1 Similarity transform 2. 2 Controllability The PBH test for controllability Observability The PBH test for observability...
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1 Contents 1 Similarity transform 2 2 Controllability 3 21 The PBH test for controllability 5 3 Observability 6 31 The PBH test for observability 7 4 Example ([1, pp121) 9 5 Subspace decomposition Invariant subspaces Reachable subspace decomposition Preliminary remarks Decomposition theorem Eigenvalues Example Observable subspace decomposition Preliminary remarks Decomposition theorem Eigenvalues Example ([1, pp 158) Some notes Kalman decomposition theorem Dimensions 20 6 Appendix 21 7 Appendix 23 8 Appendix 24 9 Appendix Appendix 26 1
2 Here we deal only with Linear Time-Invariant (LTI) systems in continuous-time (t R) ẋ(t) = Ax(t) + Bu(t) y(t) = Cx(t) (01a) (01b) discrete-time (k Z) x(k +1) = Ax(k)+Bu(k) y(k) = Cx(k) (02a) (02b) A R n n is the dynamics matrix B R n m is the input matrix C R p n is the output matrix x R n is the state vector u R m is the input (or control) vector t R denotes time in case of continuous-time systems we will assume t 0 and k = {0,1,2,} The (general) reference for this material is [1, Chapter 38 1 Similarity transform Assuming that A is diagonalizable, let Λ contain on its diagonal the eigenvalues of A Let Q be the modal matrix, ie, its columns contain the eigenvectors of A, hence AQ = QΛ We can express both systems (01) and (02) in the Q basis as follows (see [1 pp93) x(k +1) = QΛQ 1 x(k)+bu(k) Q 1 x(k +1) = ΛQ 1 x(k) +Q 1 Bu(k) }{{}}{{}}{{} z(k+1) z(k) B Essentially, above we performed a change of variable x = Qz The output equation becomes y(k) = C z, where C = CQ Hence, transforming the system with x-coordinates to the system with z-coordinates involves the following transformations A = Λ = Q 1 AQ B = Q 1 B C = CQ 2
3 2 Controllability Definition 21 Controllability (continuous-time systems) The linear system (01a) is controllable on the finite time interval [t 0,t f, if there exists an input u(t) which will drive the system from any initial state x(t 0 ) = x 0 to the zero state x(t f ) = 0 Definition 22 Reachability (continuous-time systems) The linear system (01a) is reachable on the finite time interval [t 0,t f, if there exists an input u(t) which will drive the system from any initial state x(t 0 ) = x 0 to any final state x(t f ) = x f Definition 23 Controllability (discrete-time systems) The linear system (02a) is controllable in K > 0 periods if there exists an input sequence u(k),k = 0,,K 1, which will drive the system from any initial state x(0) = x 0 to the zero state x(k) = 0 Definition 24 Reachability (discrete-time systems) The linear system (02a) is reachable in K > 0 periods if there exists an input sequence u(k),k = 0,,K 1, which will drive the system from any initial state x(0) = x 0 to any final state x(k) = x f 1 Controllability has nothing to do with the system outputs and therefore only the state equation (01a), (02a) is of interest for judging this property 2 The controllability in definitions 21 and 23 is sometimes referred to as controllability-tothe-origin 3 The concepts of controllability and reachability are equivalent for continuous-time systems [1, pp137 4 The concepts of controllability and reachability are not equivalent for discrete-time systems[1, pp137 In my opinion, the notion of controllability in the context of discrete-time systems is not very useful (see Example 21) 5 When dealing with discrete-time systems, Definition 24 is almost always used