1

Chapter 11

Compensator Design

When the full state is not available for feedback, we utilize an observer. The observer design process is

described and the applicability of Ackermann’s formula is established. The state variable compensator is

obtained by connecting the full-state feedback law to the observer.

We consider optimal control system design and then describe the use of internal model design to achieve prescribed steady-state response to selected input

commands.

2

State Variable Compensator Employing Full-State Feedback in Series with a

Full State ObserveruBAxx

.

yu ˆˆˆ.

LBxAx

C

-K

C

System Model

Observer

xu

+

-

Control Law

y

x̂

Compensator

3

Compensator DesignIntegrated Full-State Feedback and Observer

LCBKx ˆ

LCBKA

ControlGain -K

ObserverGain L+

+

u y

LCBKA

x̂

4

e

x

LC- A 0

BK BKA

e

x

BKeBK)x(Ax

exxxBKAxBAxx

CxBAxx

eLCAe

xLCLyBuxABuAxxxe

xK

LxLCBKAx

xCLBxAx

BKAI

x

xK

ˆ ;ˆ

y ;

)(

error) estimation of derivative (time ˆˆˆ

ˆ

ˆ)(ˆ

lecture) previos thesee observer, the(From )ˆ(ˆˆ

stability retain the welaw controlfeedback theusingthat when

verify toneed Wefeedback. state-full From 0))((Det

) x(of placein controlfeedback in the )(ˆ estimate stateEmploy

LawFeedback )(ˆ)(

.

.

.

.

.

.

...

.

.

u

u

u

y

yu

s

tt

ttu

5

The goal is to verify that, with the value of u(t) we retain the stability of the closed-loop system and the observer. The

characteristic equation associated with the previous equation is

)(ˆ-)( using lawfeedback state-full theoobserver t eConnect th 3.

e.performancobserver achieve toroots theplace and plane halfleft in the

roots has 0))(Det(such that Determine 2.

criteria.design meet the oproperly t roots theplace and plane halfleft in the

roots has 0))Det(such that Determine 1.

stable. is system overall then the

observer, theofdesign the tosecond and lawfeedback state-full

torelated(first equation above theof partsboth of roots theif So

))(Det( ))Det(

ttu xK

LCAsIL

BK(AsIK

Procedure Design

LCAsIBK(AsIΔ(s)

6

The Performance of Feedback Control Systems

• Because control systems are dynamics, their performance is usually specified in terms of both the transient response which is the response that disappears with time and the steady-state response which exists a long time following any input signal initiation.

• Any physical system inherently suffers steady-state error in response to certain types of inputs. A system may have no steady-state error to a step input, but the same system may exhibit nonzero steady-state error to a ramp input.

• Whether a given system will exhibit steady-state error for a given type of input depends on the type of open-loop transfer function of the system.

7

The System Performance• Modern control theory assumes that the systems engineer can

specify quantitatively the required system performance. Then the performance index can be calculated or measured and used to evaluate the system’s performance. A quantitative measure of the performance of a system is necessary for the operation of modern adaptive control systems and the design of optimum systems.

• Whether the aim is to improve the design of a system or to design a control system, a performance index must be chosen and measured.

• A performance index is a quantitative measure of the performance of a system and is chosen so that emphasis is given to the important system specifications.

8

Optimal Control Systems

• A system is considered an optimum control system when the system parameters are adjusted so that the index reaches an extreme value, commonly a minimum value.

• A performance index, to be useful, must be a number that is always positive or zero. Then the best system is defined as the system that minimizes this index.

• A suitable performance index is the integral of the square of the error, ISE. The time T is chosen so that the integral approaches a steady-state value. You may choose T as the settling time Ts

T

dtte0

2 )( ISE

9

The Performance of a Control System in Terms of State Variables

• The performance of a control system may be represented by integral performance measures [Section 5.9]. The design of a system must be based on minimizing a performance index such as the integral of the squared error (ISE). Systems that are adjusted to provide a minimum performance index are called optimal control systems.

