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

where $X = [\bar{x}_{1,2}^{\mathrm T}, \bar{x}_{2,2}^{\mathrm T}]^{\mathrm T}$ with $\bar{x}_{j,2}=[x_{j,1},x_{j,2}]^{\mathrm T},j=1,2$.

Consider the VTOL example. The following shortcuts ${\rm c}(\cdot)=\cos(\cdot),\quad {\rm s}(\cdot)=\sin(\cdot)$ are used. The $\Delta_0$ is given as

$$\begin{align*} \omega_{0}^{1} &: {\mathrm d}x-v_{x}{\mathrm d}t \\ \omega_{0}^{2} &: {\mathrm d}v_{x}-(u^{2}\epsilon {\mathrm c}(\theta)-u^{1}{\mathrm s}(\theta)){\mathrm d}t \\ \omega_{0}^{3} &: {\mathrm d}z-v_{z}{\mathrm d}t \\ \omega_{0}^{4} &: {\mathrm d}v_{z}-(u^{1}{\mathrm c}(\theta)+u^{2}\epsilon {\mathrm s}(\theta)-1){\mathrm d}t \\ \omega_{0}^{5} &: {\mathrm d}\theta-\omega {\mathrm d}t \\ \omega_{0}^{6} &: {\mathrm d}\omega-u^{2}{\mathrm d}t, \end{align*}$$

where $\epsilon$ is a constant parameter. It can be shown that $\Delta_{0,\mathrm{d}t}^{\perp} = \mathrm{span}\{\delta_{u^1},\delta_{u^2}\}$. Construct $\Delta_1 \in \Delta_0$ such that $v_0(\Delta_1)\in \Delta_1$ holds with $v_0=\Delta_{0,{\mathrm d}t}^{\perp}$, i.e. $\Delta_1$ is given as:

$\begin{align*} \omega_{1}^{1} &: {\mathrm d}x-v_{x}{\mathrm d}t \\ \omega_{1}^{2} &: {\mathrm c}(\theta){\mathrm d}v_{x}+{\mathrm s}(\theta){\mathrm d}v_{z}-\epsilon {\mathrm d}\omega+{\mathrm s}(\theta){\mathrm d}t \\ \omega_{1}^{3} &: {\mathrm d}z-v_{z}{\mathrm d}t \\ \omega_{1}^{4} &: {\mathrm d}\theta-\omega {\mathrm d}t \end{align*}$

Consider a two-link rigid robot manipulator moving a horizontal plane. The dynamic equations of this MIMO system are

$$\left[\matrix{ \ddot{q}_{1}\cr \ddot{q}_{2} }\right]=\left[\matrix{ M_{11} & M_{12}\cr M_{21} & M_{22} }\right]^{-1} \left\{\left[\matrix{ u_{1}\cr u_{2} }\right]-\left[\matrix{ -h\dot{q}_{2} & -h(\dot{q}_{1}+\dot{q}_{2})\cr h\dot{q}_{1} & 0 }\right]\left[\matrix{ \dot{q}_{1}\cr \dot{q}_{2} }\right]\right\},$$

where

$\begin{align*} M_{11}&=a_{1}+2a_{3}\cos(q_{2})+2a_{4} \sin (q_{2}),\ M_{22}=a_{2} \\ M_{12}&=M_{21}=a_{2}+\alpha_{3}\cos(q_{2})+a_{4}\sin(q_{2}) \\ h&=a_{3}\sin(q_{2})-a_{4}\cos(q_{2}) \end{align*}$

with

$\begin{align*} a_{1}&=I_{1}+m_{1}l_{c1}^{2}+I_{e}+m_{e}l_{ce}^{2}+m_{e}l_{1}^{2} \\ a_{2}&=I_{e}+m_{e}l_{ce}^{2} \\ a_{3}&=m_{e}l_{1}l_{ce}\cos(\delta_{e}) \\ a_{4}&=m_{e}l_{1}l_{ce}\sin(\delta_{e}). \end{align*}$

Consider a nonlinear system

$$\begin{align*} x_{1}(t+1) &=x_{1}(t)-x_{1}(t)x_{2}(t)+(5+x_{1}(t))u(t) \\ x_{2}(t+1) &=-x_{1}(t)-0.5x_{2}(t)+2x_{1}(t)u(t) \end{align*}$$

Consider a T-S fuzzy model

Plant Rule $i$: If $x_1(t)$ is $F_1^1(x_1(t))$

Then $x(t+1) = A_ix(t)+B_iu(t),$

where

$\begin{align*} A_1 &=\left[\matrix{-a & 2\cr -0.1 & b}\right], A_2=\left[\matrix{-a & 2\cr-0.1 & b }\right], A_3=\left[\matrix{-0.9 & 0.5\cr -0.1 & -1.7}\right] \\ B_1 &=\left[\matrix{b\cr 4}\right], B_2=\left[\matrix{b\cr 4.8}\right], B_3=\left[\matrix{3\cr 0.1}\right]. \end{align*}$

The parameters $a$ and $b$ are adjusted to compare the relaxation of stabilization conditions.