Intro. Computer Control Systems: F8
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1 Intro. Computer Control Systems: F8 Properties of state-space descriptions and feedback Dave Zachariah Dept. Information Technology, Div. Systems and Control 1 / 22 dave.zachariah@it.uu.se
2 F7: Quiz! 2 / 22 dave.zachariah@it.uu.se
3 F7: Quiz! 1) The state-space description of a system is a not unique b unique c stable 2 / 22 dave.zachariah@it.uu.se
4 F7: Quiz! 1) The state-space description of a system is a not unique b unique c stable 2) The eigenvalues of the system matrix A reveals something about a poles b zeros c the closed-loop system 2 / 22 dave.zachariah@it.uu.se
5 F7: Quiz! 1) The state-space description of a system is a not unique b unique c stable 2) The eigenvalues of the system matrix A reveals something about a poles b zeros c the closed-loop system 3) Solution to ẋ = Ax + Bu with initial condition x 0 is obtained using a a linear system of equations b the matrix exponential c the Nyquist contour 2 / 22 dave.zachariah@it.uu.se
6 Nonlinear time-invariant systems 3 / 22 dave.zachariah@it.uu.se
7 Nonlinear systems and states Most systems are nonlinear! Nonlinear differential equations: ẋ = f(x, u) y = h(x, u) Linearize around operating point x 0, u 0. Typically use a stationary point: ẋ = f(x 0, u 0 )=0 4 / 22 dave.zachariah@it.uu.se
8 Nonlinear systems and states Nonlinear differential equations: ẋ = f(x, u) y = h(x, u) Taylor series expansion around stationary point x 0, u 0 with y 0 = h(x 0, u 0 ) results in linear deviation model: x = A x + B u y = C x + D u Linear state-space description of the deviations around the operating point of system. Matrices A, B, C and D given by derivatives of f(x, u) and h(x, u) with respect to x and u. See ch. 8.4 G&L. 4 / 22 dave.zachariah@it.uu.se
9 Feedback control using states 5 / 22 dave.zachariah@it.uu.se
10 State-feedback control State space description of linear time-invariant system ẋ = Ax + Bu y = Cx Y (s) = G(s)U(s) G u x y (si A) 1 B C 6 / 22 dave.zachariah@it.uu.se
11 State-feedback control State space description of linear time-invariant system ẋ = Ax + Bu y = Cx G(s) = C(sI A) 1 B u x y (si A) 1 B C 6 / 22 dave.zachariah@it.uu.se
12 State-feedback control Idea: Feedback control using states u = Lx + l 0 r, where L and l 0 are design parameters. rl 0 u x y (si A) 1 B C + L ẋ = Ax + B ( Lx + l 0 r) }{{} =u 6 / 22 dave.zachariah@it.uu.se
13 State-feedback control rl 0 u x y (si A) 1 B C + L Closed-loop system from r to y comes: ẋ = Ax + B ( Lx + l 0 r) = (A BL)x + Bl 0 r y = Cx Is it possible to control the system to all states x in R n? design the closed-loop system s poles? (estimate the state x(t)?) 6 / 22 dave.zachariah@it.uu.se
14 Controllability 7 / 22 dave.zachariah@it.uu.se
15 Controllability A sought state x is controllable if some input u(t) can move the system from x(0) = 0 to x(t ) = x x(t) x 8 / 22 dave.zachariah@it.uu.se
16 Controllability For x 0 = 0, we can compute the state at t = T T x(t ) = e At x 0 + e Aτ Bu(T τ)dτ 0 8 / 22 dave.zachariah@it.uu.se
17 Controllability Med x 0 = 0 är tillståndet vid t = T Therefore: x(t ) = T 0 e Aτ Bu(T τ)dτ = via Cayley-Hamiltons theorem = Bγ 0 + ABγ A n 1 Bγ n 1 x(t ) is a linear combination of B, AB,..., A n 1 B. A state x is controllable if it can be expressed as such a linear combination, i.e., if x is in the column space of S [B AB A n 1 B] 8 / 22 dave.zachariah@it.uu.se
18 Controllability x(t ) x Figur : Example column space of S and non-controllable state x. Controllable system All states x are controllable S:s columns are linearly independent Note: rank(s) = n or det(s) 0 8 / 22 dave.zachariah@it.uu.se