It is not uncommon for people to use controllable and reachable interchangeably for both continuous- and discrete-time systems (even though for the latter, they are not) Example 21 ([5, pp 96, [1, pp137) Consider the second order discrete-time LTI system [ [ x(k +1) = x(k)+ u(k) Starting from x(0) = (a,b), x 2 can be set to zero in one period by using u(0) = (a+b) However, x 2 (k) = 0 (for any k), hence the profile of x 1 can not be influenced at all in general controllability need not imply reachability controllability does imply reachability if A is nonsingular ([5, pp 96) 3
4 Theorem 21 Condition for controllability/reachability (continuous-time systems) The linear system(01a) is controllable/reachable if and only if the controllability matrix (we could call it reachability matrix) has full rank, ie, rank(c) = n C = [ B AB A 2 B A n 1 B (21) Theorem 22 Condition for reachability (discrete-time systems) The linear system (02a) is reachable if and only if the reachability matrix has full rank, ie, rank(c) = n C = [ B AB A 2 B A n 1 B (22) 7 The condition in the above theorems is both necessary and sufficient for a system to be reachable 8 Controllability is preserved during similarity transformations [1, pp132 C = [ B ΛB Λ 2 B Λ n 1 B = [ Q 1 B Q 1 AQQ 1 B Q 1 AQQ 1 AQQ 1 B = [ Q 1 B Q 1 AB Q 1 A 2 B Q 1 A n 1 B = Q 1 C Since Q is full rank, so is Q 1, hence the matrices C and C have the same rank Hence, if a system is not controllable, expressing it in a different basis would not change this property 9 Controllability is unaffected by applying a linear state feedback u(t) = Kx(t)+v(t), where K R m n is a gain matrix and v(t) R m can be thought of as a new input Substituting u(t) from above in (01a) leads to ẋ(t) = (A BK)x(t)+Bv(t) (A,B) is a controllable pair if and only if (A BK,B) is controllable for all K Intuitively, this statement seems reasonable since the state feedback law can be reversed by setting v(t) = Kx(t)+u(t) Hence, the ability to control the system by using the input v should be the same as when using the input u 10 A necessary and sufficient condition for (01) to be output controllable is that the following matrix has full-rank ([4, pp 691) [ CB CAB CA 2 B CA n 1 B, where x R n Complete state controllability is neither necessary nor sufficient for controlling the output of the system So far we referred to state controllability simply as controllability 11 A system where all uncontrollable states are stable is called stabilizable 4
5 21 The PBH test for controllability PBH test stands for Popov-Belevitch-Hautus test The test can be used for both continuoustime and discrete-time LTI systems Note that we call it controllability test, although probably in the context of discrete-time systems it would be more accurate to call it reachability test Theorem 23 Consider the matrix [ λi A B R n (n+m) The pair (A,B) is controllable if and only if for all eigenvalues λ of A R n n rank ([ λi A B ) = n Or in other words, the pair (A,B) is not controllable if and only if rank ([ λi A B ) < n, for some eigenvalue λ of A Note that if λ C is not an eigenvalue of A, rank(λi A) = n rank ([ λi A B ) = n For that reason, Theorem 23 can be restated as: the pair (A,B) is controllable if and only if rank ([ λi A B ) = n, for any λ C As we will show in Appendix 6, Theorem 23 is equivalent to the following theorem (which is called PBH test as well) Theorem 24 The pair (A,B) is controllable if and only if there exists no left eigenvector w of A that is orthogonal to the columns of B, that is w T B = 0 T When w is a vector with complex entries, w T will be used to denote its complex conjugate transpose (although it is probably more common to use w ) See Appendix 6 for proofs 5
