Transcript Slide 1
Chapter 3
Laplace Transforms
1. Standard notation in dynamics and control
(shorthand notation)
2. Converts mathematics to algebraic operations
3. Advantageous for block diagram analysis
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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
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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
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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)
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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,
L1 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
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.
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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)
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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
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the area under the pulse constant at one. (i.e., let h )
tw
Let,
t impulse function
Then,
L t 1
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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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Other Transforms
e jt cos t j sin t
Note:
e jwt cos t j sin t
Chapter 3
j 1
e-jt e jt
L(cosωt) = L
2
1 1
1
=
2 s jω s jω
1 s jω
s jω
= 2
2 s ω2 s2 ω2
s
= 2
s ω2
e jt - e jt
L(sin ωt) = L
2j
ω
= 2
s ω2
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Difference of two step inputs S(t) – S(t-1)
(S(t-1) is step starting at t = h = 1)
Chapter 3
By Laplace transform
1 e s
F ( s)
s s
Can be generalized to steps of different magnitudes
(a1, a2).
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Table 3.1. Laplace Transforms
Chapter 3
See page 54 of the text.
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