Inverted pendulum
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## 1 Inverted Pendulum

The inverted pendulum is a standard controls example and it is related to rocket and missile guidance, where thrust is actuated at the bottom of a tall vehicle. Many undergraduate controls courses use the inverted pendulum as their first example plant. This classic problem in dynamics and control theory is widely used as a benchmark for testing control designs.

### 1.1 Equations of Motion[1]

Inverted Pendulum
Inverted Pendulum Free Body Diagram

Equations of motion usually start with Newton's 2nd Law and the inverted pendulum is no different. Sum of the torques must equal 0.

 $LaTeX: \sum\tau=\frac{d}{dt}\left[L\right]$

where
$LaTeX: \tau$ is torque and
$LaTeX: L$ is angular momentum.

The torques are the control torque $LaTeX: \tau$ and the gravity (positional) based torque $LaTeX: lmg\sin\left(\theta\right)$

 $LaTeX: \tau+lmg\sin\left(\theta\right)=\frac{d}{dt}\left[L\right]$

where
$LaTeX: l$ is the length of the pendulum,
$LaTeX: m$ is the mass at the end of the pendulum,
$LaTeX: g$ is gravity and
$LaTeX: \theta$ is the angle of the pendulum off vertical.

Angular torque is $LaTeX: L=I\ddot{\theta}$. The inertia of the rod is $LaTeX: I=ml^2$ so the angular torque is

 $LaTeX: \tau+lmg\sin\left(\theta\right)=ml^2\ddot{\theta}$

Rearranging the 2 sides

 $LaTeX: \ddot{\theta}=\frac{g}{l}\sin\left(\theta\right)+\frac{1}{ml^2}\tau$ IP 1

Eqn. IP 1 is nonlinear due to the sine function. However, if we can assume that $LaTeX: \theta$ is small (approx. 3 degrees or less) then the small angle approximation ($LaTeX: \sin\left(\theta\right) \approx \theta$) is valid and IP 1 becomes

 $LaTeX: \ddot{\theta}=\frac{g}{l}\theta+\frac{1}{ml^2}\tau$ IP 2

### 1.2 State-Space Formulation

The typical state-space formulation includes a description of the system via state variable

If we make the substitution $LaTeX: x=\begin{Bmatrix}\theta \\ \dot{\theta}\end{Bmatrix}$ and $LaTeX: u=\tau$ then

 $LaTeX: \dot{x}=\begin{bmatrix}0 & 1 \\ \frac{g}{l} & 0\end{bmatrix}x+\begin{bmatrix}0 \\ \frac{1}{ml^2}\end{bmatrix}u$

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