Linear System Theory

Linear System Theory Wonhee Kim Chapter 6: Controllability & Observability Chapter 7: Minimal Realizations May 2, 217 1 / 31

Recap State space equation Linear Algebra Solutions of LTI and LTV system Stability We will study Controllability & Observability Kalman Decomposition Minimal realizations 2 / 31

Controllability & Observability Controllability (informal): we want to know whether the state of the system is controllable or not from the input Analyze the system structure from the input With the input, we want to move the state to the desired point in a finite time. Observability (informal): we want to observe the initial state of the system from the output and input to quantify the behavior of the system State: position, velocity, acceleration, etc Sensors are required to measure the state. We are not able to use many sensors in real applications. Controllability & Observability Important concepts in control, estimation, and filtering problems Optimal control (LQG, Kalman filtering, etc.) 3 / 31

Controllability & Observability Example ẋ = ( ) 1 x + 3 ( b1 b 2 ) u, x = (x 1 x 2 ) T (b 1, b 2 ) T = ( 1, 1) T : can move both eigenvalues can control the state x 1 and x 2 (b 1, b 2 ) T = (1, ) T : cannot move the eigenvalue 3 cannot control state x 2 (b 1, b 2 ) T = (1, ) T : No matter input, x 2 diverges we cannot control x 2 4 / 31

Controllability & Observability Example ẋ = ( ) 1 x, y = ( ) c 3 1 c 2 x (c 1, c 2 ) = (1, 1): can observe the state x 1 and x 2 (c 1, c 2 ) = (1, ): cannot observe the state x 2 (c 1, c 2 ) = (1, ): Output is always stable, but the system is internally unstable 5 / 31

Controllability ẋ(t) = Ax(t) + Bu(t), x R n, u R m Definition (Definition 6.1) The state equation with the pair (A, B) is said to be controllable if for any initial state x() = x, any final state x 1, there exists an input that transfers x to x 1 in a finite time. Equivalent Definition: A system is controllable at time t if there exists a finite time t f such that for any initial condition x, and any final state x f, there is a control input u defined on [t, t f ] such that x(t f ) = x f. We need an input u to transfer the state from the initial to the final state Given initial and finial state conditions in R n, is it possible to steer x(t) to the final state by choosing an appropriate input u(t)? 6 / 31

Controllability: A Preview Discrete-time LTI system x(k + 1) = Ax(k) + Bu(k), x() =, x R n, u R m x(1) = Bu() x(2) = Ax(1) + Bu(1) = ABu() + Bu(1) x(3) = A 2 Bu() + ABu(1) + Bu(2). x(r) = A r 1 Bu() + A r 1 Bu(1) + + Bu(r 1) u(r 1) x(r) = ( B AB A r 1 B ) u(r 2). u() 7 / 31

Controllability: A Preview u(r 1) x(r) = ( B AB A r 1 B ) u(r 2). u() R((B AB A r 1 B)) = {z R n, z = (B AB A r 1 B)p, p R nm } If x f R((B AB A r 1 B)), then x f is reachable This implies that we can reach arbitrary x f R n at time t f = r if and only if R((B AB A r 1 B)) = R n that is equivalent to rank((b AB A r 1 B)) = n Rank of (B AB A r 1 B)) By C-H theorem, A k is a linear combination of {I, A,..., A n 1 } For r n, the rank of (B AB A r 1 B)) cannot increase Hence, if rank((b AB A n 1 B)) = n, then we can find u for an arbitrary x f R n for any finite time 8 / 31

Controllability ẋ(t) = Ax(t) + Bu(t), x R n x(t) = e At x() + e A(t τ) Bu(τ)dτ The system is controllable (page 213 of the textbook) for any x, there exists u(t) on [t, t f ] that transfers x to the origin at t f (controllability to the origin) there exists u(t) on [t, t f ] that transfers state from the origin to any final state x f at t f (reachability) Proof: Exercise!! (note that e A(t t) is always invertible!) Basically, we need the surjectivity of ea(t τ) Bu(τ)dτ 9 / 31

