1Pole Placement TechniquesStability improvement by state Feedback: The main goal of a feedback design is

to stabilize a system, if it is initially unstable or to improve its stability if transientphenomena dont die out sufficiently fast.

Consider single input single out put system with n-th order described byBu(t)AX(t)(t)X (1)

Let us assume that all the state variables are measured accurately, and then it ispossible to implement a linear control law of the form

KX(t)u(t) (2)where,

n21 k..........k[kK ] is (1xn) feedback gain matrix.Substitute Eq (2) in Eq (1) to obtain closed loop system model as

BKX(t)AX(t)(t)X BK]X(t)[A (3)

The characteristic Eqn of closed loop system will be0BK)(ASI (4)

When Eq (4) is evaluated, this yields an nth order polynomial in S containing the n-gainsK1, K2,Kn. The control Law design then consists of picking the gains so that theroots of Eq (4) are in desired locations.

In the next section it can be seen that if the state model is in controllable canonicalform, then the roots of characteristic Eqn (4) can be chosen arbitrarily. If all the givenvalues of (A-BK) are placed in left- half of s-plane, the closed loop system isasymptotically stable. And the state X(t) decays to Zero irrespective of initial state X(0)-the initial perturbation in the state . The system state can be maintained Zero value inspite of disturbances that act upon the system. System with this property are calledREGULATOR SYSTEMS. Control configuration of such system as shown

2Algorithm for Design:a) Let the desired Locations of closed loop Poles be n ........, 21b) The correspondry cha Eqn will be 0S......SS n21

0.......SSS n2n

21n

1n (1)

c) Let the controller gain matrix beK=[K1 K2----------Kn].

d) n2n21n1n ......SSSBK)(ASI e) Corresponding equation is

0......SSS n2n

21n

1n -------(2)

f) Comparing the co-efficient of like powers of s form Eqn (1) & (2) K1,K2,--------Kncan be Evaluated and this completes design of state back controller K.Example: Design a controller K for the state model

u(t)10X(t)

0010

(t)X

Such that the system shall have damping ratio as 0.707 & settling time as 1 sec.Solution:

a) Converting the desired specifications as locations of closed loop poles at 44 jb) The corresponding cha Eqn will be

(S+4+J4) (S+4-J4)=00328SS2 (1)

c) Let the controller be of the formK=[K1 K2].

d) From given state model

21 KK

000010

BKA

21 KK

10

122 KSKSBK)(ASI

e) The corresponding characteristic equation is0KSKS 12

2 (2)Comparing the co-efficient of like powers of S from Equ (1) and (2)

K2=8, K1=32.Hence controller K=[32 8].Controller gain matrix K=[32 8] in feed back will have desired specifications.Necessary and sufficient conditions:The sufficient conditions for arbitrarily pole-placement is, the system must be completelycontrollable.

3Consider the state modelBu(t)AX(t)(t)X (1)

Y(t)=CX(t) (2)The above model can be converted into controller canonical form through the

transformationPX(t)(t)X (3)

Where P is nxn non singular transformable matrix, and assumed as

nnn2n1

2n2221

1n1211

PPP

PPPPPP

P

n

2

1

PPP

; ini2i1i PPPP Taking the derivative of Eqn (3) and substituting for (t)X from Equ(1)

u(t)BXA(t)X (4)Where 1PAPA , PBB (5)

n1nn ..........01......0000......10

(6)

10.0

PBB (7)

From Equ.(3),xPxP.........xPxPx 1n1n2121111

Taking the derivativeBu(t)][Ax(t)P(t)xPx 111

Bu(t)PAx(t)P 11 (8)From Equ(4)

0xx 2 (9)Comparing (8) and (9), 0BP1 & Ax(t)Px 12 (10)Taking again derivative of Equ(10)

4Bu(t)]A[AX(t)Px 1 ABPX(t)AP 1

21 and from Equ (4)

