Chapter 3 Controllability Observability

8/13/2019 Chapter 3 Controllability Observability

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Chapter 3

Controllability and Observability

In this chapter, we study the controllability and observability concepts. Controllability is concernedwith whether one can design control input to steer the state to arbitrarily values. Observability isconcerned with whether without knowing the initial state, one can determine the state of a systemgiven the input and the output.

3.1 Some linear mapping concepts

The study of controllability and observability for linear systems essentially boils down to studyingtwo linear maps: the reachability map Lr which maps, for zero initial state, the control input u()to the final state x(t1); and the observability map, Lo which maps, for zero input, from the initialstate x(t0) to the output trajectory y(). The rangeand the nullspaces of these maps respectivelyare critical for their studies.

LetL : X(=p) Y(=q) be a linear map: i.e.

L(x1+ x2) =(Lx1) + (Lx2)

R(L) Ydenotes the range ofL, given by:

R(L) ={y Y|y= Lx for some x X }

N(L) Xdenotes the null space ofL, given by:

N(L) ={x X |Lx= 00}

Note: BothR(L) andN(L) are linear subspaces, i.e. for scalar , ,

y1, y2 R(L) (y1+ y2) R(L)

x1, x2 N(L) (x1+ x2) R(L)

L is called onto or surjective ifR(L) =Y(everything is within range).

L is called into or one-to-one (cf. many-to-one) ifN(L) ={0}. Thus,

Lx1= Lx2 x1= x2

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ME 8281:Advanced Control Systems, F01, F02, S06, S08 29

Remarks:

The followings are equivalent:

1. A system is controllable at t = t0.

2. The system can be transferred from any state att = t0 to any other state in finite time.

3. For eachx0 ,

s(, t0, x0, ) : [t0, ) u()s(t1, t0, x0, u())

is surjective (onto).

Controllability does not say that x(t) remains at x1

. e.g. for

x= Ax + Bu

to make x(t) =x1 t t0, it is necessary that Ax1 Range(B). This is generally not true.

Example - An Uncontrollable System

An linear actuator pushing 2 masses on each end of the actuator in space.

m1x1= u; m2x2= u

State: X= [x1, x1, x2, x2]T.

Action and reaction are equal and opposite, and act on different bodies.

Momentum is conserved:

m1x1+ m2x2= constant =m1x1(t0) + m2x2(t0)

E.g. if both masses are initially at rest: it is not possible to choose u() to make x1 = 0,x2= 1.

3.2.1 Effects of feedback on controllability

Consider a system

x= f(t, x(t), u(t)).

Suppose that the system is under memoryless (static) state feedback:

u(t) =v(t) g(t, x(t))

with v(t) being the new input.

A system is controllable if and only if the system under memoryless feedback withv(t) as theinput is controllable.


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Proof: Suppose that the original system is controllable. Given a pair of initial and final states,x0 and x1, suppose that the control u() transfers the state from x0 at t0 to x1 at t1. For the new

system, which has alsox(t) as the state variables, we can use the control v(t) =u(t) + g(t, x(t)) totransfer the state fromx0 at t0 to x1 at t1. Therefore, the new system is also controllable.

Suppose that the feedback system is controllable, and v() is the control that can achieve thedesired state transfer. Then for the original system, we can use the control u(t) =v(t) g(t, x(t))to achieve the same state transfer. Thus, the original system is also controllable.

Consider now the system under dynamic state feedback:

z(t) =(t, x(t), z(t))

u(t) =v(t) g(t, x(t), z(t))

where v(t) is the new input and z (t) is the state of the controller.

If a system under dynamic feedback is controllable then the original system is.

The converse is not true, i.e. the original system is controllable does not necessarily implythat the system under dynamic state feedback is.

Proof: The same proof for the memoryless state feedback works for the first statement. For thesecond statement, we give an example of the controller:

z= z; u(t) =v(t) g(t, x(t), z(t)).

The state of the new system is (x, z) but no control can affect z(t). Thus, the new system is not

controllable. These results show that feedback control cannot make a uncontrollable system become controllable,but can even make a controllable system, uncontrollable. To make a system controllable, one mustredesign the actuators.

3.3 Controllability of Linear Systems

Continuous time system:

x(t) =A(t)x(t) + B(t)u(t) x n

x(t1) = (t1, t0)x0+ t1t0

(t1, t)B(t)u(t)dt.

The reachability map on [t0, t1] is defined to be:

Lr,[t0,t1](u()) =

t1t0

(t1, t)B(t)u(t)dt

Thus, it is controllable on [t0, t1] if and only ifLr,[t0,t1](u()) is surjective (onto).Notice thatLr,[t0,t1] determines the set of states that can be reached from the origin at t = t1.The study of the range space of the linear map:

Lr,[t0,t1] : {u()} n

is central to the study of controllability.


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Proposition 3.3.1 If a continuous time, possibly time varying, linear system is controllable on[t0, t1] then it is controllable on [t0, t2] wheret2 t1.

Proof: To transfer from x(t0) =x0 to x(t2) =x2, define

x1= (t1, t2)x2

Suppose that u[t0,t1] transfers x(t0) =x0 to x(t1) =x1. Choose

u(t) =

u(t) t [t0, t1]

0 t (t1, t2]

.

This result does not hold for t0 < t2 < t1. e.g. The system

x(t) =B(t)u(t) B(t) =

0 t [t0, t2]

In t (t2, t1].

is controllable on [t0, t1] but x(t2) =x(t0) for any control u().

3.3.1 Discrete time system

x(k+ 1) =A(k)x(k) + B(k)u(k), x n, u m

x(k1) = (k1, k0)x(k0) +

k11k=k0

(k1, k+ 1)B(k)u(k)

The transition function of the discrete time system is given by:

x(k1) =s(k0, k1, x(k0), u()) = (k1, k0)x(k0) +k11k=k0

(k1, k+ 1)B(k)u(k) (3.1)

(kf, ki) = kf1k=ki

A(k) (3.2)

The reachability map of the discrete time system can be written as:

Lr,[k0,k1](u(k0), . . . , u(k1 1)) =k11

k=k0

(k1, k+ 1)B(k)u(k) =Lr(k0, k1)U

where Lr(k0, k1) n(k1k0)m, U =

u(k0)...

u(k1 1)

.

