Linear systems lecture

T. Kirubarajan

Department of Electrical and Computer EngineeringMcMaster University

Hamilton, Ontario, Canada

These viewgraphs are based on the text“Linear System: Theory and Design” by Chi-Tsong Chen

Oxford University Press, 1999.

ECE 707: Control Systems Design (10)

12. State Feedback and State Estimators

Plantu yr

Fig. 12.1In most of the control systems the plant (system) and the referencesignal r(t) are given. Our job is to find out input u(t) such that y(t)(the output) follows r(t) as closely as possible.If the input u(t) depends only on the reference signal r(t) and not onoutput y(t) then the control is called open loop control.Open loop control performance degrades if there are variations in theplant parameters (for example, properties of active components suchas inductors and capacitors may vary with temperature) and noisedisturbance (such as thermal noise).In a control system, if input u(t) depends on both r(t) as well asoutput y(t) (or state variables x(t)), then the control is called closedloop control.

Estimation, Tracking and Fusion Laboratory (ETFLab) 2 / 36

Closed loop control is more robust in noisy environment and undersystem parameter variations as it can keep track of these variationsby observing the output y(t) (or state variables x(t)).State FeedbackConsider the n-dimensional SISO state equation

x = Ax+ bu (1)

y = cTx (2)

+r+

−

kT

b

A

+s

1 cT y

plant

u xx•

Fig. 12.2


From Fig. 12.2 it is clear that

u = r − kTx = r −[

k1 k2 · · · kn]

x = r −

n∑

i=1

kixi (3)

Hence the state equation becomes

x = (A− bkT )x+ br (4)

y = cTx (5)

The next theorem shows that state feedback can not change thecontrollability of the system.Theorem: The pair (A− bkT , b), for any constant vector kT , iscontrollable if and only if (A, b) is controllable.Proof: If n = 4 (number of the state variables). For the originalsystem (plant) the controllability matrix is given by

X =[

b Ab A2b A3b]

(6)


For the feedback system the controllability matrix is given by

Xf =[

b (A− bkT )b (A− bkT )2b (A− bkT )3b]

(7)

It can be shown that

Xf = X

1 −kT b −kT (A− bkT )b −kT (A− bkT )2b

0 1 −kT b −kT (A− bkT )b

0 0 1 −kT b

0 0 0 1

(8)

Since Xf is X multiplied by a nonsingular matrix, the rank of Xf isequal to rank of X.Thus one of the systems is controllable if and only if the other systemis controllable. (EOP)The next example shows that the observability property can changeas a result of state feedback.


Example: Consider the state equation

x =[

1 23 1

]

x+[

01

]

u (9)

y = [ 1 2 ]x (10)

and state feedback

u = r − [ 3 1 ]x (11)

Find out if the initial system and the feedback system are controllableand observable.Solution: The controllability matrix and the observability matrix of theinitial system

X =[

0 21 1

]

and O =[

1 27 4

]

(12)

Both of them are of full rank. So the system is controllable andobservable. From the theorem we know that the feedback system isalso controllable.


For the feedback system the state equation is given by

x =[

1 20 0

]

x+[

01

]

r (13)

y = [ 1 2 ]x (14)

The controllability matrix and the observability matrix of the feedbacksystem

X =[

0 21 0

]

and O =[

1 21 2

]

(15)

The observability matrix does not have full row rank. So the systemwhich was initially observable becomes unobservable when we usefeedback.Consider a SISO system with transfer function

g(s) =β1s

3 + β2s2 + β3s+ β4

s4 + α1s3 + α2s2 + α3s+ α4(16)


Consider a controllable realization

x = Ax+ bu (17)

y = cTx (18)

It has controllability matrix

X =[

b Ab A2b A3b]

(19)

also consider controllable canonical form

˙x =

−α1 −α2 −α3 −α4

1 0 0 00 1 0 00 0 1 0

x+

1000

u (20)

y = [ β1 β2 β3 β4 ] x (21)


