dr.-ing. erwin sitompul president university lecture 1 feedback control systems president...
TRANSCRIPT
Dr.-Ing. Erwin SitompulPresident University
Lecture 1
Feedback Control Systems
http://zitompul.wordpress.com
President University Erwin Sitompul FCS 1/1
President University Erwin Sitompul FCS 1/2
Textbook and SyllabusTextbook:Gene F. Franklin, J. David Powell, Abbas Emami-Naeini, “Feedback Control of Dynamic Systems”, 6th Edition, Pearson International Edition.
Syllabus:1. Introduction2. Dynamic Models3. Dynamic Response4. A First Analysis of Feedback5. The Root-Locus Design Method6. The Frequency-Response Design Method
IDR 192,000
USD 112.50
Feedback Control Systems
President University Erwin Sitompul FCS 1/3
Grade Policy Final Grade = 10% Homeworks + 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 penalty10 – 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.
Feedback Control Systems
President University Erwin Sitompul FCS 1/4
Midterm and final exam schedule will be announced in time.Make up of quizzes and exams must be held within one week
after the schedule of the respective quizzes and exams. In order to maintain the integrity, the score of a make up quiz
or exam can be multiplied by 0.9 (i.e., the maximum score for a make up will be 90).
Grade Policy
• Heading of Homework Papers (Required)
Feedback Control Systems
President University Erwin Sitompul FCS 1/5
Extra points will be given every time you solve a problem in front of the class or answer a question. You will earn 1 or 2 points.
Lecture slides can be copied during class session. The updated version will be available on the lecture homepage around 2 days after class schedule. Please check regularly.
http://zitompul.wordpress.com
Grade Policy Feedback Control Systems
President University Erwin Sitompul FCS 1/6
Chapter 1
Introduction
Feedback Control Systems
President University Erwin Sitompul FCS 1/7
IntroductionControl 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 control2. Closed-loop control / Feedback control
Chapter 1 Introduction
President University Erwin Sitompul FCS 1/8
Process
Input
Performance
Measurement
Disturbance
Reference
Measurement noise
Chapter 1 Introduction
President University Erwin Sitompul FCS 1/9
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.
Open-loop vs. Feedback ControlChapter 1 Introduction
In feedback control, the system includes a sensor to measure the actual output and uses its feedback to influence the control process.
President University Erwin Sitompul FCS 1/10
Examples
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.
Open-loop control Feedback control
Example: an electric toaster, a standard gas stove.
Example: automated filling-up system, magic jar, etc.
Chapter 1 Introduction
President University Erwin Sitompul FCS 1/11
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
Chapter 1 Introduction
President University Erwin Sitompul FCS 1/12
Plus-Minus of Feedback Control
+ 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
– More complex than open-loop control– May have steady-state error– Depends on the accuracy of the sensor– May have stability problem
Chapter 1 Introduction
President University Erwin Sitompul FCS 1/13
Chapter 2
Dynamic Models
Feedback Control Systems
President University Erwin Sitompul FCS 1/14
A Simple System: Cruise Control ModelWrite 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)
Chapter 2 Dynamic Models
President University Erwin Sitompul FCS 1/15
A Simple System: Cruise Control ModelApplying the Newton’s Law for translational motion yields:
u bv ma
u bx mx
b uv vm m
u bv mv
( )V s b m U m
MATLAB (Matrix Laboratory) is the standard software used in control engineering:
By the end of this course, you are expected to be able to use MATLAB for basic applications.
