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20095ee141_1_EE141_hw6_sol

# 20095ee141_1_EE141_hw6_sol - EE141 Principles of Feedback...

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EE141 Principles of Feedback Control (Fall 2009) Solutions to Homework 6 Problem 9.2 (a) The equilibrium pairs for u = 1 is di dt = - i + v = 0 dv dt = - i + g (1 - v ) = - i + (1 - v )(1 - v - 1)(1 - v - 4) = 0 which gives v = - 1 ± 3 Thus the other two equilibrium pairs are v 2 = i 2 = 0 . 732 and v 3 = i 3 = - 2 . 732. (b) x = i v = x 1 x 2 ˙ x = ˙ i ˙ v = - i + v - i + g ( u - v ) = - x 1 + x 2 - x 1 + g ( u - x 2 ) = f 1 ( x, u ) f 2 ( x, u ) F = ∂f 1 ∂x 1 ∂f 1 ∂x 2 ∂f 2 ∂x 1 ∂f 2 ∂x 2 x 1 = x 2 =0 ,u =1 = - 1 1 - 1 3 G = ∂f 1 ∂u ∂f 2 ∂u x 1 = x 2 =0 ,u =1 = 0 - 3 Thus the linearized model here is ˙ x = - 1 1 - 1 3 x + 0 - 3 u (c) (1) For v 2 = i 2 = 0 . 732 , u = 1 F = ∂f 1 ∂x 1 ∂f 1 ∂x 2 ∂f 2 ∂x 1 ∂f 2 ∂x 2 x 1 = x 2 =0 . 732 ,u =1 = - 1 1 - 1 - 1 . 54 G = ∂f 1 ∂u ∂f 2 ∂u x 1 = x 2 =0 . 732 ,u =1 = 0 1 . 54 1

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Thus the linearized model here is ˙ x = - 1 1 - 1 - 1 . 54 x + 0 1 . 54 u (2) For v 2 = i 2 = - 2 . 732 , u = 1 F = ∂f 1 ∂x 1 ∂f 1 ∂x 2 ∂f 2 ∂x 1 ∂f 2 ∂x 2 x 1 = x 2 = - 2 . 732 ,u =1 = - 1 1 - 1 - 8 . 46 G = ∂f 1 ∂u ∂f 2 ∂u x 1 = x 2 = - 2 . 732 ,u =1 = 0 8 . 46 Thus the linearized model here is ˙ x = - 1 1 - 1 - 8 . 46 x + 0 - 8 . 46 u Problem 9.4 (a) ˙ x = - x 2 e - 1 x + sin 0 = 0 x = 0 Note: the x = 0 here is actually x = 0 + . (b) Let ˙ x = - x 2
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