SE 207: Modeling and Simulation
Introduction to Laplace Transform
Dr. Samir Al-Amer
Term 072
Why do we use them

We use transforms to transform the problem
into a one that is easier to solve then use the
inverse transform to obtain the solution to the
original problem
Original problem : compute A  1.23 *1.43 * 2.43
log A  log 1.23  log 1.43  log 2.43  0.0899  0.1553  0.3856  0.6308
A  100.6308  4.2741
Laplace Transform
f (t )  e
 2t
L
Laplace Transform
t is a real variable
f(t) is a real function
Time Domain
-1
1
F ( s) 
s2
s is complex variable
L
Inverse
Laplace Transform
F(s) is a complex
valued function
Frequency Domain
Use of Laplace Transform in solving ODE
Differential Equation
x (t )  2 x(t )  0, x(0)  1
x (t )  e 2 t
Solution of the
Differential Equation
Laplace
Transform


Inverse
Laplace
transform
Algebraic Equation
sX ( s )  1  2 X ( s )  0

1
X ( s) 
s2
Solution of the
Algebraic Equation
Definition of Laplace Transform

F ( s)  L{ f (t )} 

 st
f (t )e dt
0
Sufficient conditions for existence of the Laplace
transform
•
f (t ) is piecewise continuous
• There exist M ,  , t0 such that
f (t )  M et
for t  t0
Examples of functions of exponential order
1 t  0
f (t )  
0 t  0
Take M  1,   0, t0  0
f (t )  sin( t )
Take M  1,   0, t0  0
f (t )  2
Take M  1,   1, t0  0
x
Example
unit step
1 t  0
f (t )  
0 t  0
1
F ( s) 
s
_______________________________________
Proof :

F(s)  
0

 st 
0 1 1
f(t)e dt   1e dt 


s
s
s
0
 st
 st
e
0
Example
Shifted Step
A t  2
A e 2 s
f (t )  
F ( s) 
s
0 t  2
_______________________________________
Proof :

2

0
0
2
F(s)   f(t)e st dt   0e  st dt   Ae st dt
 st 
2 s
2 s
0

e
A
e
2
F ( s)  A
A

s
s
s
e
Integration by parts

 udv  uv
0

0

  vdu
0
Example

 st
te
 dt  ?
0
Let u  t , dv  e  st dt
 du  dt , v  
1  st
e
s
te st
1  st
uv 
, vdu 
e dt
s
s


 st 
te
0 te dt  uv 0  0 vdu   s
 st

0

 st 
1  st
 1  te

e dt  0  0  

s
s s
0
0
1
 2
s
Example
Ramp
t
f (t )  
0
t0
t0
1
F ( s)  2
s
_______________________________________
Proof :


F(s)   f(t)e dt   te dt  te
 st
0
 st
 st 
0
0
1  st
 1  te

e dt  0  0  

s
s s
0

Done Using integratio n by part
 udv  uv
0
1  st
Let u  t , dv  e dt  du  dt , v  
e
s
 st
 st 


0

  vdu
0
0
1
 2
s
Example
Exponential Function
1
f (t )  e
F (s) 
sa
_______________________________________
Proof :
 at



F(s)   f(t)e dt   e e dt   e
0
 st
0
 at  st
0
( s  a ) t
1
dt 
sa
Example
sine Function

F ( s)  2
s 2
f (t )  sin( t )
_______________________________________
Proof :

F(s)   f(t)e
0
 st

dt   sin( t )e
0
 st

dt  2
s 2
Details of the proof will be given shortly
Example
cosine Function
f (t )  cos(t )
F ( s) 
s
s2   2
_______________________________________
Proof :

F(s)   f(t)e
0
 st

dt   cos(t )e
0
 st
dt 
s
s2   2
Example
Rectangle Pulse
A
f (t )  
0
t  [0, L]
t  ( L,  ]
A
F ( s )  (1  e  sL )
s
_______________________________________
Proof :

F(s)  
0
L

A
f(t)e dt   Ae dt   0e dt  (1  e  sL )
s
0
L
 st
 st
 st
Properties of Laplace Transform
Addition
f (t )
g (t )
F (s)
G (s)
f (t )  g (t )
F (s)  G ( s)
__________________________________________________
L f (t )  g (t )  L f (t ) Lg (t )
Proof :



0
0
0
L f (t )  g (t )    f(t)  g (t ) e  st dt   f (t )e  st dt   g (t )e  st dt
Properties of Laplace Transform
Multiplication by a constant
f (t )
a f (t )
F (s)
a F ( s)
__________________________________________________
L a f (t )  a L f (t )
Proof :


