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

$$\begin{align*} y_{1}(k+1)&={{2.5y_{1}(k)y_{1}(k-1)}\over{1+y_{1}(k)^{2}+y_{2}(k-1)^{2}+y_{1}(k-2)^{2}}} \\ &+0.09u_{1}(k)u_{1}(k-1)+1.2u_{1}(k)+1.6u_{1}(k-2) \\ &+0.5u_{2}(k)+0.7\sin (0.5(y_{1}(k)+y_{1}(k-1))) \\ &\times\cos (0.5(y_{1}(k)+y_{1}(k-1))) \\ y_{2}(k+1)&=\displaystyle{{5y_{2}(k)y_{2}(k-1)}\over{1+y_{2}(k)^{2}+y_{1}(k-1)^{2}+y_{2}(k-2)^{2}}} \\ &+u_{2}(k)+1.1u_{2}(k-1)+1.4u_{2}(k-2) \\ &+0.5u_{1}(k). \end{align*}$$

The initial values are: $y_1(1)=y_1(3)=0$, $y_1(2)=1$, $y_2(1)=y_1(3)=0$, $y_2(2)=1$, $u(1)=u(2)=[0,0]^{\mathrm T}$

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 $X = [\bar{x}_{1,2}^{\mathrm T}, \bar{x}_{2,4}^{\mathrm T}]^{\mathrm T}$.