Stability, Pole Placement, Observers and Stabilization

Stability, Pole Placement, Observers and Stabilization 1 1, The Netherlands DISC Course Mathematical Models of Systems

Outline 1 Stability of autonomous systems 2 The pole placement problem 3 Stabilization by state feedback 4 State observers 5 Pole placement and stabilization by dynamic output feedback

Stability of autonomous Systems Autonomous systems Let P(ξ) R q q [ξ], with det P(ξ) 0, i.e., detp(ξ) is not the zero polynomial. Consider the system of differential equations P( d dt )w = 0. This represents the dynamical system Σ = (R, R q,b) with B = {w : R R q w satisfies P ( ) d dt w = 0 weakly}. Since det P(ξ) 0, the resulting system is autonomous. Hence B is finite-dimensional and each weak solution of is a strong one.

Stability of autonomous systems All solutions All solutions are infinitely differentiable. In fact, w B if and only if w = w 1 + w 2 + + w N, with w k s associated with one of the distinct roots λ 1,λ 2,...,λ N of det P(ξ). This w k is given by w k (t) = ( nk 1 l=0 B kl t l )e λ kt, where n k is the multiplicity of the root λ k of det P(ξ), B kl s are suitable constant complex vectors.

Stability of autnomous systems Stability definitions The linear dynamical system Σ = (R, R q,b) is said to be 1 stable if all elements of its behavior B are bounded on the half-line [0, ), i.e, if (w B) (there exists M R such that w(t) M for t 0). Of course, this bound M depends on the particular solution w B, 2 unstable if it is not stable, 3 asymptotically stable if all elements of B approach zero for t (i.e, if (w B) (w(t) 0 as t )). Important note: If Σ is stable or asymptotically stable then it must be autonomous. Hence without loss of generality B = ker P( d dt ) with P(ξ) nonsingular.

Stability of autonomous systems Definition of simple and semisimple roots Let P(ξ) R q q [ξ] be nonsingular. Then 1 The roots of P(ξ) are defined to be those of the scalar polynomial detp(ξ). Hence λ C is a root of P(ξ) if and only if rank P(λ) < q, 2 The root λ is called simple if it is a root of det P(ξ) of multiplicity one, 3 semisimple if the rank deficiency of P(λ) equals the multiplicity of λ as a root of P(ξ) (equivalently, if the dimension of ker P(λ) is equal to the multiplicity of λ as a root of det P(ξ)). Clearly, for q = 1 roots are semisimple if and only if they are simple, but for q > 1 the situation is more complicated.

Stability of autonomous systems Example λ = 0 is a root of multiplicity 2 for both the polynomial matrices ( ) ( ) ξ 0 ξ 1 and. (1) 0 ξ 0 ξ This root is semisimple in the first case, but not in the second.

Stability of autonomous systems Theorem Let P R q q [ξ] be nonsingular. )w = 0 is: P( d dt The system represented by 1 asymptotically stable if and only if all the roots of detp(ξ) have negative real part; 2 stable if and only if for each λ C that is a root of P(ξ), either (i) Re λ < 0, or (ii) Re λ = 0 and λ is a semisimple root of P(ξ). 3 unstable if P(ξ) has a root with positive real part and/or a nonsemisimple root with zero real part.

Stability of autonomous systems Examples 1 Scalar first-order system aw + d dt w = 0. Associated polynomial P(ξ) = a + ξ. Root a. Hence this system is asymptotically stable if a > 0, stable if a = 0, and unstable if a < 0. Note: behavior B = {Ae at A R}. 2 Scalar second-order system aw + d2 dt 2 w = 0. Associated polynomial P(ξ) = a + ξ 2. Roots λ 1,2 = ± a for a < 0, λ 1,2 = ±i a for a > 0, and λ = 0 is a double, not semisimple root when a = 0. Thus we have (a < 0 instability), (a > 0 stability), and (a = 0 instability). Indeed: B = {Ae at + Be at A,B R}, if a < 0, B = {A cos at + B sin ata A,B R}, if a > 0, B = {A + Bt A,B R}, if a = 0.

Stability of autonomous systems Special case: stability of state equations Autonomous state system d dt x = Ax, with A Rn n. Polynomial matrix P(ξ) = I ξ A. Roots: the eigenvalues of A. An eigenvalue λ of A is called semisimple if λ is a semisimple root of det(iξ A), i.e: the multiplicity of λ is equal to dim ker(λi A). Note: dim ker(λi A) is always equal to the number of independent eigenvectors associated with the eigenvalue λ.