0
0
L a f (t )   a f(t)e st dt a  f (t )e  st dt  a L f (t )
Properties of Laplace Transform
Multiplication by exponential
f (t )
F (s)
f (t ) e at
F ( s  a)
__________________________________________________


L f (t ) e  at  F ( s  a)
Proof :

L f (t ) e
 at

  f(t)e
0

e dt   f(t)e
 at  st
0
( a  s ) t
dt  F ( s  a)
Properties of Laplace Transform
Examples Multiplication by exponential
f (t )
F (s)
f (t ) e at
F ( s  a)
__________________________________________________

 ( s  a )

L sin( t )   2
,
s 2
L sin( t ) e
L cos(t )  
L cos(t ) e at 
L t  
1
s
2
,
s
s 
2
,
2


L te
at


at
 ( s  a )
2
( s  a)
2
( s  a)  
2
1
2
2
Useful Identities
e
j
 cos( )  j sin(  )
 j
 cos( )  j sin(  )
1 j
 j
cos( )  e  e
2
1 j
 j
sin(  ) 
e e
2j
e




Example
sin Function

F ( s)  2
s 2
f (t )  sin( t )
_______________________________________


0
0
F(s)   f(t)e st dt   sin( t )e  st dt 


1 j t
 j t
e

e
2j
0




1  jt  st
 st
 jt  st 
e dt 
e
dt   e
dt



2j0
0


1  1
1 


  2

2 j  s  j s  j   s   2
Example
cosine Function
s
f (t )  cos(t )
F ( s)  2
2
s


Inverse Laplace Transform
______________________________________________
Laplace Transform


0
0
F(s)   f(t)e st dt   cos(t )e  stdt




1 jt
1
 jt
 st
jt  st
 jt  st
0 2 e  e e dt  2  0 e e dt  0 e e dt 

1 1
1 
s
 

  2
2  s  j s  j  s   2
Properties of Laplace Transform
Multiplication by time
L t f (t )  
d
F ( s)
ds
_________________________________
1
L u (t ) 
s
d 1  1
L t u (t )      2
ds  s  s

2
L t
3

d 1 2
L t u (t )    2   3
ds  s  s
d 2 6

u (t )     
ds s
s

3

4

n
L t u (t )

n!
 n 1
s
Properties of Laplace Transform
 df (t ) 
L
  sF ( s )  f (0)
 dt 
 d 2 f (t )  2
L
  s F ( s )  sf (0)  f (0)
2
 dt 
 d 3 f (t )  3
2
L
  s F ( s )  s f (0)  sf (0)  f(0)
3
 dt 
__________________________________________________________
L x(t )  3 x (t )  2 x(t )  u (t )


 s 2 X ( s )  sx (0)  x (0)  3[ sX ( s )  x(0)]  2 X ( s ) 
1
s
Properties of Laplace Transform
Integration


1
5. L  f ( )d  F ( s )
0
s
______________________________
t
6. L f (t  a )u (t  a)  e
 sa
F ( s)
Properties of Laplace Transform
Delay
f (t )
f (t  a)u (t  a)
F ( s)
e
 sa
L f (t  a)u (t  a)  e  sa F ( s)
F ( s)
Properties of Laplace Transform
t0
0

f (t )   At 0  t  L
0
tL

f (t )  At u (t )  A(t  L)u (t  L)  ALu (t  L)
Slope =A
A
F (s)  2
s
1  Ls
A 2e
s
1  Ls
 AL e
s
L
Properties of Laplace Transform4
Slope =A
_
L
Slope =A
L
Slope =A
=
L
_
AL
Summary
SE 207: Modeling and Simulation
Lesson 3: Inverse Laplace Transform
Dr. Samir Al-Amer
Term 072
Properties of Laplace Transform
 df (t ) 
L
  sF ( s )  f (0)
 dt 
 d 2 f (t ) 
2
 ( 0)
L

s
F
(
s
)

sf
(
0
)

f

2
 dt

 d 3 f (t ) 
3
2
 (0)  f(0)
L

s
F
(
s
)

s
f
(
0
)

s
f

3
dt


 d n f (t ) 
n
n 1
L

s
F
(
s
)

s
f (0)  ...  s f

n
 dt

( n2)
(0)  f
( n 1)
(0)
Solving Linear ODE using Laplace Transform
x(t )  3 x (t )  2 x(t )  u (t ), x(0)  x (0)  0
use LaplaceTra nsform


