Laplace tranbeereerrhhhsform linear.pptx

The Laplace Transform
Assistant Professor: khaled Fawaz
Source: Modern Control Engineering
Fourth Edition
Katsuhiko Ogata

The Laplace Transform: Introduction
The Laplace transform method is an operational method that can be used advantageously for
solving linear differential equations. By use of Laplace transforms, we can convert many
common functions, such as sinusoidal functions, damped sinusoidal functions, and
exponential functions, into algebraic functions of a complex variable s.
Operations such as differentiation and integration can be replaced by algebraic operations in
the complex plane. Thus, a linear differential equation can be transformed into an algebraic
equation in a complex variable s.
If the algebraic equation in s is solved for the dependent variable, then the solution of the
differential equation (the inverse Laplace transform of the dependent variable) may be found
by use of a Laplace transform table or by use of the partial-fraction expansion technique.
An advantage of the Laplace transform method is that it allows the use of graphical
techniques for predicting the system performance without actually solving system
differential equations. Another advantage of the Laplace transform method is that, when we
solve the differential equation, both the transient component and steady-state component of
the solution can be obtained simultaneously

Introduction
Tim e dom ain
unknow n f(t), d/dt, D iff Eqs
Frequency dom ain
unknow n F(s), A lg Eqs
Laplace
Transform ation
Solve
A lgebraic
Equations
Frequency dom ain
know n F(s)
Tim e dom ain
know n f(t)
Solve
D ifferential
Equations
Inverse
Laplace
Transform

Review of complex variables and complex
functions
• Complex Variable:
• Complex Function:
• A complex function G(s) is said to be analytic in a region if G(s) and all its
derivatives exist in that region :
s j
 
 
Re( ) ,
s 
 Im( ) ,
s 
 2 2
| | ,
s  
 
1
tan
s


  
   
 
 
| | cos( ) sin( ) ,
s s s j s
    ,
s j
 
 
2
| |
s ss

( ) ( , ) ( , )
x y
G s G G j
   
 
0 0
( ) ( )
( ) lim lim
s s
d G s s G s G
G s
ds s s
   
   
 
 

Review of Complex Variables and
Complex functions
Euler’s Theorem cos( ) sin( )
j
e j

 
 
Corollaries cos( )
2
j j
e e
 



 sin( )
2
j j
e e
j
 




Proof: Consider the Taylor series expansions of the functions
2 3
1
2! 3!
x x x
e x
    
2 4 6
cos 1
2! 4! 6!
  
     
3 5 7
sin
3! 5! 7!
  
 
    
cos sin j
j e 
 
 

Laplace Transformation
• Let us define:
– f(t)= a function of time t such that f ( t ) = 0 for t < 0
– s = a complex variable
– = an operational symbol indicating that the
quantity that it prefixes is to be transformed by the
Laplace integral
– F(s) = Laplace transform of f( t )
Then the Laplace Transform of f(t) is:
L
0
( ) st
e d
f t t



0
[ ] (
) ( )
)
( st
f t f
F t
s e dt


 
L

Laplace Transform Theorems
1. Linearity:
2. Superposition:
3. Translation in time:
4. Complex differentiation:
5. Translation in the s Domain:
6. Real Differentiation:
7. Real Integration:
8. Final Value:
9. Initial Value:
10.Complex Integration:
[ ] [
( ) ( )] ( )
f t f t F s
  
 
L L
1
1 2
1 2 2
[ ] [ ]
( ) [ ] ( ) ( )
( ) ( ) ( )
f t f t f t t F s F s
f
    
L L L
1
( ) (
[ ] ( )
) s
f t u e F
t s

 
  
L
[ ] ( )
( )
d
F s
tf
ds
t 
L
( )
[ ] ( )
t
e t s
f F


 
L
[ ] ( ) (
( ) 0 )
sF s f
Df t 
 
L
1
1
0
( )
( ) )
(0
[ ]
(0
)
t
F s D f
s
f t dt f
s
D 

  

L
0
lim ( ) lim ( )
s t
sF s f t
  

0
lim ( ) lim ( )
s t
sF s f t
  

[ ]
(
)
)
(
F
f
d
t
s
t
s


L

Derivation of Laplace Transforms of Simple
Functions
• Step Function
(Echelon Unite):
• Exponential Function:
0
[1(
0, 0
1( ) ]
) ( )
0
1
,
1,
t
st
t
e d
t
t
t
t
t
t






