# Inverted pendulum with cart: unstable nonlinear system

• Oct 28th 2008, 12:35 AM
slodki
Inverted pendulum with cart: unstable nonlinear system
Hey guys,

Im really stuck when it comes to trying to solve this problem. It'll take some time to write up, so i'll upload a pdf file. I was given the following tip:

Regarding Question 2 of the assignment, you may use the following theorem to determine the stability of the nonlinear system:

Stability Theorem: Let (x_0,y_0) be a critical point of a nonlinear system

dot(x) = f(x)

and let A be the Jacobi matrix of the system, evaluated at (x_0,y_0). Then the critical point is

(i) asymptotically stable if Re(lambda_i) < 0 for all eigenvalues lambda_i of A

(ii) unstable if Re(lambda_i) > 0 for one or more eigenvalues lambda_i of A

(from Nonlinear Systems, by H. Khalil, Prentice Hall, Third edition. )

If anyone can tell me what to do i would greatly appreciate it!
• Oct 28th 2008, 05:40 AM
shawsend
You have the system:

$\displaystyle \dot{x_1}=x_2=F_1(x_1,x_2,x_3,x_4)$

$\displaystyle \dot{x_2}=\frac{u(\mathbf{x})}{v(\mathbf{x})}=F_2( x_1,x_2,x_3,x_4)$

$\displaystyle \dot{x_3}=x_4=F_3(x_1,x_2,x_3,x_4)$

$\displaystyle \dot{x_4}=\frac{w(\mathbf{x})}{lv(\mathbf{x})}=F_4 (x_1,x_2,x_3,x_4)$

(1) The critical points are called equilibrium points or fixed points. That's where all the derivatives are zero. You can plug in $\displaystyle (0,0,0,0)$ into each $\displaystyle F$ and see that this makes them all zero.

(2) You need to just start splitting out partials for the Jacobian. You know what the Jacobian is right? It's $\displaystyle \frac{\partial(F_1,F_2,F_3,F_4)}{\partial(x_1,x_2, x_3,x_4)}$

Now $\displaystyle F_1$ is just $\displaystyle x_2$ so that $\displaystyle \frac{\partial(F_1)}{\partial(x_1,x_2,x_3,x_4)}=(0 ,1,0,0)$ right? How about $\displaystyle \frac{\partial(F_2)}{\partial(x_1,x_2,x_3,x_4)}$? Can you calculate the partials with respect to each x of $\displaystyle F_2$?. Do the same with the other two and you'll get the full Jacobian matrix. Once you get it, you need to evaluate it at the fixed point $\displaystyle (0,0,0,0)$, that is $\displaystyle \textbf{J}(0,0,0,0)$. You can do that, and you get the linearized matrix $\displaystyle \textbf{A}$. Now calculate all the eigenvalues of that matrix and then determine if the critical point at $\displaystyle (0,0,0,0)$ is stable or not depending on the signs of the eigenvalues.

Oh yea, all the secrets of the Universe can be found in differential equations. :)