Thus, the system is controllable on [k0, k1] if and only ifLr,[k0,k1] is surjective, which is true ifand only ifLr(k0, k1) has rank n

Proposition 3.3.2 Suppose thatA(k) is nonsingular for eachk1 k < k2, then the discrete timelinear system is controllable on [k0, k1] implies that it is controllable on [k0, k2] wherek2 k1.

Proof: : The proof is similar to the continuous case. However for a discrete time system, we needA(k) nonsingular for k1 k < k2 to ensure that (k2, k1) is invertible. Q.E.D.


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Example

Consider a unit point mass under control of a force:

d

dt

xx

=

0 10 0

xx

+

01

u

This is the same as x= u.Suppose that x(0) = x(0) = 0, we would like to translate the mass to x(T) = 1, x(T) = 0.The reachability map L : u() x(T) is

x(T) =

T0

(T, t)B(t)u(t)dt

Let use try to solve this problem using piecewise constant control:

u(t) =

u0 0 t < T/10

u1 T /10 t


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which is non-singular (full rank). Thus, from the finite rank theorem R(Lr) = R(LrLTr ), theparticle transfer problem is solvable using piecewise constant control.

If we make the number of pieces in the piecewise constant control larger and larger, then themultiplication ofLr byL

Tr becomes an integral:

LrLTr

t1t0

(t1, t)B(t)B(t)T(t1, t)

Tdt.

Therefore for the continuous time linear system is controllable over the interval [t0, t1] if andonly if the matrix, t1

t0

(t1, t)B(t)B(t)T(t1, t)

Tdt (3.3)

is full rank.

3.4 Reachability Grammian

The matrix Wr = LrLTr

nn is called the Reachability Grammian for the interval that Lr isdefined. For the continuous time system

x= A(t)x + B(t)u

the Reachability Grammian on the time interval [t0, t1] is defined to be:

Wr,[t0,t1] =

t1t0

(t1, t)B(t)B(t)T(t1, t)

Tdt

For time invariant systems,

Wr([t1, t0]) =LrLr =

t1t0

eA(t1t)BBTeAT(t1t)dt

= 0

t1t0

eABBTeATd.

For the discrete time system,

x(k+ 1) =A(k)x(k) + B(k)u(k),

the Reachability Grammian on the interval [k0, k1] is

k11k=k0

(k1, k+ 1)B(k)B(k)T(k1, k+ 1)

T.

From the finite rank theorem, we know that the controllability of a system on [t0, t1] (or on[k0, k1]) is equivalent to Reachability Grammian on the same time interval being full rank.


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3.4.1 Reduction theorem: Controllability in terms of Controllability to Zero

Thus far, we have been deciding controllability by considering the reachability map Lr whichconcerns what states we are able to steer to from x(t0) = 0. The reduction theorem allows usto formulate the question in terms of whether a state can be steered to x(tf) = 0 in finite time.

Theorem 3.4.1 The followings are equivalent for a linear time varying differential system:

x= A(t)x + B(t)u

1. It is controllable on [t0, t1].

2. It is controllable to0 on [t0, t1], i.e. x(t0) =x0, there existsu(), [t0, t1) such that finalstate isx(t1) = 0.

3. It is reachable from x(t0) = 0, i.e. x(t1) = xf, there exists u(), [t0, t1) such that forx(t0) = 0, the final state isx(t1) =xf.

Proof:

Clearly, (1) (2) and (3).

To prove (2) (1) and (3) (1), use

x(t1) = (t1, t0)x(t0) +

t1t0

(t1, )B()u()d.

and (t1, t0) is invertible, to construct the appropriate x(t0) and x(tf).

Controllability map: Lc,[t0,t1]

The controllability (to zero) map on [t0, t1] is the map between u() to the initial statex0 such thatx(t1) = 0.

Lc,[t0,t1](u()) =

t1t0

(t0, t)B(t)u(t)dt

Proof:

0 =x(t1) = (t1, t0)x0+ t1

t0 (t1, )B()u()d

x0= (t1, t0)1

t1t0

(t1, )B()u()d

x0=

t1t0

(t0, t1)(t1, )B()u()d

x0=

t1t0

(t0, )B()u()d

Notice that

R(Lc,[t0,t1]) = (t1, t0)R(Lr,[t0,t1])

These two ranges are typically not the same for time varying systems. However, one is full rank ifand only if the other is, since (t1, t0) is invertible.


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For time invariant continuous time systems, R(Lc,[t0,t1]) =R(Lr,[t0,t1]).The above two statements are not true for time invariant discrete time systems. Can you find

counter-examples?Because of the reduction theorem, the linear continuous time system is controllable on the inter-

val [t0, tf] ifR(Lc,[t0,t1]) =n. From the finite rank theorem, R(Lc,[t0,t1]) =R(Lc,[t0,t1]L

c,[t0,t1]

) =

n. The controllability-to-zero grammian on the time interval [t0, t1] is defined to be:

Wc= LcLc =

t1t0

(t0, t)B(t)B(t)T(t0, t)

Tdt

For time invariant system,

Wc= LcLc =

t1t0

eA(t0t)BBTeAT(t0t)dt

= 0t1t0

eABBTeATd.

Since controllability of a system on [t0, t1] is equivalent to controllability to 0 which is determinedby the controllability map being surjective, it is also equivalent to controllability grammian on thesame time interval [t0, t1] being full rank.

3.5 Minimum norm control

3.5.1 Formulation

The Controllability Grammian is related to the cost of control.

Given x(t0) = x0, find a control function or sequence u() so that x(t1) = x1. Let xd =x1 (t1, t0)x0. Then we must have

xd= Lr,[t0,t1](u)

where Lr,[t0,t1] : {u()} is the reachability map which, for continuous time system is:

Lr,[t0,t1](u) =

t1t0

(t1, t)B(t)u(t)dt.

and for discrete time system:

Lr,[k0,k1](u()) =k11

k=k0 (k1, k+ 1)B(k)u(k) =C(k0, k1)U

where C n(k1k0)m, U =

u(k0)...

u(k1 1)

.

