16.31 Fall 2005 Lecture Presentation Mon 31-Oct-05 ver 1.1

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1 16.31 Fall 2005 Lecture Presentation Mon 31-Oct-05 ver 1.1 Charles P. Coleman October 31, / 40

2 : Controllability Tests Observability Tests LEARNING OUTCOMES: Perform controllability tests Perform observability tests References: Bélanger (1995), Control Engineering, 3.5, 3.6 DeRusso et al.(1998), State Variables for Engineers, 4.3 Fairman(1998), Linear Control Theory, 2.2, 2.5.3, 3.3, Furuta et al. (1988), State Variable Methods in Automatic Control, 2.1 Morris (2001), Introduction to Feedback Control, 2.2 Skelton et al. (1997), A Unified Algebraic Approach to Linear Control Design, 3.3, 3.5 Szidarovszky & Bahill (1997), Linear Systems Theory, 2nd Ed, 5.1, / 40

3 Warning! More math and proofs! You gotta see this stuff at some point in your graduate career! Why not now! Today s goal is to give you more comprehensive list of controllability and observability tests. To start off, lets revisit a result stated (no proved!) in DeRusso et al. 3 / 40

4 For a system where the system matrix A has distinct eigenvalues ẋ = Ax + Bu y = Cx Where the diagonalized transformed system x = Mq is q = Λx + Bu y = Cq with Λ = M 1 AM B = M 1 C = CM 4 / 40

5 For the diagonalized system q = Λx + Bu y = Cq Complete controllability requires no zero rows of B Complete observability requires no zero columns of C The uncontrollable modes correspond to the zero rows of B The unobservable modes correspond to the zero columns of C 5 / 40

6 Toy problem: A = ẋ 1 ẋ 2 = λ 1 x 1 + u = λ 2 x 2 + u ẋ 3 = λ 3 x 3 ẋ 4 = λ 4 x 4 y = x 1 + x 3 λ λ λ λ 4 B = C = ( ) D = ( 0 ) What states are controllable? What states are observable? What is the minimal? 6 / 40

7 DEFINITION: A state x Ō is said to be unobservable if the zero-input solution y(t) = Ce At x Ō = 0 for all t 0. By Cayley-Hamilton the state x Ō must be orthogonal to all the rows of C and all the rows of CA k for k = 0,,n 1. That is x N(M Ō O) y(t) = Ce At 1 x = 0 Ō C CA 0 (n p) 1 =. x Ō CA n 1 There are no unobservable states if rank(m O ) = n 7 / 40

8 THEOREM: The system (A,C) is unobservable iff there exists an eigenvector v of A such that Cv = 0. y(t) = Ce At 1 x Ō = 0 0 (n p) 1 = C CA. CA n 1 x Ō x Ō (t) = α 1e λ 1t v 1 + α 2 e λ 2t v α n e λ nt v n y(t) = C(α 1 e λ 1t v 1 + α 2 e λ 2t v α n e λ nt v n ) Proof? Write x Ō as a lin. combination of the eigenvectors of A. For distinct eigenvalues λ, the eigenvectors v decompose the state space R n. 8 / 40

9 - If there are repeated eigenvalues, there may be several independent eigenvectors associated with the repeated eigenvalue. v 0 i1,v 0 i2,,v 0 ik A linear combination of eigenvectors is also an eigenvector. Cv = 0 if or if v = a 1 v 0 i1 + a 2 v 0 i2 + + a n v 0 ik Cv = a 1 Cv 0 i1 + a 2 Cv 0 i2 + + a n Cv 0 ik = 0 rank( [ Cv 0 i1 Cv 0 i2 Cv 0 ik] ) < K 9 / 40

10 - If rank( [ Cv 0 i1 Cv 0 i2 Cv 0 ik] ) = K then no eigenvector of A is orthogonal to all the rows of C. REMARK: If there are more independent eigenvectors associated with some repeated eigenvalues than there are outputs (K > p) then [ Cv 0 i1 Cv 0 i2 Cv 0 ik] has fewer rows (p) than columns (K) Since the rank cannot exceed the number of rows, it meets the condition for unobservability. 10 / 40

