Consider nonlinear van der Pol oscillators coupled via a bath. The normal form of the system is expressed by

$$\begin{align*} \dot{\xi}_{1}^{1}&=\xi_{2}^{1} \\ \dot{\xi}_{2}^{1}&=-\xi_{1}^{1}+\epsilon\{1-(\xi_{1}^{1})^{2}\}\xi_{2}^{1}+k(\eta_{1}-\xi_{1}^{1})+u_{1} \\ \dot{\xi}_{1}^{2}&=\xi_{2}^{2} \\ \dot{\xi}_{2}^{2}&=-\xi_{1}^{2}+\epsilon\{1-(\xi_{1}^{2})^{2}\}\xi_{2}^{2}+k(\eta_{1}-\xi_{1}^{2})+u_{2} \\ \dot{\eta}_{1}&=k(\xi_{1}^{1}-\eta_{1})+k(\xi_{1}^{2}-\eta_{1}) \\ y_{1}&=\xi_{1}^{1} \\ y_{2}&=\xi_{1}^{2}, \end{align*}$$

where parameters $\epsilon$ and $k$ are positive constants. The system has the relative degrees $r_1 = 2, r_2 = 2$ and the following zero dynamics:

$ \dot{\eta}_{1}=-2k\eta_{1}.$

The inverted pendulum on cart to be controlled is shown in the attached image. Its structure consists of a cart and pendulum where the pendulum is hinged to the cart via a pivot and only the cart is actuated, where:

Lagrange's equations are applied with respect to $\theta$ and $x$ coordinates. The nonlinear state space model of inverted pendulum on cart system can then be obtained as:

$$\begin{align*} \dot{x}_1 &= x_2 \\ \dot{x}_2 &= \dfrac{mg \sin{x_1}+ \dfrac{ m\cos{x_1}(u-f\dot{x})-m^2lx_2^2\cos{x_1}\sin{x_1}}{(M + m)}}{\dfrac{4}{3}ml - \dfrac{m^2l\cos^2{x_1}}{(M+m)}} \\ \dot{x}_3 &= x_4 \\ \dot{x}_4 &= \dfrac{u - f x_1 + \dfrac{3}{4}mg\cos{x_1}\sin{x_1} - mlx_2^2\sin{x_1}}{(M+m) - \dfrac{3}{4}m\cos^2{x_1}}, \end{align*}$$

where the state variables are consequently assigned as $x_1 = \theta$, $x_2 = \dot{\theta}$, $x_3 = x$ an $x_4=\dot{x}$, and where the input $u$ is the applied force $F$.

Simplified Schmid pendulum:

$$\begin{align*} \ddot{\psi} + a_{21}\omega - a_{11}\sin{\psi} &= -b_1u, \\ \dot{\omega} + a_{22}\omega + a_{12}\sin{\psi} &= b_2u, \end{align*}$$

where $\psi$ is the pendulum angle; $\omega$ is the wheel angular rate; $u$ is the controlling voltage, applied to the motor; $a_{11},a_{21},a_{12},b_1,b_2$ are positive constants, depending on the design parameter of the pendulum. It is assumed that the upper (unstable) equilibrium point corresponds to $\psi=0$.

Self-excited nonlinear oscillator:

$$\begin{align*} \dot{x}_1 &=x_2,\\ \dot{x}_2 &=-\omega_1^2\sin{y_1}-\varrho x_2 + k_p \arctan(k_c y_2),\\ \dot{x}_3 &=\omega_2 \cdot (x_1 - x_3), \\ y_1 &=x_1, \\ y_2 &=x_1-x_3, \end{align*}$$

where $y(t)\in \mathbb{R}^2$ is the sensor output vector (to be transmitted over the communication channel), $\omega_1, \omega_2, \varrho, k_p, k_c$ are system parameters, $x=[x_1,x_2,x_3]^{\mathrm T}\in\mathbb{R}^3$ of the system state vector $x(t)$ based on the signals $y_1(t),y_2(t)$, transmitted over the communication channel.

The system has the form

$\dot{x}(t)=Ax(t)+\varphi(y(t)),$ $y(t)=Cx(t),$

where

$A=\begin{bmatrix}0 & 1 & 0 \\ 0 & -\varrho & 0 \\ \omega_2 & 0 & -\omega_2\end{bmatrix},$

$C = \begin{bmatrix}1, & 0, & 0\\ 1, & 0, & -1\end{bmatrix},$

$ \varphi(y) = \begin{bmatrix} 0\\ \omega_1^2 \sin{y_1} + k_p \arctan{(k_c y_2)} \\ 0 \end{bmatrix}.$

$$\begin{align*} x_1(t+1) &=\left(\dfrac{x_1(t)}{1+x_1^2(t)}+1\right)\sin{x_2(t)} \\ x_2(t+1) &=x_2(t)\cos{x_2(t)}+x_1(t)e^{-((x_1^2(t)+x_2^2(t))/8} + \dfrac{u^3(t)}{1+u^2(t)+0.5\cos{x_1(t)+x_2(t)}} \\ y(t) &=\dfrac{x_1(t)}{1+0.5\sin{x_2(t)}}+\dfrac{x_2(t)}{1+0.5\sin{x_1(t)}}+e(t), \end{align*}$$

where $e(t)$ is the noise term, has a variance of 0.1.