Stability of autonomous systems Corollary The system defined by d dt x = Ax is: 1 asymptotically stable if and only if the eigenvalues of A have negative real part, 2 stable if and only if for each λ C that is an eigenvalue of A, either (i) Reλ < 0, or (ii) Re λ = 0 and λ is a semisimple eigenvalue of A, 3 unstable if and only if A has either an eigenvalue with positive real part or a nonsemisimple one with zero real part.

The pole placement problem Consider the linear time-invariant dynamical system in state form described by d dt x = Ax + Bu, with x : R R n and u : R R m the state and the input trajectory, and with A R n n and B R n m. Allow linear state feedback controllers of the form u = Nx, with N R m n called the feedback gain matrix. Closed loop equation: d dt x = (A + BN)x. This represents an autonomous dynamical system.

Stability of Autonomous Systems d dt Plant x = Ax + Bu Feedback processor u = Nx

The pole placement problem We call the eigenvalues of A + BN the closed loop poles. Question: what closed loop pole locations are achievable by choosing the feedback gain matrix N? Closed loop characteristic polynomial: Pole placement problem χ A+BN (ξ) := det(ξi (A + BN)) Let A R n n and B R n m be given. What is the set of polynomials χ A+BN (ξ) obtainable by choosing the matrix N R m n?

The pole placement problem The pole placement theorem (W.M. Wonham, 1969) Consider the system d dt x = Ax +Bu. For any real monic polynomial r(ξ) of degree n there exists N R m n such that χ A+BN = r(ξ) if and only if the system d dt x = Ax + Bu is controllable. Monic polynomial of degree n: p(ξ) = ξ n +p n 1 ξ n 1 +...+p 1 ξ+p 0. W.M. Wonham, On pole assignment in multi-input controllable linear systems, IEEE Transactions on Automatic Control, 1967.

Proof of the pole placement theorem

Proof of the pole placement theorem System similarity Σ n,m := the family of all systems d dt x = Ax +Bu with n state and m input variables. We say: (A,B) Σ n,m. Let (A 1,B 1 ),(A 2,B 2 ) Σ n,m. We call (A 1,B 1 ) and (A 2,B 2 ) similar if there exists a nonsingular matrix S R n n such that A 2 = SA 1 S 1 and B 2 = SB 1. d If in the state space of dt x = Ax + Bu we change the coordinate basis by defining z(t) = Sx(t), then the dynamics of z are governed by d dt z = SAS 1 z + SBu. Hence similarity corresponds to changing the basis in the state space.

Proof of the pole placement theorem Lemma Assume (A 1,B 1 ),(A 2,B 2 ) Σ n,m are similar. Let r(ξ) R[ξ] be a monic polynomial. Then there exists a matrix N 1 R m n such that χ A1 +B 1 N 1 (ξ) = r(ξ) if and only if there exists a matrix N 2 R m n such that χ A2 +B 2 N 2 (ξ) = r(ξ). Proof Let A 2 = SA 1 S 1 and B 2 = SB 1. Assume χ A1 +B 1 N 1 (ξ) = r(ξ). Define N 2 = N 1 S 1. Then A 2 + B 2 N 2 = S(A 1 + B 1 N 1 )S 1, so A 1 +B 1 N 1 and A 2 +B 2 N 2 are similar, and therefore χ A2 +B 2 N 2 (ξ) = r(ξ) as well. Also the converse.

Proof of the pole placement theorem Lemma The system (A,B) Σ n,m is similar to a system (A,B ) Σ n,m with A,B of the form ( A A = 11 A ) ( ) 12 0 A, B B = 1 22 0 and with (A 11,B 1 ) controllable.

Proof of the pole placement theorem Proof Two cases: 1 if (A,B) is controllable then A 11 = A = A and B 1 = B = B, and A 12, A 22 and the 0-matrices are void. 2 If (A,B) is not controllable, then R := im(b AB A 2 B... A n 1 B) R n, so is a proper subspace of R n. Choose a basis of R n adapted to R. The matrices with respect to this new basis are of the form (A,B ) because AR R and im B R.