6 3 Observability 1 Observability has nothing to do with the input to the system but only with the way the states are interconnected and the way the output is connected to the states That is why, we will consider the following continuous- and discrete-time unforced systems ẋ(t) = Ax(t) (31a) y(t) = Cx(t), (31b) x(k +1) = Ax(k) y(k) = Cx(k) (32a) (32b) Definition 31 Observability (continuous-time systems) The linear system (31) is observable on the finite time interval [t 0,t f, if any initial state x 0 is uniquely determined by the output y(t) over the same time interval Definition 32 Observability (discrete-time systems) The linear system (32) is observable in K > 0 periods if any initial state x 0 is uniquely determined by the output sequence y(k),k = 1,,K Theorem 31 Condition for observability (continuous- and discrete-time systems) The linear systems (31), (32) are observable if and only if the observability matrix O = C CA CA 2 CA n 1 (33) has full rank, ie, rank(o) = n 2 The condition in the above theorem is both necessary and sufficient for a system to be observable 3 Observability is preserved during similarity transformations [1, pp147 C CQ C Λ CQ O = C Λ 2 CQQ 1 AQ CAQ = CQQ 1 AQQ 1 AQ = CA 2 Q = OQ C Λ n 1 CA n 1 Q 6
7 Since Q is full rank, the matrices O and O have the same rank 4 Even though the controllability property is invariant under any linear state feedback, the observability property is (in general) not (for a justification, see [3, pp 237) Example 31 Consider the state equation ([3, pp 233) [ [ ẋ = x+ u }{{}}{{} A B y = [ 1 2 x }{{} C The pair (A, B) is controllable, while (A, C) is observable Consider the linear state feedback u = [ 3 1 x+r, for r R }{{} K The state feedback equations become [ [ ẋ = x+ r }{{} A BK y = [ 1 2 x The pair (A BK,B) is controllable, however (A BK,C) is not observable 5 A system where all unobservable states are stable is called detectable 31 The PBH test for observability PBH test stands for Popov-Belevitch-Hautus test The test can be used for both continuoustime and discrete-time LTI systems Theorem 32 Consider the matrix [ λi A C R (n+p) n The pair (A,C) is observable if and only if ([ λi A rank C ) = n for all eigenvalues λ of A R n n ([ λi A Or in other words, the pair (A,C) is not observable if and only if rank C for some eigenvalue λ of A ) < n, 7
8 Note that if λ C is not an eigenvalue of A, rank(λi A) = n rank ([ λi A C ) = n For that reason, Theorem 32 can be restated as: the pair (A,C) is observable if and only if ([ ) λi A rank = n, for any λ C C Theorem 32 is equivalent to the following theorem (which is called PBH test as well) Theorem 33 The pair (A,C) is observable if and only if there exists no (right) eigenvector v of A that is orthogonal to the rows of C, that is Cv = 0 The proofs for the above conditions are similar to the ones in Appendix 6 and are left as a homework 8
9 4 Example ([1, pp121) Consider the continuous-time LTI system ẋ(t) = }{{} A y(t) = [ x(t) }{{} C x(t) }{{} B u(t) By using the similarity transformation in Section 1, it can be represented as v(t) = v(t) u(t) (41) }{{}}{{} Λ B y(t) = [ v(t) (42) }{{} C x(t) = Qv(t) Λ = Q 1 AQ B = Q 1 B C = CQ The controllability matrix C = has rank 2 < 4, hence the system is not controllable (see Fig 1) The right and left eigenvectors of A are given by Q = , W = From Fig 1 it is evident that the subsystems corresponding to λ = 2 and λ = 4 are uncontrollable We can verify this using the PBH test in Section 21 to obtain W T 2 B = 0, W T 4 B = 0 From Fig 1 it is evident that the subsystems corresponding to λ = 3 and λ = 4 are unobservable We can verify this using the PBH test in Section 31 to obtain CQ 3 = 0 and CQ 4 = 0 9
