Post on 20-Jan-2015
description
28/02/2014
DAE32103_Control System 1
Muhammad Faizal bin Ismail
Dept. of Electrical Engineering
PPD, UTHM
ifaizal@uthm.edu.my
013-7143106 1
1. Introduction
2. Laplace Transform Theorem
3. Common Time Domain Input Function
4. Transfer Function
Outline:
1. Introduction
2. Laplace Transform – Table/ Theorem/ Eg.
3. Common Time Domain Input Function
4. Transfer Function – Open/ Closed Loop & Eg.
5. Electrical Elements Modelling – Table & Eg.
6. Mechanical Elements Modelling - Table & Eg.
7. Block Diagram Reduction - Table & Eg.
8. System Response – Poles/ Zeros, Second Order, Steady State Error, Stability Analysis
2
OUTLINE
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DAE32103_Control System 2
1. Intro - Objective of this chapter
After completing this chapter you will be able to:
• Describe the fundamental of Laplace transforms.
• Apply the Laplace transform to solve linear ordinary differential equations.
• Apply Mathematical model, called a transfer function for linear time-invariant electrical, mechanical and electromechanical systems.
3
2. What is Laplace Transform?
• Laplace transform is a method or techniques used to transform the time (t) domain to the Laplace/frequency (s) domain
• What is algebra & calculus?
4
Time Domain Frequency Domain
Differential equations
Input q(t)
Output h(t)
Algebraic equations
Input Q(s)
Output H(s)
Calculus Algebra
Laplace Transformation
Inverse Laplace
Transformation
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DAE32103_Control System 3
Laplace Transform (cont.)
5
The Laplace transform solution
consists of the following three
steps:
(1) the Laplace transformation of
q1(t) and (r dhldt + h = Gq) to
frequency domain
(2) the algebraic solution for H(s)
(3) the inverse Laplace
transformation of H(s) to time
domain h(t).
(4) The calculus solution is shown as
step 4.
Definition of the Laplace Transform
• Laplace transform is defined as
• Inverse Laplace transform is defined as
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[ ])(tf )()(
0
sFdtetfst
== ∫∞
−
L
L-1 ∫∞+
∞−
==
j
j
sttfdsesF
jsF
σ
σπ
)()(2
1)]([
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DAE32103_Control System 4
Laplace Theorem
7
Laplace Table
8
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Example 1
Find the Laplace transform for
9
1)( =tf
Solution:
Example 2
Find the Laplace transform for
10
atetf
−=)(
Solution:
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DAE32103_Control System 6
Example 3
Find the inverse Laplace transform of( )( )21
32
10)(
++=
ssssF
11
Solution:
Expanding F(s) by partial fraction:
Where,
Then, taking the inverse Laplace transform
)(9
40
3
105
9
5)(
332tueteetf
ttt
++−=
−−
Example 4
Given the ,solve for y(t) if all initial conditions are
zero. Use the Laplace transform method.
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Solution:
Substitute the corresponding F(s) for each term:
Solving for the response:
Where, K1= 1 when s=0
K2=-2 when s=-4
K3= 1 when s=-8Hence
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3. Common Time Domain Input
Functions
• Unit Step Function
13
cont.
• Unit Ramp Function
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DAE32103_Control System 8
cont.
• Unit Impulse Function
15
4. Transfer Function
• Definition:
Ratio of the output to the input; with all initial conditions are zero
• If the transformed input signal is X(s) and the transformed output signal is Y(s), then the transfer function M(s) is define as;
• From this,
• Therefore the output is
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DAE32103_Control System 9
TF of Linear Time Invariant Systems
• In practice, the input-output relation of lines time-invariant system with
continuous-data input is often described by a differential equation
• The linear time-invariant system is described by the following nth-order
differential equation with constant real coefficients;
).()(
....)()(
)()(
.....)(
)(
)(
011
1
1
011
1
1
trbdt
tdrb
dt
trdb
dt
trdb
tcadt
tdca
dt
tcda
td
tcda
m
m
mm
m
m
n
n
nn
n
n
+++=
++++
−
−
−
−
−
−
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c(t) is output
r(t) is input
cont.
• Taking the Laplace transform of both sides,
• If we assume that all initial conditions are zero, hence
• Now, form the ratio of output transform, C(s) divided by input
transform. The ratio, G(s) is called transfer function.
).(___)(....)()(
)(___)(.....)()(
0
1
1
0
1
1
trofconditioninitialsRbsRsbsRsb
tcofconditioninitialsCasCsasCsa
m
m
m
m
n
n
n
n
++++=
++++
−
−
−
−
)().....()().....(01
1
101
1
1sRbsbsbsbsCasasasa m
m
m
m
n
n
n
n ++++=++++−
−
−
−
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)....(
)....(
)(
)()(
01
1
1
01
1
1
asasasa
bsbsbsb
sR
sCsG
n
n
n
n
m
m
m
m
++++
++++==
−
−
−
−
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DAE32103_Control System 10
cont.
• The transfer function can be represented as a block diagram
• General block diagram
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Block Diagram of Open Loop
System
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DAE32103_Control System 11
Block Diagram of Closed Loop
System
21
Example 1
• Problem: Find the transfer function represented by
• Solution:
Taking the Laplace transform of both sides, assuming zero initial
conditions, we have
The transfer function, G(s) is
)()(2)(
trtcdt
tdc=+
)()(2)( sRsCssC =+
22
2
1
)(
)()(
+==
ssR
sCsG
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DAE32103_Control System 12
Example 2
• Problem: Use the result of Example 1 to find the response, c(t), to an input, r(t)=u(t), a unit step and assuming zero initial conditions.
• Solution:
Since r(t)=u(t), R(s)=1/s, hence
Expanding by partial fractions, we get
Finally, taking the inverse Laplace transform of each term yields
)2(
1)()()(
+==
sssGsRsC
2
2/12/1)(
+−=
sssC
23
)(2
1
2
1)(
2tuetc
t
−=
−
Example 3
• Problem: Find the transfer function, G(s)=C(s)/R(s), corresponding to the
differential equation
• Solution:
rdt
dr
dt
rdc
dt
dc
dt
cd
dt
cd34573
2
2
2
2
3
3
++=+++
24
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Example 4
• Problem: Find the differential equation corresponding to the transfer function,
• Solution:26
12)(
2++
+=
ss
ssG
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Example 5
• Problem: Find the ramp response for a system whose transfer function is,
• Solution:
)8)(4()(
++=
ss
ssG
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