Dr.-Ing. Erwin Sitompul President University Lecture 1 Feedback Control Systems President UniversityErwin SitompulFCS 1/1

Dr.-Ing. Erwin SitompulPresident University

Lecture 1

Feedback Control Systems

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President University Erwin Sitompul FCS 1/1


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

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Grade Policy Final Grade = 10% Homeworks + 20% Quizzes +

30% Midterm Exam + 40% Final Exam + Extra Points

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Grade Policy Feedback Control Systems


Chapter 1

Introduction



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


Process

Input

Performance

Measurement

Disturbance

Reference

Measurement noise



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.


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.



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



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 2

Dynamic Models



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


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.


( ) 1

( )

V s m

U s s b m


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


( ) 1

( )

V s m

U s s b m


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



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:



A Two-Mass System: Suspension ModelEliminating X(s) yields a transfer function:

( )( )

( )

Y sF s

R s

outputtransfer function

input


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


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


o 12 o i

2

( ) 0V V

sC V VR


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,


o

i

1

2 3

V

V s


Chapter 3

Dynamic Response



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


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



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



t

( )t1

t

( )r t

1

1

unit impulse

unit step

unit ramp

Table of Laplace Transform

t

1( )t1



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


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



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



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


●Learn also the faster “Cover Up Method”


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



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



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



Homework 12.6 3.4 (b)3.5 (c)3.6 (e) all from FPE (5th Ed.)

Deadline: 11.09.2012, 07:30.


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Dr.-Ing. Erwin Sitompul President University Lecture 1 Feedback Control Systems President UniversityErwin SitompulFCS 1/1