X(t)APx0xx 2133 & 0ABP1 Similarly 0BAP&X(t)APx 2n1

1n1n

Taking again derivative1BAP 1n1

Thus,

XAP.AXPXP

X

1n1

1

1

X

AP.AP

P

1n1

1

1

(11)

With 0BA....PBAPABPBP 2n12

111 & 1BAP 1n1 From Equ (4)

BAPBAP

.ABPBP

100.00

PBB

1n1

2n1

1

1

1000B]..A....A|BA|AB|BP 1n2n21 Defining BA|....|BA|BA|BQ 1n22c 1000QP C1 Post-multiplying both sides with Qc-1 1c1 Q1000P Under this transformation, the control law will be (t)XKKX(t)u Where, n7211 K....KKKKPK Substituting (t)XKu(t) in Equation (4)

(t)XKBA(t)X

(t)X

K......KK....01......0000......10

n121n1n

(13)

5Hence the corresponding characteristic poly will be1n

1nn1

n K.......)SK(SKBA(SI (14)and characteristic equation will be

0K.......)SK()SK(S 1n2n

n21n

n1n (15)

Where n21 K,.....K,K can be chosen real numbers arbitrarily.

Let the desired characteristic equation be0a.......SaS n

1n1

n (16)Equating the coefficients of like powers of S from equation (15) and (16)

11nn11 aKKa 221n1n22 aKKa

. (17)nn11nn aKKa

Hence the controller K will be]aa[aK 111n1nnn (18)

Transferring to original system]Paa[aPKK 111n1nnn

The necessary condition is the system should be completely controllable. If the system isnot completely controllable then there are certain states that cannot be controlled bystable feedback.

Thus the necessary and sufficient condition for arbitrarily pole placement is thesystem should be completely controllable in s-plane.Steps to design the controller K:

1) For given state model, test the controllability of the system by determining therank of controllable matrix Qc.

2) If it is controllable , transfer the model into controllable canonical form throughsimilarly transformation matrix P ; where P can be obtained as.

a) P1=[00---------0 1]Qc-1P2=P1AP3=P1A2Pn=P1An-1

b)

1n1

1

1

APAP

PP

3) Compute BA & as

PBBPAPA 1

4) Assume the controller K

6 n21 K....KKK andDetermine

ni1n K-....K......1.01000..010

KBA

5) Determine the Cha equation as0K.......)SK(SKBA(SI n1

1n1n

n

6) Determine the desired cha equation as0a.......SaSaS n

2n1n1

n 7) Determine the elements of gain matrix nKKKK ,.....,, 21as nn1 aK

1n1n2 aK 11n aK and hence n21 K....KKK

8) The feedback controller K for original system isPKK

This completes the design procedureDesign of controller by Ackermanns formula:We have

1n1

1

1

APAP

PP & 01][00QP 1c1

]Paa[aK 111n1nnn

01][00A)(Q

01]A[00)(Q01][00)(Q

]aa[a1n1

c

1c

1c

111n1nnn

n)In2n221n111c (a......)A(a)A(a01]Q[00 (1)Characteristics poly of A matrix is

n2n

21n

1n .......SSSASI (2)

According to Cayley-Hammilton theorem,0I.......AA n

1n1

n I...........AA n

1n1

n --------(3)From equation (1) and (3),

7(A)01]Q[00K 1c -------------(4)

WhereIa.......AaAaAaA(A) n

3n3

2n2

1n1

n (5)& B.......A|BA|AB|BQ 1n2c (6)Design procedure:

1. For given model, determine Qc and test the controllability. If controllable,2. Compute the desired characteristics equation as

0a.........SaS n1n

1n

and determine coefficients naaa ,, 213. Compute the characteristics poly of system matrix A as

n2n

21n

1n a.......AaAaA(A)

4. Determine 1cQ and then controller k will be K=[0 0 .0 1] .1cQ )(AThis completes the design procedure.

Ex. Design controller k which places the closed loop poles at 44 j for systemu(t)

10

X(t)0110

(A)X

using Acermanns formula.