Chapter 2 Dynamic Models
( ) 1
( )
V s m
U s s b m
President University Erwin Sitompul FCS 1/16
A Simple System: Cruise Control ModelWith the parameters:
1000 kg50 Ns/m500 N
mbu
In MATLAB windows:
Response of the car velocity v to a step-shaped force u:
Time (sec)A
mp
litu
de
0 20 40 60 80 100 1200
1
2
3
4
5
6
7
8
9
10
Chapter 2 Dynamic Models
( ) 1
( )
V s m
U s s b m
President University Erwin Sitompul FCS 1/17
A Two-Mass System: Suspension Model
m1 : mass of the wheelm2 : mass of the carx,y : displacements from equilibriumr : distance to road surface
s w 1( ) ( ) ( )k x y b x y k x r m x Equation for m1:
Equation for m2:
s 2( ) ( )k y x b y x m y
Rearranging:s w w
1 1 1 1
( ) ( )k k kb
x x y x y x rm m m m
s
2 2
( ) ( ) 0kb
y y x y xm m
Chapter 2 Dynamic Models
President University Erwin Sitompul FCS 1/18
A Two-Mass System: Suspension ModelUsing the Laplace transform:
( ) ( )
( ) ( )
x t X sdx t sX s
dt
L
L
2
1
s w w
1 1 1
( ) ( ) ( )
( ) ( ) ( ) ( )
bs X s s X s Y s
mk k k
X s Y s X s R sm m m
2 s
2 2
( ) ( ) ( ) ( ) ( ) 0kb
s Y s s Y s X s Y s X sm m
to transfer from time domain to frequency domain yields:
Chapter 2 Dynamic Models
President University Erwin Sitompul FCS 1/19
A Two-Mass System: Suspension ModelEliminating X(s) yields a transfer function:
( )( )
( )
Y sF s
R s
outputtransfer function
input
Chapter 2 Dynamic Models
w s
1 2
4 3 2s w w w s
1 2 1 2 1 1 2 1 2
( )
( )
k b ks
m m bY s
R s k k k b k kb b ks s s s
m m m m m mm mm
President University Erwin Sitompul FCS 1/20
Bridged Tee Circuit
1 o1 i1 1
1 2
0V VV V
sCVR R
v1
Resistor Inductor Capacitor
dvi Cdt
v Ri div Ldt
( ) ( )V s R I s ( ) ( )V s sL I s ( ) ( )I s sC V s
Chapter 2 Dynamic Models
o 12 o i
2
( ) 0V V
sC V VR
President University Erwin Sitompul FCS 1/21
RL Circuit
1 o1 i 1 01 1
V VV V V
s
1 o o1 o
0( 1)
1
V V VV V s
s
o1 i
12
VV V
s s
oo i
11 2
VV s V
s s
o i2 3V s V
v1
Further calculation and eliminating V1,
Chapter 2 Dynamic Models
o
i
1
2 3
V
V s
President University Erwin Sitompul FCS 1/22
Chapter 3
Dynamic Response
Feedback Control Systems
President University Erwin Sitompul FCS 1/23
Review of Laplace Transform
( )f t L
( )F s
( )G s 1L( )g t
Time domain Frequency domain
Problem
Solution
easy operations
difficult operations
0
( ) ( ) ( ) stf t F s f t e dt
L
1 1( ) ( ) ( )
2
c
c
jst
j
F s f t F s e dsj
L
s j
Chapter 3 Dynamic Response
President University Erwin Sitompul FCS 1/24
Properties of Laplace Transform1. Superposition
1 2 1 2( ) ( ) ( ) ( )f t f t F s F s L
2. Time delay
( ) ( ) ( )sf t u t e F s L3. Time scaling
1( )
sf at F
a a
L
4. Shift in Frequency
( ) ( ) ( )ate f t u t F s a L
5. Differentiation in Time
2
1 2 1
( ) ( ) (0 )
( ) ( ) (0 ) (0 )
( ) ( ) (0 ) (0 ) (0 )n n n n n
f t s F s f
f t s F s s f f
f t s F s s f s f f
LL
L
Chapter 3 Dynamic Response
President University Erwin Sitompul FCS 1/25
Properties of Laplace Transform6. Integration in Time
0
1( ) ( )
tf t dt F s
s L
7. Differentiation in Frequency
( )( )
dF st f t
ds L
8. Convolution
1 2 1 2
1 2 1 2
( ) ( ) ( ) ( )
( ) ( ) 2 ( ) ( )
F s F s f t f t
F s F s j f t f t
L
L
1 2 1 2
0
( ) ( ) ( ) ( )f t f t f f t d
Chapter 3 Dynamic Response
President University Erwin Sitompul FCS 1/26
t
( )t1
t
( )r t
1
1
unit impulse
unit step
unit ramp
Table of Laplace Transform
t
1( )t1
Chapter 3 Dynamic Response
President University Erwin Sitompul FCS 1/27
Example:Obtain the Laplace transform of 2( ) ( ) 2 1( ) 3 , 0.tf t t t e t
2( ) ( ) 2 1( ) 3 tF s t t e L L L
1 11 2 3
2s s
2 4
( )( 2)
s sF s
s s
2( ) 2 1( ) 3 tt t e L L L
Laplace TransformChapter 3 Dynamic Response
President University Erwin Sitompul FCS 1/28
Laplace TransformExample:Find the Laplace transform of the function shown below.
