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Unformatted text preview: 16.30/31, Fall 2010 — Recitation # 8 Brandon Luders November 1, 2010 In this recitation, we revisit LQR to perform a bit more analysis, then use it in part of the design of a DOFB controller. 1 LQR Revisited Suppose you are asked to solve the LQR problem specified by the following four matrices: 2 1 1 q , R = [ 1 ] , A = B , Q = = , 2 2 1 q where q 1 > and q 2 > 0. Note that this is equivalent to HW 5, Problem 3, part (a), with ρ = 1 and M = 1 kg, but with a different choice of Q . Thus we’re dealing with a double integrator with state x = [ x v ], where x is position/displacement and v is velocity, and input u corresponding to the force applied. First, let’s check our assumptions on the LQR problem: 1. The matrix Q must be positive semidefinite, i.e. Q . 2 2 ≥ 0, this will always be satisfied. 2 1 Since q ≥ and q 2. The matrix R must be positive definite, i.e. R . Since 1 > 0, this is satisfied. 3. The solution P to the algebraic Riccati equation is always symmetric, such that P T = P . We will use this below. 4. If ( A, B, C z ) is stabilizable and detectable, then the correct solution of the algebraic Riccati equation is the unique solution P for which P . If ( A, B, C z ) is also observable, then P . We have that 1 M c = [ B AB ] = 1 ; this is full rank, so the system is controllable. If we choose C z = I 2 , it is immediately clear that the system is also observable. Thus, when solving for P below, we can use the fact that P 0. We can represent P symbolically as a b P = , b c where a , b , and c are scalar quantities to be found. We now plug everything into the algebraic Riccati equation to solve for P : 0 = A T P + PA + Q − P BR − 1 B T P a b a b 1 q 1 2 = + + 2 1 b c b c q 2 a b 1 a b − b c 1 1 [ 0 1 ] b c a q 1 2 = + + 2 a b b q 2 a b a b − b c 1 b c q 1 2 a a b =...
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This note was uploaded on 02/03/2012 for the course AERO 16.30 taught by Professor Ericferon during the Fall '04 term at MIT.
- Fall '04