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Unformatted text preview: EE141 HW 6 Principles of Feedback Control Instructor: Balakrishnan, A.V. Problem Set 6: Solutions For the system ˙ x = Ax + Bu with A = 1 25 0 , B = 1 1. Find a controller that steers the initial state x (0) = [1 , 2] to the origin x ( T ) = [0 , 0] in time T = 1 . Solution: Notice that the solution for this problem is not unique. There are many controllers that will steer the system to the origin. Below there is just one possible controller. If you designed a different controller (for instance, as Prof. Balakrishnan did in class), the energy will be different and therefore your solution for problem 2 will be different as well. For the cost function: C ( u ) =  x ( T ) x final  2 + λ Z t u ( t ) 2 dt The open loop controller will be given by: u opt ( t ) = B * e A * ( T t ) ( R ( T ) + λI ) 1 ( x final e AT x (0)) R ( T ) = LL * = Z T e As BB * e A * s ds In this problem energy is not a concern (it is not specified in the problem), so we set λ = 0 . We compute e At as in the previous homework using CayleyHamilton. e At = cos(5 t ) 1 / 5 sin(5 t )] 5 sin(5 t ) cos(5 t )] R ( T ) = LL * = . 021 0 . 018 . 018 0 . 473 R ( T ) 1 = 49 . 08 1 . 91 1 . 91 2 . 19 u opt ( t ) = B * e A * (1 t ) ( R ( T )) 1 e A 1 2 = = 3 . 03 sin( 5 + 5 t ) 11 . 9 cos( 5 + 5 t ) 1 In the following figures we can see how this controller indeed steers the system to the origin. In the first one, the trajectory of x 2 is plotted along the trajectory of x 1 , thus the initial point is the initial state (1 , 2) and the final point is the origin (0 , 0) . The second plot proves the same result but now time is explicit, so it is shown how the states approach the origin as time tends to the final time (1 second in this case).0.40.2 0.2 0.4 0.6 0.8 1 1.24321 1 2 x1 x2 Figure 1: Phase portrait of the state variables 2 0.2 0.4 0.6 0.8 14321 1 2 time(’secs’) x1 x2 Figure 2: Trajectories of the system 2. Now consider the case where the control energy R 1 u ( t ) 2 dt is required to be less than 1 2 the energy in problem 1. Find the optimal controller, open loop and closed loop, to minimize  x (1)  2 , formulating it first as a soft control problem. Solution For the open loop controller, the equations will be very similar to the ones we used in the first problem. Now we do care about the energy we spend, so λ will no longer be 0. We know that for problem 1 the energy was a maximum: Z 1 u ( t ) 2 dt = Z 1 ( 3 . 03 sin(...
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This note was uploaded on 03/18/2008 for the course EE 141 taught by Professor Balakrishnan during the Fall '07 term at UCLA.
 Fall '07
 Balakrishnan

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