10 u v y 1 v v 3 3 v arg Figure 1: Block diagram corresponding to the system (41), (42) subsystem with v 1 is controllable and observable subsystem with v 2 is not controllable but observable subsystem with v 3 is controllable but not observable subsystem with v 4 is not controllable and not observable 10
11 5 Subspace decomposition 51 Invariant subspaces Let A R n n, and S R n be a subspace S is called invariant under A (or A-invariant) if and only if Ax S, for all x S Or in a more compact notation we can write AS S Example 51 If S = R n, then S is A-invariant (for any matrix A R n n ) Example 52 LetS c betherangespaceofc, ie, S c = R(C), wherec isthecontrollability matrix C = [ B AB A 2 B A n 1 B The linearly independent columns of C span the reachable (controllable) subspace of (01a) (and (02a)) We can think of the controllable subspace as the subspace that contains all states which may be reached starting from state zero If Ax S c, for a state x S c, this means that by applying A we could leave the reachable subspace and reach additional points in the state space Hence, it would be reasonable to expect that S c is A-invariant We can actually demonstrate this by noting that if v S c, then there exist m-vectors a 0,a 1,,a n 1 such that Multiplying on the left by A leads to v = Ba 0 +ABa 1 + +A n 1 Ba n 1 Av = ABa 0 +A 2 Ba 1 + +A n Ba n 1 By the Cayley-Hamilton theorem, A n is a linear combination of I, A,,A n 1, hence Av R(C) Example 53 Let S o be the range space of O T, ie, S o = R(O T ), where O T is the transpose of the observability matrix O T = [ C T A T C T (A 2 ) T C T (A n 1 ) T C T The linearly independent columns of O T span the observable subspace of (31) (and (32)) We can demonstrate that S o is A T -invariant by noting that if v S o, then there exist p-vectors a 0,a 1,,a n 1 such that Multiplying on the left by A T leads to v = C T a 0 +A T C T a 1 + +(A n 1 ) T C T a n 1 A T v = A T C T a 0 +(A 2 ) T C T a 1 + +(A n ) T C T a n 1 By the Cayley-Hamilton theorem, (A n ) T is a linear combination of I, A T,,(A n 1 ) T, hence A T v R(O T ) 11
12 Example 54 Here, we demonstrate that the null space of the observability matrix (ie, N(O)) is A-invariant This implies that Av N(O) for any v N(O) Clearly, if v N(O), then hence, For k = n 1, CA k v = 0, for k = 0,,n 1, CA k (Av) = 0, for k = 0,,n 2 CA n 1 (Av) = CA n v = 0 The last equality above follows from the Cayley-Hamilton theorem The same argument can be used for any k n Note that CA k v = 0 does not imply that A k v = 0 (for example A could be full rank) If we consider the system ẋ(t) = Ax(t), S is A-invariant if and only if x(0) S x(t) S, for all t 0 The same statement holds for discrete-time systems 52 Reachable subspace decomposition 521 Preliminary remarks For a given system, the state space (set of all possible states) can always be decomposed into two subspaces a controllable subspace S c and an uncontrollable subspace S c The controllable subspace and the uncontrollable subspace are orthogonal complements, ie, the intersection of the two subspaces is the zero vector (S c S c = 0) the sum of the two subspaces equals the total state space {x c +x c : x c S c,x c S c } = R n A basis for the controllable