• The performance of a control system, written in terms of the state variables of a system, can be expressed in general as

• Where x equals the state vector, u equals the control vector, and tf equals the final time.

• We are interested in minimizing the error of the system; therefore when the desired state vector is represented as xd = 0, we are able to consider the error as identically equal to the value of the state vector. That is, we desire the system to be at equilibrium, x = xd= 0, and any deviation from equilibrium is considered an error.

ft

dttgJ0

),ux,(

10

Design of Optimal Systems using State Variable Feedback and Error-Squared Performance Indices: Consider the following control system in terms of x and u

ControlSystem

u1

u2

um

Control signals State variablesx1

x2

x3

elements. above theofaddition thefrom resultingmatrix theis ;

Kxu that sofunction linear a ofunction tfeedbak Limit the ....

..

.....................

..

...

;.. ;

as vector control thechoosemay We]. ; ;[

thereforeand x variablestate measured theoffunction some is that so controllerfeedback aSelect

:equation aldifferenti vector by the drepresente becan system The

n

2

1

mnm1

1n11

m

2

1

32222111

222111

.

nn

x

x

x

kk

kk

u

u

u

xxkuxxku

xkuxkuxku mmm

HHxBKx-Axx

k(x)u

u

BuAxx

.

.

11

Now Return to the Error-Squared Performance Index

] :used be willP symmetric[A xPxPxxx-xPxx ; xx

matrix x theof transpose theis X ; ... ...

,...,,,xx

operationmatrix use

squared, variablesstate theof sum theof integralan of in termsindex ePerformanc

operationmatrix the Utilize]. variablesstate [Two

variable]state [Single )(

.T

T.TT

0

T

T232

22

21

2

1

321T

0

22

21

2

01

pjipijdt

ddtJ

xxxx

xn

x

x

xxxx

dtxxJ

dttxJ

f

f

f

t

nn

t

t

12

To minimize the performance index J, we consider two equationsThe design steps are: first to determine the matrix P that satisfies the second

equation when H is known. Second minimize J by determining the minimum of the first equation by adjusting one or more unspecified system parameters.

1Equation of minimum thegdetermininby of value

theminimize and 2)(Equation Pmatrix theDetermine :CriteriaDesign The

2)(Equation

1)(Equation (0) (0)x

(0)Px(0)xPx)x(-

x-xPH)xP(Hx

(Hx)xPx(Hx)xPxPxxPxx

0

TT

TT

0

TTT

TT.

TT.

T

J

dtJ

dtdx

dJ

dt

d

-IPHPH

Pxxx

T

13

State Variable Feedback: State Variables x1 and x2

u1 1/s 1/s

11

x1x2

x1(0)/sx2(0)/s

1- 0

0 1

- 1-

1 0

- 1

1- 0

minimized isindex eperformanc that theso find and 1Let

. - -

1 0 have weformmatrix In

feedback negative provide tonegativesign ;)( that so system controlfeedback

1

0 ;

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1 0

)( 1

0

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222 12

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22 12

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2

21

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22112.

21.

2211

2

1

2

1

kpp

pp

pp

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k

kk

kk

xkxkxxx

xkxktu

tux

x

x

x

dx

d

-IPHPH

xHx x

Choose

BA

T

.

14

121 1

1- DetDet ;

2- 1-

1 0 3;

minimum is when 2 ;2

)42(2)22(2

2

4211

2

2

21

1

1 1

mequilibriu fromunit one displaced is stateeach Assume 1 1(0)x

index) eperformanc integral (The (0)(0)

2

2 ;

1 ;

2

1

1

0

1

2min

222

22222

2

2

222

22

22

2212112212

1211

T

T

2

22

112

2212

2221222212

2212211

1212

H-λIH

Pxx

J

Jkk

kkkk

dk

dj

k

kk

kk

kJ

ppppp

ppJ

J

k

kp

kpp

pkppkp

ppkp

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15

107.02/5.0

3/08.0

/

/, Assume;