t
( )g t4
0 1 2 3 4
( ) 4 1( 2) 4 1( 3)g t t t
2 3
( ) ( )
4 4s s
G s g t
e e
s s
L
2 34( ) ( )s sG s e e
s
Chapter 3 Dynamic Response
President University Erwin Sitompul FCS 1/29
Inverse Laplace TransformThe 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 ( 2)( 4)
( ) .( 1)( 3)
s sY s
s s s
31 2( )1 3
cc cY s
s s s
1 2 3( 1)( 3) ( 3) ( 1)
( 1)( 3)
c s s c s s c s s
s s s
Chapter 3 Dynamic Response
President University Erwin Sitompul FCS 1/30
Inverse Laplace Transform
1 2 3
1 2 3
1
14 3 6
3 8
c c cc c c
c
1
8,
3c 2
3,
2c 3
1
6c
8 1 3 1 1 1( )
3 2 1 6 3Y s
s s s
1
1 1 1
( ) ( )
8 1 3 1 1 1
3 2 1 6 3
y t Y s
s s s
L
L L L
38 3 1( ) 1( ) 1( ) 1( )
3 2 6t ty t t e t e t
1 12 32 32
1 (4 3( )( )
(
3
1)( 3)
)s s cY s
s s
c c cc c c
s
2
( 1)( 3)
6 81s s
s s s
Comparing the coefficients
Chapter 3 Dynamic Response
●Learn also the faster “Cover Up Method”
President University Erwin Sitompul FCS 1/31
Initial and Final Value Theorem
0lim ( ) lim ( )
sty t s Y s
0lim ( ) lim ( )t sy t s Y s
0( ) lim ( ) lim ( )
t sy y t s Y s
Only applicable to stable system, i.e. a system with convergent step response
Example:Find the final value of the system corresponding to
2
3( 2)( )
( 2 10)
sY s
s s s
20
3( 2)( ) lim
( 2 10)s
sy s
s s s
3 2
0.610
Chapter 3 Dynamic Response
President University Erwin Sitompul FCS 1/32
Initial and Final Value TheoremExample:Find the final value of the system corresponding to
0 0
3( ) lim ( ) lim ( ) lim
( 2)t s sy y t s Y s s
s s
3( )
( 2)Y s
s s
3
2
WRONG
Since 3 3 2 3 2
( )( 2) 2
Y ss s s s
1 2( ) ( ) 3 2 1( ) 3 2 1( )ty t Y s t e t L NOT convergentNO limit value
Chapter 3 Dynamic Response
President University Erwin Sitompul FCS 1/33
Initial and Final Value TheoremExample:Find the final value of
2
2( )
4Y s
s
20 0
2( ) lim ( ) lim ( ) lim
4t s sy y t s Y s s
s
0
WRONG
Since
2
2( ) ( ) sin 2
4Y s y t t
s
periodic signalNOT convergentNO limit value
Chapter 3 Dynamic Response
President University Erwin Sitompul FCS 1/34
Homework 12.6 3.4 (b)3.5 (c)3.6 (e) all from FPE (5th Ed.)
Deadline: 11.09.2012, 07:30.
Chapter 3 Dynamic Response