ee441spring2007final_solution

# ee441spring2007final_solution - . 2.2 J = 4 1 0 0 4 0 0 0 1...

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EE441 Final Exam Solution Spring 2007 Instructor : Dr. Jonckheere TA: Ben Raskob Due Date 5/8/2007 1 Diferential Equation 1.1 b ˙ q ( t ) ¨ q ( t ) B = b 0 1 - 5 - 6 B b ˙ q ( t ) ¨ q ( t ) B 1.2 b 0 1 - 5 - 6 B ± ²³ ´ A = b 1 1 - 1 - 5 B ± ²³ ´ S b - 1 0 0 - 5 B ± ²³ ´ D b 5 / 4 1 / 4 - 1 / 4 - 1 / 4 B ± ²³ ´ S - 1 1.3 u ( t ) = e At u (0) = b 1 1 - 1 - 5 B ± ²³ ´ S b e - t 0 0 e - 5 t B ± ²³ ´ e Dt b 5 / 4 1 / 4 - 1 / 4 - 1 / 4 B ± ²³ ´ S - 1 b 2 2 B ± ²³ ´ u (0) = 3 e - t b 1 - 1 B - e - 5 t b 1 - 5 B Yes, the system is stable. 1

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2 Jordan Form 2.1 λ 1 = 4 (multiplicity of 2), λ 2 = 1. For λ 1 = 4, there is one real eigenvector 2 1 0 and one generalized eigen- vector 1 0 0 . For λ 2 = 1, the eigenvector is 1 1 1

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Unformatted text preview: . 2.2 J = 4 1 0 0 4 0 0 0 1 S = 2 1 1 1 0 1 0 0 1 OR J = 1 0 0 0 4 1 0 0 4 S = 1 2 1 1 1 0 1 0 0 2 3 Positive Defniteness The matrix is not PD by the determinant test, v v v v 3 5 5 8 v v v v =-1. 3 4 Multiple Choice I. D II. A III. C IV. C V. B VI. D VII. D VIII. C IX. B X. A 4...
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## This note was uploaded on 09/16/2010 for the course EE 441 taught by Professor Neely during the Spring '08 term at USC.

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ee441spring2007final_solution - . 2.2 J = 4 1 0 0 4 0 0 0 1...

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