11 DEFINITION: A state x C is said to be uncontrollable if it is orthogonal to the zero-state response x(t) for all t > 0 and all input functions x T C t 0 t 0 e Aτ Bu(t τ)dτ = 0 x T Ce Aτ Bu(t τ)dτ = 0 For this to be true for all t > 0 and all u( ) x T Ce Aτ B = 0 for all t 0 11 / 40

12 By Cayley-Hamilton the state x C must be orthogonal to all the columns of C and all the columns of A k B for k = 0,,n 1. x T C x T C x T Ce Aτ B = 0 1 m [ α0 (τ)i + α 1 (τ)a + + α n 1 (τ)a n 1] B = 0 [ α0 (τ)b + α 1 (τ)ab + + α n 1 (τ)a n 1 B ] = 0 α 0 (τ)x T CB + α 1 (τ)x T CAB + + α n 1 (τ)x T CA n 1 B = 0 That is x C N(M T C ) B T e AT τ x C = 0 m 1 0 (n m) 1 = B T B T A. B T (A T ) n 1 x C There are no uncontrollable states if rank(m C ) = n 12 / 40

13 THEOREM: The system (A,B) is uncontrollable iff there exists an eigenvector w of A T such that B T w = 0. THEOREM: The system (A,B) is uncontrollable iff there exists a left eigenvector w of A such that w T B = 0. x T C x T Ce Aτ B = 0 1 m ( B AB A n 1 B ) = 0 1 (n m) x T C(t) = α 1 e λ 1t w T 1 + α 2 e λ 2t w T α n e λ nt w T n Proof? Write x T C as a linear combination of the eigenvectors w of A T. For distinct eigenvalues λ, the left eigenvectors w T decompose the state space R n. 13 / 40

14 - If there are repeated eigenvalues, there may be several independent left eigenvectors associated with the repeated eigenvalue. w 0 i1,w 0 i2,,w 0 ik A linear combination of left eigenvectors is also an eigenvector. B T w = 0 if or if w = a 1 w 0 i1 + a 2 w 0 i2 + + a n w 0 ik B T w = a 1 B T w 0 i1 + a 2 B T w 0 i2 + + a n B T w 0 ik = 0 rank( [ B T w 0 i1 B T w 0 i2 B T w 0 ik] ) < K 14 / 40

15 - If rank( [ B T w 0 i1 B T w 0 i2 B T w 0 ik] ) = K then no left eigenvector of A is orthogonal to all the columns of B. REMARK: If there are more independent eigenvectors associated with some repeated eigenvalues than there are inputs (K > m) then [ B T w 0 i1 B T w 0 i2 B T w 0 ik] has fewer rows (m) than columns (K) Since the rank cannot exceed the number of rows, it meets the condition for uncontrollability. 15 / 40

16 We can no revisit our statement about a diagonalized system now that we have discussed right eigenvectors v, and left eigenvectors w We know that the matrix M that diagonalizes A, where x = Mq contains the eigenvectors of A. (Assume distinct eigenvalues) M = ( v 1 v 2 v n ) The rows of its inverse M 1 are the left eigenvectors of A (The eigenvectors of A T ) M 1 = w1 T w2 T. w T n 16 / 40

17 The transformed distribution B and output matrix C are given by w1 T B B = M 1 w2 T B = B. wnb T C = CT = ( ) Cv 1 Cv 2 Cv n We recover the conditions we stated at the beginning of the lecture We recover the conditions of the stated theorems 17 / 40

18 PBH Test - Controllability We can combine some of the previous results into one test! Popov-Belevitch-Hautus (PBH) TEST: (A,B) is controllable iff the matrix ( (si A) B ) has rank n for all numbers s. This can be considered as a corollary to the theorem that (A,B) is completely controllable iff the matrix A T (A) has no right (left) eigenvector that is orthogonal to the columns of B (B T ). 18 / 40

19 PBH Test - Controllability PROOF: Suppose that [(si A) B] does not have full row rank Then there exists w T, a left eigenvector of A, such that w T ( (si A) B ) = 0 which says w T A = λw T w T B = 0 Then w T M C = w T ( B AB = w T ( B λb A n 1 B ) λ n 1 B ) = 0 And the system is uncontrollable 19 / 40