1
s X ( s )  sx (0)  x (0)  3[ sX ( s )  x(0)]  2 X ( s ) 
s
1
s 2 X ( s )  3sX ( s )  2 X ( s ) 
s
solve for X ( s )
2
1
s s 2  3s  2
use inverse LaplaceTra nsform
x(t )  ???
X ( s) 


Inverse Laplace Transform
Inverse Laplace Transform
f (t )  L1 F ( s)
Partial Fraction Expansion
Expand F(s)as the sum of first and second order term s
then obtain the inverse of each term and sum them.
Notation
F(s)is rational function in s , it can be expressed as
N(s)
F(s) 
where N(s) and D(s) are polynomial s
D(s)
roots of N(s) are called the zeros of F(s)
roots of D(s) are called the poles of F(s)
Notation
s2
s2

poles :  3,  4 , zero :  2
2
s  7 s  12 (s  4)(s  3)
s3
pole :  0.5 , zero : 3
2s  1
Notation
F(s) is rational function in s , it can be expressed as
N(s)
F(s) 
where N(s) and D(s) are polynomial s
D(s)
roots of N(s) are called the zeros of F(s)
roots of D(s) are called the poles of F(s)
Examples
1
s2
, 2
are strictly proper
2
s  7 s  12 s  7 s  12
1
s 2  3s  2 s  2
, 2
,
are proper
2
s  7 s  12 s  7 s  12 2s  1
Partial Fraction Expansion
We will consider t hree cases
* distict poles
* repeated poles
* complex poles
Partial Fraction Expansion
1
F (s) 
s(s  1)(s  2) 2 (s 2  2 s  5)
F ( s ) has two distict real poles at 0,-1
one repeated pole at s  -2 (double poles at - 2)
two complex poles at - 1  2j,-1 - 2j
Partial Fraction Expansion
If F(s) is strictly proper and all poles of are distict
F(s) can be expressed as
n
Ai
F(s)  
i 1 s-s i
where
Ai  s-si F ( s) s  s
i
n
inverse Laplace Transform f(t)   Ai e
i 1
si t
Example
s5
s5

s 2  5s  4 ( s  1)( s  4)
strictly proper, poles are at  1,  4 distinct
F(s) 
n
Ai
A1
A2
F(s)  


( s  1) ( s  4)
i 1 s-s i
where
A1  s-s1 F ( s) s  s  s  1F ( s ) s  1 
1
A2  s-s 2 F ( s ) s  s
2
s5
2
( s  4) s  1
s5
 s  4F ( s ) s  1 
 3
( s  1) s  4
n
inverse Laplace Transform f(t)   Ai e sit  2e t  3e  4t
i 1
for t  0
Example
s5
s5
F(s)  2

s  5s  4 ( s  1)( s  4)
strictly proper, poles are at  1,  4 distinct
s5
2
3


( s  1)( s  4) ( s  1) ( s  4)
inverse Laplace Transform
t
f(t)  2e  3e
 4t
for t  0
Alternative Way of Obtaining Ai
s5
s5
F(s)  2

s  5s  4 ( s  1)( s  4)
strictly proper, poles are at  1,  4 distinct
A1
A2
A1 ( s  4)  A2 ( s  1)
F(s) 


( s  1) ( s  4)
( s  1)( s  4)
s ( A1  A2 )  (4 A1  A2 )
s5
F (s) 

( s  1)( s  4)
( s  1)( s  4)
 A1  A2  1, 4 A1  A2  5 ; solve  A1  2, A2  3
n
inverse Laplace Transform f(t)   Ai e sit  2e t  3e  4t
i 1
for t  0
Repeated poles
5s  16
( s  2) 2 ( s  5)
strictly proper, distict pole at  5 and reperated poles are at  2
A11
A12
A2
F(s) 


( s  2) 2 ( s  2) ( s  5)
F(s) 
A11  ( s  2) 2 F ( s )
A12 


s  2
d
( s  2) 2 F ( s )
ds
5s  16
2
( s  5) s  2

A2  ( s  5) F ( s ) s  5 

s  2
 5( s  5)  (5s  16) 
d  5s  16 

1




2
ds  ( s  5)  s  2 
( s  5)
 s  2
5s  16
( s  2) 2
 1
s  5
Repeated poles
F(s) 
A11
A12
A2


distict pole at  5 and reperated poles are at  2
2
( s  2) ( s  2) ( s  5)
* The coefficien t of distict pole is obtained as before A2  ( s  5) F ( s ) s  5 
A11
A12
and
are present
( s  2) 2
( s  2)
* The coefficien t of higest order term is obtained as in distict pole case
* Note both
A11  ( s  2) 2 F ( s )
s  2

