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Lap
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CCB3013 - Chemical Process Dynamics, Instrumentation
and Control 16/16/2014
Chapter 3
Laplace Transform
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CCB3013 - Chemical Process Dynamics, Instrumentation
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Chapter Objectives
End of this chapter, you should be able to:
• Use Laplace Transform of representative functions
• Solve differential equations by Laplace Transform
techniques
• Explain other Laplace Transform properties
– Final value theorem
– Initial value theorem
– Time delay (translation in time)
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3.1 Laplace Transform of Representative
Functions
The Laplace transform of a function f(t) is defined as
0
)()]([)( dtetf =tfLsF -st
(3.1)
where F(s) is the symbol for the Laplace transform
s is a complex independent variable
f(t) is some function of time
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Transforms of some important functions
Constant function: f(t) = a
s
a
s
ae
s
adtae=aL st-st
0][0
0
(3.2)
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Transforms of some important functions
Step function:
The unit step function is defined as
0 1
0 0)(
t
ttS (3.3)
The Laplace transform of the unit step function is
obtained with a = 1
stSL
1)]([ (3.4)
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Transforms of some important functions
Exponential Functions
bteb > 0
bs
esb
dtedtee=eL tsbtsb-stbtbt
11
][0
)(
0
)(
0
(3.5)
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Transforms of some important functions
Derivatives
)0()()0(][
)()(
][
00
0
fssFffsL
etfsdtetf
dtedtdf=dtdfL
stst
-st
(3.6)
Usually define f(0) = 0
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Transforms of some important functions
Second-order derivatives
dt
dL
dt
fdL
2
2
wheredt
df
)0()0()(
)0()0()(
)0()(
2
2
2
fsfsFs
ffssFs
ss
dt
dL
dt
fdL
(3.7)
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Transforms of some important functions
Higher-order derivatives
)0()0(
...)0()0()(
)1()2(
21
nn
nnn
n
n
fsf
fsfssFsdt
fdL
(3.8)
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Transforms of some important functions
Trigonometric functions
Euler identity:2
costjtj ee
t
cos sin
cos sin
1
j t
jwt
e t j t
e t j t
j
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Transforms of some important functions
Cosine function
22
222
1
11
2
1
2cos
s
s
s
jsjs
jsjs
eeLtL
tjtj
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Transforms of some important functions
Sine function
22
222
1
11
2
1
2sin
s
s
jsjs
j
jsjsj
j
eeLtL
tjtj
(3.9)
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Rectangular pulse function
0
1 1( ) (1 )
hst hsF s e dt e
h hs
ht
htht
tf
0
0 /10 0
)( (3.10)
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Rectangular pulse function
If h = 1, unit rectangular pulse input.
Difference of two step inputs S(t) – S(t-1). (S(t-1) is a
step starting at t = h = 1)
By Laplace transform
1( )
seF s
s s
Can be generalized to steps of different magnitudes (a1, a2).
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Impulse function
Let h→0, f(t) = δ(t) (Dirac delta) L(δ) = 1
Laplace transforms can be used in process control for:
• Solution of differential equations (linear)
• Analysis of linear control systems (frequency
response
• Prediction of transient response for different inputs
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General procedure for solving differential
equations
repeated factors
and complex
factors may arise.
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Example 3.1 Solve the ODE,
5 4 2 0 1 (3-26)dy
y ydt
(3.11)
First, take L of both sides of (3-11),
2
5 1 4sY s Y ss
Rearrange,
5 2
(3-34)5 4
sY s
s s
(3.12)
Take L-1,
1 5 2
5 4
sy t
s s
L
From L Table
0.80.5 0.5 (3-37)ty t e (3.13)
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Example 3.2
00
0000
2461162
2
3
3
at t=dt
du
)=(y)=(y)=y(
udt
duy
dt
dy
dt
yd
dt
yd
To find transient response for u(t) = unit step at t > 0 :
1. Take Laplace Transform (L.T.)
2. Factor, use partial fraction decomposition
3. Take inverse L.T.
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Step 1 Take L.T. (note zero initial conditions)
3 26 11 6 ( ) 4 2s Y(s)+ s Y(s)+ sY(s) Y s = sU(s) + U(s)
Rearranging,
input) (1
1
6116
2423
stepunits
U(s)
ssss
s+Y(s)=
Step 2a. Factor denominator of Y(s)
))(s+)(s+)=s(s+s++s+s(s 3216116 23
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Step 2b. Use partial fraction decomposition
321321
24 4321
s
α
s
α
s
α
s
α
))(s+)(s+s(s+
s+
Multiply by s, set s = 0
3
1
321
2
321321
24
1
0
4321
0
α
s
α
s
α
s
αsα
))(s+)(s+(s+
s+
ss
For a2, multiply by (s+1), set s = -1 (same procedure for a3, a4)
3
531 432 , α, αα
3
3/5
2
3
1
1
3
1
ssssY(s)=
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Step 3. Take inverse of L.T.
3
1
3
53
3
1 32
) y(tt
eeey(t)= ttt
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Dynamic Analysis
2
2
3 4 1s s
Example 1:2 16
1.333 14 12
b
a
2 13 4 1 (3 1)( 1) 3( )( 1)3
s s s s s s
Transforms to e-t/3, e-t (real roots) - (no oscillation)
2
2
1
s
s s
Example 2:
2 11
4 4
b
a
2 3 31 ( 0.5 )( 0.5 )
2 2s s s j s j
Transforms to 0.5 0.53 3
sin , cos2 2
t te t e t
(oscillation)
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Remarks
• You can use this method on any order of ODE, limited only
by factoring of denominator polynomial (characteristic
equation)
• Must use modified procedure for repeated roots,
imaginary roots
One other useful feature of the Laplace transform is that one
can analyze the denominator of the transform to determine its
dynamic behavior. For example, if
23
12 ss
Y(s)=
the denominator can be factored into (s+2)(s+1).
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Remarks
Using the partial fraction technique
12
21
s
α
s
αY(s)=
The step response of the process will have exponential
terms e-2t and e-t, which indicates y(t) approaches zero.
))(s(sssY(s)=
21
1
2
12
However, if
• We know that the system is unstable and has a transient
response involving e2t and e-t. e2t is unbounded for large
time.
• We shall use this concept later in the analysis of
feedback system stability.
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Other properties of L( ):
A. Final value theorem
sY(s))=y(s 0lim
“offset”
Example 3: Step response
aτs
a
τs
asY(s)
s
a
τsY(s)
s
1lim
1
1
1
0
offset (steady state error) is a.
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Other properties of L( ):
Time-shift theorem
y(t)=0 t < Y(s)=et-yL s-
Initial value theorem
sY(s)y(t)=st limlim
0
Example 4.
)3)(2)(1(
24)(
s+s+s+s
s+=sFor Y
by initial value theorem: 0)0( y
by final value theorem:3
1)( y
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Conclusion!
• Laplace transform definition
• Laplace transform of some typical functions
• Inverse Laplace transforms
• Examples
• Dynamic Analysis