20 PBH Test - Controllability PROOF: Suppose that the system is not controllable. Then it can be transformed by a matrix T into ( x C ) ) ( ) (ĀC Ā x C = 12 )( x C BC 0 Ā C x C + u 0 Let λ be an eigenvalue of Ā C and choose left eigenvector w T C so that w T CĀ C = λw T C Define the 1 n vector w T = ( 0 w T C ) 20 / 40

21 PBH Test - Controllability PROOF: Then w T ( (si Ā) B) = ( 0 w T C w T ( (si T 1 AT) T 1 B) = 0 w T T 1 ( (si A)T B ) = 0 B C 0 λi Ā C 0 ) ( λi ĀC Ā12 T 1 is nonsingular which implies w T T 1 0 Since T is nonsingular w T T 1 (si A)T = 0 ) becomes w T T 1 (si A) = 0 21 / 40

22 PBH Test - Controllability PROOF: Putting this together implies w T T 1 ( (si A) B ) = 0 Hence, if (A,B) is not controllable, the matrix ( ) (si A) B loses rank at s = λ. 22 / 40

23 Toy problem: ( ) (si A) B s λ s λ s λ s λ 4 0 Does not have full rank for all s! Let s = λ 3 (λ 3 λ 1 ) (λ 3 λ 2 ) (λ 3 λ 4 ) 0 Column 5 is now a linear combination of columns 1 and 2! 23 / 40

24 PBH Test - Observability Similarly we have the following observability test! Popov-Belevitch-Hautus (PBH) TEST: (A,C) is observable iff the matrix has rank n for all numbers s. ( (si A T ) C T) Equivalently ( ) (si A) C has rank n for all numbers s. 24 / 40

25 PBH and Controller Pole Placement We can use the PBH test to strengthen a statement we have already made THEOREM: Whenever A has a left eigenvector w T such that w T B = 0, the corresponding eigenvalue of A, λ, is invariant to state feedback. That is λ is an eigenvalue of (A BK). PROOF: w T (A BK) = w T A w T BK = w T A 0 = λw T And we see that λ is an eigenvalue of the controlled system matrix for all feedback gain matrices K. 25 / 40

26 PBH and Observer Pole Placement A similar results holds for observer design THEOREM: Whenever A has a right eigenvector v such that Cv = 0, the corresponding eigenvalue of A, λ, is an eigenvalue of (A + LC). PROOF: (A + LC)v = Av + LCv = Av 0 = λv And we see that λ is an eigenvalue of the observer error dynamics for all observer gain matrices L We cannot control the error dynamics of unobservable states The best we can hope for is detectability 26 / 40

27 Controllability Gramian Controllability revisited x(t 1 ) e At 1 x(0) = x = x = t1 0 t1 0 t1 0 e A(t 1 σ) Bu(σ) dσ e A(t 1 σ) Bu(σ) dσ R(σ)u(σ) dσ = L(u) For the range of the linear operator L to be R n the columns of R(σ) (not square!) must be linearly independent (N(R) = 0) for σ [0,t 1 ] We form the Gramian of R (DeRusso, 3.2), and require that it be positive definite 27 / 40

28 Controllability Gramian Controllability Gramian W C (t 1 ) = = t1 0 t1 0 R(σ)R T (σ) dσ e A(t 1 σ) BB T e AT (t 1 σ) dσ For controllability we require Ẇ C (t) = W C A T + AW C + BB T W C (0) = 0 W C (t 1 ) > 0 28 / 40

29 Controllability Gramian We can rewrite the integral W C (t 1 ) = = = t1 0 0 t1 0 e A(t 1 σ) BB T e AT (t 1 σ) dσ t 1 e Aτ BB T e AT τ dτ e Aτ BB T e AT τ dτ THEOREM: (A,B) is completely controllable iff there exists t 1 < such that W C (t 1 ) = t1 0 e Aτ BB T e AT τ dτ > 0 Ẇ C (t) = W C A T + AW C + BB T W C (0) = 0 29 / 40

30 Controllability Gramian REMARK: The u that gets us to x is u(t) = B T e AT (t 1 t) W 1 C (t 1) x This is impractical to calculate, but does just nicely for proofs! x = = t1 0 t1 0 e A(t 1 σ) Bu(σ) dσ e A(t 1 σ) BB T e AT (t 1 σ) dσw 1 C (t 1) x = W C (t 1 )W 1 C (t 1) x = x 30 / 40

