Transcript feedback control

Feedback Control System
Dr.-Ing. Erwin Sitompul
Textbook and Syllabus
Textbook:
Gene F. Franklin, J. David Powell, Abbas
Emami-Naeini, “Feedback Control of Dynamic
Systems”, 6th Edition, Pearson International
Edition.
Syllabus:
1.
2.
3.
4.
5.
6.
IDR 192,000
Introduction
Dynamic Models
Dynamic Response
A First Analysis of Feedback
The Root-Locus Design Method
The Frequency-Response Design Method
USD 112.50
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Grade Policy
Final Grade = 10% Homework + 20% Quizzes +
30% Midterm Exam + 40% Final Exam +
Extra Points
 Homeworks will be given in fairly regular basis. The average
of homework grades contributes 10% of final grade.
 Homeworks are to be written on A4 papers, otherwise they
will not be graded.
 Homeworks must be submitted on time. If you submit late,
< 10 min.
 No penalty
10 – 60 min.  –20 points
> 60 min.
 –40 points
 There will be 3 quizzes. Only the best 2 will be counted.
The average of quiz grades contributes 20% of final grade.
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Grade Policy
 Midterm and final exam schedule will be announced in time.
 Make up of quizzes and exams will be held one week after
the schedule of the respective quizzes and exams, at the
latest.
 The score of a make up quiz or exam can be multiplied by 0.9
(the maximum score for a make up is 90).
 Extra points will be given every time you solve a problem in
front of the class. You will earn 1 or 2 points.
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Chapter 1
INTRODUCTION
Feedback Control System
Dr.-Ing. Erwin Sitompul
Introduction
• Control is a series of actions directed for making a system
variable adheres to a reference value (can be either
constant or variable).
• The reference value when performing control is the desired
output variable.
• Process, as it is used and understood by control engineers,
means the component to be controlled.
Fundamental structures of control are classified based on the
information used along the control process:
1. Open-loop control / Feedforward control
2. Closed-loop control / Feedback control
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Process
Reference
Measurement
Disturbance
noise
Performance
Input
Measurement
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Open-loop vs. Feedback Control
The difference:
 In open-loop control, the system does not measure the
actual output and there is no correction to make the actual
output to be conformed with the reference value.
 In feedback control, the system includes a sensor to
measure the actual output and uses its feedback to
influence the control process.
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Examples
Open-loop control
Feedback control
Example: an electric toaster,
a standard gas stove.
Example: automated filling
up system, magic jar, etc.
• The controller is constructed
based on knowledge or
experience.
• The process output is not
used in control computation.
• The output is fed back for
control computation.
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Plus-Minus of Open-loop Control
+ Generally simpler than closed-loop control
+ Does not require sensor to measure the output
+ Does not, of itself, introduce stability problem
– Has lower performance to match the desired output
compared to closed-loop control
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Plus-Minus of Feedback Control
–
–
–
–
More complex than open-loop control
May have steady-state error
Depends on the accuracy of the sensor
May have stability problem
+ Process controlled by well designed feedback control can
respond to unforeseen events, such as: disturbance,
change of process due to aging, wear, etc.
+ Eliminates the need of human to adjust the control variable
 reduce human workload
+ Gives much better performance than what is possibly given
by open loop control: ability to meet transient response
objectives and steady-state error objectives
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Chapter 2
DYNAMIC MODELS
Feedback Control System
Dr.-Ing. Erwin Sitompul
Dynamic Models
A Simple System: Cruise Control Model
Write the equations of motion for the speed and forward
motion of the car shown below, assuming that the engine
imparts a force u, and results the car velocity v, as shown.
Using the Laplace transform, find the transfer function
between the input u and the output v.
u
(Force)
x (Position)
v (Velocity)
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Dynamic Models
Applying the Newton’s Law for
translational motion yields:
u  bv  ma
u  bx  mx
u  bv  mv
b
u
v v 
m
m
MATLAB (Matrix Laboratory) is
the standard software used in
control engineering:
V (s  b m)  U m
V (s)
1m

