from which we obtain the following differential equation v R 1 5 v 3 R v R v i

From which we obtain the following differential

This preview shows page 17 - 23 out of 90 pages.

from which we obtain the following differential equation ˙ v R + 1 . 5 v 3 R + v R = v i When v i = 14, we can find a steady state solution for the circuit by considering ˙ v R = 0, and thus 1 . 5 v 3 R + v R = 14, from which we obtain v R = 2. Hence, we can linearize the differential equation of
Image of page 17
12 CHAPTER 1. the system around the solution ( v R = 2 , v i = 14). Since ˙ v R = f ( v R , v i ) = - 1 . 5 v 3 R - v R + v i we have Δ ˙ v R = ∂f ∂v R v R =2 v i =14 Δ v R + ∂f ∂v i v R =2 v i =14 Δ v i = ( - 4 . 5 v 2 R - 1) v R =2 Δ v R + (1)Δ v i = - 19Δ v R + Δ v i Exercise 1.13. Defining x 1 = φ , x 2 = ˙ φ , x 3 = s , x 4 = ˙ s , we have the following equations ˙ x 1 = x 2 ˙ x 2 = g L sin x 1 - 1 L M ( - Fx 4 + μ ( t )) cos x 1 ˙ x 3 = x 4 ˙ x 4 = - F M x 4 + μ ( t ) M We see that x i = 0, i = 1 , . . . , 4, μ ( t ) = 0 is a solution. Linearizing the system about this solution, we get Δ ˙ x 1 Δ ˙ x 2 Δ ˙ x 3 Δ ˙ x 4 = 0 1 0 0 g L 0 0 F L M 0 0 0 1 0 0 0 - F M Δ x 1 Δ x 2 Δ x 3 Δ x 4 + 0 - 1 L M 0 1 M Δ μ. Exercise 1.14. The states for the simple pendulun are shown in Figure 1.6 and 1.7, where the dashed line corresponds to the linearized system, and the solid to the nonlinear one. When θ 0 = π/ 18 and θ 0 = π/ 12 (Figure 1.6) the states are similar for the nonlinear and the linearized model. The system was linearized about the solution x = [0 , 0] T which is close to the initial condition. In the case when θ 0 = π/ 6 and θ 0 = π/ 3 the linearized system is not a good approximation of the nonlinear one. This can be observed from the evolution of the states in Figure 1.7).
Image of page 18
13 Figure 1.6: (i) θ 0 = π/ 18, (ii) θ 0 = π/ 12 Figure 1.7: (i) θ 0 = π/ 6, (ii) θ 0 = π/ 3
Image of page 19
14 CHAPTER 1.
Image of page 20
Chapter 2 Exercise 2.1. (a) In Problem 1.2(a), let x 1 = y 1 , x 2 = ˙ y 1 , x 3 = y 2 , and x 4 = ˙ y 2 . Also, let u 1 = f 1 and u 2 = f 2 , and let ( v 1 , v 2 ) = v T denote the output vector with v 1 = y 1 = x 1 and v 2 = y 2 = x 3 . From Problem 1.2(a) we obtain ˙ x 1 ˙ x 2 ˙ x 3 ˙ x 4 = 0 1 0 0 - [ ( K 1 + K ) M 1 ] - [ ( B 1 + B ) M 1 ] K M 1 B M 1 0 0 0 1 K M 2 B M 2 - [ ( K + K 2 ) M 2 ] - [ ( B + B 2 ) M 2 ] x 1 x 2 x 3 x 4 + 0 0 1 M 1 0 0 0 0 - 1 M 2 u 1 ( t ) u 2 ( t ) = A x 1 x 2 x 3 x 4 + B u 1 ( t ) u 2 ( t ) and v 1 v 2 = 1 0 0 0 0 0 1 0 x 1 x 2 x 3 x 4 = C x 1 x 2 x 3 x 4 . (b) Same as in item (a), except consider u T = ( u 1 , u 2 ) = ( f 1 , 5 f 2 ) T as the system input and (8 y 1 + 10 y 2 ) as the (scalar-valued) system output. Then ˙ x 1 ˙ x 2 ˙ x 3 ˙ x 4 = A x 1 x 2 x 3 x 4 + 0 0 1 M 1 0 0 0 0 - 1 M 2 u 1 ( t ) u 2 ( t ) = A x 1 x 2 x 3 x 4 + 0 0 1 M 1 0 0 0 0 - 5 M 2 f 1 ( t ) f 2 ( t ) = A x 1 x 2 x 3 x 4 + B f 1 ( t ) f 2 ( t ) 15
Image of page 21
16 CHAPTER 2.
Image of page 22
Image of page 23

You've reached the end of your free preview.

Want to read all 90 pages?

  • Fall '13

  • Left Quote Icon

    Student Picture

  • Left Quote Icon

    Student Picture

  • Left Quote Icon

    Student Picture