Controllability ẋ(t) = Ax(t) + Bu(t), x R n, u R m, x = Set of reachable state for a fixed time t: R t = {ξ R n, there exists u such that x(t) = ξ} Note that R t is a subspace of R n Controllability matrix and controllability subspace C AB = {ξ R n : ξ = ( B AB A n 1 B ) z, z R nm } C AB : range space of C, where C = (B AB A n 1 B) R n nm Controllability Gramian W t = e A(t τ) BB T e AT (t τ) dτ = e Aτ BB T e AT τ dτ R(W t ): the range space of W t, W t is a symmetric positive semi-definite matrix 1 / 31

Controllability Theorem: Controllability (Theorem 6.1 of the textbook) For each time t >, the following set equality holds: R t = C AB = R(W t ). C = (B AB A n 1 B): controllability matrix Hence if dim C AB = rank((b AB A n 1 B)) = n, the system is controllable Due to C AB, the controllability is independent of the time If the system is controllable, then R t = R n, all the states are reachable by an appropriate choice of the control u We will show that R t C AB, C AB R(W t ), R(W t ) R t Required tools C-H theorem (Chapter 3), R(A T ) = (N(A)) : Problem 1 in HW3 11 / 31

Controllability Theorem: R t C AB Proof: Fix t >, and choose any reachable state ξ R t. We need to show that ξ R t implies ξ C AB. We have ξ R t, which implies ξ = ea(t τ) Bu(τ)dτ. Then by C-H theorem, e At = β (t)i + + β n 1 (t)a n 1 (β i (t): scalar function); hence, ξ =B β (t τ)u(τ)dτ + + A n 1 B β n 1 (t τ)u(τ)dτ Hence, ξ C AB = ( B AB A n 1 B ) β (t τ)u(τ)dτ. } β n 1(t τ)u(τ)dτ {{ } R nm 12 / 31

Controllability Theorem: C AB R(W t ) Proof: Since C AB R(W t ) is equivalent to C AB R(W t) (proof: exercise!!), we will show that C AB R(W t). From Problem 1 in HW3, R(W t ) = (N(W t )), which is equivalent to (R(W t )) = N(W t ), and similarly, C AB = N((B AB An 1 B)). Hence we need to show that if ξ N(W t ), then ξ N((B AB A n 1 B)). Let ξ N(W t ), then W t ξ = R n, which also implies ξ T W t ξ = R. Then = ξ T e Aτ BB T e AT τ dτξ = Since y(τ) = ξ T e Aτ B =, τ [, t], we have B T e AT τ ξ 2 dτ B T e AT τ ξ =, τ [, t] ξ T ( d k dτ k eaτ ) τ= B = ξ T A k B =, k ξ T ( B AB A n 1 B ) = ξ N((B AB A n 1 B)) = C AB 13 / 31

Controllability Theorem: R(W t ) R t Proof: Let ξ R(W t ). Then there exists v R n such that ξ = W t v = Define u(τ) = B T e AT (t τ) v, τ [, t] e Aτ BB T e AT τ vdτ Then, since ẋ = Ax + Bu with x() =, we have x(t) = = e A(t τ) Bu(τ)dτ e A(t τ) BB T e AT (t τ) vdτ = W t v = ξ This means that ξ R t, since we have found the control u that steers the state to ξ from the origin. 14 / 31

Controllability Theorem (Theorem 6.1 of the textbook) If (A, B) is controllable, and A is stable (eigenvalues of A have negative real parts), then there exists a unique solution of where P = e Aτ BB T e AT τ dτ > Note that BB T AP + PA T = BB T, In Chapter 5, AP + PA T = Q where Q > Theorem (Theorem 6.1 of the textbook) (A, B) is controllable if and only if rank((a λi B)) = n for all eigenvalues, λ, of A. Hautus-Rosenbrock test 15 / 31

