CHE 185 – PROCESS
CONTROL AND DYNAMICS
LaPLACE TRANSFORM AND TRANSFER
FUNCTION FUNDAMENTALS
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
• DEFINITION

F (s)  L[ f (t )]   f (t )e dt
 st
0
• TIME (t) IS REPLACED BY A NEW INDEPENDENT
VARIABLE (s)
• WE CALL s THE LAPLACE TRANSFORM VARIABLE
LaPLACE TRANSFORMS
• WHY WORK IN THE LaPLACE DOMAIN
• OFTEN MORE CONVENIENT TO WORK IN LAPLACE
DOMAIN THAN TIME DOMAIN
• TIME DOMAIN  ORDINARY DIFFERENTIAL EQUATIONS I
Nt
• LaPLACE DOMAIN  ALGEBRAIC EQUATIONS IN s
LaPLACE TRANSFORMS
• GENERAL SOLUTION APPROACH
• LaPLACE TRANSFORMS ARE USED FOR SOLUTION OF
LINEAR ODE”S
• GIVEN A LINEAR FUNCTION f (t) WHICH IS
DEFINED FOR ALL POSITIVE VALUES OF t, t > 0.
• CONVERT MODEL TO LAPLACE DOMAIN
• TO OBTAIN THE LaPLACE TRANSFORM, MULTIPLY BY
e−st AND INTEGRATE FROM ZERO TO INFINITY
• 𝐹 𝑠 =
∞ −𝑠𝑡
𝑒 𝑓
0
𝑡 𝑑𝑡 OR L(f) = F(s)
LaPLACE TRANSFORMS
• GENERAL SOLUTION APPROACH
• SOLVE PROBLEM IN LAPLACE DOMAIN
• INVERT SOLUTION BACK TO TIME DOMAIN
• APPLY INVERSE LAPLACE TRANSFORM TO DIRECTLY
DETERMINE y(t).
• TABLES OF LAPLACE TRANSFORMS ARE AVAILABLE.
LaPLACE TRANSFORMS
• DERIVATION OF SOME COMMONLY USED
LaPLACE TRANSFORMS
• CONSTANT FUNCTION: f(t) = a
F ( s )  L( a )  

0
a  st
ae dt   e
s

 st
0
 a a
 0   
 s s
• EXPONENTIAL FUNCTION: f(t) = e-bt


F ( s )  L(e )   e e dt   e
bt
0
bt  st
0
( b  s )t

1
dt  
e ( b  s ) t
bs


0

1
sb
LaPLACE TRANSFORMS
• DERIVATION OF SOME COMMONLY USED
LaPLACE TRANSFORMS
LaPLACE TRANSFORMS
• DERIVATION OF SOME COMMONLY USED
LaPLACE TRANSFORMS
• DERIVATIVES AND INTEGRALS

 df   df  st
 st
 st 
L   
e dt   f (t )e sdt  f (t )e
 sF ( s)  f (0)
0
0
 dt  0 dt
 dn f  n
L n   s F ( s)  s n 1 f (0)  s n 2 f (1) (0)    sf ( n 2) (0)  f ( n1) (0)
 dt 
t

t
1
 st




L  f (t*)dt *    f (t*)dt * e dt  F ( s)
 0
 0  0

s
LaPLACE TRANSFORMS
• SOME COMMONLY USED
MORE LaPLACE TRANSFORMS
• IMPULSE FUNCTION - ZERO WIDTH AND TOTAL
INTEGRAL = C
• SUPERPOSITION
• FINAL VALUE THEOREM
If lim y (t ) exists
t 
 lim y (t )  lim[ sY ( s )]
• INITIAL VALUE THEOREM
lim y (t )  lim[ sY ( s )]
t 0
s 
t 
s 0
INITIAL- AND FINAL-VALUE THEOREMS
EXAMPLE
• LAPLACE TRANSFORM
OF THE FUNCTION.
• APPLY FINAL-VALUE
THEOREM
• APPLY INITIAL-VALUE
THEOREM
MORE LaPLACE TRANSFORMS
• TIME DELAY
• SUPERPOSITION
L[af (t )  bg(t )]  aF(s)  bG(s)
LaPLACE TRANSFORMS
• SOME COMMONLY USED LaPLACE TRANSFORMS
• DERIVATIVE
• OR FOR HIGHER LEVEL DERIVATIVES
LaPLACE TRANSFORMS
• SOME COMMONLY USED LaPLACE TRANSFORMS
• INTEGRALS
• DEAD TIME FOR Θ UNITS OF TIME:
LaPLACE TRANSFORMS
•
•
•
•
•
PULSE FUNCTION
𝑢 𝑡 = 0 𝑡 < 𝑡0
𝑢 𝑡 = 𝐴 𝑡0 ≤ 𝑡0 ≤ 𝑡0 + ∆𝑡
𝑢 𝑡 = 0 𝑡 ≥ 𝑡0 + 𝑡
SET UP A COMPOSITE
EQUATION TO GET THE
LaPLACIAN
• 𝑦 𝑡 = 𝐴(𝛾 𝑡 − 𝛾 𝑡 − Δ𝑡 → 𝐴
• 𝑌 𝑠 = (1 − 𝑒 −𝑠 )
𝐴
𝑠
t
A
1
𝑠
1
−