31 Observability Gramian Observability revisited y(t) = Cx(t) = Ce At x 0 For the a unique solution of x 0 over the interval t [0,t 1 ], the columns of Ce At (not square!) must be linearly independent. Again, we form a Gramian, and require that it be positive definite 31 / 40

32 Observability Gramian Observability Gramian W O (t) = t 0 e AT τ C T Ce Aτ dτ THEOREM: (A,C) is completely observable iff there exists t 1 < such that W O (t 1 ) = t1 0 e AT τ C T Ce Aτ dτ > 0 NOTE: Sometimes this is proved using the adjoint system (Szidarovsky & Bahill, 6.11) ż = A T z + C T v w = B T z 32 / 40

33 Lyapunov Stability and Gramians The controllability and observability gramians are symmetric matrices that can be positive definite. They can behave as quadratic forms Which makes them targets for use as Lyapunov functions! 33 / 40

34 Lyapunov Stability and Controllability Define THEOREM: X = W C ( ) = 0 e Aσ BB T e AT σ dσ X exists iff the controllable modes are asymptotically stable If X exists, then X > 0 iff (A,B) is controllable If X exists it satisfies 0 = XA T + AX + BB T (Skelton et al., 3.3.1) Use this result in a Lyapunov context 34 / 40

35 Lyapunov Stability and Controllability Choose the following Lyapunov function Note that X satisfies Or V (x(t)) = x T (t)x 1 x(t) 0 = XA T + AX + BB T XA T + AX = BB T < 0 V (x) > 0. We need to calculate its time derivative and show V (x) 0 and V (x) = 0 implies x = 0 to obtain the stability result we desire. 35 / 40

36 Lyapunov Stability and Controllability V (x(t)) = ẋ T (t)x 1 x(t) + x T (t)x 1 ẋ(t) = x T (t)a T X 1 x(t) + x T (t)x 1 Ax(t) = x T (t)(x 1 X)A T X 1 x(t) + x T (t)x 1 A(XX 1 )x(t) = x T (t)x 1 [XA T + AX]X 1 x(t) = x T (t)x 1 [ BB T ]X 1 x(t) = x T (t)x 1 BB T X 1 x(t) < 0 Which is the result we desire! We can now state a theorem 36 / 40

37 Lyapunov Stability and Controllability (i) (ii) (iii) (iv) THEOREM: The following are equivalent: (Skelton et al., 3.5.1) The system ẋ = Ax is asymptotically stable i.s.l The eigenvalues of A lie in the open left half plane If (A,B) is a controllable pair, then there exists X > 0 satisfying 0 = XA T + AX + BB T If (A,B) is stabilizable, then there exists X 0 satisfying 0 = XA T + AX + BB T With B = I we recover our previous Lyapunov result: ẋ = Ax is asymptotically stable if there exits X > 0 such that 0 < XA T + AX A similar results holds for observability 37 / 40

38 Lyapunov Stability and Observability Define P = W O ( ) = 0 e AT σ C T Ce Aσ dσ THEOREM: The following are equivalent: (Skelton et al., 3.5.1) (i) The system ẋ = Ax is asymptotically stable i.s.l (ii) The eigenvalues of A lie in the open left half plane (iii) If (A,C) is an observable pair, then there exists P > 0 satisfying 0 = PA + A T P + C T C 38 / 40

39 Remarks The controllability and observability gramiams are rarely used for LTI The controllability and observability test matrices are most popular! However, the PBH test can sometimes be quite slick! The link to Lyapunov stability analysis is provided to give you an introduction to the concept of matrix equalities and inequalities. We may build on this at the end of the course if we have time to look at LMI approaches to robust control 39 / 40

40 : (Done) Lyapunov stability (Done) Controller and Observer Canonical Forms, & Minimal Realizations (DeRusso, Chap 6; Belanger, 3.7.6) (Done) Kalman s Canonical Decomposition (DeRusso, 4.3, 6.8; Belanger, 3.7.4, Furuta et al ) (Some) Full state feedback & Observers (DeRusso, Chap 7; Belanger, Chap 7, How) LQR (Linear Quadratic Regulator) (Belanger, 7.4, How) Kalman Filter (DeRusso, 8.9, Belanger 7.6.4, How) Robustness & Performance Limitations (Various) 40 / 40

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