Proof of the pole placement theorem Pole placement Controllability Assume (A,B) is not controllable. Then it is similar to (A,B ) of the form ( A A = 11 A ) ( ) 12 0 A, B B = 1 22 0 with n 1, the dimension of A 11, less than n. Now apply N = (N 1 N 2 ), with N 1 Rm n 1 and N 2 Rm (n n 1), on this system. Then A + B N = ( A 11 + B 1 N 1 A 12 + B 1 N 2 0 A 22 Characteristic polynomial is given by χ A 11 +B 1 (ξ)χ N 1 A (ξ). Therefore, the characteristic polynomial χ A +B N (ξ), and hence 22 χ A+BN (ξ), has, regardless of what N is chosen, χ A 22 (ξ) as a factor. ).

Proof of the pole placement theorem Theorem Let r(ξ) be a monic polynomial of degree n. Assume (A,B) controllable, and m = 1. Let F R 1 n be the solution of the system of linear equations F(B AB A n 2 B A n 1 B) = (0 0 0 1). Then N = F r(a) yields χ A+BN (ξ) = r(ξ). Here r(a) := r 0 I + r 1 A + + r n 1 A n 1 + A n. Note: this yields a proof of the implication Controllability Pole placement for the special case m = 1!

Proof of the pole placement theorem Notation Σ cont n,m Σ n,m are all controllable systems with state space R n and input space R m. Heymann s lemma (1968) Let (A,B) Σ cont n,m. Assume K Rm 1 such that BK 0. Then there exists a N R m n such that (A + BN,BK) Σ cont 1,n. M. Heymann, Comments to: Pole assignment in multi-input controllable linear systems, IEEE Transactions on Automatic Control, 1968.

Proof of the pole placement theorem

Proof of the pole placement theorem Controllability Pole placement, m > 1: coup de grâce Choose K R m 1 such that BK 0, by controllability, B 0, hence such a K exists. Choose N R m 1 such that (A + BN,BK) controllable. We are now back in the case m = 1. Choose N R 1 n such that A + BN + BKN has the desired characteristic polynomial r(ξ). Finally, take N = N + KN. Then A + BN = A + BN + BKN so χ A+BN (ξ) = r(ξ).

The pole placement problem Algorithm Data: A R n n, B R n m, with (A,B) controllable; r(ξ) R[ξ] with r(ξ) monic and of degree n. Required: N R m n such that χ A+BN (ξ) = r(ξ). Algorithm: 1 Find K R m 1 and N R m n such that (A + BN,BK) is controllable. 2 Put A = A + BN, B = BK, and compute F from F[B,A B,...,(A ) n 1 B ] = [0 0... 0 1]. 3 Compute N = F r(a ). 4 Compute N = N + KN. Result: N is the desired feedback matrix.

Stabilization So: if (A,B) is controllable then all polynomials can be placed. Question: what polynomials can be placed if (A, B) is not controllable?. Recall (A,B) Σ n,m is similar to A = ( A 11 A 12 0 A 22 ), B = ( B 1 0 The matrix A 22 characterizes the noncontrollable behavior of the system. Its characteristic polynomial χ A22 (ξ) is called the uncontrollable polynomial of the system. Its roots are called the uncontrollable poles ( modes ). ).

Stabilization Theorem Consider the system (A,B). Let χ u (ξ) be its uncontrollable polynomial. There exists a feedback matrix N R n m such that χ A+BN(ξ) = r(ξ) if and only if r(ξ) is a real monic polynomial of degree n that has χ u (ξ) as a factor.

Stabilization Proof Take (A,B ) similar to (A,B). Partition N conformably as N = (N 1 N 2 ). Then A + B N = ( A 11 + B 1 N 1 A 12 + B 1 N 2 0 A 22 Hence χ A +B N (ξ) = χ A (ξ)χ 11 +B 1 N A 1 22 (ξ) = χ A 11 +B 1 (ξ)χ u(ξ). N 1 Let r(ξ) have χ u (ξ) as a factor. Then r(ξ) = r 1 (ξ)χ u (ξ) for some monic r 1 (ξ). Since (A 11,B 1 ) is controllable, χ A (ξ) can be 11 +B 1 N 1 made equal to r 1 (ξ). Then with N = (N 1 0) we have χ A +B N (ξ) = r 1 (ξ)χ u (ξ) = r(ξ). Also the converse. ).

Stabilization Consider the system d dt x = Ax + Bu with the control law u = Nx. The closed loop system d dt x = (A + BN)x is asymptotically stable if and only if A + BN is Hurwitz. Question: when does there exist N R m n such that A + BN is Hurwitz? Corollary There exists N such A + BN is Hurwitz if and only if the uncontrollable polynomial χ u (ξ) of (A,B) is Hurwitz.