19 Controllability x(t ) x Figur : Example column space of S and non-controllable state x. Controllable canonical form System is controllable It can be written on controllable canonical form 8 / 22 dave.zachariah@it.uu.se
20 Observability 9 / 22 dave.zachariah@it.uu.se
21 Observability Assume u(t) 0. A state x 0 is unobservable if the output y(t) 0 when system starts at x(0) = x. y(t) = Cx(t) x t 10 / 22 dave.zachariah@it.uu.se
22 Observability When u(t) 0 we obtain y(t) = Cx(t) = Ce At x +0 When y(t) 0 we do not observe any changes in the output: That is, d k dt k y(t) t=0 = CA k x =0. Cx = 0, CAx = 0,..., CA n 1 x = 0 10 / 22 dave.zachariah@it.uu.se
23 Observability When u(t) 0 and y(t) 0 we observe no changes: Cx = 0, CAx = 0,..., CA n 1 x = 0 or where O Ox = 0 C CA. CA n 1 Therefore: A state x 0 is unobservable if it belongs to the null space of O. 10 / 22 dave.zachariah@it.uu.se
24 Observability y(t) = Cx(t) x t Figur : Example null space of O and unobservable state x. Observable system All states x are observable O:s columns are linearly independent Note: rank(o) = n or det(o) 0 10 / 22 dave.zachariah@it.uu.se
25 Observability y(t) = Cx(t) x t Figur : Example null space of O and unobservable state x. Observable canonical form System is observable It can be written on observable canonical form 10 / 22 dave.zachariah@it.uu.se
26 Build intuition 11 / 22 dave.zachariah@it.uu.se
27 Build intuition from simple systems Example: controllable system System on controllable canonical form: [ 2 1 ẋ(t) = 1 0 Transfer function: y(t) = [ 1 G(s) = C(sI A) 1 B = 1 ] x(t) ] x(t) + [ ] 1 u(t) 0 s + 1 s 2 + 2s + 1 = s + 1 (s + 1) 2 = 1 s + 1 [Board: investigate observability using O] 12 / 22 dave.zachariah@it.uu.se
28 Build intuition from simple systems Example: controllable system System on controllable canonical form: [ 2 1 ẋ(t) = 1 0 Transfer function: y(t) = [ 1 G(s) = C(sI A) 1 B = 1 ] x(t) ] x(t) + [ ] 1 u(t) 0 s + 1 s 2 + 2s + 1 = s + 1 (s + 1) 2 = 1 s + 1 [Board: investigate observability using O] [ ] 1 1 O = det O = 0 unonbservable / 22 dave.zachariah@it.uu.se
29 Build intuition from simple systems Example: observable system System on observable canonical form: [ ] 2 1 ẋ(t) = x(t) Transfer function: y(t) = [ 1 G(s) = C(sI A) 1 B = 0 ] x(t) [ ] 1 u(t) 1 s + 1 s 2 + 2s + 1 = s + 1 (s + 1) 2 = 1 s + 1 [Board: investigate controllability using S] 13 / 22 dave.zachariah@it.uu.se
30 Build intuition from simple systems Example: observable system System on observable canonical form: [ ] 2 1 ẋ(t) = x(t) Transfer function: y(t) = [ 1 G(s) = C(sI A) 1 B = 0 ] x(t) [ ] 1 u(t) 1 s + 1 s 2 + 2s + 1 = s + 1 (s + 1) 2 = 1 s + 1 [Board: investigate controllability using S] [ ] 1 1 S = det S = 0 non-controllable / 22 dave.zachariah@it.uu.se
31 Build intuition from simple systems Exemple: controllable and observable system Systems in previous examples have the same transfer function G(s) = 1 s + 1. Can also be written in state-space form ẋ(t) = x(t) + u(t), y(t) = x(t). where x(t) is a scalar. [Board: investigate S and O] 14 / 22 dave.zachariah@it.uu.se
32 Build intuition from simple systems Exemple: controllable and observable system Systems in previous examples have the same transfer function G(s) = 1 s + 1. Can also be written in state-space form ẋ(t) = x(t) + u(t), y(t) = x(t). where x(t) is a scalar. [Board: investigate S and O] S = 1 O = 1 det S = 1 det O = 1 controllable and observable (1) Note: we eliminated invisible states 14 / 22 dave.zachariah@it.uu.se
33 Minimal realization 15 / 22 dave.zachariah@it.uu.se
34 Minimal realization System with transfer function G(s) and state-space form ẋ = Ax + Bu y = Cx u x y (si A) 1 B C Definition 8.2 G&L State-space form of G(s) is a minimal realization if vector x has the smallest possible dimension. 16 / 22 dave.zachariah@it.uu.se