subspace can be found by taking all linearly independent columns of the controllability matrix Clearly, only if there are n such columns, the controllable subspace becomes the whole state space, ie, the system is controllable A basis for the uncontrollable subspace is a set of vectors that span the orthogonal complement of the controllable subspace Note that this is not the null space of the controllability matrix, since R(C) is not orthogonal to N(C) For example, the range space and null space of the matrix below are the same [ The orthogonal complement of R(C) is N(C T ) 12
13 522 Decomposition theorem If the system (01) (or (02)) is not controllable (ie, rank(c) = r < n), then a non-singular matrix Q can be found, such that [ [ à = Q 1 Ã11 à AQ = 12 B1, B = Q 1 B =, 0 à 22 0 where Ã11 R r r, B1 R r m, and the pair (Ã11, B 1 ) is controllable For proof see Appendix 7 We can think of the similarity transform in the above theorem as the change of variable z = Q 1 x leading to the following system (adopting continuous-time notation) Hence, ż(t) = Ãz(t)+ Bu(t) ż 1 (t) = Ã11z 1 (t)+ã12z 2 (t)+ B 1 u 1 (t) ż 2 (t) = Ã22z 2 (t) in the z-coordinates we can only steer states of the form [ z1 z = 0 in the x-coordinates x = Qz = [ Q 1 Q 2 z = Q1 z 1, we can only steer states that belong to R(C) (see Appendix 7 for definition of Q 1 and Q 2 ) That is why we called R(C) the controllable subspace 523 Eigenvalues [ à = Q 1 Ã11 à AQ = 12 0 à 22 Since the matrices A and à are similar, they have the same eigenvalues Furthermore, because à is block triangular, the eigenvalues of A are given by the eigenvalues of Ã11 and Ã22 the eigenvalues of Ã11 are called the controllable eigenvalues the eigenvalues of Ã22 are called the uncontrollable eigenvalues 524 Example System definition A = , B =
14 u(t) B 1 z 1 (t) Ã 11 Ã 12 z 2 (t) Ã 22 + arg Figure 2: Block diagram corresponding to the decomposed system ż(t) = Ãz(t)+ Bu(t) z 1 is the controllable part of the state The controllability matrix is given by C = [ B AB A 2 B A 3 B = rank(c) = 2 (the first two columns of C are linearly independent) Define Q 1 = , Q 2 = , Q = [ Q 1 Q 2 =
15 The decomposed system becomes [ Ã = Q 1 Ã11 Ã AQ = 12 = 0 Ã , B = Q 1 B = [ B1 = B 2 The controllable eigenvalues are the eigenvalues of Ã11, while the uncontrollable eigenvalues are the eigenvalues of Ã22 In this particular example, due to the fact that A is diagonalizable, one can obtain a decomposition where Ã11 and Ã22 are diagonal matrices, while Ã12 is a zero matrix See Appendix 9 for Matlab code 53 Observable subspace decomposition 531 Preliminary remarks For a given system, the state space (set of all possible states) can always be decomposed into two subspaces an observable subspace S o and an unobservable subspace Sō The observable subspace and the unobservable subspace are orthogonal complements, ie, the intersection of the two subspaces is the zero vector (S o Sō = 0) the sum of the two subspaces equals the total state space {x o +xō : x o S o,xō Sō} = R n A basis for the observable subspace can be found by taking all linearly independent rows of the observability matrix Clearly, only if there are n such rows, the observable subspace becomes the whole state space, ie, the system is observable A basis for the unobservable subspace is a set of vectors that span the orthogonal complement of the observable subspace This is the null space of the observability matrix, since R(O T ) N(O) 532 Decomposition theorem If the system (31) (or (32)) is not observable (ie, rank(o) = r < n), then a non-singular matrix P can be found, such that [ Â = PAP 1 Â11 0 =, Ĉ = CP 1 = [ Ĉ 1 0, Â 21 Â 22 where Â11 R r r, ˆB1 R r m, and the pair (Â11,Ĉ1) is observable For proof see Appendix 8 We can think of the similarity transform in the above theorem as the change of variable z = P x leading to the following system (adopting continuous-time notation) ż(t) = Âz(t) y(t) = Ĉz(t) (51a) (51b) 15
