Control Systems 2. Lecture 7: System theory: controllability, observability, stability, poles and zeros. Roy Smith

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1 Control Systems 2 Lecture 7: System theory: controllability, observability, stability, poles and zeros Roy Smith State-space representations Idea: Transfer function is a ratio of polynomials (say n th order). Reformlate n th order differential equation as a first order matrix differential equation (with matrix dimension n). Motivation: Design & analysis methods use linear algebra (c.f. polynomial algebra). Easy to handle large systems (using Matlab). Easy to handle systems with multiple inputs and outputs. Easy to simulate systems. Numerically better than polynomial based calculations

2 State-space representations System descriptions (linear, time-invariant systems) State vector: Input vector: Output vector: x(t) R n (evolving with time) u(t) R n u y(t) R n y dx(t) = Ax(t) + Bu(t), A R n n y(t) = Cx(t) + Du(t). D R n y n u or: [ẋ(t) ] [ ] [ ] A B x(t) = y(t) C D u(t) Example: a thermal control system T 2 (t) q(t) T (t) T 1 (t) T i (t) temperature of volume i q(t) heat flux into the volume 2 m i mass of volume i c i specific heat for volume i k ij thermal conductance for i, j interface Volume 1: Volume 2: m 1 c 1 d T 1 (t) m 2 c 2 d T 2 (t) = k 12 (T 2 (t) T 1 (t)) = k 12 (T 2 (t) T 1 (t)) k 2 (T 2 (t) T (t)) + q(t)

3 Example: a thermal control system Output: T 1 (t); Inputs: e(t) and T (t); State: x(t) = [ ] T1 (t). T 2 (t) d T 1 (t) = k 12 m 1 c 1 T 1 (t) + k 12 m 1 c 1 T 2 (t) d T 2 (t) So, d T 1 (t) d T 2 (t) = k 12 m 2 c 2 T 1 (t) + k 12 k 2 m 2 c 2 T 2 (t) + k 2 m 2 c 2 T (t) + 1 m 2 c 2 q(t) [ ] = A T1 (t) T 2 (t) Exercise: find B, C and D. + B [ ] T (t), A = q(t) k 12 m 1 c 1 k 12 m 1 c 1 k 12 m 2 c 2 k 12 k 2 m 2 c Example: DC motor with rotational load L a R a τ l + θ s J l B s ω s τ m + E a I a V a Back EMF : Kirchoff s voltage law: Motor torque: Friction torque: Torque balance: E a = K dθ s V a = E a + L a di a + R ai a τ m = K τ I a τ f = B s ω s = B s dθ s J l dθ s 2 + τ f = τ m

4 State-space representations Solution (zero input case: u(t) = ) dx(t) = A x(t). Unilateral Laplace Transform: sx(s) x() = A x(s) = x(s) = (si A) 1 x(). Taking an inverse Laplace transform gives, x(t) = L 1 { (si A) 1} x() = Φ(t)x(). Φ(t), is also known as the State Transition Matrix State-space representations Solution (forced input case: u(t) ) dx(t) = A x(t) + B u(t). Taking Laplace transforms gives, sx(s) x() = A x(s) + B u(s), x(s) = (si A) 1 x() + (si A) 1 B u(s). Now the inverse Laplace gives, x(t) = Φ(t) x() }{{} zero-input solution + Φ(t τ)bu(τ) dτ. }{{} convolution of Φ(t) and B u(t)

5 State-space example First order system (α > ) y(t) α s + α u(t) Take the initial conditions to be zero. The differential equation is, dy(t) + α y(t) = α u(t). Define the state as, x(t) = y(t), then, dx(t) = α x(t) + αu(t), y(t) = x(t) = A = α B = α C = 1 D = State-space example State transition matrix Φ(t) = L 1 { (si A) 1} { } = L 1 1 = e αt (impulse reponse) s + α x(t) = e αt x() + e α(t τ) α u(τ) dτ. Step response: Intial condition: x() =, Input: u(t) = 1, t. y(t) = x(t) = e αt + e α(t τ) α dτ = e αt e ατ α dτ, = e αt [e ατ τ=t e ατ τ= ] = e αt ( e αt 1 ) = 1 e αt