Let us focus on the discrete time system.Sincexd Range(Lr,[k0,k1]) for solutions to exist, we assume Range(Lr,[k0,k1]) =

n. This impliesthat Wr,[k0,k1] is invertible.

Generally there are multiple solutions. To resolve the non-uniqueness, we find the solution sothat

J(U) = 1

2

k11

k=k0

u(k)Tu(k) =UTU

is minimized. Here U= [u(k0); u(k0+ 1); . . . ; u(k1 1)].


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3.5.2 Least norm control solution

This is a constrained optimization problem (with J(U) as the cost to be minimized, andLrUxd=0 as the constraint). It can be solved using the Lagrange Multiplier method of converting aconstrained optimization problem into an unconstrained optimization.

Define an augmented cost function, with the Lagrange multipliers n

J(U, ) =J(U) + T(LrU xd)

The optimal given by (U, ) must satisfy:

J

U(U, ) = 0;

J

(U, ) = 0

The second condition is just the constraint: LrU=xd.

This means that LTr + U = 0; LrU

=xd

Solving, we get:

=(LrLTr)

1xd= W1r xd

U =LTr =LTrW

1r xd.

Theoptimal cost of control is:

J(U) =xTd W1r LrL

TrW

1r xd= x

Td W

1r xd

Thus, the inverse of the Reachability Grammian tells us how difficult it is to perform a state transfer

from x = 0 to xd. In particular, ifWr is not invertible, for some xd, the cost is infinite.

3.5.3 Geometric interpretation of least norm solution

Geometrically, we can think of the cost as J = UTU, i.e. the inner product ofU with itself. Innotation of inner product,

J=< U, U >R

The advantage of the notation is that we can change the definition of inner product, e.g. R=U

TRV where R is a positive definite matrix. The usual inner (dot) product has R = I.We say that U and Vare normal to each other if< U, V >R= 0.Any solution that satisfies the constraint must be of the form

(U Up) Null(Lr)

where LrUp =xf is any particularsolution.

Claim: LetU be the optimal solution, and Uis any solution that satisfies the constraint. Then,(U U) U, i.e.

R= 0

which is the normal equation for the least norm solution problem.

Proof: Direct substitution.

(U U)TRUR1U L

Tpc LrR

1U L

Tr

1xf = (Lr(U U

))TLrR1U L

Tr

1xf = 0.


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3.5.4 Open-loop least norm control

Applying the above result to linear continuous time system using standard norms, i.e.

J=

t1t0

uT()u()d.

(Lr,[t0,t1])(t) =BT(t)(t, t1)

T

Wr,[t0,t1] = LrLr =

t1t0

(t1, )B()BT()(t1, )

Td

uopt(t) =B(t)T(t, t1)TW1r,[t0,t1](x1 (t1, t0)x0) (3.4)

Eq, (3.4) solves uopt() in a batch form given x0, i.e. it is openloop.

3.5.5 Recursive formulation

Openloop control does not make use of feedback. It is not robust to disturbances or modelunceratinty. Suppose now that x(t) is measured. Conceptually, we can solve (3.4) for uopt(t)and lett0= t.

Computing Wr,[t,t1] is expensive. Try recursion:

uopt(t) =B(t)T(t, t1)

TW1r,[t,t1]

(x1 (t1, t)x(t))

=

B(t)T(t, t1)TW1r,[t,t1]

x1

B(t)T(t, t1)

TW1r,[t,t1](t1, t)

x(t)

For any invertible matrix Q(t) nn, Q(t)Q1(t) =I, therefore

d

dtQ1(t) =Q1(t)Q(t)Q1(t)

Since

Wr,[t,t1] =

t1t

(t1, )B()B()(t1, )

d

d

dtWr,[t,t1] = (t1, t)B(t)B(t)

(t1, t)

Thus we can solve for W1r,[t,t1]

dynamically by integrating on-line

d

dtW1

r,[t,t1]= W1

r,[t,t1](t1, t)B(t)B(t)

(t1, t)W1

r,[t,t1]

Issue: Does not work well when t t1 because Wr,[t,t1] can be close to singular.


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3.5.6 Optimal cost

Optimal cost as a function of initial time t0

Since

Wr,[t0,tf] =

tft0

(tf, )B()B()(tf, )

d

and the integrand is a positive semi-definite term, for each v n, vTW1r,[t0,tf]v decreases as t0decreases, i.e. given more time, the cost of control decreases.

To see this, fix the final time tf, and consider the minimum cost function,

Emin(xf, t0) =xTfW

1r,[t0,tf]

xf

and considerd

dt0Emin(xf, tf) =v

T ddtf

W1r,[t0,tf]

xf

Since

d

dtfW1

r,[t0,tf]= W1

r,[t0,tf]

d

dtfWr,[t0,tf]W

1r,[t0,tf]

=

W1r,[t0,tf]

(tf, t0)B(t0)B(t0)(tf, t0)

W1r,[t0,tf]

is negative semi-definite, ddt0 Emin(v, tf) 0. Hence, the optimal control increases as the initialtime decreases, increasing the allowable time interval.

In other words, the set of states reachable with a cost less than a given value grows as t0decreases.

If the system is unstable, in that, (tf, t0) becomes unbounded as t0 , then one cansee that Wr,[t0,tf] also becomes unbounded. This means that for some xf, Emin(xf, t0) 0 andtf . This means that some directions do not require any cost of control. Which direction dothese correspond to?

For time invariant system, decreasing t0 is equivalent to increasing tf. Thus for time-invariantsystem, increasing the time interval increases Wr, thus decreasing cost.

