The system under study consists of two sets of single shell heat exchangers filled with water, placed in parallel and cooled by a liquid saturated refrigerant flowing through a coil system, as it is illustrated in the attached image. The saturated vapour generated in the coil system is separated from the liquid phase in the stages $S1$ and $S2$, both of neglected volumes. This vapour, withdrawn in $S1$ and $S2$, reduces the refrigerant mass flow rate along the cooling system, and only the saturated liquid portion is used for cooling purposes. Table below provides the fluid properties and equipment dimensions. The temperature of the refrigerant remains constant at $T_C$ as the liquid is saturated, and the energy exchanged with water is used to vapourise a small portion of the refrigerant fluid. The idividual heat exchanger energy balances can be expressed in terms of deviation variables to define the following LTI system:

$$\begin{align*} A &= \begin{bmatrix} -\dfrac{1+\nu_A}{\tau_1} & \dfrac{1}{\tau_1} & 0 & 0\\ 0 & -\dfrac{1+\nu_A}{\tau_2} & 0 & 0 \\ 0 & 0 & -\dfrac{1+\nu_B}{\tau_3} & \dfrac{1}{\tau_3} \\ 0 & 0 & 0 & -\dfrac{1+\nu_B}{\tau_4} \\ \end{bmatrix} \\ B &= \begin{bmatrix} 0 & 0 & 0 & 0\\ \dfrac{k_1}{\tau_1} & 0 & \dfrac{k_3}{\tau_2} & 0\\ 0 & 0 & 0 & 0\\ 0 & \dfrac{k_2}{\tau_t} & 0 & \dfrac{k_4}{\tau_4}\\ \end{bmatrix}, C = \begin{bmatrix} \mu & 0 & \mu & 0\\ 0 & \mu & 0 & \mu\\ \mu & 0 & \mu & 0\\ 0 & \mu & \mu & 0\\ \end{bmatrix}, D = \begin{bmatrix} 0 \end{bmatrix}, \end{align*}$$

where: $\nu_A \triangleq (hA/C_p(\dot{m}_1 + \dot{m}_3)), \\ \nu_B \triangleq (hA/C_p(\dot{m}_2 + \dot{m}_4)),\\ \tau_1 \triangleq (M_1/(\dot{m}_1 + \dot{m}_3)),\\ \tau_2 \triangleq (M_2/(\dot{m}_1 + \dot{m}_3)),\\ \tau_3 \triangleq (M_3/(\dot{m}_2 + \dot{m}_4)),\\ \tau_4 \triangleq (M_4/(\dot{m}_4 + \dot{m}_3)),\\ k_1 \triangleq(\dot{m}_1/(\dot{m}_1 + \dot{m}_3)),\\ k_2 \triangleq(\dot{m}_2/(\dot{m}_2 + \dot{m}_4)),\\ k_3 \triangleq(\dot{m}_3/(\dot{m}_1 + \dot{m}_3)),\\ k_4 \triangleq(\dot{m}_4/(\dot{m}_1 + \dot{m}_3)),\\ \mu \triangleq(hA/h_{lv}).\\$

This system was discretised with a sampling time of $T = 1 s$ and was discretised while rounding the input and output delays to the closest integer-multiples of $T$.

The dynamic equations governing the behavior of a single link robot with flexible joint are traditionally obtained from Lagrangian dynamics considerations. The simple robot system under study is shown in the attached image. Let $x_1=\theta_m$ be the motor angular position, the corresponding angular velocity $x_2 = d\theta/dt$, the elastic force $x_3 = k_s(\theta_t - \theta_m)$ and $x_4 = \{ d\theta_l/dt - d\theta_m/dt\}/\rho$, where $\rho^2=1/k_s$. Then the state variable representation is:

$$\begin{align*} \dot{x}_1(t) &= x_2(t)\\ \dot{x}_2(t) &= -a_5x_2(t)+a_1x_3(t)+a_1u(t)\\ \dot{x}_3(t) &= x_4(t)/\rho\\ \dot{x}_4(t) &= \{ -a_2a_3\sin{[\rho^2x_3(t)+x_1(t)]}-a_4x_3(t)-a_7x_2(t)-a_6\rho x_4(t) - a_1u(t)\}/\rho \end{align*}$$

with $a_1=1/J_m$, $a_2=1/J_l$, $a_3=mgl$,$a_4=a_1+a_2$,$a_5=B_m/J_m;a_6=B_l/J_l$,$a_7=a_6-a_5$ and $u(t)=\tau(t)$.

Attached image shows the principal structure of the three-tank system. The plant consists of three cylinders $T_1$, $T_2$, and $T_3$ with the cross section $S_A$. These tanks are connected serially with each other by pipes with the cross section $S_n$. A single outflow valve with the cross section $S_n$ is located at tank 2. The outflowing liquid (usually distilled water) is collected in a reservoir, which supplies the pumps 1 and 2. $H_{max}$ denotes the highest possible liquid level. The control input signals are the pump liquid flow rates $Q_1$ and $Q_2$ , the output signals are the liquid levels $h_1$ and $h_2$ .