16 Hence, ż 1 (t) = Â11z 1 (t) ż 2 (t) = Â21z 1 (t)+â22z 2 (t) y 1 (t) = Ĉ1z 1 (t) y 2 (t) = 0 z 1 is observable because the pair (Â11,Ĉ1) is observable in the z-coordinates the state is unobservable [ 0 z = z 2 in the x-coordinates x = P 1 z = [ M N [ 0 z 2 = Nz 2 N(O) Any stateinn(o)isunobservable That iswhy we call N(O)theunobservable subspace 533 Eigenvalues [  = PAP 1 Â11 0 =  21  22 Since the matrices A and  are similar, they have the same eigenvalues Furthermore, because  is block triangular, the eigenvalues of A are given by the eigenvalues of Â11 and Â22 the eigenvalues of Â11 are called the observable eigenvalues the eigenvalues of Â22 are called the unobservable eigenvalues 534 Example ([1, pp 158) System definition A = The observability matrix is given by O = C CA CA 2 =, C = [ rank(o) = 2 (the first two rows of O are linearly independent) Define [ P 1 =, P = [ [ P1, P = = P 2 16,
17 y 1 (t) Ĉ 1 z 1 (t)  12 z 2 (t)  11  22 + arg Figure 3: Block diagram corresponding to the decomposed system (51) y 1 is the observable part of the output The decomposed system becomes [  = PAP 1 Â11 0 = =  21  , Ĉ = CP 1 = [ Ĉ 1 Ĉ 2 = [ The observable eigenvalues are the eigenvalues of Â11, while the unobservable eigenvalues are the eigenvalues of Â22 Again, as in the example in Section 524, due to the fact that A is diagonalizable one could obtain a decomposition where Â11 and Â22 are diagonal matrices, while Â21 is a zero matrix The matrices  and Ĉ below are obtained by setting P equal to the inverse of the modal matrix  = Ĉ = [ = Note that the eigenvalues of Â11 are λ = 2 and λ = 1 See Appendix 10 for Matlab code 17
18 54 Some notes 1 Any left eigenvector w of A corresponding to a uncontrollable eigenvalue is orthogonal to the columns of B, ie, w T B = 0 2 Any (right) eigenvector v of A corresponding to a unobservable eigenvalue is orthogonal to the rows of C, ie, Cv = 0 55 Kalman decomposition theorem First, consider the special case when the matrix A is diagonalizable In this case we can find a set of coordinates in which A is diagonal and, possibly with some additional reordering of the states, the system can be written as (we adopt continuous-time notation) ẋ = A co A cō A co A cō y = [ C co 0 C co 0 x x+ B co B cō 0 0 u the subsystem Σ co = (A co,b co,c co ) is controllable and observable the subsystem Σ cō = (A cō,b cō,0) is controllable but not observable the subsystem Σ co = (A co,0,c co ) is observable but not controllable the subsystem Σ cō = (A cō,0,0) is neither controllable nor observable See Fig 4 (it is adopted from [2, pp 223) In the general case, when A is not diagonalizable, the state space can still be decomposed in four parts, however, coupling between the subsystems appears The equations have the following form ẋ = A co 0 0 A cō 0 0 A co A cō y = [ C co 0 C co 0 x, x+ B co B cō 0 0 u where * denotes block matrices with appropriate dimensions (see Fig 5) The pair ([ Aco 0 A cō [ Bco, B cō ), is controllable The pair ([ Aco, [ ) C 0 co C co, is observable A co 18
19 u Σ co + y Σ cō Σ co Σ cō Figure 4: Block diagram illustrating the Kalman decomposition when A is diagonalizable u Σ co + y Σ cō Σ co Σ cō Figure 5: Block diagram illustrating the Kalman decomposition in the general case 19
