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 L1 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)
3
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,
L1 aX s bY s ax t b y t
4
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)
6
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
n2
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
n1
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.
8
4. Exponential Functions
Consider f t ebt where b > 0. Then,
Chapter 3
b s t
L ebt ebt e st dt e dt
0
0
1 b s t
1
e
0
bs
sb
(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)
9
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).
11
Table 3.1. Laplace Transforms
Chapter 3
See page 54 of the text.
12
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 0.5e0.8t
(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
s5
s 1 s 4
Perform a partial fraction expansion (PFE)
1
2
s5
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
s5
1
s4
4
s 1 3
Chapter 3
To find 2 : Multiply both sides by s + 4 and let s = -4
s5
2
s 1
s 4
1
3
A General PFE
Consider a general expression,
Y s
N s
Ds
N s
n
s bi
(3-46a)
i 1
15
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.
16
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
17
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)
18
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
s0
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 eθsY s
20