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EE 580 Linear Control Systems

X. Structural Properties of Linear Systems (Part 1)

Department of Electrical Engineering

Pennsylvania State University

Fall 2010

10.1 Introduction

Recall that a linear map, or its matrix representation, between finite-dimensional vector spaces aredecomposed into Jordan blocks, which completely characterize the stability properties of the linear

map. Similar things can be done to linear dynamical systems, or their state-space representations.Consider the linear time-invariant system

x(t) =Ax(t) + Bu(t),

y(t) =Cx(t) + Du(t), (10.1)

where A Rnn, B Rnm, C Rln, and D Rlm are given matrices. If the system (10.1)is not controllable (i.e., not reachable), then it is possible to separate the controllable part of thesystem from the uncontrollable part. Similarly, if the system is not observable, then it is possibleto separate the observable part from the unobservable part.

10.2 Standard Form for Uncontrollable Systems

Lemma 10.1 Let C be the controllability matrix of the pair (A, B). If rankC = nr, then thereexists a nonsingular matrixQ Rnn such that

Q1AQ=

A1 A120 A2

and Q1B=

B10

, (10.2)

whereA1 Rnrnr , B1 R

nrm, and the pair(A1, B1) is controllable.

Proof. Let {v1, . . . , vnr} be a basis for the controllable subspace R(C). Since R(C) is in-variant under A, there exists a unique matrix A1 Rnr

nr , whose columns are the coordinaterepresentations ofAv1, . . . , Avnr with respect to {v1, . . . , vnr}, such that

A v1 vnr= v1 vnrA1.On the other hand, if an x Rn is such that x= Bu for some u Rm, then

x= [B AB An1B][uT 0 0]T,

so R(B) R(C). Thus there exists a unique B1 Rnrm , whose columns are the coordinate

representations of the columns ofBwith respect to {v1, . . . , vnr}, such that

B=

v1 vnr

B1.

1

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EE 580 Linear Control Systems 2

Now, choosing any linearly independent vectors qnr+1, . . . , qn such that

Q=

v1 vnr qnr+1 qn

is nonsingular, we have

A v1 vnr= Q A10 and A qnr+1 qn= Q A12A2 for some A12 Rnr

(nnr) and A2 R(nnr)(nnr); similarly

B= Q

B10

.

This leads to (10.2). It remains to show that (A1, B1) is controllable. The controllability matrixC of the pair Q1AQ, Q1B is given by

C=

B1 A1B1 A

n11 B1

0 0 0

= Q1C. (10.3)

Here, the first equation in (10.3) implies that the rank ofC is equal to the rank ofB1 A1B1 A

n11 B1

,

where n nr. By the Cayley-Hamilton theorem, however, the columns of this matrix are linearcombinations of the columns of the controllability matrix of (A1, B1) given by

C1=

B1 A1B1 Anr11 B1

Rnrmnr ,

whose rank is at most nr. On the other hand, the second equation in (10.3) says that rankC =rankC= nr. Therefore, we have rank C1= nr, and hence the pair (A1, B1) is controllable.

IfQ is as in Lemma 10.1, then the change of state variables given by Qx(t) = x(t) results inan equivalent system of the form x1(t)

x2(t)

=A1 A12

0 A2

x1(t)x2(t)

+B1

0

u(t),

y(t) =

C1 C2 x1(t)

x2(t)

+ Du(t),

(10.4)

where x(t) = [x1(t)T x2(t)]

T with x1(t) Rnr and x2(t) Rnnr , and where (A1, B1) is con-trollable. This form of state-space representations is called the standard form for uncontrollablesystems. The eigenvalues ofA1 (resp. A2) are called controllable eigenvalues(resp. uncontrollableeigenvalues) of the system (10.1). Recall that the inverse Laplace transform of (sI A)1 yields

e

At

=

p

i=1mi1

k=0

Aikt

k

e

it

with

Aik = 1

k!(mi 1 k)! limsi

(s i)

mi(sI A)1mi1k ,

where 1, . . . , p are distinct eigenvalues of A and m1, . . . , mp are their algebraic multiplicities,and where the terms tkeit are called the modesof the system (10.1). In particular, the modes ofthe system corresponding to controllable eigenvalues (resp. uncontrollable eigenvalues) are calledcontrollable modes (resp. uncontrollable modes) of the system.

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Lemma 10.2 The input-output description of the system(10.1)is determined solely by its control-lable part; that is, if (10.1) and (10.4) are equivalent, then the transfer function matrix of (10.1)is given by

H(s) =C(sI A)1B + D= C1(sI A1)

1B1+ D.

