Nonlinear State Space Models

Summary

Motivation: Sampled Data Digital Control Systems are not Linear Time Invariant, so we need to move beyond LTI models to be able to analyze sampled data systems.

See Drawing 2025-10-24 11.54.34.excalidraw for examples for this concept.

Rendered Excalidraw

For an LTI model $f(x,u)=Ax+Bu$ and $g(x,u)=Cx+Du$…

Definition

A non-linear state space model of a system is a model of the form…

• $x^{\prime }=f(x,u)$ (*)
• $y=g(x,u)$

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

Note that this is time-invariant since neither functions depend on t

This is a general framework for modelling the dynamics of a system which gives us a well defined solution.

For regular linear State Space Models, we have 2 options as we approach infinity. We either approach 0 or infinity. In non-linear systems, there are other states that we can approach rather than just this binary choice.

Definition (Equilibrium Point)

What does steady state look like in non-linear systems?

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

$x^{\prime }=f(x_e,\bar{u})=0$ which implies that $x(t)=x_e\text{, x(t) is a constant}\forall t\ge 0$

For an LTI system…
$x=f(x_e,\bar{u})=A{x}_e+\bar{B}u=0⟹x_e=-A^{-1}B\bar{u}$
If $\bar{u}=0⟹x_e=0$
LTI systems only have 1 equilibrium point for a fixed $u(t)$

Example

We can apply Linearization about $x^e,\bar{u}$. as follows…

• $\bar{x}=f(x,u)$
• $y=g(x,u)$

$f(x,u)=f(x^e,\bar{u})+\frac{\partial f}{\partial x}{|}_{x^e,u}(x-x^e)+\frac{\partial f}{\partial u}{|}_{x^e,u}(u-\bar{u})$

$g(x,u)=g(x^e,\bar{u})+\frac{\partial g}{\partial x}{|}_{x^e,u}(x-x^e)+\frac{\partial g}{\partial u}{|}_{x^e,u}(u-\bar{u})$

Where $f$ is vector valued…
$f(x,u)=\left[\begin{array}{c}{f}_1(x,u)\\ \dots \\ f_n(x,u)\end{array}\right]$

$x=\left[\begin{array}{c}\frac{\partial f_1}{\partial u}\\ \dots \\ \frac{\partial f_n}{\partial u}\end{array}\right]$

$\frac{\partial f_1}{\partial x_1}:=\left[\begin{array}{ccccc}\frac{\partial f_1}{\partial x_1}& & \dots & & \frac{\partial f}{\partial x_n}\\ \dots & & \dots & & \dots \\ \frac{\partial f_n}{\partial x_1}& & \dots & & \frac{\partial f_n}{\partial x_n}\end{array}\right]$

$\frac{\partial f}{\partial u}:=\left[\begin{array}{c}\frac{\partial f_1}{u}\\ \dots \\ \frac{\partial f_n}{u}\end{array}\right]$

$\frac{\partial g}{x}:=[\frac{\partial g}{\partial x}\dots \frac{\partial g}{\partial u}]$

$\frac{\partial g}{\partial u}=\frac{\partial g}{\partial u}$

Definition

An equilibrium point $x^e$ is stable if all of the eigenvalues of the linearization at $x^e\left(A=\frac{\partial f}{\partial x}\right){|}_{x^e,\bar{u}}$ are stable in $C^{-}1$

An equilibrium point which is not stable is called unstable.

Note

Note that $x^{e,u}$ is an unstable equilibrium point because it has an unstable eigenvalue ${\lambda }^u$.

Consider the linearization about $x^{e,s}=\left[\begin{array}{c}0\\ 0\end{array}\right]$.

$\Delta \dot{x}=\left[\begin{array}{ccc}0& & 1\\ -\frac{g}{l}& & -\frac{0}{ml}\end{array}\right]\Delta x+\left[\begin{array}{c}0\\ \frac{1}{ml}\\ B\end{array}\right]\Delta u$

$\Delta y=[10]\Delta x$

Assume $\Delta u=0$ $\forall t{\ge }_0$
$\Delta \dot{x}=A\Delta x$ and it can be shown that A has 2 stable eigenvalues

${\lambda }^s=\mathrm{Re}({\lambda }^s)+j\mathrm{Im}({\lambda }^s)$
$v^s$ eigenvector
$e^e,s$ is a stable equilibrium point

Linear systems are a pretty good approximation of non-linear systems around an equilibrium point.

Definition

The region of attraction of $x^{e,s}$ is the set of initial conditions $x^0$ such that $x(t=0)=x^0$ then ${\lim }_{t\to \infty }x(t)=x^{e,s}$. Where your stable equilibrium point is your desired operating point