 
 

 

L
0 0
1( ) 1( )
[ ]
t t
st st
t t
e
t t dt e dt

 

 
 
 
 
L
0
1
t
st
t
e
s





 
 
 
1
s

0, 0
( )
, 0
at
t
f t
Ae t


 
 

 
0
[ ]
a t
t at s A
e dt
Ae A
s a
e

 

 


L

Derivation of Laplace Transforms of Simple
Functions
• Ramp Function:
• Sinusoidal Function:
0, 0
( )
, 0
t
f t
At t

 
 

 
0 0
0
2
0
)
[ ]
st st
st
st
e Ae
e dt At dt
s s
A A
At A
e d
s s
t
t

 
 



  
 
 
 

L
0, 0
( )
sin , 0
t
f t
A wt t

 
 

 
 
0
2 2
2
1 1
2 2
sin ( ) st
j t j t
A
e dt
j
A A Aw
j s jw j
A wt
s jw s w
e e
 




  
 



L

Inverse Laplace Transformation
The inverse Laplace transform can be obtained by use of the inversion integral
given by:
• Partial-Fraction Expansion Method for Finding Inverse Laplace
Transforms:
For problems in control systems analysis, F(s), the Laplace transform off (t),
frequently occurs in the form:
where A(s) and B(s) are polynomials in s. In the expansion of F(s) = B(s)/A(s)
into a partial-fraction form, it is important that the highest power of s in A(s)
be greater than the highest power of s in B(s).
If F(s) is broken up into components:
-1 1
[ ] (
( ) ( )
2
)
c j
st
c j
f t F s e ds
j
F s

 
 
  
L
( ) ( ) / ( )
F s B s A s

1 2
( ) ( ) ( ) ( )
n
F s F s F s F s
   

1 2
-1 -1 -1 -1
( ) ( ) ( ) ( )
[ ] [ ] [ ] [ ]
n
F s F s F s F s
  
L L L L

• Partial-Fraction Expansion when F(s) Involves Distinct Poles Only
If F(s) involves distinct poles only, then it can be expanded into a sum of
simple partial fractions as follows:
where ak (k = 1,2,. . . , n) are constants. The coefficient ak is called the residue at the
pole at s = -pk. The value of a, can be found by multiplying both sides of the equation by
( s + pk) and letting s = -pk, which gives
1 2
1 2
( )( ) ( )
( )
( ) ,
( ) ( )( ) ( )
m
n
K s z s z s z
B s
F s for m n
A s s p s p s p
  
  
  


1 2
1 2
( )
( )
( )
n
n
a
a a
B s
F s
A s s p s p s p
   
  

     
   
1 2
1 2
( )
( ) k
k
k k k
s p
k n
k k k
k n s p
a a
B s
s p s p s p
A s s p s p
a a
s p s p a
s p s p



 
     

   
  

     

  



• Example:
The partial-fraction expansion of F(s) is:
3
( )
( 1)( 2)
s
F s
s s


 
3
( )
( 1)( 2) 1 2
s
F s
s s s s
B
A

  
   
1
1
3 3
( 1) 2
( 1)( 2) 2 s
s
s s
A s
s s s 

 
 
 
   
   
  
 
 
2
2
3 3
( 2) 1
( 1)( 2) 1 s
s
s s
B s
s s s 

 
 
 
   
   
  
 
 
Thus -1 -1 -1
2
2 1
( ) [ ]
(
2
2
)
1
t t
f t
s s
e e
F s
 

   
  
   
 
   
 
L L L

• Example:
Here, since the degree of the numerator polynomial is higher than that of the
denominator polynomial, we must divide the numerator by the denominator.
3 2
5 9 7
( )
( 1)( 2)
s s s
F s
s s
  

 
3
( ) 2
( 1)( 2)
s
F s s
s s

  
 
2
1 2
A B
s
s s
   
 
2
( ) '( ) 2 ( ) 1( )
t t
f t t t Ae Be t
   
 
   
 
2
( ) ( ) 2 ( ) 2 , 0
t t
d
f t t t e e t
dt
   
    
A and B same as in previous problem.

