• For cart-pole

$$ \dot{x} = Ax+Bu $$

So, then

$$ x = \begin{bmatrix} x \\ v \\ \theta \\ \omega \end{bmatrix} $$

and

$$ \dot{x} = \begin{bmatrix} v \\ a \\ \omega \\ \alpha \end{bmatrix} $$

we know

$$ u=F $$

now for A and B which are linearized about vertical position fixed point.

$$ x^{\star} = \begin{bmatrix} 0 \\ 0 \\ 0 \\ 0 \end{bmatrix} $$

new state

$$ x=x-x^\star $$

now for A and B we get the following equations:

$$ A = \begin{bmatrix} 0 & 1 & 0 & 0 \\ 0 & 0 & -mg/M & 0 \\ 0 & 0 & 0 & 1\\ 0 & 0 &(M+m)g/(lM)& 0 \end{bmatrix} , B = \begin{bmatrix} 0 \\ 1/M\\ 0 \\ -1/Ml \end{bmatrix} $$

now numerical solution for

$$ x_{t+1}=? $$

$$ (x_{t}-x_{t-1})/\delta t= \dot{x} $$

for cart velocity

$$ v_{t+1}=v_{t}+(-mg/M\theta_{t}+F/M)*\delta t $$