This preview has intentionally blurred sections. Sign up to view the full version.View Full Document
Unformatted text preview: EE363 Prof. S. Boyd EE363 homework 2 1. Derivative of matrix inverse. Suppose that X : R → R n × n , and that X ( t ) is invertible. Show that d dt X ( t )- 1 = − X ( t )- 1 parenleftBigg d dt X ( t ) parenrightBigg X ( t )- 1 . Hint: differentiate X ( t ) X ( t )- 1 = I with respect to t . 2. Infinite horizon LQR for a periodic system. Consider the system x t +1 = A t x t + B t u t , where A t = braceleftBigg A e t even A o t odd B t = braceleftBigg B e t even B o t odd In other words, A and B are periodic with period 2. We consider the infinite horizon LQR problem for this time-varying system, with cost J = ∞ summationdisplay τ =0 parenleftBig x T τ Qx τ + u T τ Ru τ parenrightBig . In this problem you will use dynamic programming to find the optimal control for this system. You can assume that the value function is finite. (a) Conjecture a reasonable form for the value function. You do not have to show that your form is correct. (b) Derive the Hamilton-Jacobi equation, using your assumed form. Hint: you should get a pair of coupled nonlinear matrix equations. (c) Suggest a simple iterative method for solving the Hamilton-Jacobi equation. You do not have to prove that the iterative method converges, but do check your method on a few numerical examples. (d) Show that the Hamilton-Jacobi equation can be solved by solving a single (bigger) algebraic Riccati equation. How is the optimal u related to the solution of this equation? Remark: the results of this problem generalize to general periodic systems. 3. LQR for a simple mechanical system. Consider the mechanical system shown below: d 1 d 2 d 3 d 4 m 1 m 2 m 3 m 4 k 1 k 2 k 3 k 4 u 1 u 1 u 2 u 2 u 3 1 Here d 1 ,...,d 4 are displacements from an equilibrium position, and u 1 ,...,u 3 are forces acting on the masses. Note that u 1 is a tension between the first and second masses, u 2 is a tension between the third and fourth masses, and u 3 is a force between the wall (at left) and the second mass. You can take the mass and stiffness constants to all be one: m 1 = ··· = m 4 = 1, k 1 = ··· = k 4 = 1....
View Full Document
This document was uploaded on 10/05/2011.
- Winter '09