POLE PLACEMENT. Sadegh Bolouki. Lecture slides for ECE 515. University of Illinois, Urbana-Champaign. Fall S. Bolouki (UIUC) 1 / 19

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1 POLE PLACEMENT Sadegh Bolouki Lecture slides for ECE 515 University of Illinois, Urbana-Champaign Fall 2016 S. Bolouki (UIUC) 1 / 19

2 Outline 1 State Feedback 2 Observer 3 Observer Feedback 4 Reduced Order Observers S. Bolouki (UIUC) 2 / 19

3 General objective: Stabilization of LTI systems via feedback. Example. Consider the following LTI system: ẋ = Ax + Bu = x + 1 u [ ] y = Cx = x System modes are 1, 2, 1 (check the eigenvalues of A). Modes 1, 2 are controllable (check the rank of [ λi A B ] ). Modes 1, 1 are observable (check the rank of [ λi A T C T] ). Consider a linear state feedback u = Kx + r, where K = [ k 1 k 2 k 3 ]. Thus, {ẋ = (A BK)x + Br = Acl x + Br y = Cx Challenge: Find K such that A cl is asymptotically stable.

4 (Example cont.) ẋ = A cl x + Br = y = Cx = [ ] x 1 k 1 k 2 k 3 1 k 1 k 2 k 3 x + 1 r Controllable subspace (therefore controllability) is invariant. Uncontrollable modes are invariant. Controllable modes are not invariant. Observability is not invariant. Stability is not invariant.

5 State Feedback State Feedback S. Bolouki (UIUC) 5 / 19

6 State Feedback State Feedback General objective: Stabilization via state feedback. {ẋ = Ax + Bu y = Cx Considering the state feedback u = Kx, we have ẋ = A cl x = (A BK)x Determine K to stabilize the system, i.e., to place all of eigenvalues of A cl in the left half-plane. Theorem Any controllable LTI system can be stabilized. Controllability is not necessary for stabilizability. Like controllability, stabilizability is usually associated with the pair (A, B). S. Bolouki (UIUC) 6 / 19

7 State Feedback State Feedback Theorem Given the eigenvalues of A are distinct, a pair (A, B) is stabilizable if and only if all uncontrollable modes of A are in the left half-plane. sketch of proof: We show that (i) Controllable modes of A can be placed arbitrarily via state feedback, respecting complex conjugate constraints. (ii) Uncontrollable modes of A are invariant under state feedback. We use the KCCF to prove (i) and (ii). ] ] [Āc [ B Ā =, B = c 0 Theorem 0 Ā c A pair (A, B) is stabilizable if and only if all eigenvalues of Ā c are in the left half-plane. Stabilizability Unstable subspace Controllable subspace. S. Bolouki (UIUC) 7 / 19

8 Observer Observer S. Bolouki (UIUC) 8 / 19

9 Observer Observer General objective: Estimating the states (x) via partial observation (y). Why do we estimate the states? What is a good estimation? How do we do it? S. Bolouki (UIUC) 9 / 19

10 Observer Observer {ẋ = Ax + Bu y = Cx + Du Example. Ignoring the observations y (terrible idea!), let ˆx = Aˆx + Bu, where ˆx is an estimation of x. Will e(t) x(t) ˆx(t) converge to 0 as t grows? ė = Ae e(t) = exp{at}e(0). Thus, e(t) will vanish if A is stable. S. Bolouki (UIUC) 10 / 19

11 Observer Observer Better idea: where ŷ = Cˆx + Du. ˆx = Aˆx + Bu + L(y ŷ), Will e(t) = x(t) ˆx(t) converge to 0 as t grows? } ẋ = Ax + Bu ˆx = LCx + (A LC)ˆx + Bu ė = (A LC)e lim t e(t) = 0 A LC is stable Convergence occurs at an exponential rate. A LC is stable for some L if and only if (A T, C T ) is stabilizable. Given the eigenvalues of A are distinct, (A T, C T ) is stabilizable if and only if all unobservable modes of A are in the left half-plane. (A T, C T ) is stabilizable if and only if all eigenvalues of Āō of the KOCF are in the left half-plane. S. Bolouki (UIUC) 11 / 19

12 Observer Feedback Observer Feedback S. Bolouki (UIUC) 12 / 19

13 Observer Feedback Observer Feedback General objective: Stabilization via observer feedback. ẋ = Ax BKˆx ˆx = LCx + (A BK LC)ˆx [ẋ ] ˆx = [ [ ] A BK LC A BK LC] xˆx S. Bolouki (UIUC) 13 / 19

14 Observer Feedback Observer Feedback [ ] [ [ ] x xˆx] I 0 = P, where P = e I I ẋ = Ax BKˆx = (A BK)x + BKe ė = (A LC)e [ẋ ] [ ] [ A BK BK x = ė 0 A LC e] }{{} A cl ( ) si (A LC) det(si A cl ) = det ( si (A BK) ) }{{} arbitrary pp if (A, B) is controllable Seperation Principle }{{} arbitrary pp if (A, C) is observable S. Bolouki (UIUC) 14 / 19

15 Reduced Order Observers Reduced Order Observers S. Bolouki (UIUC) 15 / 19

16 Reduced Order Observers Reduced Order Observers {ẋ = Ax + Bu Let C p n be full-row-rank. Define [ P = y = Cx C n p rows to make P non-singular ] ] [ x1, x = = Px x 2 S. Bolouki (UIUC) 16 / 19

17 Reduced Order Observers Reduced Order Observers Let C p n be full-row-rank. Define [ P = {ẋ = Ax + Bu y = Cx C n p rows to make P non-singular ] ] [ x1, x = = Px x 2 Therefore x = ] [ x 1 = x 2 ] ] ] [Ā11 Ā 12 [ x1 [ B + 1 u Ā 21 Ā 22 x 2 B 2 }{{}}{{} PAP 1 PB y = [ I 0 ] [ x ] 1 x 2 S. Bolouki (UIUC) 16 / 19

18 Reduced Order Observers Reduced Order Observers x 2 = Ā 22 x 2 + Ā 21 y + B 2 u }{{} ū ẏ Ā 11 y B 1 u = Ā }{{} 12 x 2 ȳ S. Bolouki (UIUC) 17 / 19

19 Reduced Order Observers Reduced Order Observers Thus, to estimate x 2, we employ x 2 = Ā 22 x 2 + Ā 21 y + B 2 u }{{} ū ẏ Ā 11 y B 1 u = Ā }{{} 12 x 2 ȳ observer ˆx2 = Ā 22ˆx 2 + ū + L(ȳ ŷ) where ŷ = Ā 12ˆx 2 S. Bolouki (UIUC) 17 / 19

20 Reduced Order Observers Reduced Order Observers Thus, to estimate x 2, we employ x 2 = Ā 22 x 2 + Ā 21 y + B 2 u }{{} ū ẏ Ā 11 y B 1 u = Ā }{{} 12 x 2 ȳ observer ˆx2 = Ā 22ˆx 2 + ū + L(ȳ ŷ) Then, if e = x 2 ˆx 2, we have where ŷ = Ā 12ˆx 2 ė = (Ā 22 LĀ 12 )e Remember: Arbitrary pole placement Observability of (Ā 22, Ā 12 ). S. Bolouki (UIUC) 17 / 19

21 Reduced Order Observers Reduced Order Observers Reduced Order Observer Estimation of x S. Bolouki (UIUC) 18 / 19

22 Reduced Order Observers Reduced Order Observers Reduced Order Observer No derivative of y is required! S. Bolouki (UIUC) 19 / 19