U (s) s  b m
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In the end of this course, you are
expected to be able to know how to
use MATLAB for basic applications.
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Dynamic Models
m  1000 kg
b  50 Ns/m
u  500 N
In MATLAB windows:
Response of the car velocity v
to a step-shaped force u:
10
9
8
7
Amplitude
With the parameters:
6
5
4
3
2
1
0
0
V (s)
1m

U (s) s  b m
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20
40
60
80
100
120
Time (sec)
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Dynamic Models
A Two-Mass System: Suspension Model
m1 : mass of the wheel
m2 : mass of the car
x,y : displacements from equilibrium
r : distance to road surface
Equation for m1:
ks ( x  y)  b( x  y)  kw ( x  r )  m1x
Equation for m2:
ks ( y  x)  b( y  x)  m2 y
Rearranging:
ks
kw
kw
b
x  ( x  y)  ( x  y) 
x
r
m1
m1
m1
m1
ks
b
y
( y  x) 
( y  x)  0
m2
m2
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Dynamic Models
Using the Laplace transform:
L
x (t ) 
 X (s)
d
L
x (t ) 
 sX ( s )
dt
to transfer from time domain to frequency domain yields:
b
s X ( s)  s  X ( s)  Y ( s)  
m1
ks
kw
kw
 X ( s )  Y ( s )   X ( s )  R( s )
m1
m1
m1
2
ks
b
s Y ( s) 
s Y ( s )  X ( s )  
Y ( s )  X ( s )   0
m2
m2
2
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Dynamic Models
Eliminating X(s) yields a transfer function:
kw b 
ks 
s 
m1m2 
b
Y ( s)

R( s )
 b
kw ks
b  3  ks k kw  2  kw b 
4
s  
 s 
s  
s
m1m2
 m1 m2 
 m1 m2 m1 
 m1m2 
Y (s)
output
 F (s) 
 transfer function
R( s)
input
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Dynamic Models
Bridged Tee Circuit
v1
v  Ri
V1  Vi V1  Vo

 sC1V1  0
R1
R2
v  L di
dt
i  C dv
dt
V (s)  R  I (s) V (s)  sL  I (s) I (s)  sC V (s)
Resistor
Inductor Capacitor
Vo  V1
 sC2 (Vo  Vi )  0
R2
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Dynamic Models
RL Circuit
v1
V1  Vi V1 V1  Vo
 
0
1
1
s
V1  Vo Vo  0

 V1  Vo ( s  1)
s
1
Further calculation and
eliminating V1,
1  Vo

V1  2     Vi
s s

1  Vo

Vo  s  1  2     Vi
s s

Vo  2s  3  Vi
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Vo
1

Vi 2s  3
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Chapter 3
DYNAMIC RESPONSE
Feedback Control System
Dr.-Ing. Erwin Sitompul
Review of Laplace Transform
Time domain
Problem
f (t )
Frequency domain
L

F (s)
difficult
operations
easy
operations
1
L
g (t )

G ( s)
Solution
L  f (t )  F (s) 


f (t )e st dt
s    j

1
L
F ( s) 
f (t ) 
0   j
c
1
2 j  
c
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st
F ( s )e ds
j
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Properties of Laplace Transform
1. Superposition
L   f1 (t )    f2 (t )    F1 (s)    F2 (s)
2. Time delay
L  f (t  )  u(t  )  es  F (s)
3. Time scaling
1 s
L  f (at )   F  
a a
4. Shift in Frequency
L e  at  f (t )  u (t )  F ( s  a)


5. Differentiation in Time
L  f (t )  s  F ( s )  f (0 )
L  f (t )  s 2  F ( s )  s  f (0 )  f (0  )
L  f n (t )  s n  F ( s )  s n 1  f (0 ) 
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 s  f n  2 (0 )  f n 1 (0  )
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Properties of Laplace Transform
6. Integration in Time
t
1
L  f (t )dt   F ( s )
0
s
7. Differentiation in Frequency
dF ( s )
L t  f (t )  
ds
8. Convolution
F1 ( s )  F2 ( s )  L  f1 (t )  f 2 (t )


