Chapter 3 (10-5

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Transcript Chapter 3 (10-5

Laplace Transforms
Chapter 3
• Important analytical method for solving linear ordinary
differential equations.
- Application to nonlinear ODEs? Must linearize first.
• Laplace transforms play a key role in important process
control concepts and techniques.
- Examples:
• Transfer functions
• Frequency response
• Control system design
• Stability analysis
1
Definition
The Laplace transform of a function, f(t), is defined as
Chapter 3

F ( s)  L  f (t )   f  t  e st dt
0
(3-1)
where F(s) is the symbol for the Laplace transform, L is the
Laplace transform operator, and f(t) is some function of time, t.
Note: The L operator transforms a time domain function f(t)
into an s domain function, F(s). s is a complex variable:
s = a + bj, j
1
2
Inverse Laplace Transform, L-1:
Chapter 3
By definition, the inverse Laplace transform operator, L-1,
converts an s-domain function back to the corresponding time
domain function:
f  t   L1  F  s  
Important Properties:
Both L and L-1 are linear operators. Thus,
L  ax  t   by  t   aL  x  t   bL  y  t 
 aX  s   bY  s 
(3-3)
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where:
- x(t) and y(t) are arbitrary functions
Chapter 3
- a and b are constants
- X  s  L  x  t  and Y  s  L  y  t 
Similarly,
L1  aX  s   bY  s    ax  t   b y  t 
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Laplace Transforms of Common
Functions
Chapter 3
1. Constant Function
Let f(t) = a (a constant). Then from the definition of the
Laplace transform in (3-1),

L  a    ae
0

 st
a  st
dt   e
s
0
 a a
 0  
 s s
(3-4)
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2. Step Function
Chapter 3
The unit step function is widely used in the analysis of process
control problems. It is defined as:
S t 
0 for t  0

1 for t  0
(3-5)
Because the step function is a special case of a “constant”, it
follows from (3-4) that
1
L  S  t   
s
(3-6)
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3. Derivatives
Chapter 3
This is a very important transform because derivatives appear
in the ODEs we wish to solve. In the text (p.53), it is shown
that
 df 
L    sF  s   f  0 
 dt 
(3-9)
initial condition at t = 0
Similarly, for higher order derivatives:
dn f
L n
 dt

n
n 1
n  2 1

s
F
s

s
f
0

s
f  0 






n2
n 1
...  sf   0  f   0
(3-14)
 
 
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where:
- n is an arbitrary positive integer
Chapter 3
- f  k   0
dk f
dt k
t 0
Special Case: All Initial Conditions are Zero
Suppose
1
n1
f  0   f    0   ...  f    0  . Then
dn f
L n
 dt
 n
  s F s

In process control problems, we usually assume zero initial
conditions. Reason: This corresponds to the nominal steady state
when “deviation variables” are used, as shown in Ch. 4.
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4. Exponential Functions
Consider f  t   ebt where b > 0. Then,


Chapter 3
 b s t
L ebt    ebt e st dt   e   dt

 0
0
1   b s t  
1

e

0
bs 
sb
(3-16)
5. Rectangular Pulse Function
It is defined by:
0 for t  0

f  t   h for 0  t  t w
0 for t  t
w

(3-20)
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h
Chapter 3
f t 
tw
Time, t
The Laplace transform of the rectangular pulse is given by
F s 

h
1  e t w s
s

(3-22)
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Chapter 3
6. Impulse Function (or Dirac Delta Function)
The impulse function is obtained by taking the limit of the
rectangular pulse as its width, tw, goes to zero but holding
1
the area under the pulse constant at one. (i.e., let h  )
tw
Let,
  t  impulse function
Then,
L   t   1
Solution of ODEs by Laplace Transforms
Procedure:
1. Take the L of both sides of the ODE.
2. Rearrange the resulting algebraic equation in the s domain to
solve for the L of the output variable, e.g., Y(s).
3. Perform a partial fraction expansion.
4. Use the L-1 to find y(t) from the expression for Y(s).
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Table 3.1. Laplace Transforms
Chapter 3
See page 54 of the text.
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Example 3.1
Chapter 3
Solve the ODE,
dy
5  4y  2
y 0  1
dt
First, take L of both sides of (3-26),
2
5  sY  s   1  4Y  s  
s
Rearrange,
5s  2
Y s 
s  5s  4 
Take L-1,
(3-26)
(3-34)
1 
5s  2 
y t   L 

