

1
2
1
2
1
2


iq 2 



(e
1
F [ h ( z )]  F [ 

 ( x  a )e
 iqa
b
e
 iqx
b
(e
)
iqa
dx 
 iqb
e
iqb
dx 
q 2
2
2
1
2
)
[
e
 iqx
 iq
q 2
b
] b
2 sin qb
~
2  f ( q ) g~ ( q )
4 cos qa sin qb
1
2 cos qa
f ( x ) g ( z  x ) dx ]
 F [ f * g] 

e
 iqx



 ( x  a )e
 iqx
dx
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.
~
h (k ) 
~
2  f ( k ) g~ ( k ) the Fourier transform of the measured
~
1 h (k )
~
 f (k ) 
~
2 g ( k )
~
1
h
(k )
1
 f (x) 
F [~
] extract the true distribution
g
(
k
)
2
distribution
Chapter 13 Integral transforms
 Correlation functions and energy spectra
The cross-correlation of two functions f and g is defined by
c( z ) 



f ( x ) g ( x  z ) dx  c  f  g
*
It provides a quantitative measurement of the similarity of two
functions f and g as one is displaced through a distances z
relative to the other.
[ f  g ]( z )  [ g  f ] (  z )
*
c~ ( k ) 
~
* ~
2  [ f ( k )] g
(k )
c~ ( k ) 
1


2
1
2
1
2




dze
 ikz


{

theorem
*



dxf ( x ){ 
*
e
f ( x ) g ( x  z ) dx }
*


Wiene r - Kinchin

dx f ( x ){ 
non - commutativ
g ( x  z )e


g ( u )e
 ikz
 ik ( u  x )
dz } set u  x  z
du }
Chapter 13 Integral transforms
1
c~ ( k ) 
2
1

2


*


f ( x )e
~
*
2  [ f ( k )] 
 F [ f ( x ) g ( x )] 
*
a(z) 



dx 


g ( u )e
 iku
du
2  g~ ( k )
~
*
2  [ f ( k )] g~ ( k )

a(z) 
ikx


~
~(k )
2  [ f ( k )] g
f ( x ) f ( x  z ) dx auto - correlatio n function
1
2
1
2
1
2
*









ikz
a~ ( k ) e dk
~
* ~
ikz
2  [ f ( k )] f ( k ) e dk
~
2
ikz
2  | f ( k ) | e dk
~
2
2  | f ( k ) | : the energy
spectrum
of
f
of
f (x)
Chapter 13 Integral transforms
Parseval’s theorem:
c( z ) 



f ( x ) g ( x  z ) dx 
*
set z  0 



f ( x ) g ( x ) dx 
*


let g  f 



| f ( x ) | dx 
2


~
*
ikz
[ f ( k )] g~ ( k ) e dk





~
*
[ f ( k )] g~ ( k ) dk
~
2
| f ( k ) | dk
Ex: The displacement of a damped harmonic oscillator as a
function of time is given by
f ( t )  0 for t  0
e
 t/τ
sin  0 t for t  0
Find the Fourier transform of this function and so give a physical
interpretation of Parseval’s theorem.
Sol: ~f ( )  

0
e
 t /
sin  0 te
~
2
| f (  ) | : the energy
| f ( t ) |  total energy
2
 i t
dt 
1
[
1
2   0  i /

1
  0  i /
spectrum
(K  V) of the oscillator
]
Chapter 13 Integral transforms
Fourier transforms in higher dimensions:
~
f (k x , k y , k z ) 
~ 
 f (k ) 
( 2 )
1
( 2 )
f ( x, y, z) 

 f (r ) 
1
3/2

1
( 2 )
3/2
1
( 2 )
3/2

3/2

f ( x , y , z )e
1
( 2 )
3

e
e
 ik y y
e
 ik z z
dxdydz
  i k  r 3 
f ( r )e
d r

~
ik x ik y ik z
f ( k x , k y , k z ) e x e y e z dk x dk y dk z
~  i k  r 3 
f ( k )e d k
three dimensional δ-function:

 (r ) 
 ik x x
 
ik  r

3
d k
Chapter 13 Integral transforms
Ex: In three-dimensional space a function f ( r ) possesses spherical 
symmetry, so that f ( r )  f ( r ) . Find the Fourier transform of f ( r ) as
a one-dimensional integral.
Sol:
d r  r sin  drd  d 
3
2
~ 
f (k ) 



