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 da f u zt z bc v −+ ⎛⎞ =⇔ = ⎜⎟ ⎝⎠ If the system matrix is invertible, then 1 * () 1 f u c u a f v z v b u d v ba 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 f dcb a f c d b a f c d b a f c d λ λλ + +=+ + −−= −− ± − − − − ⇒= Assuming that b(a-f) > cd, If , then two eigenvalues are real and negative. Therefore, the system is asymptotically stable and has no oscillation. 2 4 ( ( ( 4 ( b d b a f −≥ +≥ If , then two eigenvalues are complex numbers with negative real part. Hence, the system is asymptotically stable and exhibits oscillation.
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This note was uploaded on 06/16/2010 for the course MS&E 201 taught by Professor Edisontse during the Spring '08 term at Stanford.

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

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