/

/

is system optimalan ofy sensitivit The

system) (optimum 1.0 y Accordingl .2

form theof isequation sticcharacteri The system.order second a is This

222

22

2

kk

JJSkk

kk

JJS

nss

optk

optk

n

R(s)=0 Y(s)U(s)x2 x1

k2

k1

--

1/s 1/s

16

Compensated Control System

1 1/s 1/s

11

x1x2

u

x1(0)/sx2(0)/s

-k1= -1

-k2 = -2

17

Design Example: Automatic Test SystemA automatic test system uses a DC motor to move a set of test probes as shown in Figure 11.23 in the textbook. The system uses a DC motor with an encoded disk to

measure position and velocity. The parameters of the system are shown with K representing the required power amplifier.

5

1

s 1

1

s

2.5)(Section )/)(/(

)(ff LRsJbss

KsG

u

K 1/s

x1

x2x3

Amplifier Field Motor

x1 = ; x2 = d / dt; x3 = If

State variables

Vf

5/;1/ ff LRJb

33

2323

1)(;

)5)(1()(

Ks

K

KKsKsH

sss

KsG

18

State Variable FeedbackThe goal is to select the gains so that the response to a step command

has a settling time (with a 2%criterion) of less than 2 seconds andan overshoot of less than 4%

5

1

s 1

1

s

rK

1/s

x1

x2x3

Amplifier Field Motor

k3

k2

k1

-

+

rxkxkxkrxkkku 332211321

19

00.369.3;62.10 are roots The

240;76.0;12When

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120 ;8

plane.- in theleft the tolocus thepull order toin

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1 ;

0)5)(1(

)(1)()(1

3

321

33

32

33

32

23

jss

kkk

kkk

kk

kk

s

ba

kb

k

kka

sss

basskksHsG

20

To achieve an accurate output position, we let k1 = 1 and determine k, k2 and k3. The aim is to find the characteristic equation

056

)(5s -

1- 1s 0

0 1-

Det

0

0

)(5- - -

1 1- 0

0 1 0

0 0 1

0

0

5- 0 0

1 1- 0

0 1 0

232

323

32

32

.

.

kskkkskksksss

kkkkk

s

r

kkkkkk

y

u

k

xBuAxx

x

xBuAxx

21

3.003.69- and -10.62

240at roots The

240 ;05.0 ;35.0 ;1

201

8

24at zeros place We

20 ;8 ;1Let

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)5)(1(

)(1

321

33

32

1

33

322

3

jss

k

kkkk

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kk

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kb

k

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22

P11.1:

1/30 is index eperformanc theof value then the31, Select

realizablenot is this,Physically minimum. is them, If

1

1)0(

yields 1 assuming of value theCalculate

1. if stable is System )0()(

1

0

2)1(2

)1(

.

.

Jk

Jk

kdtxeJ

kJ

kxetx

xkkxxx

kxu

uxx

tk

tk

23

P11.3: An unstable robot system is described below by the vector differential equation. Design gain k so that the performance index is minimized.

unstable. is system The 2; and -19 where

03817A]-[SIDet by determined are roots system loop-closed The

Ax x8- 11-

10- 9-x ;

19

1 10,Select

feedback without unstable is system The 0. togoes , togoes As .1] [1When x(0)

12

12 (0) (0):be tocomputed isindex eperformanc The

)384(4

)362( ;

)384(4

122 ;

)384(4

762

1)2(22;0)1()23(;1)1(2)1(2

)-(2 )(1-

- )-(1

is system theapplied,feedback with Then,

1]x 1[)( );( 1

1

2 1-

0 1

21

2

.

T

221211T

2

2

222

2

122

2

11

22122211121211

212

1

2

1

ss

ss

Jk

Jk

kpppJ

kk

kkp

kk

kkp

kk

kkp

kpkpkpkpkpkpkp

kk

kk

kxxktutux

x

x

x

dt

d

Pxx

-IPHPH

xHxx

T

.