Inverse of the controllability matrix for this form

X−1 =

1 α1 α2 α3

0 1 α1 α2

0 0 1 α1

0 0 0 1

(22)

Thus the equivalence transform x = Px in this case is given by

P−1 = XX−1 =[

b Ab A2b A3b]

1 α1 α2 α3

0 1 α1 α2

0 0 1 α1

0 0 0 1

(23)

This transform is important as we will see in the next theorem.Theorem: If a n-dimensional state equation is controllable, then bystate feedback u = r − kTx, where k is real vector, the eigenvalues ofA− bkT can arbitrarily assigned provided that complex conjugateeigenvalues are assigned in pairs.


Proof: Again we prove the theorem for n = 4.Let us consider a controllable system equation

x = Ax+ bu (24)

y = cTx (25)

Let us assume that the transfer function is given by

g(s) =β1s

3 + β2s2 + β3s+ β4

s4 + α1s3 + α2s2 + α3s+ α4(26)

Then as we have already shown, the equivalence transform x = Px

will form controllable canonical form of the state equation

˙x =

−α1 −α2 −α3 −α4

1 0 0 00 1 0 00 0 1 0

x+

1000

u (27)

y = [ β1 β2 β3 β4 ] x (28)


where the transform matrix P is given by

P−1 =[

b Ab A2b A3b]

1 α1 α2 α3

0 1 α1 α2

0 0 1 α1

0 0 0 1

(29)

Now consider that a state feedback u = r− kTx is used in the originalsystem. The feedback for the transformed system (that gives thesame effect) is given by

u = r − kTx = r − kTP−1x△= r − k

Tx (30)

where kT= kTP−1. Since

A− bkT= P (A− bkT )P−1 (31)

A− bkT and A− bkT have the same eigenvalues.

If kT is chosen as

kT=

[

α1 − α1 α2 − α2 α3 − α3 α4 − α4

]

(32)


then the feedback equation becomes

˙x =

−α1 −α2 −α3 −α4

1 0 0 00 1 0 00 0 1 0

x+

1000

r (33)

y = [ β1 β2 β3 β4 ] x (34)

because of the companion form the characteristic equation ofA− bk

T and, consequently, of A− bkT becomes equal to

∆f (s) = s4 + α1s3 + α2s

2 + α3s+ α4 (35)

Thus, we can use state feedback to achieve any characteristicequation of A− bkT or, we can place eigenvalues (zeros of (35))place in the complex plane we want.If we have complex eigenvalues they will appear in pairs as thecoefficients of ∆f (s) are real. (EOP)


The Procedure of Placing EigenvaluesLet us consider a n-dimensional system equation

x = Ax+ bu (36)

y = cTx (37)

Let us consider that we want to place the eigenvalues (at λi withmultiplicity mi) such that the characteristic equation of A− bkT

becomes

∆f (s) = sn + α1sn−1 + · · ·+ αn−1s+ αn =

p∏

i=1

(s− λi)mi (38)

To do this we have to find out characteristic equation of A

∆(s) = sn + α1sn−1 + · · ·+ αn−1s+ αn (39)

The feedback gain vector is given by

kT = [α1 − α1 α2 − α2 · · · αn − αn]P (40)


where P can be obtain from

P−1 =[

b Ab · · · An−1b]

1 α1 · · · αn−1

0 1 · · · αn−2

...... · · ·

...0 0 · · · 1

(41)

Notice that if the initial system has a transfer function

g(s) =β1s

n−1 + β2sn−2 + · · ·+ βn

s4 + α1sn−1 + · · ·+ αn−1s+ αn(42)

Then after state feedback the transfer function (from r to s) becomes

gf (s) =β1s

n−1 + β2sn−2 + · · ·+ βn

s4 + α1sn−1 + · · ·+ αn−1s+ αn(43)

So the poles of the transfer function changes however there is nochange in the zeros.Thus if any of the replaced poles coincide with zeros, then numeratorand denominator of gf (s) are not coprime. Hence the feedbacksystem becomes unobservable.