Stabilization Definition We call the system d dt x = Ax +Bu, or, equivalently, the pair (A,B), stabilizable if its uncontrollable polynomial χ u (ξ) is Hurwitz. Theorem (Hautus test) The system d dt x = Ax + Bu is stabilizable if and only if rank ( λi A B ) = n for all λ C + := {λ C Re(λ) 0}.

Stabilization

State observers Consider the following plant d dt x = Ax + Bu, y = Cx x is the state, u the input, and y the output. System parameters: A R n n, B R n m, and C R p n. Denote the class of such systems by Σ n,m,p. We also write (A,B,C) Σ n,m,p. The external (manifest) signals u and y are measured. Aim: deduce the internal (latent) signal x from these measurements.

State observers D. Luenberger, 1963

State observers An algorithm that estimates x from u and y is called a state observer. Denote the estimate of x by ˆx, and define the estimation error as e := x ˆx. A state observer is a dynamical system with u and y as input, ˆx as output, and that makes e = x ˆx small in some sense. Here we focus on the asymptotic behavior of e(t) for t.

State observers x + e u Plant y u Observer ˆx

State observers Proposed state observer: Combining this equation with d dt ˆx = (A LC)ˆx + Bu + Ly, d dt x = Ax + Bu, y = Cx yields the following equation for e = x ˆx: d d e = (A LC)e. dt dt e = d dt x d dt ˆx = (Ax + Bu) ((A LC)ˆx + Bu + Ly) = (Ax+Bu) ((A LC)ˆx+Bu+LCx) = (A LC)(x ˆx) = (A LC)e.

State observers We want that e(t) 0 as t for all e(0) = x(0) ˆx(0), i.e., we want A LC to be Hurwitz. Often we need a certain rate of convergence for e(t). This leads to the following question: What eigenvalues ( observer poles ) can we achieve for A LC by choosing the observer gain matrix L? In linear algebra terms: Let A R n n and C R p n be given. What is the set of polynomials χ A LC (ξ) obtainable by choosing the matrix L R n p?

State observers Theorem Consider the system d dt x = Ax + Bu, y = Cx. There exists for every real monic polynomial r(ξ) of degree n a matrix L such that χ A LC (ξ) equals r(ξ) if and only if the system is observable. Proof (A,C) observable pair if and only if (A T,C T ) controllable pair. Note: for any real square matrix M, χ M (ξ) = χ M T(ξ). Assume (A,C) observable. By the pole placement theorem, there exists for all r(ξ) a matrix N R p n such that χ A T +C T N(ξ) = r(ξ). Thus χ A LC (ξ) = r(ξ) with L = N T. Conversely: Assume there exists for all r(ξ) a matrix L R n p such that χ A LC (ξ) = r(ξ). Then χ A T +C T ( L) T (ξ) = r(ξ) so (AT,C T ) controllable, whence (A,C) observable.

State observers What if (A,C) is not observable? The dynamical systems (A 1,B 1,C 1 ) Σ n,m,p and (A 2,B 2,C 2 ) Σ n,m,p are called similar if there exist a nonsingular S such that A 1 = SA 2 S 1, B 1 = SB 2, C 1 = C 2 S 1. Theorem The system (A,B,C) Σ n,m,p is similar to a system of the form (A,B,C ) in which A and C have the following structure: ( ) A A11 0 =, C = (C A 12 A 1 0), 22 with (A 11,C 1 ) observable. All such decompositions lead to matrices A 22 that have the same characteristic polynomial.

State observers The polynomial χ A22 (ξ) is called the unobservable polynomial of (A,C). Notation: χ 0 (ξ) Its roots are called the unobservable eigenvalues (modes). Theorem Consider the system (A,B,C) Σ n,m,p. Let χ 0 (ξ) be its unobservable polynomial. There exists L R n p such that χ A LC (ξ) = r(ξ) if and only if r(ξ) is a real monic polynomial of degree n that has χ 0 (ξ) as a factor. Proof ( A11 0 A 12 A 22 ) ( L1 + L 2 ) ( A11 + L (C 1 0) = 1 C 1 0 A 12 + L 2 C 1 A 22 )

State observers Corollary There exists an observer d dt ˆx = (A LC)ˆx + Bu + Ly such that for all initial states x(0) and ˆx(0) lim ˆx(t) x(t) = 0, t i.e. such that A LC is Hurwitz, if and only if the unobservable polynomial χ 0 (ξ) of (A,C) is Hurwitz. Definition The system (A,B,C) Σ n,m,p is called detectable if the the unobservable polynomial χ 0 (ξ) of (A,C) is Hurwitz.

