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 Observer

uBAxx +=.

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

−=

+−=

−=−=+=

=+=

−=

+−−−+=−=

−=+−−=

−++=

=−−

−=

ex

LC- A 0BK BKA

e

x

BKeBK)x(Ax

exxxBKAxBAxx

CxBAxx

eLCAe

xLCLyBuxABuAxxxe

xKLxLCBKAx

xCLBxAx

BKAIx

xK

ˆ ;ˆ

y ;

)(

error) estimation of derivative (time ˆˆˆ

ˆˆ)(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

uy

yu

stt

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(AsIKProcedure Design

LCAsIBK(AsI∆(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 theperformance 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 modernadaptive 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 performanceof 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 systemparameters 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

xx

kk

kk

u

uu

xxkuxxkuxkuxkuxku mmm

×==

=

=

+−=+−=−=−=−==

+=

HHxBKx-Axx

k(x)uu

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 )(

.TT.TT

0

T

T232

22

21

2

1

321T

0

22

21

2

01

pjipijdtddtJ

xxxx

xn

xx

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

dtdxdJ

dtd

-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- 00 1

- 1-1 0

- 11- 0

minimized isindex eperformanc that theso find and 1Let

. - -

1 0 have weformmatrix In

feedback negative provide tonegativesign ;)( that so system controlfeedback 10

; 0 01 0

)( 10

0 01 0

222 12

1211

22 12

1211

2

21

21

22112.

21.

2211

2

1

2

1

kpppp

pppp

k

kkkk

xkxkxxx

xkxktu

tuxx

xx

dxd

-IPHPH

xHx x

Choose

BA

T

.

14

[ ]

[ ]

( )

[ ] 121 11-

DetDet ;2- 1-1 0

3;

minimum is when 2 ;2

)42(2)22(2

24211

22

211

1 1

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

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

22 ;1 ;

21

10

1

2min

222

22222

2

2

222

22

22

2212112212

1211

T

T2

22

112

2212

2221222212

2212211

1212

++=

+

=

==

=++−+

=

++=++

+=

++=

=

=

=

+===

−=−+−=−−

−=−−

λλλ

λH-λIH

Pxx

J

Jkk

kkkkdkdj

kkk

kkkJ

ppppppp

J

J

kkp

kpp

pkppkpppkp

pp

15

107.02/5.03/08.0

//, Assume;

//

is system optimalan ofy sensitivit Thesystem) (optimum 1.0 y Accordingl .2

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

222

22

2==

∆∆

==∆∆

=

=++

kkJJSkk

kkJJS

nss

optk

optk

n ξωξω

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 Krepresenting the required power amplifier.

51+s 1

1+s

2.5)(Section )/)(/(

)(ff LRsJbss

KsG++

=

u

K 1/s

x1θ

x2x3

Amplifier Field Motor

x1 = θ; x2 = dθ / dt; x3 = If

State variables

Vf

5/;1/ == ff LRJb

+

++=

++=

33

2323

1)(;)5)(1(

)(K

sK

KKsKsH

sssKsG

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%

51+s 1

1+s

rK

1/s

x1θ

x2x3

Amplifier Field Motor

k3

k2

k1

-

+

[ ] rxkxkxkrxkkku =−−−=+−−−= 332211321

19

00.369.3;62.10 are roots The240;76.0;12When

05.0;35.0;1

120 ;8

plane.- in theleft the tolocus thepull order toin j2-4sat zerosget will we20 and 8set weIf

1 ;

0)5)(1(

)(1)()(1

3

321

33

32

33

32

23

jsskkk

kkkkk

kks

bak

bk

kka

sssbasskk

sHsG

±−=−====

===

=+

=

±===

=+

=

=++++

+=+

ξ

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

00

)(5- - -1 1- 0 0 1 0

0 0 1

00

5- 0 01 1- 00 1 0

232

323

32

32

.

.

=++++++

+++

+

+=+=

=

+

=+=

kskkkskksksss

kkkkk

s

rkkkkkk

y

uk

xBuAxx

x

xBuAxx

21

3.003.69- and -10.62240at roots The

240 ;05.0 ;35.0 ;1

201 8

24at zeros place We20 ;8 ;1Let

1 ; ;0)5)(1(

)(1

321

33

32

1

33

322

3

jssk

kkkkkk

kkjs

bakk

bk

kka

sssbasskk

±===

====

==+

±−====

=+

==++++

+

22

P11.1:

( )

1/30 is index eperformanc theof value then the31, Select realizablenot is this,Physically minimum. is them, If

11)0(

yields 1 assuming of value theCalculate1. if stable is System )0()(

1

0

2)1(2

)1(

.

.

JkJk

kdtxeJ

kJkxetx

xkkxxx

kxuuxx

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 where03817A]-[SIDet by determined are roots system loop-closed The

Ax x8- 11-

10- 9-x ;

191 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(4122 ;

)384(4762

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

)-(2 )(1-

- )-(1is system theapplied,feedback with Then,

1]x 1[)( );( 11

2 1-

0 1

21

2

.

T

221211T

2

2222

2122

211

22122211121211

212

1

2

1

===−+=

=

===

∞=

−=++==

+−

+−−=

+−

−−=

+−

+−−=

−=−+−=+−−−−=+−−=+

+

==

−=+−=

+

=

ssss

Jk

Jkk

pppJ

kkkkp

kkkkp

kkkkp

kpkpkpkpkpkpkp

kkkk

kxxktutuxx

xx

dtd

Pxx

-IPHPH

xHxx

T

.