20 551 Dimensions Let n c = rank(c) = dim(r(c)) n c = n rank(c) = dim(n(c T )) n o = rank(o) = dim(r(o T )) nō = n rank(o) = dim(n(o)) n cō = dim(r(c) N(O)) is the dimension of the subspace that contains all state vectors x R n that are controllable but unobservable [ [ Aco 0 R nc nc, A co 0 R n c n c 0 A cō 0 A cō [ [ Aco 0 Acō 0 R no no, R nō nō 0 A co 0 A cō One can determine the dimensions of all blocks above, for example the dimension of A co is (n c n cō ) (n c n cō ) 20
21 6 Appendix See PBH test in Section 21 Proof First, we recall the following facts Let S R n n, then if λ C is an eigenvalue of S, and v 0 is its corresponding eigenvector, then Sometimes v is called a right eigenvector a vector w that satisfies Sv = λv w T S = λw T, is called a left eigenvector of S, corresponding to the eigenvalue λ of S Note that by transposing both sides we obtain S T w = λ T w, hence, w is a (right) eigenvector of S T (λ T is the complex conjugate of λ) In Matlab the left eigenvectors of S can be computed using: [W,D = eig(s ); W = conj(w) S and S T have the same eigenvalues S and S 2 have the same eigenvectors if λ is an eigenvalue of S, then λ 2 is an eigenvalue of S 2 (i) First, we show that rank ([ λi A B ) < n implies that (A,B) is not controllable If rank ([ λi A B ) < n, then at least one of the rows of [ λi A B is linearly dependent on the others, hence there exists a vector x (possibly complex) such that From direct observation we have x T [ λi A B = 0 T λx T = x T A, and x T B = 0 T Clearly, x is a left eigenvector of A (corresponding to λ) and it is orthogonal to the columns of B It follows that This leads to hence, x T AB = λx T B = 0 T x T A 2 B = λ 2 x T B = 0 T x T A n 1 B = λ n 1 x T B = 0 T x T [ B AB A 2 B A n 1 B = 0 T R(x T ) [ B AB A 2 B A n 1 B = 0 T, which implies that the controllability matrix does not have full-rank, or in other words, (A, B) is not controllable To summarize: rank ([ λi A B ) < n, there exists a left eigenvector w of A, such that w T B = 0 T, (A, B) is not controllable 21
22 (ii) Second, we show that (A,B) is not controllable implies that rank ([ λi A B ) < n forsomeeigenvalueλofa When(A,B)isnotcontrollable, wecanperformthefollowing change of coordinates (see Section 522) [ à = Q 1 Ã11 à AQ = 12 0 à 22 [ B1, B = Q 1 B = 0 Let λ be an eigenvalue of à 22 (ie, an uncontrollable eigenvalue), and w 2 0 be an associated left eigenvector w T 2Ã22 = λw T 2 Define w T = [ 0 T w T 2 Q 1 w is a left eigenvector of A (corresponding to the eigenvalue λ) w T A = [ 0 T w T 2 Q 1 QÃQ 1 = [ [ 0 T w T à 11 à 12 2 Q 1 0 à 22 = [ 0 T w T 2Ã22 Q 1 = [ 0 T λw T 2 Q 1 = λ [ 0 T w T 2 Q 1 = λw T w is orthogonal to the columns of B w T B = [ 0 T w T 2 = [ 0 T w T 2 = 0 T Q 1 Q B [ B1 0 Since the vector w satisfies the conditions λw T = w T A, and w T B = 0 T we conclude that w [ T λi A B = 0 T, and hence, rank ([ λi A B ) < n To summarize: (A, B) is not controllable, there exists a left eigenvector w of A, such that w T B = 0 T, rank ([ λi A B ) < n 22