Answer: Check for controllability

I

0110

AB|BQc Its controllable as f(Qc)=2.

0110

Q 1c

II. Desired characteristics equation is (S+4+j4)(S+4-j4)

32.a8,a0328SS

21

2

III.

80032

IaAaA(A) 212

IV. 83280032

0110

10K

State regulator design: The state Regulator system is a system whose non zeroinitial state can be transferred to zero initial state by the application of the input u(t).Design of such system requires a control lawU(t)=-KX(t) which transfers the non-zero initial state to zero initial state (Rejection ofdisturbances) by properly designing controller K.The following steps give design of such controller. For a system given by

8]......KK[KKKX(t)whereu(t)

CX(t)withY(t)Bu(t)Ax(t)(t)X

n21

From above relations, the closed loop system isBK]X(t)[A(t)X (1)

1. From the characteristics poly of matrix An

2n2

1n1

n .......SSSASI Determine n ,......., 21

2. Determine the transformation matrix P, which transfers the given state model intocontrollable caunomical form.

P=

1n1

1

1

APAP

P; 1P [0..001 ] 1cQ

B.......A|BA|AB|BQ 1n2c 3. Using the desired location of closed loop poles (desired Egen values) obtain the

characteristics polyn

2n2

1n1

nn21 .....aSaSaS)).....(S)(S(S

determine naaa ,......., 214. The required state feed back controller K is P.....aaaK 111n1nnn

Example: A regulator system has the plant

u(t)100

X(t)6116

100010

(t)X

X(t)001Y Design a state feed back controller which place the closed loop poles at 464.32 j and-5. Give the block diagram of the control configurationSolution: By observing the structure of A and B matrices, the model is in controllableconmical form. Hence controller K can be designed.

1. 1711S6SS6S116

1S001S

ASI 23

Comparing with 322

13 SSS , we obtain 1711,6, 321

92. Desired characteristics poly is(S+2+j3.464)(S+-j3.464)(S+5)

6020S9SS 23 Comparing with 32

21

3 asasas we obtain1a =9, 2a =20, 3a =60

3. Hence the required controller K isK=[a3-3 a2-2 a1-1]

= [ K1 K2 K3]= [43 9 3]

Block Diagram:

Design of state observer:Design of observer: The device that estimates the state variables with measurement ofoutput along with input u(t) is called state observer.If all the state variables are estimated, it is called Full-order state observer. If only fewstatevariable are estimated (which cant be measured accurately), the observer is calledReduced-order state observer.Design of Full-order observer: In this case all the state variables are to be estimated.

a) Open loop observer: Here it is assumed that all the state variables are estimatedcorrectly and initial state are also estimated correctly.i.e For State model

Cx(t);Y(t)Bu(t)Ax(t)(t)X

initial state X(0) are known and)(tX is estimated states and)0(X is estimated initial states the resulting observer in open loop observer.

K1=43K2=9K3=3

10

Draw backs of open loop observer.1. Precise Lack of information of x(0)2. If )0(X X(0), the estimated state )(tX obtain from open loop estimator will have

continuously growing error and hence the output.3. Small errors in A,B, and disturbances that enters the system but not in estimator

will cause slow estimate process.Hence better to go for closed loop estimator.Design of closed loop Full-order state observer: Figure shows closed loop Full orderstate observer (Luen berger state observer)

State model is Bu(t)Ax(t)(t)X (1)Cx(t)Y(t) (2)

Estimator output is (t)XC(t)Y (2a)

Error in output is (t)XX(t)c(t)YY(t) (2b) Error in state vector

(t)XX(t)(t)X~ (3)

From figure (t)Ym(Y(t)Bu(t)(t)XA(t)X (4)

m is nx1 constant gain matrix.