Example: Consider the state equation

x =

0 1 0 00 0 −1 00 0 0 10 0 5 0

x+

010−2

u (44)

y = [ 1 0 0 0 ]x (45)

Place the eigenvalues at −1.5± 0.5j and −1± j by using statefeedback.Solution: We can show that the characteristic equation of A-matrix(How?) is given by

∆(s) = s2(s2 − 5) = s4 + 0 · s3 − 5 · s2 + 0 · s+ 0 (46)

The desired characteristic function of the feedback system

∆f (s) = (s+ 1.5− 0.5j)(s+ 1.5 + 0.5j)(s+ 1− j)(s+ 1 + j)

= s4 + 5s3 + 10.5s2 + 11s+ 5 (47)


Thus we have

kT = [5− 0 10.5 + 5 11− 0 5− 0]P (48)

where P is given by

P−1 = XX−1 =

0 1 0 21 0 2 00 −2 0 −10−2 0 −10 0

1 0 −5 00 1 −5 00 0 1 00 0 0 1

(49)

=

0 1 0 −31 0 −3 00 −2 0 0−2 0 0 0

(50)

Hence

P =

0 0 0 −1

2

0 0 −1

20

0 −1

30 −

1

6

−1

30 −

1

60

(51)


So

kT = [−1.6667 − 3.6667 − 8.5833 − 4.3333] (52)

Regulation and TrackingThe problem to find a state feedback gain so that the response due tononzero initial conditions die out at a desired rate is called aregulation problem.For example, bringing an aircraft to zero deviation from its desiredaltitude is a regulation problem.In this case the reference signal r is zero.Consider another problem of designing an overall system so that y(t)approaches r(t) = a at t → ∞. This is called asymptotic tracking of astep reference input.Consider a state equation

x = Ax+ bu (53)

y = cTx (54)


The regulation problem is has a simple solution, use state feedback toshift the eigenvalues of A− bkT so that the system becomes stable.Then the state feedback equation becomes

x = (A− bkT )x+ br (55)

y = cTx (56)

Since r(t) is zero the response caused by x0 is

y(t) = cT e(A−bkT )tx0 (57)

If all eigenvalues of (A− bkT ) are in the left s plane then the outputwill decay rapidly to zero.For the tracking problem, in addition to the state feedback we need afeedforword gain p

u(t) = pr(t)− kTx(t) (58)


Then we have (for n = 4)

g(s) = pβ1s

3 + β2s2 + β3s+ β4

s4 + α1s3 + α2s2 + α3s+ α4(59)

As r(t) = a then the output approaches gf (0) · a as t → ∞.Thus y(t) will track the input asymptotically if

1 = gf (0) = pβ4

α4or p =

α4

β4(60)

We can see that tracking is not possible if β4 = 0, i.e., if the transferfunction has a zero at s = 0.StabilizationAgain, consider a system equation

x = Ax+ bu (61)

y = cTx (62)

If (A, b) is controllable, all eigenvalues can be assigned arbitrarily.Estimation, Tracking and Fusion Laboratory (ETFLab) 19 / 36

The system can be stabilized by shifting all eigenvalues of A to theleft s-plane.However this is not true for a uncontrollable system as we see in thefollowing.Any uncontrollable state equation can be transformed into

[

xcxc

]

=

[

Ac A12

0 Ac

] [

xcxc

]

+

[

bc0

]

u (63)

If we introduce the state feedback

u = r − kTx = r − [kT1 kT2 ]

[

xcxc

]

(64)

Then the state update equation becomes[

xcxc

]

=

[

Ac − bckT1 A12 − bck

T2

0 Ac

] [

xcxc

]

+

[

bc0

]

r (65)

We can see that Ac and consequently its eigenvalues are not effectedby the state feedback. Thus the state equation is still uncontrollable.


This also shows that the system can be stabilized by state feedbackonly if Ac is stable.State EstimatorIn order to apply state feedback, we need the state variables of asystem.

