### Chapter 4

```Chapter 4
Laplace Transforms
Overall Course Objectives
• Develop the skills necessary to function as an
industrial process control engineer.
– Skills
•
•
•
•
Tuning loops
Control loop design
Control loop troubleshooting
Command of the terminology
– Fundamental understanding
• Process dynamics
• Feedback control
Laplace Transforms
• Provide valuable insight into process
dynamics and the dynamics of feedback
systems.
• Provide a major portion of the terminology
of the process control profession.
Laplace Transforms
L  f (t )   f (t ) e dt  F ( s)

 st
0
• Useful for solving linear differential equations.
• Approach is to apply Laplace transform to
differential equation. Then algebraically solve for
Y(s). Finally, apply inverse Laplace transform to
directly determine y(t).
• Tables of Laplace transforms are available.
Method for Solving Linear
ODE’s using Laplace Transforms
sY(s) - y(0) =
F(s,Y)
Y(s) = H(s)
Lapl ace Domain
Ti m e Dom ai n
dy/dt = f(t,y)
y(t) = h(t)
Some Commonly Used Laplace
Transforms
f (t   )  F ( s ) e  s
Unit Step  1 / s
t
n
n!
 n 1
s
e  at
1

sa

sin( t )  2
s  2
d f (t )
 s F ( s )  f (0)
dt
d 2 f (t )
2

s
F ( s )  s f (0)  f (0)
2
dt
e
 at
sin( t ) 

( s  a) 2   2
Final Value Theorem
limt   f (t )  lims0 s F (s)
• Allows one to use the
Laplace transform of a
function to determine
value of the function.
• A good consistency
check.
Initial-Value Theorem
limt 0  f (t )  lims s F (s)
• Allows one to use the
Laplace transform of a
function to determine
the initial conditions
of the function.
• A good consistency
check
Apply Initial- and Final-Value
Theorems to this Example
2
Y ( s) 
s ( s  2) ( s  4)
2 (0)
1
limt   f (t ) 

(0) (0  2) (0  4) 4
limt 0  f (t ) 
2 ( )
0
() (  2) (  4)
• Laplace
transform of the
function.
• Apply final-value
theorem
• Apply initialvalue theorem
Partial Fraction Expansions
s 1
A
B


(s  2) ( s  3) s  2 s  3
s 1

( s  2) ( s  3)
A  B 1
• Expand into a term for
each factor in the
denominator.
A( s  3)  Bs  2
• Recombine RHS
( s  2) ( s  3)
3 A  2B 1
s 1
1
2


( s  2) ( s  3) s  2 s  3
• Equate terms in s and
constant terms. Solve.
• Each term is in a form so
that inverse Laplace
transforms can be applied.
Heaviside Method
Individual Poles
N ( s)
N ( s)


Y (s) 
D( s ) ( s  a1 )( s  a2 )  ( s  an )
Cn
C2
C1
 

s  an
s  a1 s  a2
N (s)
Equation 4.4.2 : Ci 
D( s) /( s  ai ) s  a
i
Heaviside Method
Individual Poles
s3
C1
C2
Y ( s) 


( s  1)( s  2) s  1 s  2
N ( s)
Equation 4.4.2 : Ci 
D( s) /( s  ai ) s  a
i
s3
s3
C1 
 2; C2 
 1
s  2 s 1
s  1 s 2
s3
2
1
Y ( s) 


( s  1)( s  2) s  1 s  2
Heaviside Method
Repeated Poles
N ( s)
N ( s)
Y (s) 


n
D( s ) ( s  a)
Cn
C1
C2


2
s  a (s  a)
(s  a)n
1 d i  ( s  a)n N ( s) 
Equation 4.4.2 : Ci 

i 
i ds 
D( s )
 s  ai
Heaviside Method
Example with Repeated Poles
C3
N (s)
s3
C1
C2
Y (s) 




2
2
D( s ) ( s  1) ( s  2) s  1 ( s  1) s  2
s3
C2 
 2;
s  2 s 1
s3
C3 
2
( s  1)
1
s 2
n

1 d ( s  a) N (s) 
Equation 4.5.3 : Ci 

i ds i 
D( s )
 s  ai
i
 1
1 d  s  3
s3 
C1 


 1
2


1 ds  s  2  s 1  s  2 ( s  2)  s 1
s3
1
2
1
Y (s) 



2
2
( s  1) ( s  2) s  1 ( s  1) s  2
Example of Solution of an ODE
d2y
dy

6
 8 y  2 y(0)  y' (0)  0 •
2
dt
dt
s 2 Y (s)  6s Y (s)  8Y (s)  2 / s
2
Y ( s) 
s ( s  2) ( s  4)
1
1
1
Y ( s) 


4s 2 ( s  2) 4 ( s  4)
1 e 2t
e 4t
y (t )  

4
2
4
ODE w/initial conditions
• Apply Laplace transform
to each term
• Solve for Y(s)
• Apply partial fraction
expansions w/Heaviside
• Apply inverse Laplace
transform to each term
Overview
• Laplace transforms are an effective way to
solve linear ODEs.
```