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, $\theta$ , its angular speed, ${\stackrel{\dot{}}{\theta }}^{\prime }$ , 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:

$\frac{d}{dt}\delta \text{x}=\text{A}\delta \text{x}+\text{B}\delta u$

However, since x₀= 0 and u₀= 0, then let: ${\text{x=x}}_{\text{0}}\text{+δx=δx}$ and $u=u_0+\text{δ}u=\text{δ}u$ . Thus the state equation may be rewritten as (Prasad, 2012):

$\text{<mover accent="true"> <mo> x </mo> <mo> ˙ </mo> </mover> =Ax+B}u$

where

$\text{A}=\left[\begin{array}{cccc}0& 1& 0& 0\\ \frac{\left(M+m\right)g}{Ml}& 0& 0& 0\\ 0& 0& 0& 1\\ -\frac{mg}{M}& 0& 0& 0\end{array}\right]$ and $\text{B}=\left[\begin{array}{c}0\\ \frac{-}{Ml}\\ 0\\ \frac{1}{M}\end{array}\right]$

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

$y=l\theta +x=\text{Cx=}\left[\begin{array}{cccc}l& 0& 1& 0\end{array}\right]\left[\begin{array}{c}\theta \\ \stackrel{\dot{}}{\theta }\\ x\\ \stackrel{\dot{}}{x}\end{array}\right]$

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) = X_m(s)/U(s).

FIGURE P3.17 Motor-driven inverted pendulum can system¹⁵