• Example:
Notice that the denominator polynomial can be factored as:
2
2 12
( )
2 5
s
F s
s s


 
2
2 5 ( 1 2)( 1 2)
s s s j s j
      
If the function F(s) involves a pair of complex-conjugate poles, it is convenient not to expand
F(s) into the usual partial fractions but to expand it into the sum of a damped sine and a damped
cosine function.
Noting that s2
+ 2s + 5 = (s + 1)2
+ 22
and referring to the Laplace transforms of e-at
sinwt
and e-at
cos wt, rewritten thus,
2 2 2 2 2 2 2
2 12 10 2( 1) 2 1
( ) 5 2
2 5 ( 1) 2 ( 1) 2 ( 1) 2
s s s
F s
s s s s s
   
   
       
It follows that
( ) 5 sin 2 2 cos2 , 0
t t
f t e t e t t
 
  

• Partial-Fraction Expansion when F(s) Involves Multiple Poles
Consider the following F(s):
2
3
2 3
( )
( 1)
s s
F s
s
 


The partial-fraction expansion of this F(s) involves three terms
2
3
2 3
( )
( 1)
s s
F s
s
 


2 3
1 ( 1) ( 1)
A B C
s s s
  
  
By multiplying both sides of this equation by ( s + 1 )3
, we have
2
3 3
3 2 3
2 3
( 1) ( 1)
( 1) 1 ( 1) ( 1)
s s A B C
s s
s s s s
 
 
    
 
   
 
2 2
2 3 ( 1) ( 1)
s s s A s B C
      

 
2
1
1
2 3 2 2 0
s
s
d
B s s s
ds 

 
 
     
 
 
 
2
3 3
3 2 3
1
1
2 3
( 1) ( 1) 2
( 1) 1 ( 1) ( 1) s
s
s s A B C
s s C
s s s s 

   
 
      
   
   
 
 
Then letting s = -1
2
2
2
1
2 3
s
d
A s s
ds 
 
 
  
 
 
 
2
1
2 2
s
d
s
ds 
 
 
 
 
 
 
2

2 2
( ) 2
2
t t t t t
C
f t Ae Bte t e e t e
    
    
Thus

Solving Linear, Time-Invariant, Differential
Equations
In this section we are concerned with the use of the Laplace transform
method in solving linear, time-invariant, differential equations.
• The Laplace transform method yields the complete solution
(complementary solution and particular solution) of linear, time-invariant,
differential equations. Classical methods for finding the complete solution
of a differential equation require the evaluation of the integration
constants from the initial conditions. In the case of the Laplace transform
method, however, this requirement is unnecessary because the initial
conditions are automatically included in the Laplace transform of the
differential equation.
• If all initial conditions are zero, then the Laplace transform of the
differential equation is obtained simply by replacing d/ dt with s, d2
/ dt2
with s2
, and so on.
• In solving linear, time-invariant, differential equations by the Laplace
transform method, two steps are involved:

Equations
1. By taking the Laplace transform of each term in
the given differential equation, convert the
differential equation into an algebraic equation
in s and obtain the expression for the Laplace
transform of the dependent variable by
rearranging the algebraic equation.
2. The time solution of the differential equation is
obtained by finding the inverse Laplace
transform of the dependent variable.

Equations
Example: Find the solution x(t) of the differential equation
3 2 0, (0) ,
x x x x a x b
    
  
By writing the Laplace transform of x(t) as X(s), we obtain:
2
[ ] ( ) (0)
[ ] ( ) (0) (0)
x
x
sX s x
s X s sx x
 
  

 

L
L
And so the given differential equation becomes
 
2
( ) (0) (0) 3 ( ) (0) 2 ( ) 0
s X s sx x sX s x X s
 
     
 

By substituting the given initial conditions into this last equation, we obtain
2
3 2 ( ) 3
s s X s sa b a
 
    
 

Equations
2
3
( )
3 2
sa b a
X s
s s
 

 
3
( 1)( 2)
sa b a
s s
 

 
The inverse Laplace transform of X(s) gives
1 2
A B
s s
 
 
1
3
2
2 s
sa b a
A b a
s 
 
 
  
 

  2
3
1 s
sa b a
B b a
s 
 
 
  
 

 
2
( ) (2 ) ( ) , 0
t t
x t a b e a b e t
 
    

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