2006 MIDTERM SOLUTIOINS

2006 MIDTERM SOLUTIOINS - MS&E201 Dynamic Systems Professor...

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MS&E201 Dynamic Systems page 1 of 4 Professor Edison Tse May 26, 2006 Midterm Solutions Problem 1. 1) F 2) F 3) F 4) F 5) T 6) a)F b)F c)T d)T e)T f)F Problem 2. (a) * ( ) 0 d a f u z t z b c v + = = If the system matrix is invertible, then 1 * ( ) 1 ( ) d a f u cu a f v z b c v bu dv b a f cd + + = = + If the system matrix is singular, then there is no equilibrium point. Multiply the first row by c and the second row by -(a-f) and sum up two rows yields: 0 ( , which is a contradiction. ( )) cu v a f = − + < (b) 2 2 det ( )( ) ( ) ( ) ( ) 0 ( ) ( ) 4( ( ) ) 2 d a f d c b a f c d b a f cd b c c d c d b a f cd λ λ λ λ λ λ λ + = + = + + = ± = Assuming that b(a-f) > cd, If , then two eigenvalues are real and negative. Therefore, the system is asymptotically stable and has no oscillation.
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