F1 ( s )  F2 ( s )  2 j  L  f1 (t )  f 2 (t )

f1 (t )  f 2 (t ) 
 f ()  f (t  )d 
1
2
0
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Table of Laplace Transform
 (t )
1
unit impulse
t
1(t )
1
unit step
t
r (t )
1
1
unit ramp
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t
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Laplace Transform
Example:
2t
Obtain the Laplace transform of f (t )   (t )  2 1(t )  3e , t  0.
F ( s )  L  (t )  L 2 1(t )  L 3e 2t 
 L  (t )  2  L 1(t )  3  L e 2t 
1
1
 1 2   3
s
s2
s2  s  4

s ( s  2)
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Laplace Transform
Example:
Find the Laplace transform of the function shown below.
g (t )
4
0
1
2
3
4
t
g (t )  4 1(t  2)  4 1(t  3)
G ( s)  L  g (t )
e2 s
e3s
 4
 4
s
s
4 2 s 3s
 (e  e )
s
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Inverse Laplace Transform
The steps are:
1. Decompose F(s) into simple terms using partial-fraction
expansion.
2. Find the inverse of each term by using the table of Laplace
transform.
Example:
Find y(t) for Y ( s) 
( s  2)( s  4)
.
s( s  1)( s  3)
c3
c1
c2
Y (s)  

s s 1 s  3
c1 ( s  1)( s  3)  c2  s( s  3)  c3  s( s  1)

s( s  1)( s  3)
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Inverse Laplace Transform
(c1  c2  c3 )s 2  (4c1  3c2  c3 )s  3c1
1s 2  6s  8

Y ( s) 
s( s  1)( s  3)
s(s  1)(s  3)
Comparing the c1  c2  c3  1
coefficients
4c1  3c2  c3  6
3c1  8
8
3
1
c1  , c2   , c3  
3
2
6
81 3 1  1 1 
Y ( s)     
 

3  s  2  s 1  6  s  3 
y (t )  L1 Y ( s )
8 1  1  3 1  1  1 1  1 
 L   L 
 L 

3
s 2
 s  1 6
 s  3
8
3 t
1 3t
 1(t )  e 1(t )  e 1(t )
3
2
6
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Initial and Final Value Theorem
lim y (t )  lim s  Y ( s )
s 
t 0
lim y (t )  lim s  Y ( s )
t 
Only applicable to stable
system, i.e. a system with
convergent step response
s 0
Example:
Find the final value of the system corresponding to
3( s  2)
Y ( s) 
s( s 2  2s  10)
y ()  lim y (t )  lim s Y ( s )
t 
s 0
3( s  2)
3 2
y()  lim s 

 0.6
2
s 0
s( s  2s  10)
10
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Initial and Final Value Theorem
Example:
Find the final value of the system corresponding to
3
Y (s) 
s( s  2)
3
3
y ()  lim y (t )  lim s Y ( s)  lim s 

t 
s 0
s 0
s( s  2) 2
WRONG
Since
3
3 2 3 2
Y ( s) 


s( s  2)
s
s2
y(t )  L1 Y (s)   3 2 1(t )  3 2 e2t 1(t ) NOT convergent
NO limit value
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Initial and Final Value Theorem
Example:
Find the final value of
2
Y ( s)  2
s 4
2
y ()  lim y (t )  lim s Y ( s)  lim s  2
0
t 
s 0
s 0
s 4
WRONG
Since
2
Y (s)  2
 y (t )  sin 2t
s 4
Erwin Sitompul
periodic signal
NOT convergent
NO limit value
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Homework 1
 2.6
 3.4 (b)
 3.5 (c)
 3.6 (e)
 Deadline: 10.05.2011, 7:30 am.
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