Proof. IfQ is as in Lemma 10.1, then we have

eQ1AQ =Q1eAQ and e

A1 A12

0 A2

=

eA1

0 eA2

,

where the symbol denotes the block that we do not care. Hence

CeAB=

CQ

Q1eAQ

Q1B

=

C1 C2 eA1

0 eA2

B10

= C1e

A1B1,

from which the result follows. As a simple example, consider A = [ 1 10 0 ], B = [

10 ], and C = [ 1 2 ]. Then C = [

1 10 0 ], so the

reachable subspace R(C) = span{v1} with v1= [ 10 ]. Choose v2= [01 ], and let Q= [ v1 v2] = [

1 00 1 ]

to obtain Q1AQ= A1 A120 A2 = [ 1 10 0 ], Q1B = B10 = [ 10 ], and CQ = [ C1 C2] = [ 1 2 ]. Thus,among the two eigenvalues (namely, 1 and 0) of the system, the eigenvalue 1 is controllable and theeigenvalue 0 is uncontrollable. Indeed, the transfer function of the system is

H(s) =C(sI A)1B=

1s1

2s1s(s1)

10

=

1

s 1=C1(sI A1)

1B1.

Since rankO = rank[ 1 21 1 ] = 2, both eigenvalues are observable. However, the uncontrollableeigenvalue 0 is never excited by the input, and hence does not appear as a system pole.

10.3 Standard Form for Unobservable Systems

Lemma 10.3 LetO be the observability matrix of the pair (A, C). If rankO = no, then thereexists a nonsingular matrixQ Rnn such that

Q1AQ=

A1 0

A21 A2

and CQ=

C1 0

, (10.5)

whereA1 Rnono, C1 Rln0, and the pair(A1, C1) is observable.

Proof. Let {vno+1, . . . , vn} be a basis for the unobservable subspace N(O). Since N(O) isinvariant under A, there exists a unique matrix A2 R

(nno)(nno), whose columns are thecoordinate representations ofAvno+1, . . . , Avn with respect to{vno+1, . . . , vn}, such that

A vno+1 vn= vno+1 vnA2.On the other hand, ifOx= 0, then Cx= 0, so N(O) N(C). Thus we have

C

vno+1 vn

= 0.

Chooseq1, . . . , qno be linearly independent vectors such that

Q=

q1 . . . qno vno+1 vn

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is nonsingular. Then we have

A

q1 qno

= Q

A1

A21

and A

vno+1 vn

= Q

0

A2

for some A1 R

nono and A21 R(nno)no . Similarly, CQ= [C1 0] for some C1 R

lno. This

gives (10.5). Moreover, the observability matrixO of the pair Q1AQ, CQ is given byO= CT1 AT1 CT1 An11 TCT1

0 0 0

T= OQ.

Here, the first equation, along with the Cayley-Hamilton theorem, implies that rankO is equal tothe rank of the observability matrix

O1 =

CT1 AT1 C

T1

An11

TCT1

T Rlnono

of (A

1,C

1), which is at most no; however, the second equation gives rankO = rankO = no.Therefore, we have rankO1= no, and hence (A1, C1) is observable. IfQ is as in Lemma 10.3, then the change of state variables given by Qx(t) = x(t) results in

an equivalent system of the formx1(t)x2(t)

=

A1 0

A21 A2

x1(t)x2(t)

+

B1B2

u(t),

y(t) =

C1 0 x1(t)

x2(t)

+ Du(t),

(10.6)

wherex(t) = [x1(t)T x2(t)]

T withx1(t) Rno andx2(t) Rnno, and where (A1, C1) is observable.This form of state-space representations is called the standard form for unobservable systems. The

eigenvalues ofA1 (resp. A2) and the corresponding modes are called observable eigenvalues(resp.unobservable eigenvalues) and observable modes (resp. unobservable modes) of the system (10.1).

Lemma 10.4 The input-output description of the system (10.1)is determined solely by its observ-able part. That is, if (10.1) and (10.6) are equivalent, then the transfer function matrix of (10.1)is given by H(s) =C(sI A)1B + D= C1(sI A1)1B1+ D.

Proof. The proof parallels that of Lemma 10.2. For example, suppose A = [ 1 10 0 ], B = [

11 ], and C= [ 1 1 ]. Since O = [

1 11 1 ], the unobservable

subspace N(O) = span{v2} with v2 =

11

. Choose a v1 = [ 10 ], and let Q = [ v1 v2] = 1 10 1 to obtain Q1AQ= A1 0A21 A2 = [ 1 00 0 ], Q1B = B1B2 = 21 , and CQ = [ C1 0 ] = [ 1 0 ]. Thus,

among the two eigenvalues of the system, the eigenvalue 1 is observable and the eigenvalue 0 isunobservable. Indeed, the transfer function of the system is

H(s) =C(sI A)1B=

1 1 s+1

s(s1)1s

=

2

s 1=C1(sI A1)

1B1.