Optimal cost as a function final state xf

Since Wr,[t0,tf]is symmetric and positive semi-definite, has n orthogonal eigenvectors,V = [v1, . . . , vn],

and associated eigenvalues = diag(1, . . . , n), such that V VT =I. Then

Wr,[t0,tf] = VVT W1

r,[t0,tf]= V diag(11 , . . . ,

1n )V

T

This shows that most expensive direction to transfer to is the eigenvector associated with theminimumi, and the least expensive direction is the eigenvector associated with the maximum i

3.6 Linear Time Invariant (LTI) Continuous Time Systems

Consider first the LTI continuous time system:

x= Ax + Buy= Cx (3.5)


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Note: Because of time invariance, the system is controllable (observable) at t0 means that itis controllable (observable) at any time.

We will develop tests for controllability and observability by using directly the matrices A andB.

The three tests for controllability of system (3.5) is given by the following theorem.

Theorem For the continuous time system (3.5), the followings are equivalent:

1. The system is controllable over the interval [0, T], for some T >0.

2. The system is controllable over any interval [t0, t1] with t1> t0.

3. The reachability grammian,

Wr,T := T0

(T, t)B(t)B(t)T(T, t)Tdt

=

T0

eAtBBeAtdt

is full rank n. (This test works for time varying systems also)

4. The controllability matrix

C :=

B AB A2B An1B

has rank n. Notice that the controllability matrix has dimension n nm where m is thedimension of the control vector u.

5. (Belovich-Popov-Hautus test)For each s C, the matrix,sI A B

has rank n.

Note that rank of (sI A, B) is less than n only ifs is an eigenvalue ofA.

We need the Cayley-Hamilton Theorm to prove this result:

Theorem 3.6.1 (Cayley Hamilton) LetA nn. Consider the polynomial

(s) =det(sI A)

Then(A) = 0.

Notice that ifi is an eigenvalue ofA, then(i) = 0.

Proof: This is true for any A, but is easy to see using the eigen decomposition ofA when A issemi-simple.

A= TT1

where is diagonal with eigenvalues on its diagonal. Since (i) = 0 by definition,

(A) =T ()T1 = 0.


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Proof: Controllability test(1) (3): We have already seen before that controllability over the interval [0, T] means that

the range space of the reachability map,

Lr(u) =

T0

eA(T)Bu()d

must be the whole state space, n. Notice that

Wr,T :=

T0

eAtBBeAtdt= LrL

r

where Lr is the adjoint (think transpose) ofLr.

From the finite rank linear map theorem, R(Lr

) =R(Lr

Lr

) but, Lr

Lr

is nothing but Wr,T

.

(3) (4): We will show this by showing not (4) not (3). Suppose that (4) is not true.Then, there exists a 1 n vector vT so that,

vTB = vTAB = vTA2B = vTAn1B = 0.

Consider now,

vTWr,Tv=

T0

vTeAtB22dt.

Since

eAt =I+ At +A2t2

2

+An1tn1

n 1!

+

and by the Cayley Hamilton Theorem, Ak is a linear combination ofI, A, ... An1, therefore,

vTeAtB = 0

for all t. Hence, vTWr,Tv= 0 or Wr,T is not full rank.

(3) (2): We will show this by showing not (2) not (3). If (2) is not true, then thereexists 1 n vector vT so that

vTWr,Tv= T

0

vTeAtB22dt= 0.

BecauseeAt is continuous in t, this implies that vTeAtB = 0 for all t [0, T] .Hence, the all time derivatives ofvTeAtB = 0 for all t [0, T]. In particular, at t = 0,

vTeAtB|t=0= vTB = 0

vT d

dteAtB|t=0= v

TAB = 0

vT d2

dt2eAtB|t=0= v

TA2B = 0

...

vT dn1

dtn1eAtB|t=0= v

TAn1B = 0


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Hence,

vTB AB A2B An1B= 0 0 0i.e. the controllability matrix of (A, B) is not full rank.

(4) (5): We will show not (5) implies not (4). Suppose (5) is not true so that there exists a1 n vector v , and C,

vT(I A) = 0, vTB = 0.

for some . Then vTA = vT, vTA2 = 2vT etc. Because vTB = 0 by assumption, we havevTAB = vTB = 0, vTA2B = 2vTB = 0 etc. Hence,

vTB = vTAB = vTA2B = vTAn1B = 0.

Therefore the controllability is not full rank.

The proof of (4) (3) requires the 2nd representation theorem, and we shall leave it afterward.Note that (2) (1) is obvious. To see that (1) (2), notice that the controllability matrix C

does not depend on T, so controllability does not depend on duration of time interval. Since thesystem is time invariant, it does not depend on the initial time (t0).

Proof of (4) (3) - Optional

To prove (4) (3), we need the so called 2nd Representation Theorem. It is included here forcompleteness. We will not go through this part.

Definition Let A : n n and W n a subspace with the property that for each w W,Aw W. We say that W is Ainvariant.

Theorem(2nd Representation theorem) Let

A: n n is a linear map (i.e. a matrix)

W n be anAinvariantk dimensional subspace ofn.

There exists a nonsingularT = e1 e2 en s.t. e1, , ek form a basis ofW,

in the basis ofe1, e2, , en, A is represented by:

T1AT =

A11 A12

0 A22

A11 kk wherek= dim W.

For any vectorB W, the representation ofB is given by

B = TB

0

.


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42 cPerry Y.Li

The point of the theorem is that A can be put in a form with 0 in strategic locations ( A21 arezeros), similarly for any vector B W (B2 are zeros).

Proof: Take e1, , ek to be a basis ofW, and choose ek+1, en to complete the basis ofn.BecauseAei W for each i= 1, k,

Aei =k

l=1

liel.

Therefore, li are the coefficients ofA11 and the coefficients ofA21 are all zero.LetB W, then

B =k

l=1

lel

so that l are simply the coefficients in B1 and the coefficients ofB2 are zeros.

Proof of (4) (3) in controllability test: Notice that

A

B AB A2B An1B

=

AB A2B An1B AnB

therefore, by the Cayley Hamilton Theorem, the range space of the controllability matrix is Ainvariant.