+r+

−

kT

b

A

+s

1 cT y

plant

u xx•

Fig. 12.2The state variables may not be accessible for direct connection assensing devices may be either not available or very expansive.We need to design a state estimator for our feedback system to work.


Consider the state equationx = Ax+ bu (66)

y = cTx (67)

where A,b and cT are known and the input and output are alsoavailable.We can duplicate the original system as

˙x = Ax+ bu (68)

where x be an estimate of x.

b

A

+s

1 cT yu x

b

A

+s

1 x

Estimator

Fig. 12.3Estimation, Tracking and Fusion Laboratory (ETFLab) 22 / 36

Note that the original system and the duplicated system can be madeof different types of components.For example, original system can be electromechanical system andthe duplicate system can be an op-amp circuit that copies thebehavior of the original system.Fig. 12.3 shows such an estimator. This kind of duplication is calledopen-loop estimator.Here if the initial state same in the original and the duplicate circuit,then estimation is perfect.The problem here is that we need perfect knowledge of the initialstate of the original system for the estimator to work.Next we will discuss an estimator which uses input as well as theoutput of the original system.


b

A

+s

1 cT yux

A

+s

1x

Estimator

b +

cT

+

−

b

+

+

Fig. 12.4Fig. 12.4 shows a closed-loop estimator. Here we have gain vector lthat amplifies the difference of the output of the original system andthe estimated output.The vector obtained by this operation is added to the contribution ofinput to get state update.


Duplicate system equation of the closed-loop estimator

˙x = Ax+ bu+ l(y − cT x) (69)

which can be written as

˙x = (A− lcT )x+ bu+ ly (70)

Let us define

e(t)△= x(t)− x(t) (71)

Hence

e(t) = x(t)− ˙x(t) = Ax+ bu− (A− lcT )x− bu− ly

= (A− lcT )x− (A− lcT )x

= (A− lcT )e(t) (72)

If all eigenvalues of (A− lcT ) can be placed arbitrarily then the rate atwhich e(t) approaches zero can be controlled.The following theorem tells the condition under which this can bedone.


Theorem: All eigenvalues of (A− lcT ) can be assigned arbitrarily byselecting real constant vector l if and only if (A, cT ) is observable.Proof: We know the pair (A, cT ) is observable if and only if the pair(AT , c) is controllable.Now if (AT , c) is controllable, the eigenvalues of AT

− clT can beplaced arbitrarily by the choice of l.Which means if (A, cT ) is observable, the eigenvalues of A− lcT canbe placed arbitrarily by the choice of l. (EOP)The procedure for computing state feedback gain can be used tocompute the gain l.Feedback from Estimated StatesConsider a system equation

x = Ax+ bu (73)

y = cTx (74)


Consider that x is not available. We construct a close-loop stateestimator as

˙x = (A− lcT )x+ bu+ ly (75)

we apply the feedback gain to the estimated state to get

u = r − kT x (76)

Substituting this into (73) and (75) we get

x = Ax− bkT x+ br (77)˙x = (A− lcT )x+ b(r − kT x) + ly (78)

+ yu

x

kT

Plant

Estimator

+r

−

Fig. 12.5Fig. 12.5 shows the corresponding system.


The combined state equation of the controller-estimator is given by[

x˙x

]

=

[

A −bkT

lcT A− lcT − bkT

] [

x

x

]

+

[

b

b

]

r (79)

= y = [cT 0T ]

[

x

x

]

(80)

Let us consider the following equivalence transform[

x

e

]

=

[

x

x− x

]

=

[

I 0I −I

] [

x

x

]

△= P

[

x

x

]

(81)

Computing P−1, same as P , we can get the correspondingA, B, C, D matrices. This gives

[

x

e

]

=

[

A− bkT bkT

0 A− lcT

] [

x

e

]

+

[

b

0

]

r (82)


and

y = [cT 0T ]

[

x

e

]

(83)