Controllability Theorem (Theorem 6.1 of the textbook) (A, B) is controllable if and only if W t >, that is, the controllability Gramian is non-singular Theorem (Theorem 6.2 of the textbook) Let Ā = PAP 1 and B = PB. Then (A, B) is controllable if and only if (Ā, B) is controllable Controllability is invariant under the similarity transformation Fact: The state space equation with the controllable canonical form is always controllable. 16 / 31

Controllability ẋ = Ax + Bu, G(s) = X (s) U(s) = (si A) 1 B Kalman Decomposition Theorem (Theorem 6.6 of the textbook): Suppose that C AB = r < n. Let P = (v 1,..., v r, v r+1,..., v n ) where v i, i = 1, 2,..., r is eigenvectors of C, and v r+1,..., v n are arbitrary vectors that guarantees P being nonsingular. Let z = Px. Then ż = PAP 1 z + PBu Ā = PAP 1 = (Ā11 Ā 12 Ā 22 Ā 11 R r r, B1 R r m ) ) ( B1, B = PB = Also, (Ā11, B 1 ) is controllable, and G(s) = (si Ā11) 1 B1. 17 / 31

Observability ẋ = Ax + Bu, y = Cx, x R n, y R p Definition (Definition 6.O1) The state-space equation is said to be observable if for any unknown initial condition, there exists a finite t 1 such that the knowledge of the input and the output over [, t 1 ] is suffices to determine uniquely the initial condition x(). W.L.G., u =, (since u is completely known) Note that y(t) = Ce At x() Hence, if N(Ce At ) =, i.e., dim(n(ce At )) = nullity(ce At ) =, then the system is observable. N(Ce At ): unobservable subspace 18 / 31

Observability Let O = C CA. CA n 1 O: Observability matrix, O R pn n Theorem: N(Ce At ) = N(O) Proof: We will show that N(Ce At ) N(O) and N(Ce At ) N(O). If x N(Ce At ), then ( d = Ce At x = C dt eat) t= x = CA k x, k Hence, x N(O). If x N(O). then x N(Ce At ), since by C-H Theorem, we have Ce At = Cβ (t)i + + CA n 1 β n 1 (t) 19 / 31

Observability If N(Ce At ) =, i.e., dim(n(ce At )) = nullity(ce At ) =, then the system is observable. We need N(Ce At ) = N(O) = Hence, by the rank-nullity theorem, the system is observable if rank(o) = n We say that the system is observable if and only if the pair (C, A) is observable Observability also does not depend on the time (by C-H Theorem) 2 / 31

Observability Duality Theorem (Theorem 6.5 of the textbook) The following are equivalent: (C, A) is observable (A T, C T ) is controllable Proof: (A T, C T ) is controllable if and only if O T = (C T A T C T (A T ) n 1 C T ) rank(o T ) = n = rank(o) C CA O =. CA n 1 21 / 31

Observability Theorem (Theorem 6.O1) If (A, C) is observable, and A is stable, then there exists a unique solution of A T P + PA = C T C, where P = e AT τ C T Ce Aτ dτ >. Theorem (Theorem 6.O1) (C, A) is observable if and only if ( ) C rank = n A λi 22 / 31

Observability Theorem (Theorem 6.O1) (C, A) is observable if and only if the observability Gramian Q t = e AT τ C T Ce Aτ dτ > Theorem (Theorem 6.O3) Let Ā = PAP 1 and C = CP 1. Then (C, A) is observable if and only if ( C, Ā) is observable. 23 / 31

Observability ẋ = Ax + Bu, y = Cx, G(s) = Y (s) U(s) = C(sI A) 1 B Kalman Decomposition Theorem (Theorem 6.O6 of the textbook): Suppose that rank(o) = q < n. Let P = v 1. v q v q+1. v n, v 1,..., v q : eigenvectors. Let z = Px. Then ż = PAP 1 z + PBu, y(t) = CP 1 z, and ) ) (Ā11 Ā = PAP 1 ( B1 ( =, B = PB =, C = C Ā 21 Ā 22 B 1 ) 2 Ā 11 R q q, C 1 R p q Also, ( C 1, Ā 11 ) is observable, and G(s) = C 1 (si A 11 ) 1 B 1. 24 / 31