6 Matrix exponential State transition matrix The zero-input solution (via repeated differentiation) is: x(t) = x() + At x() + A2 2 t2 x() + A3 3! t3 x() + ] = [I + At + A2 2 t2 + A3 3! t3 + x(). }{{} define this as e At x(t) = e At x() and so Φ(t) = e At. Properties: e A = I, e A(s+t) = e As e At, e At e At = I, d e At = Ae At Solution via matrix exponential We can begin by guessing a solution of the form, x(t) = e At v(t), where v(t) is a time-varying vector. Differentiate, dx(t) = Ae At v(t) + e At dv(t) = A x(t) + B u(t) = Ae At v(t) + B u(t). So e At dv(t) = B u(t) = dv(t) Solve this by integrating to get: = e At Bu(t). v(t) v() = e Aτ Bu(τ) dτ

7 Solution via matrix exponential v(t) v() = e Aτ Bu(τ) dτ. and our assumed solution is: x(t) = e At v(t) so v(t) = e At x(t) and v() = x(). Substituting these gives, or, e At x(t) x() = e Aτ Bu(τ) dτ. x(t) = e At x() + e At e Aτ Bu(τ) dτ, = e At x() + e A(t τ) Bu(τ) dτ, Again we have Φ(t) = e At Impulse response x(t) = Φ(t)x() + = e At x() + Output equation: y(t) = Cx(t) + Du(t) = Ce At x() + C Φ(t τ)bu(τ) dτ e A(t τ) Bu(τ) dτ e A(t τ) Bu(τ) dτ + Du(t) For the impulse response, x() = and u(t) = δ(t), { t < g(t) = Ce At B + Dδ(t) t

8 Example P (s) = (s 1) (s + 1)(s + 2). This system has a state-space representation: [ ] [ ] A =, B =, C = [ 1 1 ], D =. 1 { ([s ] e At = L 1 s { = L 1 1 s(s + 3) + 2 = 2e 2t e t e 2t + e t [ ]) } [ ]} s 2 1 s + 3 2e 2t 2e t e 2t + 2e t Transfer function representation Take x(t) t= = x() =. Notation sx(s) = AX(s) + BU(s) = X(s) = (si A) 1 BU(s) Y (s) = CX(s) + DU(s) = ( C(sI A) 1 B + D ) }{{} G(s) = 1 (C adj(si A) B) + D det(si A) U(s) [ A B G(s) = C D ]

9 Example (revisited) P (s) = (s 1) (s + 1)(s + 2) = From before we have, so, (si A) 1 = 1 s(s + 3) + 2 C(sI A) 1 B + D = = = [ ] s 2, 1 s [ ] [ ] [ ] s s(s + 3) s s(s + 3) + 2 s 1 s 2 + 3s + 2. [ ] [ ] s Transfer function representations Controllable canonical form One state-space representation of a transfer function. For example, G(s) = b 1 s 2 + b 2 s + b 3 s 3 + a 1 s 2 + a 2 s + a 3. Controllable canonical form: a 1 a 2 a 3 1 A = 1, B = 1 C = [ b 1 b 2 b 3 ], D =

10 System poles For a system, G(s), with a (simple) pole at s = p i, G(s) = a(s) b(s) = (s z 1 )... (s z m ) (s p 1 )... (s p i )... (s p n ) Impulse response; = E 1 (s p 1 ) + + E i (s p i ) + + E n (s p n ). p(t) = E 1 e p 1t + + E i e p it + + E n e p nt. The zero-input solutions of the corresponding differential equation will have terms of the form, y(t) = k i e p it System poles Zero-input case: dx(t) = A x(t) x(t) = e p it x() form of candidate solution. Differentiating, dx(t) Choosing t = gives, = p i e p it x = p i x(t) = Ax(t). A x() = p i x() eigenvalue equation The eigenvalues of A are the poles of P (s). The poles (eigenvalues) are also called natural frequencies or modes of G(s)

11 System zeros If G(s) has a zero at s = s, There exists u(s ) and x(s ), such that G(s )u(s ) =. State-space Laplace domain (with s = s ), s x(s ) = A x(s ) + B u(s ) = C x(s ) + D u(s ) In matrix form: [ ] [ ] (s I A) B x(s ) =, with the constraint: x(s C D u(s ) ), u(s ). Or equivalently, [ ] (s I A) B det = with the constraint: x(s C D ), u(s ) System zeros At a zero s = z i, the rank of G(s) s=zi Examples G 1 (s) = [ (s+2) (s+1) (s+1) (s+2) ] drops from it s normal value. G 1 (s) has poles at -1 and -2 G 1 (s) has zeros at -1 and -2 Poles and zeros do not cancel (different directions) G 2 (s) = (s+2) (s+1) (s+1) (s+2) G 3 (s) = [ (s+2) (s+1) ] (s+1) (s+2) G 2 (s) and G 3 (s) have poles at -1 and -2 G 2 (s) has no MIMO zeros. G 3 (s) has no MIMO zeros