11

Differentiating equation (3)(t)X(t)XX ~

= (t)XmcA ~ .. by equations (2 to 3) (5)The characteristics equation of the error is given by

0mc)(ASI (6)If m is chosen properly, (A-mc) have reasonably fast roots, then )(~ tX will decay tozero irrespective of )0(~X i.e )(~)( tXtX , regardless of )0(X . Further it can be seen thatequation (5) is independent of control input u(t).The observer gain matrix m can be obtained in exactly the same fashion as controllergain matrix K

Design procedure:1. Let the desired pole locations of observer error roots be n ,......., 21 then the

corresponding poly equation will be0)).....(S)(S(S n21 (1)

0.....aSaSaS n2n

21n

1n (2)

determine co-effients naaa ,......., 212. Let the observer gain matrix be m

Tn21 ],.......mm,[mm

obtain the characteristics poly equation of the closed loop system with observer as0mc)(ASI

0)m(.......)Sm()Sm(S 1n2n

1n21n

n1n (3)

3. Comparing the co-efficient of like powers of S from equations (2) and (3)11nn11 amma nn11nn amma

where the co-efficients n ,......., 21 are obtained from poly equation0 ASI

n1n

1n .......SS =0.

4. The observer gain matrix m will be thenT

n21 ],.......mm,[mm Example : Consider the state model

u100

X(t)6116

100010

(t)X

X(t)100Y(t) Design an observer; where observer-error poles are required to be located at 64.32 jand -5

12

Solution: Model in oberserver conamical form will be

100

X(t)6101100610

(t)X

X(t)100Y(t) 1) Desired characteristic equation will be

(S+2+j 3 .64) (S+2-j3.64) (S+5)=0.=> 06020S9SS 23 hence comparing with 0asasaS 32

21

3 1a =9, 2a =20, 3a =60 (I)

2) Let the observer be Tnmmmm ],.......,[ 21

3

2

1

m00m00m00

6101101600

mcA

)m(610)m(1101)m(600

mcA

3

2

1

0)m(ASI c 0)m(6)Sm(11)Sm(6S 12

23

3 Comparing with

0sss 322

13

)m(6),m(11),m(6 132231 (II)3) From I and II,

3mm69 33 9mm1120 22

54mm660 11 4) Hence T]3954[m

13

Reduced-order state observerRequired when few of state variables cannot be measured accurately.Design of Reduced-order state observer :In this design procedure, let us assume one measurement Y(t). Which measures one stateX1(t). The output equation is given byY(t)=CX(t) (1)Where C=[1 0 -----0]Partitioning the states X(t) into states directly measured X1(t) and states directly nonmeasurable Xe(t), which are to be estimated.i.e

(t)eX(t)1XX(t) (2)

Partionionly the matrixes accordly & state equation can be written as

u(t)bb

(t)X(t)X

aaaa

(t)X

(t)Xe

1

e

1

eee1

1e11

e

1

(3)

and

(t)X(t)X

01Y(t)e

1 (4)

from equation (3)

inputKnown

eue1eeee bX(t)a(t)Xa(t)X (5)

from (4) (t)XY(t) 1=> X1(t) Y(t) Also fromEquation (3) u(t).b(t)Xa(t)Xa(t)X 1e1e1111 Collecting the known terms.

(t)Xau(t)b-(t)Xa(t)Y e1etermsKnown

1111 (6)Comparing equation (5) & (6) with

CX(t) Y(t)andbu(t)AX(t)(t)X

we can have similarity as

u(t)b(t)Xa(t)YY

acu(t)b(t)Xabu

aA(t)(t)XX(t)

1111

1e

e1e1

ee

e

(7)

Using Equation (7), the state model for reduced order state observer can be obtained as e ee e e1 e 1 11 1 1e eX (t) a X (t) a y b u(t)X (t) + m(Y(t) (Y a y b u a X (t)) (8)

14

(same as (t))ym(y(t)bu(t)(t)XA(t)X where (t)Xa(t)y e1e Defining eee XX(t)X

~ (t)X(t)X(t)X eee

~ Substituting for (t)X&(t)X ee

in above equation & on simplification,

ee 1e eX(t) (a ma )X (9)Corresponding characteristic equation is

0)ma(aSI 1eee (10)Comparing with desired characteristic equation m can be evaluated same as m in full-order observer.To implement the reduced-order observer,Equation (8) can be written as

(t)ym)umb(be)Y(t)ma(a(t)X)ma(a(t)X 111e1e1eeee (11)

As (t)Y

is difficult to realize,Define

my(t)X(t)X ee

(t)ym(t)X(t)X ee

)u(t)mb(b)y(t)ma(a(t)X)ma(a 1e11e1e1eee (12)Equation (12) can be represented by block diagram as shown below.

