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State space linearization of an inverted pendulum

Inverted pendulum

An inverted pendulum on a cart is as shown in the figure.

inverted-pendulum.png

The system dynamics defined by Newton's second law of motion can be written as the following set of equations.

\begin{align} \ddot{p} &= \frac{\dot{\theta}^{2} m L \sin \theta-\frac{3}{4} m g \sin \theta \cos \theta+u}{M+m-\frac{3}{4} m \cos ^{2} \theta} \\ \ddot{\theta} &= \frac{\frac{3 g}{4 L}(M+m) \sin \theta-\frac{3 m}{4} \dot{\theta}^{2} \sin \theta \cos \theta-\frac{3}{4 L} u \cos \theta}{M+m-\frac{3}{4} m \cos ^{2} \theta} \end{align}

Given the dynamics, the state vector can be chosen to be the positions & velocities.

\(\mathbf{x}=\left[ \begin{array}{c}{\theta} \\ {\dot{\theta}} \\ {p} \\ {\dot{p}}\end{array}\right]=\left[ \begin{array}{c}{x_{1}} \\ {x_{2}} \\ {x_{3}} \\ {x_{4}}\end{array}\right]\)

The nonlinear state equations as functions of the states and inputs

\begin{aligned} \dot{\mathbf{x}} &=\mathbf{f}(\mathbf{x}, u) \\ y &=g(\mathbf{x}, u) \end{aligned}

can be written as,

\begin{align} \mathbf{\dot{x}} = \left[ \begin{array}{c}\dot{\theta} \\ {\ddot{\theta}} \\ \dot{p} \\ {\ddot{p}}\end{array}\right] &= \left[ \begin{array}{c}{x_{2}} \\ \frac{\frac{3 g}{4 L}(M+m) \sin x_{1}-\frac{3 m}{4} \sin x_{1} \cos x_{1} \cdot x_{2}^{2}-\frac{3}{4 L} \cos x_{1} \cdot u}{M+m-\frac{3}{4} m \cos ^{2} x_{1}} \\ x_{4} \\ \frac{m L \sin x_{1} \cdot x_{2}^{2}-\frac{3}{4} m g \sin x_{1} \cos x_{1}+u}{M+m-\frac{3}{4} m \cos ^{2} x_{1}} \end{array}\right] \\ y = p &= x_{3} \end{align}

The state equations above are description of a nonlinear system due to the sine and cosine values. The stationary state is selected as \(\mathbf{x}_{0}=\left[ \begin{array}{llll}{0} & {0} & {0} & {0}\end{array}\right]^{T}\) and these nonlinearities can be linearly approximated by assuming that \(\sin \theta \cong \theta, \cos \theta \cong 1\) and \(\dot{\theta}^{2} \cong 0\)

Substituting these values in the state equations gives,

\begin{align} \dot{x}_{2} &= \ddot{\theta} \cong \frac{\frac{3 g}{4 L}(M+m) \theta-\frac{3}{4 L} u}{M+\frac{m}{4}} \\ \dot{x}_{4} &= \ddot{p} \cong \frac{-\frac{3}{4} m g \theta+u}{M+\frac{m}{4}} \end{align}

At the stationary point calculating the Jacobian matrix, the partial derivatives of the state equations with respect to the states and inputs, will give the linearized state equation where \(\Delta\) variables are perturbation from the stationary point.

\begin{align} \dot{\Delta x} &= \left[ \begin{array}{cccc} 0 & 1 & 0 & 0 \\ \frac{3 g(M+m)}{\left(M+\frac{m}{4}\right) 4 L} & 0 & 0 & 0 \\ 0 & 0 & 0 & 1 \\ \frac{-\frac{3}{4} m g}{M+\frac{m}{4}} & 0 & 0 & 0 \end{array} \right] \Delta \mathbf{x}+ \left[ \begin{array}{c} 0 \\ \frac{-\frac{3}{4 L}}{M + \frac{m}{4}} \\ 0 \\ {\frac{1}{M+\frac{m}{4}}} \end{array}\right] \Delta \mathbf{u} \\ \Delta y &= \left[ \begin{array}{llll}{0} & {0} & {1} & {0}\end{array}\right] \Delta \mathbf{x} \end{align}

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