𝑠
−𝑒
𝑠
= 𝑌(𝑠)
LaPLACE TRANSFORMS
• EXAMPLE, FOR LINEAR ODE:
dy
5  4y  2
dt
y (0)  1
• LAPLACE TRANSFORM
2
5[ sY ( s )  y (0)]  4Y ( s) 
s
LaPLACE TRANSFORMS
• SUBSTITUTE Y(0) & REARRANGE:
5s  2
s  0.4
Y ( s) 

s(5s  4) s( s  0.8)
• INVERSE LAPLACE TRANSFORM
 s  0.4 
y(t )  L [Y (s)]  L 

s
(
s

0
.
8
)


1
1
PARTIAL FRACTION EXPANSIONS
•
EXPAND INTO A TERM FOR
EACH FACTOR IN THE
DENOMINATOR.
•
RECOMBINE RHS
•
EQUATE TERMS IN s AND
CONSTANT TERMS. SOLVE.
•
EACH TERM IS IN A FORM SO
THAT INVERSE LaPLACE
TRANSFORMS AN BE APPLIED.
PARTIAL FRACTION EXPANSIONS
• HEAVISIDE METHOD INDIVIDUAL POLES.
PARTIAL FRACTION EXPANSIONS
• HEAVISIDE METHOD INDIVIDUAL POLES.
PARTIAL FRACTION EXPANSIONS
• HEAVISIDE METHOD REPEATED POLES.
PARTIAL FRACTION EXPANSIONS
•
GENERAL FORMAT FOR THE EXAMPLE FROM TABLE 4.1.

 a3  a1 a1t a3  a2 b2t
s  a3
L 
e 
e

a1  a2
 (s  a1 )(s  a2 )  a2  a1
1
•
EXAMPLE VALUES
 s  0.4 
y (t )  L 
 a1  0 a2  0.8 a3  0.4

 s(s  0.8) 
1
•
SUBSTITUTE AND SIMPLIFY

 0.4  0 0t 0.4  0.8 0.8t
s  0.4
0.8t
L 

e

e

0
.
5

0
.
5
e
 0.8  0
(
s

0
)(
s

0
.
8
)
0  0.8


1
PARTIAL FRACTION EXPANSIONS
•
SECOND EXAMPLE
•
CALCULATE COEFFICIENTS
s5
s5
1
2
Y (s )  2



s  5s  4 (s  1)(s  4) s  1 s  4
s 5
4
s 5
1
1 

2 

s  4 s 1 3
s  1 s 4
3
•
INVERSE LaPLACE TRANSFORM
 4 / 3 1 / 3  4 t 1  4 t
y(t )  L [Y ( s)]  L 

 e  e

3
 s 1 s  4  3
1
1
PARTIAL FRACTION EXPANSIONS
•
REPEATED FACTOR EXAMPLE
•
CALCULATE COEFFICIENTS USING HEAVISIDE METHOD
3
s 1
1
2
Y (s ) 



2
2
s (s  2 )
s  2 (s  2 )
s
s 1
1
s 1
2 

3 
s s  2 2
(s  2 ) 2
•
s 0
1
1

1  
4
4
SUBSTITUTE AND DETERMINE INVERSE LaPLACE TRANSFORM
 1/ 4
1/ 2
1/ 4
Y (s ) 


2
s  2 (s  2 )
s
1  2t 1  2t 1
y (t ) L [Y (s )]   e  te  S(t )
4
2
4
1
QUADRATIC FACTOR
1
Y (s )  2 2
s (s  as  b )
1

2
2



a

a

4
b
a

a

4
b
 s 

s s 



2
2



• EVALUATION OF ROOTS DEPENDS ON THE
VALUES OF a2 – 4b
QUADRATIC FACTOR EXAMPLE
 3s   4
s 1
1  2
Y (s )  2 2

 2  2
s (s  4s  5) s s
s  4s  5
𝑎 = 4 𝑏 = 5 𝑎2 − 4𝑏 = −4
• EQUATING AROUND LIKE POWERS OF s:
s  1  1s(s 2  4s  5)   2 (s 2  4s  5)  ( 3s   4 )s 2
(1   3 )s 3  ( 41   2   4 )s 2  (51  4 2  1)s  (5 2  1)  0
• SOLVING FOR EACH TERM:
1
1
∝2 = = 0.2 , 5 ∝1 +4 ∝2 = 1 , ∝1 =
= 0.4,
5
25
∝3 = −∝1 =
1
−
25
= −0.4 ∝4 = −4 ∝1 −∝2 =
9
25
= 0.36
QUADRATIC FACTOR EXAMPLE
• SUBSTITUTING VALUES INTO EQUATION:
Y (s ) 
 3s   4
s 1
1  2
0.04 0.2  0.04s  0.36