Suppose that (3) is not true, so that the rank of

B AB A2B An1Bis less than n. Since the controllability matrix is A-invariant, by the 2nd representation theorem,there is an invertible matrix Tso that in the new coordinates, z = T1x,

z=

A11 A12

0 A22

z+

B0

u (3.6)

where B = T

B0

, and

A= T

A11 A12

0 A22

T1.

and the dim ofA22 is non-zero (it is n - rank(C) ).LetvT2 be a left eigenvector of

A22 so that

vT2(I A22) = 0

for some C. Define vT :=

0 vT2

T1. Evaluating vTB, vT(I A) gives

vTB =

0 vT2

T1T

B0

= 0

vT(I A) =

0 vT2

T1(T T1 TAT1)

= 0 vT2 (I

A)T1 = 0.

This shows that [I A, B] has rank less than n.Remarks:


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1. Notice that the controllability matrix is not a function of the time interval. Hence, if a LTIsystem is controllable over some interval, it is controllable over any (non-zero) interval. c.f.

with result of linear time varying system.

2. Because of the above fact, we often say that the pair (A, B) is controllable.

3. Controllability test can be done by just examiningAandB without computing the grammian.The test in (4) is attractive in that it enumerates the vectors in the controllability subspace.However, numerically, since it involves power ofA, numerical stability needs to be considered.

4. The PBH Test in (5), or the Hautus test for short, involves simply checking the condition atthe eigenvalues. It is because for (sIA, B) to have rank less than n,smust be an eigenvalue.

5. The range space of the controllability matrix is of special interests. It is called the control-

lable subspace and is the set of all states that can be reached from zero-initial condition.This is A invariant.

6. Using the basis for thecontrollable subspaceas part of the basis forn, the controllabilityproperty can be easily seen in (3.6).

3.7 Linear Time Invariant (LTI) discrete time system

For the discrete time system (3.7),

x(k+ 1) =Ax(k) + Bu(k)

y(k) =C x(k) (3.7)

where x n, u m, the conditions for controllability is fairly similar except that we need toconsider controllability over period of length greater than n.

First, recall that the reachability map for a linear discrete time system Lr,[k,k1] : u() s(k, k0, x= 0, u()) is given by:

Lr,[k0,k1][u()] =k1i=k0

k1j=i+1A(j)B(i)u(i) (3.8)

Theorem For the discrete time system (3.7), the followings are equivalent:

1. The system is controllable over the interval [0, T], with T n (i.e. it can transfer from anyarbitrary state at k = 0 to any other arbitrary state at k = T).

2. The system is controllable over the interval [k0, k1], with k1 k0 n

3. The controllability grammian,

Wr,T :=T1l=0

AlBBAl

is full rank n.

4. The controllability matrix

B AB A2B An1Bhas rank n. Notice that the controllability matrix has dimension n nm where m is thedimension of the control vector u.


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44 cPerry Y.Li

5. For eachz C, the matrix,

zI A Bhas rank n.

Proof: The proof of this theorem is exactly analogous to the continuous time case. However, inthe case of the equivalence of (2) and (3), we can notice that

x(k) =Akx(0) + [B,AB,A2B, An1B]

u(k 1)u(k 2)

...u(0)

so clearly one can transfer to arbitrary states at k = n iff the controllability matrix is full rank.Controllability does not improve forT > nis again the consequence of Cayley-Hamilton Theorem.

3.8 Observability

The observability question deals with the question of while not knowing the initial state, whetherone can determine the state from the input and output. This is equivalent to the question ofwhether one can determine the initial state x(0) when given the input u(t), t [0, T] and the

output y (t), t [0, T].

Definition 3.8.1 A state determined dynamical system is called observable on [t0, t1] if for allinputs, u(), and outputsy(), [t0, t1], the statex(t0) =x0 can be uniquely determined.

For observability, the null space of the observability map on the interval [t0, t1], given by:

Lo : x0y() =C()(, t0)x0, y() =C()(, t0)x0(, 0)x, [t0, t1]

must be trivial (i.e. only contains 0). Otherwise, ifLo(xn)(t) =y(t) = 0, for all t [0, T], then for

any ,y(t) =C(t)(t, 0)x= (t, 0)(x + xn).

Recall from the finite rank theorem that if the observability map Lo is a matrix, then the nullspace ofLo,N(Lo) =N(LToLo). For continuous time system, Lo can be thought of as an infinitelylong matrix, then the matrix product operation in LToLo becomes an integral, and is written asLoLo where L

denotes the adjoint of the L.

The observability Grammian is given by:

Wo,[t0,t1] = LoLo=

t1

t0

(t, t0)TCT(t)C(t)(t, t0)

nn

where Lo is the adjoint (think transpose) of the observability map.


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3.9 Least square estimation

3.9.1 Formulation and solution

Consider the continuous time varying system:

x(t) =A(t)x(t) y(t) =C(t)x(t).

The observability map on [0, t1] is given by: [0, t1]

Lo: x(0)y() =C()(, 0)x(0) (3.9)

The so-called Least squares estimate ofx(0) given the output y (), [0, t1] is:

x(0|t1) = arg minx0

t10

eT()e()d.

The least squares estimation problem can be cast into a set of linear equations. For example, wecan stack all the outputs y(t), t [0, t1] into a vector Y = [Y(0), . . . , Y (0.1), . . . , Y (0.2), . . . , Y (t1)]T pp, and represent Lo as a matrix ppn, then,

Y =Lox0+ E;

xo= argminx0ETE= (LToLo)

1LToY

Here, E = [E(0), . . . , E (t1)]T = pp is the modeling error or noise. The optimal estimatex0 usesthe minimum amount of noise Eto explain the data Y.

Notice that the resulting E satisfies ETLo = 0, i.e. E is normal to Range(Lo).

Using the observability map for the LTI continuous time system (3.9) as Lo,

x(0|t1) =W1o (t1)

t10

T(, 0)CT()y()d

Wo(t1) =LoLo=

t10

T(, 0)CT()C()(, 0)d.

One can also find estimates of the states at time t2 given data up to time t1:

x(t2|t1) = (t2, 0)x(0|t1)

This refers to three classes of esimation problems:

t2< t1: Filtering problem

t2> t1: Prediction problem

t2= t1: Observer problem

The estimate x(t|t) is especially useful for control systems as it allows us to effective do statefeedback control.


