Nonlinear State Space Models

Take for example a pendulum following Newton’s 2nd Law:

$$ml\ddot{\theta }=-mg\sin \theta -D\theta +u$$

• $u$ is torque applied to pendulum
• $D$ is damping
• $l$ is the length of the rod

Choosing states:

$$\begin{array}{r}{x}_1:=\theta \\ x_2:=\dot{\theta }\end{array}$$

Then we have:

$$\begin{array}{rl}\dot{x}=\left[\begin{array}{c}{\dot{x}}_1\\ {\dot{x}}_2\end{array}\right]& =\left[\begin{array}{c}\dot{\theta }\\ -\frac{g}{l}\sin \theta -\frac{D}{ml}\dot{\theta }+\frac{1}{ml}u\end{array}\right]\\ & =\left[\begin{array}{c}{x}_2\\ -\frac{g}{l}\sin x_1-\frac{D}{ml}{x}_2+\frac{1}{ml}u\end{array}\right]\\ & :=f(x,u)\end{array}$$

and

$$\begin{array}{rl}y=\theta & =x_1\\ & =:g(x,u)\end{array}$$

Coming to our standard non-linear state space representation:

$$\boxed{\begin{array}{rl}\dot{x}& =f(x,u)\\ y& =g(x,u)\end{array}}$$

In general, this is not LTI!

For an LTI model, we can choose:

$$\begin{array}{rl}f(x,u)& =Ax+Bu\\ g(x,u)& =Cx+Du\end{array}$$

(Nonlinear) state space model

A (nonlinear) state space model is a model of the form

$$\begin{array}{r}\dot{x}=f(x,u)\\ y=g(x,y)\end{array}(\ast )$$

such that for any initial condition $x(t=0)=x_0$ and any input signal $u(t)$, there exists a unique solution to Equation $(\ast )$, and it is equal to the system’s output.

Equilibrium Point

Given a constant control signal $u(t)=\overline{u}\forall t\ge 0$ for some $\overline{u}\in \mathbb{R}$, an equilibrium point of a state space model is any state $x_e$ that satisfies $f(x_e,\overline{u})=0$.

If we start at $x=x_e$, we have:

$$\begin{array}{rl}& \dot{x}=f(x_e,\overline{u})=0⟹x(t)=x_e\forall t\ge 0\\ & \therefore x(t)\text{ is constant in time}\\ & y(t)=g(x(t),u(t))=g(x_e,\overline{u})\\ & ⟹y(t)\text{ is also constant }\forall t\ge 0\end{array}$$

LTI example:

$$\begin{array}{rl}\dot{x}=f(x_e,\overline{u})& =0\\ A{x}_e+B\overline{u}& =0\\ x_e& =-A^{-1}B\overline{u}\end{array}$$

• If $\overline{u}=0$, then $x_e=0$
• LTI systems only have 1 equilibrium point for fixed $u(t)=\overline{u}\forall t\ge 0$

Pendulum example continued:

• Fix $u(t)=\overline{u}\forall t\ge 0$ for some $\overline{u}\in \mathbb{R}$
• Find all equilibria of the system for $u(t)=\overline{u}\forall t\ge 0$

To do so, we solve $f(x_e,\overline{u})=0$ for $x_e$.

$$\begin{array}{rl}& ⟹f(x_e,\overline{u})=\left[\begin{array}{c}{x}_2^e\\ -\frac{g}{l}\sin x_1^e-\frac{D}{ml}{x}_2^e+\frac{1}{ml}\overline{u}\end{array}\right]=0\\ & ⟹x_2^e=0\\ & ⟹-\frac{g}{l}\sin x_1^e+\frac{1}{ml}\overline{u}=0[x_2^e=0]\\ & ⟹x_1^e={\sin }^{-1}\left(\frac{\overline{u}}{mg}\right)\end{array}$$