HW1, MAE 200A. Fall 2005. UCI

Nasser Abbasi

Problem 1

Given this simple pendulum, compute the equilibrium points and determine the linearized dynamics at each equilibrium point


sys.png

Answer

part(1)

The system equation is given by MATH

Since this is a second order ODE, there are 2 state variables. Convert this to state space formulation:

Let $x_{1}=\theta $ and $x_{2}=\dot{\theta}$, hence

MATH

At the equilibrium points we must have,MATH

Hence we obtain 2 equations

MATH

Solving these equations, we obtain MATH or $x_{1}=n\pi $ for $n=0,1,2,...$

Since the period is $2\pi $, then $x_{1}=0$ or $\pi $, but $x_{1}$ is state variable that represents the angle $\theta $ hence

equilibrium occurs at $\theta =0$ and $\theta =\pi $

note: equilibrium at $\theta =0$ is stable, while at $\theta =\pi $ is unstable.

part(2)

starting with the nonlinear system equation

MATH

Near the equilibrium points, we express the nonlinear term in taylor series.

Suppose the penulium is at angle $\theta $ near the angle $\theta _{eq}$ so it is a distance MATH $\theta _{eq}$

Hence now MATH

For the first equilibrium point, $\theta _{eq}=0$ so the above becomes

MATH

For the first equilibrium point, $\theta _{eq}=\pi $ so equation (1) becomes

MATH

Hence near $\theta _{eq}=0$, the linearized system equation is

MATH

and near $\theta _{eq}=\pi $ the system equation is

MATH