feedback control system

Feedback Control System

Dr.-Ing. Erwin Sitompul

1/2Erwin Sitompul Feedback Control System

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. Introduction2. Dynamic Models3. Dynamic Response4. A First Analysis of Feedback5. The Root-Locus Design Method6. The Frequency-Response Design Method

IDR 192,000

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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 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.


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.

Grade Policy


INTRODUCTION


Chapter 1


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 control2. Closed-loop control / Feedback control


Process

Input

Performance

Measurement

DisturbanceReference

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.

In feedback control, the system includes a sensor to measure the actual output and uses its feedback to influence the control process.

Open-loop vs. Feedback Control


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


DYNAMIC MODELS


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)


Dynamic ModelsApplying 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

( ) 1

( )

V s m

U s s b m

MATLAB (Matrix Laboratory) is the standard software used in control engineering:

In the end of this course, you are expected to be able to know how to use MATLAB for basic applications.


Dynamic ModelsWith the parameters:

1000 kg50 Ns/m500 N

mbu

( ) 1

( )

V s m

U s s b m

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


Dynamic Models

A Two-Mass System: Suspension Modelm1 : 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


Dynamic ModelsUsing 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:


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

Dynamic ModelsEliminating X(s) yields a transfer function:

( )( )

( )

Y sF s

R s

outputtransfer function

input


Dynamic Models

Bridged Tee Circuit

1 o1 i1 1

1 2

0V VV V

sCVR R

o 12 o i

2

( ) 0V V

sC V VR

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


Dynamic Models

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

o

i

1

2 3

V

V s

RL Circuit

v1

Further calculation and eliminating V1,


DYNAMIC RESPONSE


Chapter 3


Review of Laplace Transform

( )f t L

( )F s

( )G s 1L( )g t

Time domain Frequency domainProblem

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


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 L

3. 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 s

s s

2( ) 2 1( ) 3 tt t e L L L

Laplace Transform


Laplace Transform

Example: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 se es


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

( 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 11( ) 1( ) 1( )

3 2 6t tt 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


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



Example: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

WRONGSince

3 3 2 3 2( )

( 2) 2Y s

s s s s

1 2( ) ( ) 3 2 1( ) 3 2 1( )ty t Y s t e t L NOT convergentNO limit value



Example: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 12.63.4 (b)3.5 (c)3.6 (e)

Deadline: 10.05.2011, 7:30 am.

feedback control system

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