Define the following variables and the parameters: $az_i$ : outflow coefficients of tank $i$ ; $h_1$ , $h_2$ , $h_3$ : liquid levels (m); $Q_{13}$ : flow rate from tank 1 to tank 3 $(m^3/sec)$ ; $Q_{32}$ : flow rate from tank 3 to tank 2 $(m^3/sec)$ ; $Q_{20}$ : flow rate from tank 2 to reservoir $(m^3/sec)$ ; $Q_1$ , $Q_2$ : supplying flow rates $(m^3/sec)$ ; $S_A$ : section of cylinder $(m^2)$ ; $S_1$ : section of leak opening $(m^2)$ ; $S_n$ : section of connection pipe $(m^2)$. Then, the dynamics of the three-tank system is expressed by a set of differential equations

$$\eqalignno{S_{A}\displaystyle{{dh_{1}}\over{dt}}=&\,Q_{1}(t)-Q_{13}(t)\cr S_{A}\displaystyle{{dh_{3}}\over{dt}}=&\, Q_{13}(t)-Q_{32}(t)\cr S_{A}\displaystyle{{dh_{2}}\over{dt}}=&\, Q_{2}(t)+Q_{32}(t)-Q_{20}(t)\cr Q_{13}(t)=&\,\alpha z_{1}S_{n}sgn(h_{1}(t)-h_{3}(t))\sqrt{(2g\left\vert h_{1}(t)-h_{3}(t)\right\vert}\cr Q_{32}(t)=&\,\alpha z_{3}S_{n}sgn(h_{3}(t)-h_{2}(t))\sqrt{2g\left\vert h_{3}(t)-h_{2}(t)\right\vert}\cr Q_{20}(t)=&\,\alpha z_{2}S_{n}sgn(h_{2}(t))\sqrt{2g\left\vert h_{2}(t)\right\vert},}$$

where $\alpha_1$, $\alpha_2$, $\alpha_3$: outflow coefficients (dimensionless, real values ranging from 0 to 1), $g$: earth acceleration $(m/s^2)$, $sgn(z)$: sign of the argument $z$.

In the simulation, the discretized model of three-tank system is obtained by first-order Euler's method, the sampling period is 1 s and the simulation time is 1500 s. The parameters of three-tank system are given in the table below. The initial conditions are $h_1(1)=0$, $h_2(1)=0$, $h_3(1)=0$, $Q_1(1)=0$, and $Q_2(1)=0$.

Consider the nonlinear system

$$\begin{align*} x_{11}(k+1) &=\frac{x_{11}^2(k)}{1+x_{11}^2(k)}+0.3x_{12}(k), \\ x_{12}(k+1) &=\frac{x_{11}^2(k)}{1+x_{12}^2(k)+x_{21}^2(k)+x_{22}^2(k)}+a(k)u_{1}(k), \\ x_{21}(k+1) &=\frac{x_{21}^2(k)}{1+x_{21}^2(k)}+0.2x_{22}(k), \\ x_{22}(k+1) &=\frac{x_{21}^2(k)}{1+x_{11}^2(k)+x_{12}^2(k)+x_{22}^2(k)}+b(k)u_{2}(k), \\ y_1(k+1) &= x_{11}(k+1)+0.005 \mathrm{rand}(1), \\ y_2(k+1) &=x_{21}(k+1)+0.005 \mathrm{rand}(1), \end{align*}$$

where $a(k)=1+0.1\sin{(2\pi k/1500)}$, $b(k)=1+0.1\cos{(2\pi k/1500)}$are two time-varying parameters. This example is controlled by using neural network without time-varying parameters $a(k)$, $b(k)$, and the noise.

The initial values are: $x_{1,1}(1)=x_{1,1}(2)=x_{2,1}(1)=x_{2,1}(2)=0.5$, $x_{1,2}(1)=x_{1,2}(2)=x_{2,2}(1)=x_{2,2}(2)=0$, $u(1)=u(2)=[0,0]^{\mathrm T}.$

$$\Sigma _{S_{2}}: \cases{\begin{align*} \dot{x}_{1,1} &=f_{1,1}(\bar {x}_ {1,1},\bar {x}_{2,3})+g_{1,1}(\bar {x}_{1,1},\bar{x}_{2,3})x_{1,2} \\ \dot{x}_{1,2} &=f_{1,2}(X)+g_{1,2}(\bar {x}_{1,1},\bar{x}_ {2,3})u_{1} \\ \dot{x}_{2,1} &=f_{2,1}(\bar {x}_{2,1})+g_{2,1} (\bar {x}_{2,1})x_{2,2} \\ \dot{x}_{2,2} &=f_{2,2}(\bar {x}_{2,2})+g_{2,2}(\bar {x}_{2,2})x_{2,3} \\ \dot{x}_{2,3} &=f_{2,3} (\bar {x}_{1,1},\bar {x}_{2,3})+g_{2,2}(\bar {x}_{1,1},\bar{x}_{2,3})x_{2,4} \\ \dot{x}_{2,4} &=f_{2,4}(X, u_{1})+g_{2,4}(\bar {x}_ {1,1},\bar{x}_{2,3})u_{2} \\ y_{j} &=x_{j,1}, \quad j=1,2, \end{align*}}$$

where $\bar{x}_{j,i_j}=[x_{j,1},\dots,x_{j,i_j}]^{\mathrm T},j=1,2, i_1=1,2, i_2=1,\dots,4$, and $X = [\bar{x}_{1,2}^{\mathrm T}, \bar{x}_{2,4}^{\mathrm T}]^{\mathrm T}$.