 s  5s  4  
From Table 3.1,
y  t   0.5  0.5e0.8t
(3-37)
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Partial Fraction Expansions
Chapter 3
Basic idea: Expand a complex expression for Y(s) into
simpler terms, each of which appears in the Laplace
Transform table. Then you can take the L-1 of both sides of
the equation to obtain y(t).
Example:
Y s 
s5
 s  1 s  4 
Perform a partial fraction expansion (PFE)
1
2
s5


 s  1 s  4  s  1 s  4
(3-41)
(3-42)
where coefficients 1 and  2 have to be determined.
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To find 1 : Multiply both sides by s + 1 and let s = -1
s5
 1 
s4
4

s 1 3
Chapter 3
To find  2 : Multiply both sides by s + 4 and let s = -4
s5
 2 
s 1
s 4
1

3
A General PFE
Consider a general expression,
Y s 
N s
Ds

N s
n
  s  bi 
(3-46a)
i 1
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Here D(s) is an n-th order polynomial with the roots  s  bi 
all being real numbers which are distinct so there are no repeated
roots.
The PFE is:
Chapter 3
Y s 
N s
n
  s  bi 
n

i 1
i
(3-46b)
s  bi
i 1
Note: D(s) is called the “characteristic polynomial”.
Special Situations:
Two other types of situations commonly occur when D(s) has:
i) Complex roots: e.g., bi  3  4 j
ii) Repeated roots (e.g., b1  b2  3 )
j
1

For these situations, the PFE has a different form. See SEM
text (pp. 61-64) for details.
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Example 3.2 (continued)
Chapter 3
Recall that the ODE, y  6 y  11y  6 y  1, with zero initial
conditions resulted in the expression
Y s 

1
s s  6s  11s  6
3
2

(3-40)
The denominator can be factored as


s s3  6s 2  11s  6  s  s  1 s  2  s  3
(3-50)
Note: Normally, numerical techniques are required in order to
calculate the roots.
The PFE for (3-40) is
Y s 




1
 1  2  3  4 (3-51)
s  s  1 s  2  s  3 s s  1 s  2 s  3
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Solve for coefficients to get
1
1
1
1
1  ,  2   ,  3  ,  4  
6
2
2
6
Chapter 3
(For example, find  , by multiplying both sides by s and then
setting s = 0.)
Substitute numerical values into (3-51):
1/ 6 1/ 2 1/ 2 1/ 6
Y ( s) 



s
s 1 s  2 s  3
Take L-1 of both sides:
1
1 1/ 6 
1  1/ 2 
1  1/ 2 
1  1/ 6 
L Y  s    L 
L 
L 
L 



 s 
 s  1
s  2
 s  3 
From Table 3.1,
y t  
1 1 t 1 2t 1 3t
 e  e  e
6 2
2
6
(3-52)
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Important Properties of Laplace Transforms
1. Final Value Theorem
Chapter 3
It can be used to find the steady-state value of a closed loop
system (providing that a steady-state value exists.
Statement of FVT:
lim y  t   lim  sY  s 
t 
s0
providing that the limit exists (is finite) for all
Re  s   0, where Re (s) denotes the real part of complex
variable, s.
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Example:
Suppose,
Chapter 3
Y s 
5s  2
s  5s  4 
(3-34)
Then,
 5s  2 
y     lim y  t   lim 
 0.5

t 
s 0  5s  4 
2. Time Delay
Time delays occur due to fluid flow, time required to do an
analysis (e.g., gas chromatograph). The delayed signal can be
represented as
y  t  θ  θ  time delay
Also,
L  y  t  θ    eθsY  s 
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