The A-matrix is block triangular. Thus the eigenvalues of the mainsystem (corresponding to states x) remains unchanged.So inserting the state estimator does not effect the eigenvalues ofmain system and vice versa.Thus the design of state feedback and state estimator can be doneindependently (separation property).State Estimator–Multivariate CaseConsider the n-dimensional p-input and q-output system equation

x = Ax+Bu (84)

y = Cx (85)


In state feedback, the input u is given by

u = r −Kx (86)

where K is p× n constant matrix and r is a reference signal. Thestate equation of the feedback system

x = (A−BK)x+Br (87)

y = Cx (88)

Theorem: The pair (A−BK,B) is controllable if and only if (A,B) iscontrollable.Theorem: All eigenvalues of (A−BK) can be assigned arbitrarily(complex conjugate eigenvalues in pairs) if and only if (A,B) iscontrollable.


Cyclic DesignA square matrix A is called cyclic if its characteristic polynomial andminimal polynomial are the same i.e., if only one Jordan block isassociated with each distinct eigenvalue.Theorem: If the pair (A,B) is controllable then for almost any vectorv the pair (A,Bv) is controllable.We argue the validity of this theorem by an example.Controllability is invariant under equivalence transform, so withoutloss of generality we can assume A to be in Jordan form. Considerthe example

A =

2 1 0 0 00 2 1 0 00 0 2 0 00 0 0 −1 10 0 0 0 −1

B =

0 10 01 24 31 0

(89)

A is cyclic and (A,B) is controllable.


Let us consider an arbitrary v

Bv = B[

v1v2

]

=

x

x

α

x

β

(90)

It is clear that if α and β are nonzero then (A,Bv) is controllable.We have

α = v1 + 2v2 and β = v1 (91)

Thus v1 and v2 can’t have the following values

v1 = 0 or v1 = −2v2 (92)


v1=-2v2

v1

v2

v1=0

Fig. 12.6Fig. 12.6 shows that the probability of an arbitrary v to give acontrollable (A,Bv) is 1.Theorem: If the pair (A,B) is controllable, then for almost any realconstant matrix K, the matrix (A−BK) has only distinct eigenvalues(cyclic).Again the theorem is shown intuitively for n = 4. Let the characteristicpolynomial of (A−BK) be

∆f (s) = s4 + a1s3 + a2s

2 + a3s+ a4 (93)


where ai are functions of entries of K. Differentiation of ∆f (s) w.r.t. sgives

∆′

f (s) = 4s3 + 3a1s2 + 2a2s+ a3 (94)

If ∆f (s) has repeated roots, then ∆f (s) and ∆′

f (s) are not coprime, inthat case

∣

∣

∣

∣

∣

∣

∣

∣

∣

∣

∣

a4 a3 0 0 0 0 0 0a3 2a2 a4 a3 0 0 0 0a2 3a1 a3 2a2 a4 a3 0 0a1 4 a2 3a1 a3 2a2 a4 a31 0 a1 4 a2 3a1 a3 2a20 0 1 0 a1 4 a2 3a10 0 0 0 1 0 a1 40 0 0 0 0 0 1 0

∣

∣

∣

∣

∣

∣

∣

∣

∣

∣

∣

= 0 (95)

It is obvious that the above equation is satisfied by a small subset ofall possible K. Hence arbitrary K has probability 1 of not satisfyingthis or eigenvalues of (A−BK) are distinct.Next we discuss the cyclic design.


Let us consider that A is not cyclic, but (A,B) is controllable. Weintroduce u = w −K1x as shown in Fig. 12.7.

A

+s

1y

u x

+ B + Cw

r

K1

K2

+ ++

+− −

x•

Fig. 12.7Then the update equation becomes

x = (A−BK1)x+Bw = Ax+Bw (96)

For arbitrary choice of K1 A is cyclic. Now because (A,B) iscontrollable so is (A, B).Hence there exists vector v such that (A, Bv) is controllable.


Next we introduce another feedback w = r −K2x, where K2 = vkT .Then the update equation becomes

x = (A−BvkT )x+Bw (97)

Now the eigenvalues can be assigned arbitrarily by selecting kT

(SISO case).


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