Example: System is described by

X(t)100Y(t)u(t)

12

50X(t)

10102401000

(t)X

To design the reduced-order observer for un measurable states such that the given valueswill be at 10, -10.

15

Solution:From output Equation it follows that states X1(t) & X2(t) cannot be measured ascorresponding elements of C matrix are zero. Let us write the state equation as

u(t)12

50

(t)X(t)X(t)X

10102401000

X

X

X

3

2

1

3

2

1

10a,240

a,0100

a 1ee1ee

Let

2

1

mm

m

Characteristic equation of an observer is0)ma(aSI 1eee 0mSmS 12

2 Desirable characteristic equation is

010)10)(S(S 010020SS2

From two characteristic equations, equating the co-efficient of like powers of S20=m2 & 100=m1.

20

100m

Hence state model of observer will be

1}u(t)20

1002

50{10}]}Y(t){

20100

240

{(t)X}1020

1000100

{(t)X ee

Design of compensator by the principle of separation:Consider the completely controllable and observable system described by

Bu(t)AX(t)(t)X (1a)CX(t)Y(t) (1)

Let the control law bekX(t)u(t) (2)

Assume the full-order observer model be

(t))xcM(yBu(t)XA(t)X

(3)Hence the state feedback control law using the estimated states will be

(t)XKU(t) (4)

1b2

50be;10a 111

16

Using equation (4) in (1a)(t)XBKAX(t)(t)X

(t))XBK(XBK)X(t)(A (5)Defining observer state error vector as

(t)XX(t)(t)X ~ (6)using (6) in (5)

(t)XBKBK)X(t)(A(t)X ~ (7)WKT observer error-vector model is

(t)XmC)(A(t)X ~~

(8)Equation (7) & (8) can be obtained in vector model as

(t)X

X(t)mC)(A0

BKBK)(A

(t)X

(t)X~

~

(9)

Equation (9) represents the model of 2n dimensional system. The correspondingcharacteristic equation is 0mC)(ASIBK)(ASI (10)Equation (10) represents the poles (given values) of the system, which is union of polesof controller & observers. The design of controller & observer can be carried byseparately, & when used together, the system is unchanged. This special case ofseparation principle controller can be designed using

0BK)(ASI and observer by 0mC)(ASI .The corresponding block diagram is as shown

Transfer function of compensator:

Now (t))xcm(yBu(t)(t)XA(t)X

(1)(t)XKU(t) (2)

From (1) & (2)mY(t)(t)XmC)BK(A(t)X

(3)

17

Taking L.T.mY(S)(S)XmC)BK(A(s)XS

(S)XmC)]BK(A[SImY(s) (4)Taking L.T. of equation (2)

(S)XK.U(S) (5)From equations (4) & (5)

]mmC)BK(AK[SIY(S)

U(S) 1D(S)

Example: A System is described by

u(t)100

X(t)1165100010

(t)X

X(t)001Y(t) Design controller to take place closed loop poles at 1 j1, -5. Also design an observersuch that observer poles are at 6, -6, -6. Derive the Un compensated & compensatedT.F.Solution: System is observable & Controllable. Using the procedure of controller design& observer design, the controller will beK=[4 1 1] and observer will beM=[12 25 -72]T

Transfer function of uncompensated system will beMu(S)= C[SI-A]-1B.

611S6SS1M 23u(S)

Transfer function of compensated system will be

mmC)]BK(AK[SIY(S)U(S)(s)M 1C

257121S19SS618119SS(S)M 23

2

C