1
( 2 )
3/2
1
( 2 )
3/2
1
( 2 )
3/2
1
( 2 )
3/2

dr
0

and
  i k  r 3 
f ( r )e
d r


 
k  r  kr cos 



0
d

2
d  f ( r ) r sin  e
2
0
dr 2  f ( r ) r [
2
e
 ikr cos 
ikr
0


0

4 r f ( r )
sin kr
k
dr
]0
 ikr cos 
Chapter 13 Integral transforms
13.2 Laplace transforms:
~
f  0 as t    f diverge
a function
f ( t ) : if
 Fourier
transform
does not exist
 Laplace transform of a function f(t) is defined by
L [ f ( t )]  f ( s ) 


f ( t )e
 st
dt
0
 define a linear transformation of L
L[ af 1 ( t )  bf 2 ( t )]  aL [ f 1 ( t )]  bL [ f 2 ( t )]  a f 1 ( s )  b f 2 ( s )
Chapter 13 Integral transforms
Ex: Find the Laplace transforms of the functions:
(i) f ( t )  1 (ii) f ( t )  e
(i)

L [1 ] 

e
 st
1
dt 
(ii)
L[ e ] 
 st


at
e e
 st
dt 
0
(iii) L [ t ]  f n ( s ) 
 0

n

L[t
n
t e
 st
n1
] 
for s  0
n
s
f2(s) 
1
for n  0 ,1 , 2 ..
for s  0
dt 
e
t e
(a  s )t
1

as
 st
s
0
2
(a  s )t
dt 
n
s
0
for t  1  f 0 ( s ) 
s
e
1
n

s
1

(iii) f ( t )  t
0
n
f1 ( s ) 

|0 
s
0
at
e
at

|0 
|0 
n
s

for s  a
sa

t
n1
e
 st
dt
0
f n  1 (s) for s  0
for s  0
s
2
s
f1 ( s ) 
2!
s
21
f3(s) 
3!
s
31
.......
fn (s) 
n!
s
n1
Chapter 13 Integral transforms
Standard
(1)
Laplace
transforms
c
f (t )  c  f ( s) 
s
f (s) 


ce
 st
dt 
c
f ( t )  ct
n
 f (s) 
n1
f ( t )  sin bt  f ( s ) 
f (s) 


sin bte
 st
b
s b
2
dt  Im
0
(4)

2

e
( ib  s ) t
dt  Im
0
f ( t )  cos bt  f ( s ) 
f (s) 

|0  c / s
cn !
s
(3)
 st
s
0
(2)
e


0
cos bte
 st
e
( ib  s ) t
1

ib  s
| 0  Im
ib  s
s
s b
2
dt  Re

2

0
e
( ib  s ) t
dt  Re
1
ib  s

s
s b
2
2

b
s b
2
2
Chapter 13 Integral transforms
(5)
f (t )  e
f (s) 
 f (s) 
at


at
e e
 st
dt 
0
1
sa
e
(a  s )t
as
f (t )  t e
(7)
f ( t )  sinh at  f ( s ) 
f (s) 

at
 f (s) 

sinh ate
 st
(s  a)
dt 
a
s a
2
2
f ( t )  cosh bt  f ( s ) 
(9)
f (t )  e
(10) f ( t )  e
at
n1
s a
2
1

2
(a  s )t

[e
(a  s )t
e
(a  s )t
0
for s  a  0
(8)
at
for s  a
a
2
0

sa
n!
(6)
n
1

|0 
s
s a
sin bt  f ( s ) 
cos bt  f ( s ) 
2
2
b
[( s  a )  b ]
2
2
sa
[( s  a )  b ]
2
2
(a  s )t
1 e
e

]dt  [

]0
2 as
a s
Chapter 13 Integral transforms
(11) f ( t )  t
using
1  1/2
( 3)
2 s
 f (s) 
1/2
gamma
:  ( n  1 )  n!  n  ( n ) 
function
1
! (
2
3
1
1
1
( )  ( ) 
2
2
2
2