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46 cPerry Y.Li

3.9.2 Recursive Least Squares Observer

The least squares solution computed above is expensive to compute:

x(t|t) = (t, 0)W1o (t)

t0

T(, 0)CT()y()d

Wo(t) =LoLo=

t0

T(, 0)CT()C()(, 0)d.

It is desirable if the estimate x(t|t) can be obtained recursively via a differential equation.

d

dtx(t|t) =A(t) (t, 0)W1o (t)

t0

T(, 0)CT()y()d

x(t|t)+ (t, 0)

d

dtW1o (t)

t0

T(, 0)CT()y()d

+ (t, 0)W1o (t)T(t, 0)CT(t)y(t)

Now by differentiating Wo(t)W1o (t) =I, we have:

d

dtW1o (t) =W

1o (t)

dWodt

(t)W1o (t)

whered

dtWo(t) =

T(t, 0)CT(t)C(t)(t, 0)

d

dtx(t|t) =A(t)x(t|t) (t, 0)W1o (t)

T(t, 0)CT(t)C(t)x(t|t) + (t, 0)W1o (t)T(t, 0)CT(t)y(t)

=A(t)x(t|t) (t, 0)W1o (t)T(t, 0)

P(t)

CT(t)[C(t)x(t|t) y(t|t)

y(t)] (3.10)

P(t) satisfies:P(t) =A(t)P(t) + P(t)AT(t) P(t)CT(t)C(t)P(t)

Typically, we would initialize P(0) to be a large non-singular matrix, e.g. P(0) = I, where islarge.

In the recursive least squares observer in (3.10), the first term is an openloop prediction term,

and the second term is called an output injection where an output prediction error is used to correctthe estimation error.We can think of the observer feedback gain as:

L(t) =P(t)CT(t)

3.9.3 Covariance Windup

Let us rewrite P(t) as follows:

P(t) = (t, 0)

t0

T(, 0)CT()C()(, 0)d

1T(t, 0)

= t

0T(, t)CT()C()(, t)d

1=:K1(t)


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The matrix K(t) :=P1(t) is known as the Information Matrixand satisfies:

K=CT(t)C(t) AT(t)K(t) K(t)A(t)

Notice that Wo(t) increases with time. Thus, Wo1(t) decreases with time, as would K1(t).This means that the observer gain L(t) will decrease. The observer will asymptotically rely moreand more on open loop estimation:

d

dtx(t|t) =A(t)x(t|t)

Forgetting factor

Forget old information ( far away from current time): Choose >0:

K(t) =

t0

T(, t)CT()C()(, t)e(t)d;

K= CT(t)C(t) AT(t)K(t) K(t)A(t) K

Covariance reset

IfK(t) becomes large, reset it to small number.

Check maximum singular value ofK(t), K(t).

If K(t) max, K(t) =I, i.e. P(t) =I /.

3.10 Observability Tests

The observability question deals with the question of whether one can determine what the initialstate x(0) is, given the input u(t), t [0, T] and the output y (t), t [0, T].

For observability, the null space of the observability map,

Lo: x y() = (, 0)x, [0, T],

must be trivial (i.e. only contains 0). Otherwise, ifLo(xn)(t) =y(t) = 0, for all t [0, T], then forany ,

y(t) = (t, 0)x= (t, 0)(x + xn).

Hence, one cannot distinguish the various initial conditions from the output.

Just like for controllability, it is inconvenient to check the rank (and the null space) of theLo which is tall and thin. We can instead check the rank and the null space of the observabilitygrammian given by:

Wo,T =LoLo

where Lo is the adjoint (think transpose) of the observability map.

Proposition 3.10.1 Null(Lo) =Null(LoLo).

Proof:


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48 cPerry Y.Li

Let x Nul(Lo) so, Lox = 0. Then, clearly, LToLox = LT0. Therefore, Null(Lo)

Null(LoLo).

Let x Nul(LoLo). Then, xTLoLox = 0. Or, (Lox)

T(Lox) = 0. This can only be true ifLox= 0. Therefore, Null(L

oLo) Null(Lo).

3.11 Observability Tests for Continuous time LTI systems

Theorem For the LTI continuous time system (3.5), the followings are equivalent:

1. The system is observable over the interval [0, T]2. The observability grammian,

Wo,T :=

T0

eAtCCeAtdt

is full rank n.

3. The observability matrix

CCA

CA2

..

.CAn1

has rank n. Notice that the observability matrix has

dimension np n where p is the dimension of the output vector y .

4. For eachs C, the matrix, sI A

C

has rank n.

Proof: The proof is similar to the ones as in the controllable case and will not be repeated here.Some differences are that instead of considering 1 n vectorvT multiplying on the left hand sidesof the controllability matrix and the grammians, we consider n 1 vector multiplying on the RHS

of the observability matrix and the grammian etc.Also instead of considering the range space of the controllability matrix, we consider the NULL

space of the observability matrix. Its null space is also A-invariant. Hence if the observabilitymatrix is not full rank, then using basis for its null space as the last k basis vectors ofn, thesystem can be represented as:

z =

A11 0

A21 A22

z+

B1B2

u

y=

C 0

z

where C= C 0T1, andA= T

A11 A12

0 A22

T1.


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and the dim ofA22 is non-zero (and is the dim of the null space of the observability matrix).

Remark

1. Again, observability of a LTI system does not depend on the time interval. So, theoreticallyspeaking, if observing the output and input for an arbitrary amount of time will be sufficientto figure out x(0). In reality, when more data is available, one can do more averaging toeliminate effects of noise (e.g. using the Least square ie. Kalman Filter approach).

2. The subspace of particular interest is the null space of the controllability matrix. An initialstate lying in this set will generate identically 0 zero-input response. This subspace is calledthe unobservable subspace.

3. Using the basis of the unobservable subspaceas part of the basis ofn, the observability

property can be easily seen.

3.12 Observability Tests for Discrete time LTI systems

The tests for observability of the discrete time system (3.7) is given similarly by the followingtheorem.

