Control Systems Engineering
Control Systems Engineering
7th Edition
ISBN: 9781118170519
Author: Norman S. Nise
Publisher: WILEY
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Chapter 3, Problem 30P

Figure P3.17 shows a free-body diagram of an inverted pendulum, mounted on a cart with a mass, M. The pendulum has a point mass, m, concentrated at the upper end of a rod with zero mass, a length, I, and a frictionless hinge. A motor drives the cart, applying a horizontal force, u(t). A gravity force, mg, acts on m at all times. The pendulum angle relative to the y-axis, θ , its angular speed, θ ˙ , the horizontal position of the cart, x, and its speed, x’, were selected to be the state variables. The state-space equations derived were heavily nonlinear.14 They were then linearized around the stationary point, x0= 0 and u0= 0, and manipulated to yield the following open-loop model written in perturbation form:

d d t δ x = A δ x + B δ u

However, since x0= 0 and u0= 0, then let: x=x 0 +δx=δx and u = u 0 + δ u = δ u . Thus the state equation may be rewritten as (Prasad, 2012):

x ˙ =Ax+B u

where

A = 0 1 0 0 M + m g M l 0 0 0 0 0 0 1 m g M 0 0 0 and B = 0 M l 0 1 M

Assuming the output to be the horizontal position of m = x m = x + l sin θ = x + l θ for a small angle, θ , the output equation becomes:

y = l θ + x = Cx= l 0 1 0 θ θ ˙ x x ˙

Given that: M = 2.4 kg, m = 0.23 kg, MATLAB ML l = 0.36 m, g = 9.81 m/s2, use MATLAB to ?nd the transfer function, G(s) = Y(s)/U(s) = Xm(s)/U(s).

Chapter 3, Problem 30P, Figure P3.17 shows a free-body diagram of an inverted pendulum, mounted on a cart with a mass, M.

FIGURE P3.17 Motor-driven inverted pendulum can system15

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