# Finite-Time Control of Singular Linear Semi-Markov Jump Systems
## 4 Numerical Examples  
**Example 1**. : _Consider the SSMJS in (1) with \(\omega(t)=0\), which operates in two modes. Some detailed system parameters are given as follows._  
\[E= \begin{bmatrix}1&0\\ 0&0\end{bmatrix} ,A_{1}= \begin{bmatrix}0.2&-0.5\\ 0.3&0.7\end{bmatrix} ,A_{2}= \begin{bmatrix}-0.5&-1.2\\ 1&2\end{bmatrix} ,B_{1}= \begin{bmatrix}1\\ 1\end{bmatrix} ,B_{2}= \begin{bmatrix}1\\ -1\end{bmatrix} .\]  
_To study the finite-time stabilization problem, the following parameters are given:_  
\[R= \begin{bmatrix}1&0\\ 0&1\end{bmatrix} ,H= \begin{bmatrix}0\\ 1\end{bmatrix} ,c_{1}=1,c_{2}=2,T=4.\]Meanwhile, the transition rate matrix is bounded with_  
\[ \begin{bmatrix}\underline{\mu}_{ij}\end{bmatrix} = \begin{bmatrix}-0.1&0.1\\ 0.3&-0.3\end{bmatrix} ,\quad \begin{bmatrix}\overline{\mu}_{ij}\end{bmatrix} = \begin{bmatrix}-0.9&0.9\\ 0.7&-0.7\end{bmatrix} .\]  
_It can be shown that the matrix inequalities in Theorem 2 are feasible, and a solution can be given as_  
\[ \begin{array}{c}\alpha=0.1,\theta=0.7818,Z_{1}=Z_{2}=0.8909,\\ Y_{1}=\begin{bmatrix}-1.1335&-1.8287\end{bmatrix},Y_{2}=\begin{bmatrix}-2.6346&2. 9817\end{bmatrix},\\ P_{1}=\begin{bmatrix}0.8994&0.8625\\ 0.8625&1.0564\end{bmatrix},P_{2}=\begin{bmatrix}0.8748&-0.8619\\ -0.8619&0.9297\end{bmatrix},S_{1}=\begin{bmatrix}0.3159\\ 0.8888\end{bmatrix},S_{2}=\begin{bmatrix}-0.5487\\ 0.8030\end{bmatrix}.\end{array} \]  
_Therefore, the desired controller gain can be obtained as_  
\[K_{1}= \begin{bmatrix}1.4353&-2.0574\end{bmatrix} ,K_{2}= \begin{bmatrix}2.9757&3. 7133\end{bmatrix} ,\]  
_and the corresponding closed-loop system matrices are_  
\[\mathcal{A}_{1}= \begin{bmatrix}1.6353&-2.5574\\ 1.7353&-1.3574\end{bmatrix} ,\mathcal{A}_{2}= \begin{bmatrix}2.4757&2.5133\\ -1.9757&-1.7133\end{bmatrix} .\]  
_With this control law, when the semi-Markov jump mode is as Figure 1a, the state trajectories of closed-loop SSMJS are as in Figure 1b, and the evolution of \(x^{T}(t)E^{T}REx(t)\) is presented in Figure 1c._  
**Example 2**. : _The second example is concerned with the SSMJS in (1) with the following parameters:_  
\[E= \begin{bmatrix}1&0\\ 0&0\end{bmatrix} ,A_{1}= \begin{bmatrix}0.9&-0.4\\ 0.5&0.8\end{bmatrix} ,A_{2}= \begin{bmatrix}-0.2&-0.4\\ 0.3&-0.5\end{bmatrix} ,\] \[B_{1}= \begin{bmatrix}1\\ 1\end{bmatrix} ,B_{2}= \begin{bmatrix}1\\ -1\end{bmatrix} ,C_{1}= \begin{bmatrix}0.1\\ 0\end{bmatrix} ,C_{2}= \begin{bmatrix}0.1\\ 0.1\end{bmatrix} ,\]  
_with \(\omega(t)\) shown in Figure 2a._  
Figure 1: The closed-loop SSMJS with \(\omega(t)=0\). _Similarly, the following parameters are also given:_  
\[H = \begin{bmatrix}0\\ 1\end{bmatrix} ,R= \begin{bmatrix}1&0\\ 0&1\end{bmatrix} ,c_{1}=1,c_{2}=1.5,T=4,d=2,\] \[\left[\underline{\mu}_{ij}\right]= \begin{bmatrix}-0.4&0.4\\ 0.2&-0.2\end{bmatrix} ,\left[\overline{\mu}_{ij}\right]= \begin{bmatrix}-1.2&1. 2\\ 1.3&-1.3\end{bmatrix} .\]  
_From Theorem 3, it follows that there is a state feedback controller for the closed-loop SSMJS to be stochastically finite-time bounded. A feasible solution to the matrix inequalities is_  
\[\theta=0.5418,\eta=0.8998,\alpha=0.3,Z_{1}=Z_{2}=0.7709,\] \[Y_{1}= \begin{bmatrix}-0.3833&-0.4818\end{bmatrix} ,Y_{2}= \begin{bmatrix} -0.0196&0.0027\end{bmatrix} ,W_{1}=0.6198,W_{2}=0.6259,\] \[P_{1}= \begin{bmatrix}0.6391&0.4055\\ 0.4055&0.3644\end{bmatrix} ,P_{2}= \begin{bmatrix}0.6571&0.0038\\ 0.0038&0.5808\end{bmatrix} ,S_{1}= \begin{bmatrix}0.2095\\ 0.3636\end{bmatrix} ,S_{2}= \begin{bmatrix}0.0077\\ 0.5778\end{bmatrix} .\]  
_With this solution, we can get the desired controller gain as_  
\[K_{1}= \begin{bmatrix}0.6751&-1.3250\end{bmatrix} ,K_{2}= \begin{bmatrix}-0.0299&0.0047\end{bmatrix} ,\]  
_and the corresponding closed-loop system matrices are_  
\[\mathcal{A}_{1}= \begin{bmatrix}1.5751&-1.7250\\ 1.1751&-0.5250\end{bmatrix} ,\mathcal{A}_{2}= \begin{bmatrix}-0.2299&-0.3953\\ 0.3299&-0.5047\end{bmatrix}