 2 
2
2
2
2
s (s  4s  5) s s
s  4s  5
s
s
s 2  4s  5
• IN TERMS SIMILAR TO TABLE 4.1
 0.04s  0.36  0.04(s  2)
 0.28


s 2  4s  5
(s  2 ) 2  1 (s  2 ) 2  1
• INVERSE LaPLACE SUBSTITUTION:
y (t )  L1 [Y (s )]  0.04S(t )  0.2t  0.04e  2t cos t  0.28e  2t sin t
LaPLACE TRANSFORMS
• LINEARITY PROPERTIES INCLUDE ADDITIVITY, SO
L(af(t)+𝑏𝑔(𝑡)) = 𝑎𝐿(𝑓 𝑡 + 𝑔(𝑡)) = 𝑒 −𝑠𝑡 (𝑎𝑓 𝑡 + 𝑏𝑔(𝑡))𝑑𝑡
• EXAMPLE USING THE HYPERBOLIC COSINE
FUNCTION: cosh 𝑎𝑡 =
(𝑒 𝑎𝑡 +𝑒 −𝑎𝑡 )
2
= 𝑓(𝑡)
• TRANSFORMING:
• 𝐿(cosh 𝑎𝑡) =
1
2
𝐿 𝑒
𝑎𝑡
+
1
𝐿
2
• WHEN s > a: 𝐿(cosh 𝑎𝑡) =
𝑒
−𝑎𝑡
=
𝑠
( 2 2)
𝑠 −𝑎
1 1
(
2 𝑠−𝑎
+
1
)
𝑠+𝑎
LaPLACE TRANSFORMS
• PARTIAL FRACTION EXPANSION OBJECTIVE
• THIS METHOD IS USED TO ALGEBRAICALLY
CONVERT THE LaPLACE TRANSFORM INTO
TERMS THAT ARE IN THE TABLE
• EXAMPLE INVERT THE LaPLACE TRANSFORM:
𝑠+1
−1
𝐿 ( 3
)
2
𝑠 + 𝑠 − 6𝑠
• EXPAND INTO A TERM FOR EACH FACTOR IN
THE DENOMINATOR
PARTIAL FRACTION EXAMPLE
• THE RESULTING EQUATION HAS THE FORM:
𝑠+1
𝑠+1
𝑐1
𝑐2
𝑐3
=
= +
+
3
2
𝑠 + 𝑠 − 6𝑠
𝑠(𝑠 − 2)(𝑠 + 3) 𝑠 𝑠 − 2 𝑠 + 3
• CLEARING FRACTIONS YIELDS
𝑠 + 1 = 𝑐1 𝑠 − 2 𝑠 + 3 + 𝑐2 𝑠 𝑠 + 3 + 𝑐3 𝑠(𝑠 − 2)
• EQUATE THE COEFFICIENTS WITH LIKE
POWERS OF s: s 2 : (𝑐1 + 𝑐2 + 𝑐3 ) = 0;
s: 𝑐1 + 3𝑐2 − 2𝑐3 = 1; −6𝑐1 = 1
RESULTING VALUES:
𝑐1 =
1
−
6
𝑐2 =
3
10
𝑐3 =
2
−
15
LaPLACE TRANSFORMS
• SUBSTITUTION BACK INTO THE ORIGINAL
EQUATION
• 𝐿−1
𝑆+1
𝑆 3 +𝑆 2 −6𝑆
1
6
= − 𝐿−1
1
𝑆
+
3 −1
1
𝐿
10
𝑆−2
−
2 −1 1
𝐿 ( )
15
𝑆+3
• USING TABLE 4.1, THE SUBSTITUTIONS YIELD:
−1
• 𝐿
𝑆+1
𝑆 3 +𝑆 2 −6𝑆
=
1
−
6
3 2𝑡
+ 𝑒
10
−
2 −3𝑡
𝑒
15
EXAMPLE OF ODE SOLUTION
• ODE WITH INITIAL
CONDITIONS
• APPLY LAPLACE
TRANSFORM TO EACH
TERM
• SOLVE FOR Y(S)
• APPLY PARTIAL
FRACTION EXPANSIONS
WITH HEAVISIDE
• APPLY INVERSE LaPLACE
TRANSFORM TO EACH
TERM
MIX TANK TRANSIENT RESPONSE
• STEP INPUT INTO MIX TANK
• V = 4 m3 q = 2 m3/min
• STEP CONCENTRATION IS
5%
• COMPONENT BALANCE
dc1
dc1
V
 q(ci  c1 )  4
 2c1  2ci
dt
dt
• STEP INPUT LaPLACIAN
0 t  0
5
ci (t )  
 Ci (s ) 
s
5 t  0
c1 (0)  0
MIX TANK TRANSIENT RESPONSE
• LaPLACE TRANSFORM FOR COMPOSITION
2
4[ sC1 ( s)  0]  2C1 ( s)  2Ci ( s)  C1 ( s) 
Ci ( s)
4s  2
• ADD INPUT SUBSTITUTION
2 5
5
C1 ( s) 

4s  2 s s(2s  1)
• INVERSE LaPLACE TRANSFORM


5
t / 2
c1(t )  L 

5
1

e

 s(2s  1) 
1