5s  16
2
( s  5) s  2
* New formula for the coefficien t of other term s
A12 

d
( s  2) 2 F ( s )
ds


s  2
 5( s  5)  (5s  16) 
d  5s  16 

1




2
ds  ( s  5)  s  2 
( s  5)
 s  2
5s  16
( s  2) 2
 1
s  5
Repeated poles
If F(s) has repeated poles the formula used to obtain A1 is modified
5s  16
strictly proper, poles are at  2,2 ,  5
2
( s  2) ( s  5)
A11
A12
A2
F(s) 


( s  2) 2 ( s  2) ( s  5)
F(s) 
A11  ( s  2) 2 F ( s )
A12 
s  2

d
( s  2) 2 F ( s )
ds
A2  ( s  5) F ( s ) s 5

5s  16
2
( s  5) s  2


s  2
 5( s  5)  (5s  16) 
d  5s  16 

1
2




ds  ( s  5)  s  2 
( s  5)
 s  2
5s  16

( s  2) 2
 1
( distict pole )
s  5
inverse Laplace Transform f(t)  2te2 t  1e 2 t  1e 5t
for t  0
Repeated poles
If F(s) has repeated poles the formula used to obtain A1 is modified
s3
F(s) 
strictly proper, poles are at  1,1,  1,  4
( s  1)3 ( s  4)
A11
A12
A13
A2
F(s) 



( s  1)3 ( s  1) 2 ( s  1) ( s  4)
A11  ( s  1)3 F ( s )
s  1

d
A12 
( s  1)3 F ( s )
ds

s  1
1 d2
3
A13 
(
s

1
)
F ( s)
2
2! ds

A2  ( s  4) F ( s ) s  4

s  1
( distict pole )
inverse Laplace Transform f(t)  0.5 A11t 2e t  A12tet  A13e t  A2e 4 t
for t  0
Common Error
s3
Some may expand F(s) 
as
3
( s  1) ( s  4)
A11
A2
F(s) 

3
( s  1) ( s  4)
This is not valid in general. It should be expanded as
A13
A11
A12
A2
F(s) 



3
2
( s  1) ( s  1) ( s  1) ( s  4)
Complex Poles
4s  8
F(s)  2
has two complex poles at - 1  j2 and - 1 - j2
s  2s  5
k1
k2
F(s) 

( s  1  j 2) ( s  1  j 2)
k1  ( s  1  j 2) F ( s ) s  1 j 2
k 2  ( s  1  j 2) F ( s ) s  1 j 2  k 1
f (t )  k1e ( 1 j 2 )t  k 2 e ( 1 j 2 )t  ?
Complex Poles
Alternativ e Way
4s  8
Bs  C
F(s)  2
can be expressed as
s  2s  5
( s  a) 2   2
s 2  2 s  5  s 2  2 s  1  4  ( s  1) 2  2 2
 B  4, C  8, a  1,   2
f (t )  Be
 at
 C  aB   at
cos(t )  
e sin( t )
  
What do we do if
F(s) is not strictly proper
If F(s) is proper but not strictly proper 
use long division t o express it as F(s)  k  G(s)
where k is a real number and G(s) is strictly proper.
F(s) 
s 2  4s  6
 1
s4
s  3s  2
s 2  3s  2
s4
3
2
F(s)  1 
 1

( s  2)( s  1)
( s  1) ( s  2)
2
f (t )   (t )  3e t  2e 2t
for t  0
Solving for the Response
 df (t ) 
L
  sF ( s )  f (0)
 dt 
 d 2 f (t )  2
 (0)
L

s
F
(
s
)

sf
(
0
)

f

2
dt


__________________________________________________________
Solve x(t )  3 x (t )  2 x(t )  0, x(0)  1, x (0)  2
s X (s)  sx(0)  x(0) 3[sX (s)  x(0)]  2 X (s)  0
s X (s)  s  2 3[sX (s)  1]  2 X (s)  0
2
2
X ( s) 
s5
s 2  3s  2
Final value theorem
2
2

2
3
F(s) 
 3 
( s  1)( s  4) ( s  1) ( s  4)
2 t 2  4 t
f(t)  e  e
for t  0
3
3
2  2  4 
f ( )  e  e  0
3
3
2s
0
f ()  lim

