I am dealing with obtaining an space state representation of a nonlinear differential equation that arises from an inverted pendulum. It includes some terms that reflect the fact pendulum is controlled in a PID similarly fashion.
The original EOM would be like this:
$$ \ddot{\theta}(t)=a_1\sin{\theta(t)} -a_2\theta(t)-a_3\dot{\theta}(t)-a_4\int_0^\tau{\theta(\tau)d\tau} +b_1u(t) $$
So if I derive it once to get rid of the integral term, the equation takes the form:
$$ \dddot{\theta}(t)=a_1\cos{\theta(t)}\dot{\theta}(t) -a_2\dot{\theta}(t)-a_3\ddot{\theta}(t)-a_4\theta(t) +b_1\dot{u}(t) $$
And form here I'm kinda stuck and second guessing myself a little bit. If I want to express the equation in state space form, $\vec{x}(t)=\mathbf{f}\{\theta(t),\dot{\theta}(t),\ddot{\theta}(t),u(t)\}$ to then linearize it to obtain the classic form, $\dot{\vec{x}}(t) = A(\sigma)\vec{x}(t) + B(\sigma)\vec{u}(t)$...
Kinda my question would be how to deal with that input derivative.. Could a change of variable be the right thing to do? Like set $v(t)=\dot{u}(t)$ and go from there.. or consider a vector input $\vec{u}(t)=[u(t), \dot{u}(t)]$?
I mean, for instance if I would want to obtain the TF of that system for a particular operation point, I could get back from the change of variable knowing that $V(s)=sU(s)$, but not sure if I would get the same results with the extended input vector.
Here is one way:
Let $x_1 = \dot{\theta}$, $x_2 = \theta$ and $x_3(t) = x_3(0) + \int_0^t \theta(\tau) d \tau$.
Then we obtain $\dot{x} = \begin{bmatrix} a_1 \sin x_2 - a_2 x_2 - a_3 x_1 - a_4 x_3 +b_1 u \\ x_1 \\ x_2\end{bmatrix} $.
Note however, the behaviour of this system matches the original system iff $x_3(0) = 0$. From a transfer function perspective they are of course equivalent.