Kalman Decomposition Theorem Theorem (Theorem 6.7 of the textbook) We can extract the state that is controllable and observable. Fact: The state space equation with the observable canonical form is always observable Discrete-time LTI system x(k + 1) = Ax(k) + Bu(k), y(k) = Cx(k) (A, B) is controllable if and only if rank(c) = n (C, A) is observable if and only if rank(o) = n 25 / 31

Minimum Energy Control (page 189) ẋ = Ax + Bu, x() = x, x(t 1 ) = x f, (A, B) controllable Let W t1 = 1 e Aτ BB T e AT τ dτ >, invertible u (t) = B T e AT (t 1 t) Wt 1 1 (e At1 x x f ) ( 1 ) x(t 1 ) = e At1 x e A(t1 τ) BB T e AT (t 1 τ) dτ Wt 1 1 (e At1 x x f ) = x f } {{ } W t1 We can show that the controller u is the minimum energy controller in the sense that for any controller u that transfers the state from x to x f, we have 1 u(t) 2 dt 1 u (t) 2 dt, u 26 / 31

Stabilizability & Detectability Weaker notions of controllability and observability A system is stabilizable if and only if Ā22 is stable and (Ā11, B 1 ) is controllable A system is detectable if and only if Ā22 is stable and ( C 1, Ā11) is observable How about the example on pages 4-5. Is it stabilizable? Is it detectable? 27 / 31

Controllability & Observability: LTV system ẋ(t) = A(t)x(t) + B(t)u(t), y(t) = C(t)x(t) W t = Q t = Φ(t, τ)b(τ)b T (τ)φ T (t, τ)dτ, t Φ T (t, τ)c T (τ)c(τ)φ(t, τ)dτ, t The LTV system is is controllable if and only if there exists t f > such that W tf > is observable if and only if there exists t f > such that Q tf > W t : controllability Graminan Q t : observability Graminan 28 / 31

Minimal Realizations We have seen that the realization of the state-space equation is not unique. ẋ = Ax + Bu, y = Cx + Du, x() = ẋ = A 1 x + B 1 u, y = C 1 x + D 1 u, x() = y(t) = C Lemmas (not in the textbook) e A(t τ) Bu(τ)dτ = C 1 e A1(t τ) B 1 u(τ)dτ Two system realizations (A, B, C, D) and (A 1, B 1, C 1, D 1 ) are equivalent if and only if D = D 1 and Ce At B = C 1 e A1t B 1, t Two system realizations (A, B, C, D) and (A 1, B 1, C 1, D 1 ) are equivalent if and only if D = D 1 and CA k B = C 1 A k 1B, k 29 / 31

Minimal Realizations In view of the Kalman decomposition, we have the following result: Suppose (A, B, C, D) is a system realization. If either (C, A) is not observable or (A, B) is not controllable, then there exists a lower-order realization (A 1, B 1, C 1, D 1 ) for the system Definition (page 233 of the textbook) Realizations with the smallest possible dimension are called minimal realizations Theorem (Theorem 7.M2 (page 254)) (A, B, C, D) is a minimial realization of the transfer function G(s) if and only if (A, B) is controllable and (C, A) is observable If the system is not controllable or not observable (or not controllable and observable), then there are pole-zero cancellations in a transfer function. 3 / 31

MATLAB Commands controllability matrix: ctrb(a, B) observability matrix: ctrb(a T, C T ) minimal realization: minreal(a, B, C, D) reduce the system order that has only controllable and observable state Mostly, we use the balanced realization (Chapter 7.4) related to controllability and observability Gramians (robust control, advanced control topic) 31 / 31