12 System poles and zeros Poles are given by: det(si A) = Zeros are given by: [ ] [ ] (si A) B x (s) =, x C D u (s) (s) and u (s), so [ ] (si A) B det = must also be satisified. C D Similarity transformations dx(t) = Ax(t) + Bu(t) A R n n y(t) = Cx(t) + Du(t) D R n y n u Define a new state, z = T x, with T invertible. T 1 dz(t) = AT 1 z(t) + Bu(t) y(t) = CT 1 z(t) + Du(t) Giving the equivalent system with state z(t) dz(t) = T AT 1 z(t) + T Bu(t) y(t) = CT 1 z(t) + Du(t)

13 Other domains Continuous time, time invariant: Discrete time, time invariant: Nonlinear, time invariant: Nonlinear, time varying: d x(t) = A x(t) + B u(t), y(t) = C x(t) + D u(t) x(k + 1) = A x(k) + B u(k), d x(t) y(k) = C x(k) + D u(k) = f(x(t), u(t)), y(t) = g(x(t), u(t)) d x(t) = f(t, x(t), u(t)), y(t) = g(t, x(t), u(t)) State controllability Formal definition: A system is controllable (or state controllable) if and only if for any given initial state, x(t ) = x, and any given final state, x 1, specified at an arbitrary future time, t 1 > t, there exists a control input u(t) that will drive the system from x(t ) = x to x(t 1 ) = x 1. Controllability matrix: C = [ B AB A 2 B A n 1 B ] R n n un. A system is called controllable if and only if C has rank n (equivalently for n u = 1, C 1 exists)

14 State controllability Controllability Grammian W c (t) := e Aτ BB T e AT τ dτ R n n. To get from state x(t ) to x(t 1 ), (with t 1 > t ), use, ) u(t) = B T e AT (t 1 t) Wc 1 (t 1 ) (e At 1 x x 1. (A,B) is controllable if and only if W c (t) has rank n for all t >. Lyapunov equation P := W c ( ) = e Aτ BB T e AT τ dτ. (for stable systems) We can show that P solves the Lyapunov equation: AP + P A T = BB T State controllability Controllability as a design tool Consider a two-input, two-output system, [ ] [ ] [ d x1 (t) 1 x1 (t) = x 2 (t) x 2 (t) ] + [ 3 2 ] [ ] u1 (t) u 2 (t) [ y1 (t) y 2 (t) ] = [ 5 4 ] [ x1 (t) x 2 (t) ], This system is unstable (where are the poles?) Suppose that implementing u 1 (t) costs CHF 5 and implementing u 2 (t) costs CHF 1. What is the minimum cost to stabilize the system: CHF 5, CHF 1 or CHF 15?

15 State observability Formal definition: The dynamical system pair (A,C) is state observable if and only if, for any time t 1 >, the initial state, x() can be determined from the time history of the input, u(t), and output, y(t), over the interval [, t 1 ]. Observability matrix (A, C) is observable rank C CA. CA n 1 = n State observability Observability Grammian: Observability (for a stable system) is equivalent to Q := e AT τ C T Ce Aτ dτ, having full rank (positive definite). Lyapunov equation Q solves, A T Q + QA = C T C

16 Kalman decomposition G(s) = A 11 A 12 B 1 A 22 A 31 A 32 A 33 A 34 B 3 A 42 A 44 C 1 C 2 D States x 1 x 2 x 3 x 4 Decomposition controllable and observable uncontrollable and observable controllable and unobservable uncontrollable and unobservable Kalman decomposition D y + C 1 x 1 (si A 11 ) 1 + B 1 u A 12 C 2 x 2 (si A 22 ) 1 A 31 A 32 x 3 (si A 33 ) B 3 A 34 A 42 x 4 (si A 44 )

17 Stability Internal stability A system is internally stable if for all initial conditions, and all bounded signals injected at any place in the system, all states remain bounded for all future time. Stabilizability A system is stabilizable if all unstable modes are state controllable. Dectectability A system is detectable if all unstable modes are state observable Internal stability v + G(s) y u K(s) + r Closed-loop input-output relationship: [ ] [ ] [ ] y N11 (s) N 12 (s) v = u N 21 (s) N 22 (s) r Are all four transfer functions stable? Example: G(s) = s 1 s + 1 K(s) = k(s + 1) s(s 1) Work out S(s), T (s), and all four transfer functions, N ij (s)

18 Notes and references Skogestad & Postlethwaite (2nd Ed.) State space systems: section 4.1 Poles and zeros: sections 4.4 & 4.5 Controllability and observability: section 4.2 Stability: sections 4.3 &

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