0
s 0 ( s  1)( s  4)
(0  1)(0  4)
Final value theorem
2
2

2
5
F(s) 
 5 
( s  1)( s  4) ( s  1) ( s  4)
2
2
f(t)  et  e  4t for t  0
5
5
2  2 4
f ( )  e  e  
5
5
2s
0
f ()  lim

 0 Not valid
s 0 ( s  1)( s  4)
(0  1)(0  4)
Remember w e can apply final value theorem if F has
no poles with positive real parts
Step function
A t  0
u(t)  
0 t  0
A
A
U(s) 
s
impulse function
(t)  0 for t  0



(t) dt  1,
F ( s)  1
 0
impulse function
(t)  0 for t  0



(t) dt  1,   0
 f (c )
a (t  c) f (t )dt   0
sampling property
b
L(t)  1
if c  [a, b]
otherwise
Initial Value& Final Value Theorems
Initial Value f (0 ) and Final Value f ()
can be obtained directly from F(s)
with out the need to obtain
inverse Laplace Transfrom
Initial Value Theorem
f (0)  lim s F ( s)
s 
the value of the function at the initial time
is obtained by taking the limit lim s F ( s )
s 
Final Value Theorems
f ()  lim s F ( s )
s 0
provided F(s) has no poles on the in the right half
of the complex plane and with a possible exception
of single pole at the origin.
Examples :
s5
s4
G(s) 
, F (s) 
(s  2)(s - 3)
s(s  2)(s  3)
We can obtain f () but not g ().
SE 207: Modeling and Simulation
Lesson 4: Additional properties of Laplace
transform and solution of ODE
Dr. Samir Al-Amer
Term 072
Outlines




What to do if we have proper function?
Time delay
Inversion of some irrational functions
Examples
Step function
A t  0
u(t)  
0 t  0
A
A
U(s) 
s
impulse function
(t)  0 for t  0



(t) dt  1,
F ( s)  1
 0
impulse function
You can consider the unit
impulse as the limiting case
for a rectangle pulse with
unit area as the width of the
pulse approaches zero
1

Area=1

impulse function
(t)  0 for t  0



L(t)  1
(t) dt  1,   0
sampling property

b
a
 f (c )
(t  c) f (t )dt  
 0
if c  [a, b]
otherwise
Sample property of impulse function



5
1
5
1
2
5
(t  2 ) cos(3t )dt  cos(6)
(t  3 ) e dt  e
t
3
(t  3 ) e dt  0
t
Time delay
f(t)
F(s)
g(t)
G(s)
g (t )  f (t  2)
G ( s )  F ( s )e
2 s
What do we do if
F(s) is not strictly proper
1
G(s) 
(s  3)(s  4)
strictly proper, We can apply the
techniques discussed earlier.
s 2  5s  6
H(s)  2
s  5s  4
not strictly proper, We cannot apply
the technique s discussed earlier.
s6
F (s)  2
e 2 s
s  5s  4
Not rational
What do we do if
F(s) is not strictly proper
If F(s) is proper but not strictly proper 
use long division t o express it as F(s)  k  G(s)
where k is a real number and G(s) is strictly proper.
s 2  7s  8
4s  6
F(s)  2
 1 2
s  3s  2
s  3s  2
4s  6
2
2
F(s)  1 
 1

( s  2)( s  1)
( s  1) ( s  2)
f (t )   (t )  2e t  2e  2t
for t  0
Example
2s 2  s  3
F(s)  2
s  4s  4
 2
2
s 2  4s  4 2s 2  s  3
−
2 s 2 − 8s − 8
 7s  5
 7s  5
s 2  4s  4
Example
2s 2  s  3
 7s  5
 7s  5
F(s)  2
 2 2
2
s  4s  4
s  4s  4
s  22
2
A
B

s  22 s  2
A  s  2 
2
 7s  5
9
2
s  2 s 2
d 
2  7s  5 

B   s  2 
2 
ds 
s  2 

s  2
d
 7 s  5
ds
 7
s  2

9
7 
2t
 2t
f (t )  L 2 


2

(
t
)

9
te

7
e

2
s  2  


s

2

1
Solving for the Response
 df (t ) 
L
  sF ( s )  f (0)
 dt 
 d 2 f (t )  2
 (0)
L

s
F
(
s
)

sf
(
0
)

f

2
dt


__________________________________________________________
Solve x(t )  3 x (t )  2 x(t )  0, x(0)  1, x (0)  2
s X (s)  sx(0)  x(0) 3[sX (s)  x(0)]  2 X (s)  0
s X (s)  s  2 3[sX (s)  1]  2 X (s)  0
2
2
X ( s) 
s5
s 2  3s  2