# Robust Stabilization of Linear Switched Systems with Unstable Subsystems
## 2 Preliminaries
With initial condition \(x(0)\in\mathbb{R}^{n}\), Lemma 1 yields:  
\[k\in\mathbb{K}_{p}^{s}=\{\,m\in\mathbb{I}_{n}|\,Re\left(\lambda_{m}\left(\sum_{ j=1}^{2^{n}}\theta_{j}(t)A_{[p,j]}\right)\right)\geq 0,p\in I,\theta\in\Theta,T\geq t \geq 0\}.\]6  
Therefore, using (4) and Lemma 2:  
\[k\in\mathbb{K}_{p}^{s}=\{\,m\in\mathbb{I}_{n}|\,Re\left(\lambda_{m}\left(\sum_{ j=1}^{2^{n}}\theta_{j}(t)A_{[p,j]}\right)\right)\geq 0,p\in I,\theta\in\Theta,T\geq t \geq 0\}.\]7 \[k\in\mathbb{K}_{p}^{s}=\{\,m\in\mathbb{I}_{n}|\,Re\left(\lambda_{m}\left(\sum_{ j=1}^{2^{n}}\theta_{j}(t)A_{[p,j]}\right)\right)\geq 0,p\in I,\theta\in\Theta,T\geq t \geq 0\}.\]8 \[k\in\mathbb{K}_{p}^{s}=\{\,m\in\mathbb{I}_{n}|\,Re\left(\lambda_{m}\left(\sum_{ j=1}^{2^{n}}\theta_{j}(t)A_{[p,j]}\right)\right)\geq 0,p\in I,\theta\in\Theta,T\geq t \geq 0\}.\]9 \[\lambda_{k}\left(\sum_{j=1}^{2^{n}}\theta_{j}(t)A_{[p,j]}\right),\]0 \[\lambda_{k}\left(\sum_{j=1}^{2^{n}}\theta_{j}(t)A_{[p,j]}\right),\]1 \[\lambda_{k}\left(\sum_{j=1}^{2^{n}}\theta_{j}(t)A_{[p,j]}\right),\]2  
where \(\Phi_{1},\Phi_{2}\) denote the sets of \(q\) satisfying \(\sigma(t_{q})\in I_{1},I_{2}\), respectively. Therefore, if:  
\[\lambda_{k}\left(\sum_{j=1}^{2^{n}}\theta_{j}(t)A_{[p,j]}\right),\]3  
\[\lambda_{k}\left(\sum_{j=1}^{2^{n}}\theta_{j}(t)A_{[p,j]}\right),\]4  
is used, then:  
\[\lambda_{k}\left(\sum_{j=1}^{2^{n}}\theta_{j}(t)A_{[p,j]}\right),\]5  
which means that the system is GUES under MDADT, satisfying (19). \(\square\)  
It is worth mentioning that this theorem can be used even if all of the subsystems are stable; in such a case, \(I_{1}=\emptyset\), \(I_{2}=I\), \(\Omega_{1}=\emptyset\), \(\Omega_{2}=\mathbb{R}^{n}\). Even more, the result can be used even if all of the subsystems are unstable; in such a case, \(I_{2}=\emptyset\), \(I_{1}=I\), \(\Omega_{2}=\emptyset\), \(\Omega_{1}=\mathbb{R}^{n}\). **4. Simulations**  
In this section is illustrated the main result of this paper through MATLAB simulations. For simplicity and demonstrative reasons, the following system is proposed:  
\[\lambda_{k}\left(\sum_{j=1}^{2^{n}}\theta_{j}(t)A_{[p,j]}\right),\]6with vertexes:  
\[A_{1,1}=\left[ \begin{array}{cc}0.1074&0.0154\\ -0.0154&0.2026\end{array} \right],\ A_{1,2}=\left[ \begin{array}{cc}0.0966&0.0206 \\ -0.0206&0.2234\end{array} \right],\ A_{1,3}=\left[ \begin{array}{cc}0.1069&0.0189 \\ -0.0189&0.2231\end{array} \right],\]  
\[A_{1,A}=\left[ \begin{array}{cc}0.0971&0.0171\\ -0.0171&0.2029\end{array} \right],\ A_{2,1}=\left[ \begin{array}{cc}-4.9701&-0. 1995\\ 0.1995&-6.3299\end{array} \right],\ A_{2,2}=\left[ \begin{array}{cc}-5.4816&-0. 1228\\ 0.1228&-6.3184\end{array} \right],\]  
\[A_{2,3}=\left[ \begin{array}{cc}-4.9556&-0.2962\\ 0.2962&-6.9744\end{array} \right],\ A_{2,4}=\left[ \begin{array}{cc}-5.4671&-0. 2194\\ 0.2194&-6.9629\end{array} \right],\]  
In Figure 1 is shown the dynamic behavior for the system (27), with nominal parameter values and the switching law plotted in Figure 2. Note that while the first state converges to zero, the second state is not GUES, and a switching law is designed based on the main result of this paper; the objective is to design a switching signal \(p=\sigma(t)\in F_{MDADT}[N_{[0,p]},\tau_{p}]\) such that (27) is GUES. From Lemma 2, \(\lambda_{1}^{M}=0.22\), \(\lambda_{1}^{m}=\emptyset\), \(\lambda_{2}^{M}=\emptyset\), \(\lambda_{2}^{m}=-5.00\),  
\[T_{[U,1]}=\left[ \begin{array}{cc}0.90&0.15\\ 0.15&0.90\end{array} \right], \tag{28}\]  
\[T_{[U,2]}=\left[ \begin{array}{cc}1&0.15\\ 0.15&1\end{array} \right], \tag{29}\]  
\(\epsilon_{p}=0.6931\ \forall p\). Set \(\mathcal{T}_{p}(T,t)=26s\), \(N_{[\sigma,1]}(T,t)=N_{[\sigma,2]}(T,t)=26\). For (4) and using \(\tau_{1},\tau_{2}\leq 1\), then \(N_{[0,1]}=N_{[0,2]}=1\). From (17)-(19):  
\[-5+\frac{0.6931}{\tau_{2}}<\beta_{2}<0 \tag{30}\]  
\[0<0.22+\frac{0.6931}{\tau_{1}}<\alpha_{1} \tag{31}\]  
\[26\alpha_{1}<-26\beta_{2}-1.3862 \tag{32}\]  
With \(\tau_{1}=\tau_{2}=0.5s\), \(\beta_{2}=3.5\) and \(\alpha_{1}=1\) are selected, and all the conditions of Theorem 1 are satisfied.