Theorem 3.12.1 For the discrete time system (3.7), the followings are equivalent:

1. The system is observable over the interval [0, T], for someT n.

2. The observability grammian,

Wo,T :=T1k=0

AkCCAk

is full rankn.

3. The observability matrix

CCA

CA2

...CAn1

has rank n. Notice that the observability matrix has

dimensionnp n wherep is the dimension of the output vectory.4. For eachz C, the matrix,

zI AC

has rankn.

The controllability matrix can have the following interpretation: zero-input response is givenby:

y(0)y(1)y(2)

...y(n 1)

=

C

CACA2

...CAn1

x(0).


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50 cPerry Y.Li

Thus, clearly if the observability matrix is full rank, one can reconstruct x(0) from measurementsofy(0), , y(n 1). One might think that by increasing the number of output measurement times

(e.g. y(n)) the system can become observable. However, because of Cayley-Hamilton theorem,y(n) =C Anx(0) can be expressed as

n1i=0 aiCA

ix(0) consisting of rows already in the observabilitymatrix.

3.13 PHB test and Eigen/Jordan Forms

Considerx= Ax + Bu, y= C x

Suppose first that A is semi-simple, then A= T DT1 and D = diag 1, . . . , n where i might berepeating.

The PBH test in the transformed eigen-coordinates is to check

rank(iD, T1B)

for each i if it is n. This shows thatT1B must be non-zero in each row. Also, ifi = j for somei=j , then a single input system cannot be controllable.

Now suppose that A = T JT1 and J is in Jordan form,

J=

1 1 0 00 1 0 00 0 1 00 0 0 2

This Jordan form has 3 generalized eigenvector chains.

Controllability

From homework, we saw that if the entries ofB 4 at the beginning of some chain is 0,i.e..

T1B =

b1b2b30

, T1B=

b1b20b4

, T1B =

b10

b3b4

then the system is uncontrollable.

Thus, T1B being nonzero at the beginning of each chain is a necessary condition for con-trollability.

PBH test shows that it is in fact a sufficient condition.

Observability

Similarly, ifC T 1n is a zero entry at the end of any chain (these correspond to thecoordinates for the eigenvector), i.e.

CT =

0 c2 c3 c4

, CT =

c1 c2 0 c4

,

or, CT = c1 c2 c3 0then we saw from the homework that the system is not observable.


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Hence, C Thaving non-zero entries at the end of each chainis necessary for observability.

PHB test shows that it is also sufficient.

Note If the system is either uncontrollable or unobservable, then the order of the transferfunction will be less than the order ofA.

However, for a multi-input-multi-output system, the fact that the order of each transfer functionin the transfer function matrix has order less than the order ofA, does NOT imply that the systemis uncontrollable or unobservable. e.g.

A=

1 00 2

B = C=

1 00 1

gives,

Y(s) = 1

s+1 00 1s+2

U(s).

3.14 Kalman Decomposition

3.14.1 Controllable / uncontrollable decomposition

Suppose that the controllability matrix C nn of a system has rank n1 < n. Then there existsan invertible transformation, T nn such that:

z= T1x,

z=

A11 A12

0 A22

z+

B0

u (3.11)

where B = T

B0

, and

A= T

A11 A12

0 A22

T1.

and the dim ofA22 is n n1.

3.14.2 Observable / unobservable decompositionHence if the observability matrix is not full rank, then using basis for its null space as the last kbasis vectors ofn, the system can be represented as:

z =

A11 0

A21 A22

z+

B1B2

u

y=

C 0

z

where C=

C 0

T1, and

A= TA11 A12

0 A22T1.

and the dim ofA22 is non-zero (and is the dim of the null space of the observability matrix).


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52 cPerry Y.Li

3.14.3 Kalman decomposition

The Kalamn decomposition is just the combination of the observable/unobservable, and the con-trollable/uncontrollable decomposition.

Theorem 3.14.1 There exists a coordinate transformationz= T1x n such that

z =

A11 0 A13 0

A21 A22 A23 A240 0 A33 0

0 0 A43 A44

z+

B1B200

u

y =

C1 0 C2 0

z

LetT = T1 T2 T3 T4 (with compatible block sizes), then T2 (C O): T2= basis for Range(C) Null(O).

T1 (C O): T1 T2 = basis for Range(C)

T4 (C O): T2 T4 = basis for Null(O).

T3 (C O): T1 T2 T3 T4 = basis forn.

Note: Only T2 is uniquely defined.

Proof: This is just combination of the obs/unobs and cont/uncont decomposition.

3.14.4 Stabilizability and Detectability

If the uncontrollable modes A33, A44 are stable (have eigenvalues on the Left Half Plane), then,system is called stabilizable.

One can use feedback to make the system stable;

Uncontrollable modes decay, so not an issue.

If the unobservable modes A22, A44 are stable (have eigenvalues on the Left Half Plane), then,

system is called detectable.

The states z2, z4 decay to 0

Eventually, they will have no effect on the output

State can be reconstructed by ignoring the unobservable states (eventually).

The PBH tests can be modified to check if the system is stabilizable or detectable, namely,check

rank

sI A B

rank

sI A

C

for alls with non-negative real parts if they lose rank. Specifically, one needs to check onlys whichare the unstable eigenvalues ofA.


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3.15 Realization

3.15.1 Degree of controllability / observability

Formal definitions for controllability and observability are black and white. In reality, some statesare very costly to control to, or have little effect on the outputs. They are therefore, effectivelyuncontrollable or unobservable.

Degree of controllability and observability can be evaluated by the sizes of Wr and Wo withinfinite time horizon:

Wr(0, ) = limT

T0

eA(T)BBTeAT(T)d

=

0

eAtBBTeATtdt

Wo(0, ) = limT

T0

eATCTCeAd

Notice that Wr satisfies differential equation:

d

dtWr(0, t) =AWr(0, t) + Wr(0, t)A

T + BBT

Thus, if all the eigenvalues ofA have strictly negative real parts (i.e. Ais stable), then, as T ,

0 =AWr+ WrAT + BBT

which is a set of linear equations in the entries ofWr, and can be solved effectively. The equationis called a Lyapunov equation. Matlab can solve for Wr using gram.

