Question

Problem 2 and 3

A simplified model of a magnetic levitation system has the dynamic model 1 2 (a) Find the transfer function G(s) of the system. (b) Find the poles and zeros of the system. (c) The plant is unstable. Explain why Problem 2 The plant in Problem 1 is to be stabilized by use of "proportional plus derivative" control: U(s)-(Kis + K2)Y(s) Find and sketch the region in the Ki, K2 plane for which the closed loop system, using this control law, is stable.. Problem3 In Problem 2. Ki - K, K2-3K. Sketch the root locus of the poles of the the closed-loop system as varies from 0 to oo. Show important features, such as Imaginary axis ctossings Asymptotes as K -oo . Segments on the real axis.

Answer 1

Given The following is the dynamic model x_1 = x_2 x_2 = 4 x_1 + x_3 x_3 = -6 x_3 + 4 y = x_1 In matrix form [x_1 x_2 x_3] = [0 4 0 1 0 0 0 1 -1] [x_1 x_2 x_3] + [0 0 1] u It is in the form of X = Ax + Bu A = [0 4 0 1 0 0 0 1 -2] B = [0 0 1] y = x_1 y = [1 0 0] [x_1 x_2 x_3] + [0] u \r\n Formula for transfer function is y (s)/u (s) = C (SI - A)^-1 B + 0 (SI - A) = [s u 0 -1 s 0 0 -1 s + 1] (SI - A)^-1 = Adj (SJ - A)/SJ - A Adj (SI - A) = [s (s + 1) u (s + 1) 0 (s + 1) s (s + 1) 0 1 s s^2 - 1] (SI - A) = s [s (s + 1) + 1 - 4 (s + 1) - 0 + 0] = s^3 + s^2 - 4s - 4 Therefore y (s) u (s) = [1 0 0] 1/s^3 + s^2 - 4s - 4 [s (s