I 
set

e

1
2  1
2

y
e
y
2
1
dy 

2
0

e
y
1
3
 1)   ( )
2
2
2
dy
0
2
y
dy
0
I 
2


u
e
2
du 
0



e
v
2
dv 
0
d
0
I 




e

2
 
0
 d   2 (
1 / 2

2
e
e

2
3

2
2

1
2

)
dudv

) |0  
! f ( s) 
2
 f (s)  (
2
(u v )
0
1
0
  ( ) 
(12) f ( t )  t

( 1 / 2 )!
s
3

1  1/2
( 3)
2 s
1/2
s
f (s) 


0
t
1 / 2
e
 st
dt 
(  1 / 2 )!
s
1/ 2
, (
-1
2
)!   (
1
1
 1)   ( ) 
2
2
  f (s)  (

s
)
1/2
Chapter 13 Integral transforms
(13) f ( t )   ( t  t 0 )  f ( s )  e
f (s) 


0
 ( t  t 0 )e
 st
 st 0
dt  e
 st 0
(14) f ( t )  H ( t  t 0 )  1 for t  t 0  f ( s ) 
e
 st 0
s
 0 for t  t 0
f (s) 


0
H ( t  t 0 )e
 st
dt 


e
t0
 st
dt 
e
 st 0
s
The inverse Laplace transform is unique and linear
1
L [ a f 1 ( s )  b f 2 ( s )]  af 1 ( t )  bf 2 ( t )
f (s) 
f (s) 
s3
s( s  1)
3
s

2
 f (t )  ?
2
1 3
1
t
 f (t )  L [ ]  L [
]  3  2e
for s  0
s1
s
s1
Chapter 13 Integral transforms
Laplace transforms of derivatives and integrals
(1) L [
df
]
dt


0
df
e
 st
dt  [ f ( t ) e
dt
 st

]0  s 

 st
f ( t )e
dt
0
  f ( 0 )  s f ( s ) for s  0
2
L[
d f
dt
2
]


0

 df
d df
df  st 
 st
 st
(
) e dt  [
e ]0  s 
e dt
0 dt
dt dt
dt
 df
( 0 )  s (  f ( 0 )  s f ( s ))  s f ( s )  sf ( 0 ) 
2
dt
 L[
dt
n
]  s f (s)  s
n
n1
f (0)  s
n2
df
( 0 )  .......... ... 
dt
t
(2) L [  f ( u ) du ] 
0

( 0 ) for s  0
dt
n
d f
df


dte
 st
0
1
s

t
f ( u ) du  [
s
0


0
e
 st
f ( t ) dt 
1
1
s
L[ f ]
df
dt
e
 st

t
0

f ( u ) du ] 0 
n1
n1


0
( 0 ) for s  0
1
s
e
 st
f ( t ) dt
Chapter 13 Integral transforms
Other properties of Laplace transforms:
(1) L [ e
at

f ( t )] 

at
f ( t )e e
 st
dt 
0
(2) for b  0  e


f ( t )e
( sa )t
 bs
f (s) 


 s(tb)
e
f ( t ) dt 
0
 e
 bs
dt  f ( s  a )
0
f ( s ) is the Laplace
g(t )  0


e
 sz
f ( z  b ) dz
for t  b  z
0
transform
of g ( t ) defined
by
for 0  t  b
 f (t  b)
for t  b
1
s
f( )
a
a
(3) L [ f ( at )] 
n
(4) L [ t f ( t )]  (  1 )
n
n
d f (s)
ds
(5) L [
f (t )
]
t


e
 st
dt 
t
0

f (t )
n

 {
s
for n  1 , 2 , 3 ....


0

0
f ( t )e
 ut

f ( t ){  e
dt } du 
 ut
du } dt
s


f ( u ) du for lim [ f ( t ) / t ] exists
s
t 0
Chapter 13 Integral transforms
Ex: Find the expression for the Laplace transform of t ( d 2 f / dt 2 )
Sol:
2
L[ t
d f
dt
2
]

2

e
 st
t
d f
dt
0
 
d
2
dt  
d
ds

2

e
 st
dt
0
[ s f ( s )  sf ( 0 )  f ( 0 )]
2
'
ds
 s
2
d f
d
ds
f ( s )  2 sf ( s )  f (0 )
2