Since rankC = rank[ 1 21 0 ] = 2, both eigenvalues are controllable. However, the unobservable eigen-value 0 is never observed from the system output, and hence does not appear as a system pole.

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10.4 Kalman Canonical Form of LTI Systems

The following theorem is called the Kalman decomposition theorem. The theorem unifies the de-composition lemmas for uncontrollable and unobservable linear time-invariant systems.

Theorem 10.5 LetC andO be the controllability and observability matrices of the triple(A, B, C).

IfrankC=nr, rankO= no, anddim R(C) N(O) =nro, then there exists a nonsingular matrixQ Rnn such that

Q1AQ=

A11 0 A13 0

A21 A22 A23 A240 0 A33 0

0 0 A43 A44

, Q1B=

B1B20

0

, CQ= C1 0 C3 0 , (10.7)where A11 R

(nrnro)(nrnro), A22 Rnronro, A33 R

(no(nrnro))(no(nrnro)), and A44 R

((nno)nro)((nno)nro), and such that the following hold:

(a) The pair(Ac, Bc) with

Ac= A11 0A21 A22

and Bc= B1B2

is controllable.

(b) The pair(Ao, Co) with

Ao=

A11 A13

0 A33

and Co=

C1 C3

is observable.

(c) The triple(A11, B1, C1) is controllable and observable.

Proof. Choose a basis {v1, . . . , vnr} for R(C) such that {vnrnro+1, . . . , vnr} is a basis forR(C) N(O). Choose vectors vno+nro+1, . . . , vn such that

{vnrnro+1, . . . , vnr ,vno+nro+1, . . . ,vn n no vectors

}

is a basis for N(O). Finally, choose qnr+1, . . . , qno+nro such that

Q=

v1 vnr qnr+1 qno+nro vno+nro+1 vn

is nonsingular. Then, since the firstnr columns ofQ span R(C), (the proof of) Lemma 10.1 impliesthat Q1AQ, Q1B, and CQare of the following partitioned forms:

Q1AQ=

A11

A12

A13

A14

A21 A22 A23 A240 0 A33 A340 0 A43 A44

, Q1B= B

1B20

0

, CQ= C1 C2 C3 C4 . (10.8)Let

T=

Inrnro 0 0 0

0 0 Inro 0

0 In0(nrnro) 0 0

0 0 0 I(nno)nro

,

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where Ik are the k-by-k identity matrices. Then

QT=

v1 vnrnro qnr+1 qno+nro vnrnro+1 vnr vno+nro+1 vn

,

where the lastn nocolumns ofQT spanN(O). Thus (the proof of) Lemma 10.3, along with thefact that T1 =TT, implies that Q1AQ, Q1B, and CQare of the following partitioned forms:

Q1AQ= T(QT)1A(QT)T1 =T

F11 F12 0 0

F21 F22 0 0

F31 F32 F33 F34F41 F42 F43 F44

T1 = F11 0 F12 0

F31 F33 F32 F34F21 0 F22 0

F41 F43 F42 F44

,

Q1B= T(QT)1B= T

G1G2G3G4

=

G1G3G2G4

,CQ= C(QT)T1

H1 H2 0 0

T1 =

H1 0 H2 0

.

Comparing these equations with those in (10.8), we conclude that (10.7) holds. The pair (Ac, Bc)

is controllable by Lemma 10.1, and the pair (Ao, Co) is observable by Lemma 10.3. It follows from

rank

Bc AcBc Anr1c Bc

= rank

B1 A11B1 A

nr111 B1

= nr,

along with the Cayley-Hamilton theorem, that

rank

B1 A11B1 Anr111 B1

= rank

B1 A11B1 A

nrnro111 B1

= nr nro.

Thus the triple (A11, B1, C1) is controllable. Since (Ac, Bc, [C1 0]) is already in the standard formfor unobservable systems, we conclude that (A11, B1, C1) is observable as well.

IfA11, A22, A33, and A44 are as in Theorem 10.5, then we have

det(sI A) = det(sI A11)det(sI A22)det(sI A33)det(sI A44).

The eigenvalues of A11 are controllable and observable, those of A22 are controllable and unob-servable, those ofA33 are uncontrollable and observable, and those ofA44 are uncontrollable andunobservable.

Theorem 10.6 The input-output description of the system (10.1) is determined solely by its con-trollable and observable part. That is, if Q is as in (10.7), then the transfer function matrixof (10.1) is given byH(s) =C(sI A)1B + D= C1(sI A11)1B1+ D.

