lect10 - EL 625 Lecture 10 1 EL 625 Lecture 10 Pole...

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Unformatted text preview: EL 625 Lecture 10 1 EL 625 Lecture 10 Pole Placement and Observer Design Pole Placement Consider the system x = A x (1) The solution to this system is x ( t ) = e At x (0) (2) If the eigenvalues of A all lie in the open left half plane, x ( t ) asymp- totically approaches the origin as time goes to infinity (the system is asymptotically stable ). If the eigenvalues of A lie in the closed left half plane (with some eigenvalues possibly on the imaginary ax- is), the state x ( t ) remains bounded (the system is stable ). If some eigenvalues of A lie in the right half plane, the system is unstable . Consider the system x = A x + B u (3) The state x is measured and can be used to calculate a suitable value of u to make the system stable. The simplest feedback is a static EL 625 Lecture 10 2 linear state feedback, u = K x (where K is a constant matrix). With this feedback, x = A x + B u = A x + BK x = ( A + BK ) x (4) = A x (5) A = A + BK (6) If it is possible to make the eigenvalues of A lie in the open left half plane by using the gains K , then, choosing u = K x would make the resulting closed loop system stable. It can be proved that the eigenvalues of ( A + BK ) can be arbitrarily assigned through a suitable choice of K if the pair ( A, B ) is controllable. If(A,B) is a controllable pair, a similarity transformation, T can be found such that the transformed matrices, A = T- 1 AT and B = EL 625 Lecture 10 3 T- 1 B are in the controller canonical form , shown below. A = 1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1- a- a 1 . . . . . . . . .- a n- 1 (7) B = . . . . . . 1 (8) a n- 1 , a n- 2 , . . . , a 1 , a are the coefficients of the characteristic poly- nomial (the characteristic polynomial is n + a n- 1 n- 1 + . . . + a 1 + a ). The similarity transformation to achieve this controller canon- ical structure for the single-input case is described below. Characteristic polynomial , p ( ) = n + a n- 1 n- 1 + . . . + a 1 + a (9) EL 625 Lecture 10 4 = ( n- 1 + a n- 1 n- 2 + . . . + a 1 ) | {z } p 1 ( ) + a (10) p 1 ( ) = n- 1 + a n- 1 n- 2 + . . . + a 1 (11) = ( n- 2 + a n- 1 n- 3 + . . . + a 2 ) | {z } p 2 ( ) + a 1 (12) ....
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lect10 - EL 625 Lecture 10 1 EL 625 Lecture 10 Pole...

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