Homework6_2016_sol.pdf - ESE 551 Linear Dynamic Systems...

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ESE 551: Linear Dynamic Systems Assignment 6 Due: December 12, 2016 1. Consider the linear dynamical system ˙ x = - 3 x + 3 u and the cost functional η = Z 5 0 2 x 2 + u 2 dt. Minimizing η under the assumption that the final end point is not constrained. Solution: This is a typical free Endpoint optimal control problem, LQR gives us the optimal control law as b u ( t ) = - 3 k ( t ) x ( t ) which minimizes η , where k ( t ) is the solution to the Riccati equation ˙ k ( t ) = 6 k ( t ) + 9 k 2 ( t ) - 2 with boundary condition k (5) = 0. We solve this equation by change of variable, p = k + 1 3 . ˙ k = 6 k + 9 k 2 - 2 = 9 ( k 2 + 2 3 k - 2 9 ) = 9 h ( k + 1 3 ) 2 - 1 3 i ˙ p = 9 ( p 2 - 1 3 ) d p p 2 - 1 3 = 9d t - 3 2 · d p p + 1 3 + 3 2 d p p - 1 3 = 9d t - 3 2 · ln p + 1 3 + 3 2 ln p - 1 3 = 9 t + c 3 2 ln p - 1 3 p + 1 3 = 9 t + c Because p (5) = k (5) + 1 3 = 1 3 then p - 1 3 < 0, so plugging in the exponential results p - 1 3 p + 1 3 = - e 6 3 t + c p = 1 3 1 - Ae 6 3 t 1 + Ae 6 3 t ! for A > 0 . Solving for A using p (5) = 1 3 results in A = 3 - 1 1 + 3 e - 30 3 and therefore k ( t ) = 1 3 1 - Ae 6 3 t 1 + Ae 6 3 t ! - 1 3 = 1 3 3 1 + 3 - ( 3 - 1) e 6 3( t - 5) 1 + 3 + ( 3 - 1) e 6 3( t - 5) ! - 1 ! = 2 3 1 - e 6 3( t - 5) 1 + 3 - (1 - 3) e 6 3( t - 5) ! Then k (5) = 0 as required, and the optimal control law b u that minimizes η is given by b u ( t ) = - 2 1 - e 6 3( t - 5) 1 + 3 - (1 - 3) e 6 3( t - 5) ! · x ( t ) . 1
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The minimum value of η is given by min u η = ( x (0)) 2 k (0) = 2 3 ( x (0)) 2 1 - e - 30 3 1 + 3 - (1 - 3) e - 30 3 ! 2 3 1 1 + 3 ( x (0)) 2 . 2. It is often useful to think about second order scalar systems using the phase plane. This is nothing more than a plot of the trajectories on the plane whose horizontal axis is x and whose vertical axis is ˙ x . Consider the linear quadratic problem min η = Z 0 x 2 + u 2 dt s . t . ¨ x + x = u. Use the Riccati equation formalism to find the optimal control in feedback form. Sketch the optimal trajectories in the phase plane. Solution: Let y = ˙ x so that ˙ z = ˙ x ˙ y = 0 1 - 1 0 x y + 0 1 u = Az + Bu and x = 1 0 x y = Cz By LQR, the control law b u ( t ) = - B 0 Π z ( t ) minimizes η where Π is the solution to the stationary Riccati equation for the triple [ A, B, C ], given by A 0 K + KA - KBB 0 K = - C 0 C . Hence let K = a b b c and thus 0 - 1 1 0 a b b c + a b b c 0 1 - 1 0 - a b b c 0 0 0 1 a b b c = - 1 0 0 0 - b - c a b + - b a - c b - b 2 bc bc c 2 = - 1 0 0 0 - 2 b - b 2 = - 1 a - c - bc = 0 2 b - c 2 = 0 The unique real solution to this system of equations is a = 2 p 2 - 1, b = 2 - 1, and c = q 2( 2 - 1). We conclude that the solution to the stationary Riccati equation associated with [ A, B, C ] is Π = " 2 p 2 - 1 2 - 1 2 - 1 q 2( 2 - 1) # The minimizing control is b u ( t ) = - B Π z ( t ) = - ( 2 - 1) x ( t ) - q 2( 2 - 1) y ( t ).
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