# set9 - Chapter 4 State Space Design Part 1 Pole Placement \$...

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a39 a38 a36 a37 Chapter 4 – State Space Design Part 1 – Pole Placement P. M. Van Dooren Department of Mathematical Engineering Catholic University of Louvain K. A. Gallivan School of Computational Science Florida State University Spring 2006 1

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a39 a38 a36 a37 Pole Placement Definition 9.1 Let λ stands for d/dt or E then redefining the input u ( · ) as Fx ( · ) + u ( · ) is using state feedback to alter the system to the form: 8 < : λx ( · ) = ( A + BF ) x ( · ) + Bu ( · ) y ( · ) = ( C + DF ) x ( · ) + Du ( · ) . Problem 9.1 Given a linear time invariant system use state feedback to place the poles of the modified system at specified points in the complex plane. 2
a39 a38 a36 a37 Similarity Transformations Note 9.1 A similarity transformation T does not affect the solution of this problem. This is seen from the feedback F t = FT applied to the transformed pair ( A t , B t ) . = ( T - 1 AT, T - 1 B ) A t + B t F t = T - 1 ( A + BF ) T which has the same eigenvalues and Jordan structure as A + BF T can be chosen to: simplify the construction of F t (and F ) to make explicit its possible solution(s). 3

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a39 a38 a36 a37 Limits of State Feedback Assume ( A, B ) has been transformed by a unitary T into the form [ A t | B t ] = 2 4 A r × B r 0 A ¯ r 0 3 5 where A ¯ r contains the unreachable modes and ( A r , B r ) is reachable. If F t = [ F r | F ¯ r ] then A t + B t F t = 2 4 A r + B r F r × 0 A ¯ r 3 5 . So F t can only affect the spectrum of A r . 4
a39 a38 a36 a37 Controller Canonical Form Ackerman’s method for pole placement depends upon controller canonical form. A similarity transformation that puts an arbitrary system into that form can be deduced analytically. Lemma 9.1 If ( A, b ) is given and T 1 = [ b Ab · · · A n - 1 b ] is the controllability matrix then h T - 1 1 AT 1 T - 1 1 b i = h A 1 b 1 i = h e 2 · · · e n a e 1 i where a T = h α 0 · · · α n - 1 i defines the characteristic polynomial of A . 5

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a39 a38 a36 a37 Controller Canonical Form Proof By definition we have that T 1 e i = A i - 1 b and therefore T - 1 1 b = e 1 and e i = T - 1 1 A i - 1 b Now consider the columns of T - 1 1 AT 1 for 1 i n 1 . T - 1 1 AT 1 e i = T - 1 1 A ( A i - 1 b ) = T - 1 1 A i b = e i +1 The last column T - 1 1 AT 1 e n has n components we label α i for 0 i n 1 which are the coefficients of the characteristic polynomial. square . 6
a39 a38 a36 a37 Controller Canonical Form Definition 9.2 The system h A 2 b 2 i is in controller canonical form if b 2 = e 1 and A 2 e i = e i +1 α n - i e 1 A 2 e n = α 0 e 1 where 1 i n 1 and P ( ξ ) = ξ n + P n - 1 i =0 α i ξ i is the characteristic polynomial of A 2 . 7

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a39 a38 a36 a37 Controller Canonical Form Lemma 9.2 Let h A b i , h A 1 b 1 i and T 1 be as in Lemma 9.1. If T 2 = 2 6 6 6 6 6 6 6 4 1 α n - 1 . . . α 1 . . . . . . . . . . . . α n - 1 1 3 7 7 7 7 7 7 7 5 then h A 2 b 2 i = h T - 1 2 A 1 T 2 T - 1 2 b 1 i is in controller canonical form with Λ( A 2 ) = Λ( A 1 ) = Λ( A ) .
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