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

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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

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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 0 x (5) A 0 = A + BK (6) If it is possible to make the eigenvalues of A 0 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, ˆ A = T - 1 AT and ˆ B =
EL 625 Lecture 10 3 T - 1 B are in the controller canonical form , shown below. ˆ A = 0 1 0 . . . . . . 0 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 0 . . . 0 1 - a 0 - a 1 . . . . . . . . . - a n - 1 (7) ˆ B = 0 . . . . . . 0 1 (8) a n - 1 , a n - 2 , . . . , a 1 , a 0 are the coefficients of the characteristic poly- nomial (the characteristic polynomial is λ n + a n - 1 λ n - 1 + . . . + a 1 λ + a 0 ). The similarity transformation to achieve this ‘controller canon- ical’ structure for the single-input case is described below. Characteristic polynomial , p ( λ ) = λ n + a n - 1 λ n - 1 + . . . + a 1 λ + a 0 (9)

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EL 625 Lecture 10 4 = λ ( λ n - 1 + a n - 1 λ n - 2 + . . . + a 1 ) | {z } p 1 ( λ ) + a 0 (10) p 1 ( λ ) = λ n - 1 + a n - 1 λ n - 2 + . . . + a 1 (11) = λ ( λ n - 2 + a n - 1 λ n - 3 + . . . + a 2 ) | {z } p 2 ( λ ) + a 1 (12) . . . p n - 1 ( λ ) = λ + a n - 1 (13) p n ( λ ) = 1 (14) Define the set of vectors ( v 1 , . . . , v n ) as v 1 = p 1 ( A ) B v 2 = p 2 ( A ) B . . . v n = p n ( A ) B (15) p i ( A ) is the polynomial p i ( λ ) evaluated at λ = A . To prove that this coordinate transformation achieves the desired transformed form, we have to compute the effect of A on the coordi- nate vectors, v 1 , v 2 , . . . , v n . Av 1 = A ( A n - 1 + a n - 1 A n - 2 + . . . + a 2 A + a 1 I ) B
EL 625 Lecture 10 5 = ( A n + a n - 1 A n - 1 + . . . + a 2 A 2 + a 1 A ) = ( A n + a n - 1 A n - 1 + . . . + a 2 A 2 + a 1 A + a 0 I - a 0 I ) B = - a 0 IB = - a 0 B = - a 0 v n (16) (By Cayley Hamilton theorem, p ( A ) = A n + a n - 1 A n - 1 + . . . + a 2 A 2 + a 1 A + a 0 I = 0).

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