23 7 Appendix See decomposition theorem in Section 522 Proof Let Q = [ Q 1 Q 2, where Q 1 R n r contains r linearly independent columns of C (ie, R(Q 1 ) = R(C)) Q 2 R n (n r) is chosen so that Q is non-singular Let the inverse of Q be partitioned as Q 1 = [ M N, where M R r n and N R (n r) n, hence [ [ M [ Q 1 MQ1 MQ Q = Q1 Q N 2 = 2 NQ 1 NQ 2 [ Ir 0 = 0 I (n r) From the fact that NQ 1 = 0 we can conclude that R(C) N(N) [ M Q 1 AQ = A [ [ [ MAQ1 MAQ Q N 1 Q 2 = 2 MAQ1 MAQ = 2 NAQ 1 NAQ 2 0 NAQ 2, since R(C) is A-invariant (see Example 52), R(AQ 1 ) = R(Q 1 ) = R(C) NAQ 1 = 0 [ [ [ M MB MB B = Q 1 B = B = =, N NB 0 since the columns of B belong to R(C) (and hence belongs to N(N)) Let C = [ B AB A 2 B A n 1 B, [ C = B Ã B Ã2 B n 1 B Ã = Q 1 C [ = B1 Ã 11 B1 Ã r 1 11 B 1 Ã n 1 11 B (71) Since controllability is preserved during similarity transformations, rank(c) = rank( C) = r From the Cayley-Hamilton theorem we can conclude that ([ ) rank B1 Ã 11 B 1 Ã r 1 11 B 1 = r, and hence, that (Ã11, B 1 ) is controllable Furthermore, from the fact that B 2 = 0, it is clear that the pair (Ã22, B 2 ) is not controllable 23
24 8 Appendix See decomposition theorem in Section 532 [ P1 Proof Let P =, where P 2 P 1 R r n contains r linearly independent rows of O (ie, R(P T 1) = R(O T )) P 2 R (n r) n is chosen so that P is non-singular Let the inverse of P be partitioned as P 1 = [ M N, where M R n r and N R n (n r), hence [ [ PP 1 P1 [ P1 M P = M N = 1 N P 2 M P 2 N P 2 [ Ir 0 = 0 I (n r) From the fact that P 1 N = 0 we can conclude that R(N) N(P 1 ) = N(O) [ PAP 1 P1 = A [ M N [ [ P1 AM P = 1 AN P1 AM 0 = P 2 P 2 AM P 2 AN P 2 AM P 2 AN, since R(O T ) is A T -invariant (see Example 53), R(A T P T 1) = R(P T 1) = R(O T ), or in words, the space spanned by the rows of the matrix P 1 A is equal to R(O T ) Since R(O T ) N(O), it follows that any linear combination of the rows of P 1 A is orthogonal to any linear combination of the columns of N, ie, (P 1 A)N = 0 O = C CA CA 2 CA n 1, Ô = Ĉ ĈÂ ĈÂ2 ĈÂn 1 = Ĉ 1 0 Ĉ 1 Â 11 0 Ĉ 1 Â Ĉ 1 Â r Ĉ 1 Â n Since observability is preserved during similarity transformations, rank(o) = From the Cayley-Hamilton theorem we can conclude that rank Ĉ 1 Ĉ 1 Â 11 Ĉ 1 Â 2 11 Ĉ 1 Â r 1 11 = r, rank(ô) = r and hence, that (Â11,Ĉ1) is observable Furthermore, from the fact that Ĉ2 = 0, it is clear that the pair (Â22,Ĉ2) is not observable 24
25 9 Appendix Matlab code for Example 524 % system definition A = [ ; ; ; ; B = [-1;1;1;-1; % controllability matrix (rank(c) = 2) Ctr = [B,A*B,A^2*B,A^3*B; % form a full-rank matrix Q Q1 = Ctr(:,[1,2); Q2 = [1 0; 0 1; 0 0; 0 1; Q = [Q1, Q2; iq = inv(q); % form decomposed system At = iq*a*q; Bt = iq*b; % verify M = iq(1:2,:); N = iq(3:4,:); N*Q1 % should be equal to zero 25
26 10 Appendix Matlab code for Example 534 % % example observable subspace (Example 321, pp 158) % clear;clc % system definition A = [-1-4 4; ; ; C = [0 1 0; 1 3-2; % % observability matrix (rank(o) = 2) O = [C;C*A;C*A^2; % form a full-rank matrix P P1 = O([1,2,:); P2 = [1 0 0; P = [P1;P2; ip = inv(p); % form decomposed system Ah = P*A*iP; Ch = C*iP; % verify M = ip(:,1:2); N = ip(:,3); P1*A*N % should be equal to zero % using the fact that A is diagonalizable [Q,L=eig(A); P = inv(q); Ad = P*A*inv(P); Cd = C*inv(P); % Note that even if A is diagonalizable, when using Q as a basis, we might % have to rearrange the states in order to obtain the desired decomposition % (in this particular example we were lucky) %%%EOF 26
27 References [1 E Hendricks, O Jannerup, and P Sørensen, Linear Systems Control - Deterministic and Stochastic Methods, Springer, 2010 [2 K Åström, and R Murray, Feedback Systems: An Introduction for Scientists and Engineers, Princeton University Press, 2008 [3 C-T Chen, Linear system theory and design, Oxford University press, 1999 [4 K Ogata, Modern control engineering, 5 th edition, Pearson, 2010 [5 T Kailath, Linear systems, Prentice-Hall,
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