Similarly, ifA is stable, as T , Wo satisfies the Lyapunov equation

0 =ATWo+ WoA + CTC

which can be solved using linear method (or using Matlab via gram).

Wr is related to the minimum norm control. To reach x(0) = x0 from x() = 0 in infinitetime,

u= LT

rW1

c x0

minu()J(u) = minu()

0

uT()u()d =xT0 W1c x0.

If statex0is difficult to reach, thenxT0 W

1c x0is large meaning thatx0is practically uncontrollable.

Similarly, if the initial state is x(0) =x0, then, with u() 0 0

yT()y()d=xT0 Wox0.

Thus, xT0 Wox0 measures the effect of the initial condition on the output. If it is small, the effect ofx0 in the output y() is small, thus, it is hard to observe, or practically unobservable.

Generally, one can look at the smallest eigenvalues Wc and Wo, the span of the associatedeigenvectors will be difficult to control, or difficult to observe.


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54 cPerry Y.Li

3.15.2 Relation to Transfer Function

For the system

x= Ax + Bu

y= C x + Du

The transfer function from U(s) Y(s) is given by:

G(s) =C(sI A)1B+ D

=Cadj(sI A)B

det(sI A) + D.

Note that transfer function is not affected by similarity transform.From Kalman decomposition, it is obvious that

G(s) = C1(sI A11)1B1+ D

where A11, B1, C1 correspond to the controllable and observable component.

Poles

they are values ofs C s.t. G(s) becomes infinite.

poles ofG(s) are clearly the eigenvalues ofA11.

Zeros

For the SISO case, z C(complex numbers) is a zero if it makes G(z) = 0.

Thus, zeros satisfy

Cadj(zI A)B+ Ddet(zI A) = 0

assuming system is controllable and observable, otherwise, need to apply Kalman decompo-sition first.

In the general MIMO case, a zero implies that there can be non-zero inputsU(s) that produce

output Y(s) that is identically zero. Therefore, there exists X(z), U(z) such that:

zI A X(z) =B U(z)

0 =C X(z) + D U(z)

This will be true if and only if at s = z

rank

sI A B

C D

is less than the normal rank of the matrix at other s.

We shall return to the multi-variable zeros when we discuss the effect of state feedback on the zeroof the system.


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3.15.3 Pole-zero cancellation

Cascade system with input/output (u1, y2):

System 1 with input/output (u1, y1) has a zero at and a pole at .

x1= A1x1+ B1u1

y1= C1x1

System 2 with input/output (u2, y2)has a pole at and a zero at.

x2= A2x2+ B2u2

y2= C2x2

Cascade interconnection: u2= y1.

Then

1. The system pole at is not observable from y2

2. The system pole at is not controllable from u1.

Combined system:

A=

A1 0B2C1 A2

; B =

B10

; C=

0 C2

Considerx =

x1x2

.

Use PBH test with = .

I A1 0B2C1 I A2

0 C2

x1x2

= 0???

Letx1 be eigenvector associated with for A1.

Letx2 = (I A2)1B2C2x1 (i.e. solve second row in PBH test).

Then since

Cx= C2x2= (C2(I A2)1B2)C2x1

and is a zero of system 2, PHB test shows that the mode is not observable.

Similar for the case when mode in system 2 not controllable by u.

These example shows that when cascading systems, it is important not to do unstable pole/zerocancellation.


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56 cPerry Y.Li

3.16 Balanced Realization

The objective is to reduce the number of states while minimizing effect on I/O response (transferfunction). This is especially useful when the state equations are obtained from distributed param-eters system, P.D.E. via finite elements methods, etc. which generate many states. If states areunobservable or uncontrollable, they can be removed. The issue at hand is if some states are lightlycontrollable while others are lightly observable. Do we cancel them?

In particular,

Some states can be lightly controllable, but heavily observable.

Some states can be lightly observable, but heavily controllable.

Both contribute to significant component to input/output (transfer function) response.

Idea: Exhibit states so that they are simultaneously lightly (heavily) controllable and observ-able.

Consider a stable system:

x= Ax + Bu

y= C x + Du

IfA is stable, reachability and observability grammians can be computed by solving the Lya-punov equations:

0 =AWr+ WrAT + BBT

0 =ATWo+ WoA + CTC

If state x0 is difficult to reach, then xT0 W

1r x0 is large practically uncontrollable.

IfxT0 Wox0 is small, the signal ofx0 in the output y() is small, thus, it is hard to observe practically unobservable.

Generally, one can look at the smallest eigenvalues Wr and Wo, the span of the associatedeigenvectors will be difficult to control, or difficult to observe.

Transformation of grammians: (beware of which side T goes !):

z = T x

Minimum norm control should be invariant to coordinate transformation:

xT0 W1r x0= z

TW1r,zz = xTTTW1r,zT x

ThereforeW1r =T

TW1r,zT Wr,z=T WrTT.

Similarly, the energy in a state transmitted to output should be invariant:

xTWox= zWo,zz Wo,z =TTWoT

1.

Theorem 3.16.1 (Balanced realization) If the linear time invariant system is controllable andobservable, then, there exists an invertible transformationT nn s.t. Wo,z =Wr,z.


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58 cPerry Y.Li

However, this does not retain the D.C. (steady state) gain from u to y (which is important froma regulation application point of view).

To maintain the steady state response, instead of eliminating z2 completely, replace z2 by itssteady state value:

In steady state:

z2= 0 =A21z1+ A22z2+ B2u

z2 =A122(A21z1+ B2u)

This is feasible ifA22 is invertible (0 is not an eigenvalue).A truncation approach that maintains the steady state gain is to replace z2 by its steady state

value:

z1= A11z1+ A12z2+ B1uy = C1z1+ C2z

2+ Du

so that,

z1=

A11 A12A122A21

Ar

+

B1 A12A122B2

Br

u

y=

C1 C2A122A21

Cr

z1+

D C2A122B2

Dr

u

The truncated system with state z1 n1 is

z1 = Arz1+ Bru

y= Crz1+ Dru

which with have the same steady state response as the original system.

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Chapter 3 Controllability Observability

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