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problemset6_Soln - The University of Texas at Austin...

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The University of Texas at Austin Department of Electrical and Computer Engineering EE362K: Introduction to Automatic Control—Fall 2009 Problem Set Six Solution C. Caramanis Due: Friday, October 30, 2009. This problem set focuses on the new concepts introduced in the last three classes: reachability and state feedback, and integral action, and also observability and state estimation. As usual, we also work in some exercise with important concepts from linear algebra. 1. Integral Action: (a) Consider the system: A = 1 6 4 0 , B = 0 1 , C = [ 1 0 ] Compute a feedback control policy, u = Kx , so that the closed loop eigenvalues are both at 1. (b) Let the reference signal r be equal to 1. Compute the feedforward term so that the steady state output is equal to the reference signal, r . (c) Plot the trajectory of the output (i.e., plot y ( t )) to verify that you have chosen K and k r appropriately so that the system is stable, and so that the steady state output is equal to r . (d) As in the previous exercise, compute a perturbation matrix, and choosing a value of p so that the closed loop system is still stable, plot the behavior of the perturbed system. (e) Now back to the nominal system: suppose we have a constant disturbance, so that the system dynamics are now: ˙ x = Ax + Bu + Fd, where d is an unknown but constant disturbance, and F is the 2 × 1 matrix (1 , 1) . Pick a constant value for d , and plot the output trajectory for the same values of K and k r . (f) Write down the closed loop augmented system, where you add a state z that is the integral of the error, y r , and where you use feedback control with integral action: u = Kx k i z. In particular, write down the augmented system matrix (this should be a 3 × 3 matrix). (g) Find values for K and k i so that the augmented closed loop system matrix is stable. (You can use the same K you computed above, if you like). (h) Plot the output trajectory ( y ( t )) for different values of the feedforward gain, k r . 1
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(i) Fix your control parameters, and plot the output trajectory for different values of con- stant disturbance, d . What happens? Solution : (a) First, we compute the reachable canonical form as: Φ A ( λ ) = | λI A | = λ 2 λ + 24 ˜ A = 1 24 1 0 , ˜ B = 1 0 , and the change of coordinates which leads to it: W r = 0 6 1 0 , ˜ W r = 1 1 0 1 T = ˜ W r W 1 r = 1 6 1 1 6 0 The desired closed loop eigenvalues give us the closed loop characteristic function as: Φ A,cl ( λ ) = i ( λ λ cl,i ) = ( λ + 1)( λ + 1) = λ 2 + 2 λ + 1 therefore, the desired ˜ K and K would be: ˜ K = 2 ( 1) 1 24 = 3 23 K = ˜ KT = 10 3 3 (b) In order to have the steady state output to be equal to the reference point, which is equal to having zero frequency gain equal to one, we should have kr as given in 6.13 in the textbook: k r = 1 C ( A BK ) 1 B = 0 . 1667 . (c) In order to see the output trajectory we should compute the response of the closed loop system. The model of the closed loop system is: ˙ x = Ax + Bu = Ax + B ( Kx + k r r ) = ( A BK ) x + Bk r r = ˆ Ax + ˆ Br y = Cx.
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