Proof. The result is an immediate consequence of Lemmas 10.2 and 10.4. For example, consider

A=1 3 21 3 3

1 1 0

, B= 1 01 10 1

, C= 1 1 02 2 1

.

The controllability and observability matrices are

C=

1 0 2 1 4 11 1 2 0 4 20 1 0 1 0 1

, O=1 2 0 1 1 21 2 0 1 1 2

0 1 1 2 0 1

T .

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The controllable(or reachable) subspace Rr = R(C), the unobservable subspace Ro = N(O),and their intersection Rro= R(C) N(O) are

Rr= span

01

1

,

110

, Ro = span

110

, and Rro= span

110

,

respectively. Then

v1 =

011

, v2= 11

0

, q3=11

1

Q= v1 v2 q3= 0 1 11 1 1

1 0 1

.This Qgives

Q1AQ=

1 0 11 2 70 0 1

, Q1B=

0 1

1 00 0

, CQ=

1 0 21 0 5

.

Among the three eigenvalues of A, the eigenvalue 1 is both controllable and observable, theeigenvalue 2 is controllable but unobservable, and the eigenvalue 1 is observable but uncontrollable.Indeed, the transfer function matrix shows a single pole at s= 1:

H(s) =C(sI A)1B= 11

(s (1))1

0 1

=

0 1/(s+ 1)0 1/(s+ 1)

.

10.5 Popov-Belevitch-Hautus (PBH) Tests

We know the system (10.1) is controllable if and only if its controllability matrix has full rowrank. Similarly, the system is observable if and only if its observability matrix has full rank. Thedecomposition results presented above reveal the structural properties of linear systems, and provideadditional tests for controllability and observability. These tests facilitate further insights into thestructural properties of linear systems, and will play an important role in controller synthesisproblems.

Theorem 10.7 (PBH Eigenvector Tests)

(a) The pair(A, B) is controllable if and only if no eigenvectorv ofAT satisfiesBTv= 0.

(b) The pair(A, C) is observable if and only if no eigenvectorv ofA satisfiesCv= 0.

Proof. Suppose there exists an eigenvector v Cn of AT such that BTv = 0. Let Cbe the eigenvalue associated with v. Then we have vTB = 0, vTAB = vTB = 0, vTA2B =

vTAB = 0, . . . , an d s o vTC = 0, where C is the controllability matrix of (A, B). That is,(v)TC =(v)TC = 0. Since v= 0, we have (v)= 0 or (v)= 0, so rank C < n. This provesthat the pair (A, B) being controllable implies BTv= 0 for any eigenvector v ofAT. Conversely,suppose (A, B) is not controllable, so that there exists a nonsingular Q satisfying (10.2), whereA2 R

(nnr)(nnr) with nr < n. Choose an eigenvalue ofAT2, and let v be the corresponding

eigenvector ofAT2 . Then putting v=

Q1T

[0vT]T Rn gives

ATv=

Q1T

Q1AQT 0

v

=

Q1T AT1 0

AT12 AT2

0v

=

Q1T 0

v

= v

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and

BTv= BT

Q1T 0

v

=

Q1BT 0

v

=

BT1 0 0

v

= 0.

This shows that the pair (A, B) is controllable whenever no eigenvector vofAT satisfiesBTv= 0,and hence that part (a) of the theorem holds. The proof of part (b) is similar.

Corollary 10.8 (PBH Rank Tests)

(a) The pair(A, B) is controllable if and only if

rank

I A B

= n

for all eigenvalues ofA (and hence for all C).

(b) A number C is an uncontrollable eigenvalue ofA if and only if

rank

I A B

< n.

(c) The pair(A, C) is observable if and only if

rank

I A

C

= n

for all eigenvalues ofA (and hence for all C).

(d) A number C is an unobservable eigenvalue ofA if and only if

rank

I A

C

< n.

10.6 Kalman Decomposition of Discrete-Time Systems

For discrete-time LTI systems of the form

x(t+ 1) =Ax(t) + Bu(t), t= 0, 1, . . . ;

y(t) =Cx(t) + Du(t), t= 0, 1, . . . ,

even though reachability implies controllability, controllability does not imply reachability. Simi-larly, observability implies constructibility, but the converse does not necessarily hold. Thus, wewill speak of the reachability (as well as observability) of discrete-time systems. Otherwise, allthe theorems in Sections 10.4 and 10.5, with controllability replaced by reachability and theLaplace transformH(s) by the z-transformH(z), carry over to discrete-time LTI systemswithout further change.References

[1] P. J. Antsaklis and A. N. Michel, A Linear Systems Primer. Boston, MA: Birkhauser, 2007.

[2] , Linear Systems, 2nd ed. Boston, MA: Birkhauser, 2006.

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