35 Minimal realization System with transfer function G(s) and state-space form ẋ = Ax + Bu y = Cx u x y (si A) 1 B C Definition 8.2 G&L State-space form of G(s) is a minimal realization if vector x has the smallest possible dimension. Result 8.11(+8.12) G&L A state-space form is minimal realization controllable and observable A:s eigenvalues = G(s):s poles 16 / 22 dave.zachariah@it.uu.se
36 Design of state-feedback control 17 / 22 dave.zachariah@it.uu.se
37 State-feedback control State-space model with controller u = Lx + l 0 r where L = [ l 1 l 2 l n ] 18 / 22 dave.zachariah@it.uu.se
38 State-feedback control State-space model with controller u = Lx + l 0 r where L = [ l 1 l 2 l n ] Closed-loop system ẋ = (A BL)x + Bl 0 r y = Cx 18 / 22 dave.zachariah@it.uu.se
39 State-feedback control State-space model with controller u = Lx + l 0 r where L = [ l 1 l 2 l n ] Closed-loop system as a transfer function Output is Y (s) = G c (s)r(s), where G c (s) = C(sI A + BL) 1 Bl 0 18 / 22 dave.zachariah@it.uu.se
40 State-feedback control State-space model with controller u = Lx + l 0 r where L = [ l 1 l 2 l n ] System matrix of closed-loop system: (A BL) Eigenvalues/poles given by polynomial equation det(si A + BL) = 0 which we can design via L! 18 / 22 dave.zachariah@it.uu.se
41 State-feedback control Design of the gain l 0 Y (s) = G c (s)r(s) where G c (s) = C(sI A + BL) 1 Bl 0. It is desirable to have at least G c (0) = 1 19 / 22 dave.zachariah@it.uu.se
42 State-feedback control Design of the gain l 0 Y (s) = G c (s)r(s) where G c (s) = C(sI A + BL) 1 Bl 0. It is desirable to have at least G c (0) = 1 G c (0) = C( A + BL) 1 Bl 0 = 1 and so l 0 = 1 C( A + BL) 1 B 19 / 22 dave.zachariah@it.uu.se
43 State-feedback control Design of the gain l 0 Y (s) = G c (s)r(s) where G c (s) = C(sI A + BL) 1 Bl 0. It is desirable to have at least G c (0) = 1 G c (0) = C( A + BL) 1 Bl 0 = 1 and so l 0 = 1 C( A + BL) 1 B More generally, replace l 0 r with F r (s)r(s) How to design L? 19 / 22 dave.zachariah@it.uu.se
44 Build intuition from simple systems Exemple: state-vector in R 2 y u Figur : Force u(t) and position y(t). State-space form: [ ] [ ] ẋ = x + u k/m 0 1/m y = [ 1 0 ] x [Board: design L so that closed-loop system has poles -2 and -3] 20 / 22 dave.zachariah@it.uu.se
45 Pole placement State-feedback control rl 0 u x y (si A) 1 B C + L Result 9.1 State-space form is controllable L can be designed to yield arbitrarily placed poles (real and complex-conjugated) of the closed-loop system 21 / 22 dave.zachariah@it.uu.se
46 Pole placement State-feedback control rl 0 u x y (si A) 1 B C + L Result 9.1 State-space form is controllable L can be designed to yield arbitrarily placed poles (real and complex-conjugated) of the closed-loop system L solved by det(si A + BL) = 0 with desired roots L very simple to solve for system on controllable canonical form 21 / 22 dave.zachariah@it.uu.se
47 Pole placement State-feedback control rl 0 u x y (si A) 1 B C + L Result 9.1 State-space form is controllable L can be designed to yield arbitrarily placed poles (real and complex-conjugated) of the closed-loop system L solved by det(si A + BL) = 0 with desired roots L very simple to solve for system on controllable canonical form What to do when we can t measure x directly? 21 / 22 dave.zachariah@it.uu.se
48 Summary and recap Linearization of nonlinear system models Properties: Controllable Observable Minimal realization State-feedback control Pole placement for the closed-loop system 22 / 22 dave.zachariah@it.uu.se
Intro. Computer Control Systems: F9
Intro. Computer Control Systems: F9 State-feedback control and observers Dave Zachariah Dept. Information Technology, Div. Systems and Control 1 / 21 dave.zachariah@it.uu.se F8: Quiz! 2 / 21 dave.zachariah@it.uu.se
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