This preview shows pages 1–5. Sign up to view the full content.
This preview has intentionally blurred sections. Sign up to view the full version.
View Full DocumentThis preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
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 singleinput 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) ....
View Full
Document
 Spring '10
 khorami

Click to edit the document details