State observers Theorem (Hautus test) The system (A,B,C) Σ n,m,p is detectable if and only if ( λi A rank C for all λ C + := {λ C Re(λ) 0}. ) = n

Dynamic output feedback Plant: d dt x = Ax + Bu, y = Cx. Linear time-invariant feedback controller: d z = Kz + Ly, u = Mz + Ny, dt with z : R R d the state of the controller, K R d d, L R d p, M R m d, and N R m p the parameter matrices specifying the controller. State dimension d N is called the order of the controller. Is a design parameter: typically, we want d to be small.

Dynamic output feedback By combining the equations of the plant and controller we obtain the closed loop system ( ) ( )( ) d x A + BNC BM x dt =,y = Cx, u = Mz + Ny z LC K z Compact form: with x e := col(x,z) (the extended state), A e, C e, and H e defined in the obvious way, this yields the closed loop system equations d dt x e = A e x e, y = C e x e, u = H e x e. This is an autonomous dynamical system. We call the eigenvalues of A e the closed loop poles and χ Ae (ξ) the closed loop characteristic polynomial.

Dynamic output feedback Question What closed loop pole locations are achievable by choosing (K,L,M,N)? More precisely: Let A R n n, B R n m, and C R p n be given. Determine the set of polynomials χ Ae (ξ) obtainable by choosing d N and K R d d, L R d p, M R m d, N R m p, and where A e is given by ( ) A + BNC BM A e =. LC K Full solution to this problem is unknown. We will describe a very useful partial result.

Dynamic output feedback We have already seen how to proceed if C = I, i.e., if the full state vector is measured. Let u = N x be a memoryless state feedback control law obtained this way. We have also seen how we can estimate the state x of from (u,y). Let d dt ˆx = (A L C)ˆx + Bu + L y be a suitable observer. Separation principle: combine an observer with a state controller and use the same controller gains as in the case in which the state is measured.

Dynamic output feedback This yields the feedback controller: d dt ˆx = (A L C)ˆx + BN ˆx + L y, u = N ˆx. This is, of course, a feedback processor with order d = n, K = A L C + BN, L = L, M = N, and N = 0. Closed loop system obtained by using this feedback controller: ( ) ( ) ( ) d x A BN x dt = ˆx L C A L C + BN, ˆx u = N ˆx, y = Cx.

Dynamic output feedback u Plant N ˆx Observer Controller

Dynamic output feedback We are interested in the characteristic polynomial of A e. Define a similarity transformation S R 2n 2n by ( ) I 0 S := I I Note: S 1 = S. Also ( A + BN BN 0 A L C ) ( ) A BN = S L C A L C + BN S 1 Hence the closed loop characteristic polynomial equals the product of χ A+BN (ξ) and χ A L C(ξ).

Dynamic output feedback Theorem (Pole placement by dynamic output feedback) Consider the system (A,B,C) and assume that (A,B) is controllable and that (A,C) is observable. Then for every real monic polynomial r(ξ) of degree 2n, factorizable into two real polynomials of degree n, there exists a feedback controller (K, L, M, N) of order n such that the closed loop system matrix A e has characteristic polynomial r(ξ). Proof Take d = n, K = A L C + BN, L = L, M = N, and N = 0. Choose N such that χ A+BN (ξ) = r 1 (ξ) and L such that χ A L C(ξ) = r 2 (ξ), where r 1 (ξ) and r 2 (ξ) are real factors of r(ξ) such that r(ξ) = r 1 (ξ)r 2 (ξ).

Dynamic output feedback Theorem (Stabilization by dynamic output feedback) Consider the system (A,B,C) and let χ u (ξ) be its uncontrollable polynomial, χ 0 (ξ) its unobservable polynomial. Then 1 For any real monic polynomials r 1 (ξ) and r 2 (ξ) of degree n such that r 1 (ξ) has χ u (ξ) as a factor and r 2 (ξ) has χ 0 (ξ) as a factor, there exists a feedback controller (K,L,M,N) of order n such that the closed loop system matrix A e has characteristic polynomial r(ξ) = r 1 (ξ)r 2 (ξ). 2 There exists a feedback controller (K,L,M,N) as such that the closed loop system is asymptotically stable if and only if both χ u (ξ) and χ 0 (ξ) are Hurwitz, i.e., if and only if (A,B) is stabilizable and (A,C) is detectable.