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Unformatted text preview: EE141 Principles of Feedback Control (Fall 2009) Solutions to Homework 6 Problem 9.2 (a) The equilibrium pairs for u = 1 is di = −i + v = 0 dt dv = −i + g (1 − v ) = −i + (1 − v )(1 − v − 1)(1 − v − 4) = 0 dt which gives v = −1 ± √ 3 Thus the other two equilibrium pairs are v2 = i2 = 0.732 and v3 = i3 = −2.732. (b) x i =1 v x2 x= x= ˙ ˙ −x1 + x2 f (x, u) −i + v i = =1 = f2 (x, u) −i + g (u − v ) −x1 + g (u − x2 ) v ˙ ∂f1 ∂x1 F = ∂f 2 ∂x1 ∂f1 ∂u G = ∂f 2 ∂u ∂f1 ∂x2 ∂f2 ∂x2 =
x1 =x2 =0,u=1 −1 1 −1 3 =
x1 =x2 =0,u=1 0 −3 Thus the linearized model here is x= ˙ −1 1 0 x+ u −1 3 −3 (c) (1) For v2 = i2 = 0.732, u = 1 ∂f1 ∂x1 F = ∂f 2 ∂x1 ∂f1 ∂u G = ∂f 2 ∂u ∂f1 ∂x2 ∂f2 ∂x2 =
x1 =x2 =0.732,u=1 −1 1 −1 −1.54 =
x1 =x2 =0.732,u=1 0 1.54 1 Thus the linearized model here is x= ˙ −1 1 0 x+ u −1 −1.54 1.54 (2) For v2 = i2 = −2.732, u = 1 ∂f1 F = ∂x1 ∂f2 ∂x1 ∂f1 ∂u G = ∂f 2 ∂u ∂f1 ∂x2 ∂f2 ∂x2 =
x1 =x2 =−2.732,u=1 −1 1 −1 −8.46 =
x1 =x2 =−2.732,u=1 0 8.46 Thus the linearized model here is x= ˙ 0 −1 1 u x+ −8.46 −1 −8.46 Problem 9.4 (a) x = −x2 e− x + sin 0 = 0 ⇒ x = 0 ˙ Note: the x = 0 here is actually x = 0+ . (b) 1 Let x = −x2 e− x + sin u = f (x, u). ˙ F= G== ∂f ∂x ∂f ∂u = −(2xe− x + e− x )
x=0+ ,u=0
1 1 1 x=0+ ,u=0 =0 x=0+ ,u=0 = cos ux=0+ ,u=0 = 1 Thus the linearized model is x=u ˙ Problem 9.6 (a) x=0⇒x=0 ˙ 2 Thus the equilibrium point is x = 0 (b) Assume α = 1, and there is a small variation > 0 at x = 0.
3 x = x3 x= = ˙ >0 which will make x further away from 0. Therefore x = 0 is not a true equilibrium point and the linearized model is not valid. (c) Assume α = 1, and there is a small variation > 0 at x = 0.
3 x = −x3 x= = − ˙ <0 which will make x get back to 0. Therefore x = 0 is a true equilibrium point and the linearized model is valid. 3 ...
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This note was uploaded on 04/04/2010 for the course EE 141 taught by Professor Balakrishnan during the Fall '07 term at UCLA.
 Fall '07
 Balakrishnan

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