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Unformatted text preview: MASSACHUSETTS INSTITUTE OF TECHNOLOGY Department of Aeronautics and Astronautics 16.060: Principles of Automatic Control Fall 2003 PROBLEM SET 6 Solutions Problem 1 1. ⎣ ⎤ ⎦ = B AB = 1 − 1 − 1 1 There are two states, so we need two independent columns to show controllability. But the control lability matrix has only one independent column, so the system is uncontrollable. 2. 1 1 1 ⎤ ⎦ = B AB A 2 B = 1 ⎡ 1 1 There are three states, so we need to find three independent columns to show controllability. The 1 controllability matrix includes the vectors ⎡ , 1 ⎡ , and ⎡ , which are independent and span the 1 statespace ⇒ 3 , so the system is controllable. Problem 2 1. (a) The characteristic equation of the openloop system is: det( sI − A ) = s 3 + 9 s 2 + 23 s + 15 = ( s + 5)( s + 3)( s + 1) = So the poles of the openloop system are at s = − 1, s = − 3, and s = − 5. (b) We are given that the closedloop poles are at s = − 3 and s = − 4 ± 4 j , so we want the characteristic equation for the closedloop system to be: ( s + 3)( s 2 + 8 s + 32) = s 3 + 11 s 2 + 56 s + 96 = In terms of the matrices A , B , and K , the characteristic equation is given by det( sI − ( A −...
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This note was uploaded on 11/06/2011 for the course AERO 100 taught by Professor Willcox during the Fall '03 term at MIT.
 Fall '03
 willcox
 Aeronautics, Astronautics

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