Figure P3.17 shows a free-body diagram of an inverted

Chapter 3, Problem 30

(choose chapter or problem)

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, l, 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, \(\dot{\theta}^{\prime}\), the horizontal position of the cart, x, and its speed, \(x^{\prime}\), were selected to be the state variables. The state-space equations derived were heavily nonlinear. \({ }^{14}\) They were then linearized around the stationary point, \(\mathbf{x}_{\mathbf{0}}=\mathbf{0}\) and \(u_0=0\), and manipulated to yield the following open-loop model written in perturbation form:

\(\frac{d}{d t} \delta \mathbf{x}=\mathbf{A} \delta \mathbf{x}+\mathbf{B} \delta u\)

However, since \(\mathbf{x}_{\mathbf{0}}=\mathbf{0}\) and \(u_0=0\), then let: \(\mathbf{x}=\mathbf{x}_0+\delta \mathbf{x}= \delta \mathbf{x}\) and \(u=u_0+\delta u=\delta u\). Thus the state equation may be rewritten as (Prasad, 2012):

\(\dot{\mathbf{x}}=\mathbf{A} \mathbf{x}+\mathbf{B} u\)

where

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

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:

\(y=l \theta+x=\mathbf{C x}=\left[\begin{array}{llll} l & 0 & 1 & 0 \end{array}\right]\left[\begin{array}{l} \theta \\ \dot{\theta} \\ x \\ \dot{x} \end{array}\right]\)

Given that: \(M=2.4 \mathrm{~kg}, m=0.23 \mathrm{~kg}\), \(I=0.36 \mathrm{~m}, g=9.81 \mathrm{~m} / \mathrm{s}^2\), use MATLAB to find the transfer

function, \(G(s)=Y(s) / U(s)=X_m(s) / U(s)\)