## 4 Simulation Results
### Stabilization  
The elementary equations describing the behavior of the boost converter depicted in Figure 1 are given by  
\[\left\{ \begin{array}{l}L\frac{di_{L}(t)}{dt}=V_{in}-(1-u_{1}(t))y(t)-r_{L}i_{L}( t)\\ C\frac{d\
u_{C}(t)}{dt}=(1-u_{1}(t))i_{L}(t)-\frac{y(t)}{R}=\frac{y(t)-v_{C}(t)}{r_{C}} \end{array} \right. \tag{75}\]  
Taking the state vector, \(x(t)=\left[ \begin{array}{cc}i_{L}(t)&v_{C}(t)\end{array} \right]\), we have  
\[\left\{ \begin{array}{l}\frac{dx(t)}{dt}=\left[\underbrace{\left[ \begin{array}{cc}-\frac{r_{L}+ar_{C}}{L}&-\frac{\alpha}{L}\\ \frac{\alpha}{\alpha}&-\frac{\alpha}{RC}\end{array}\right]}_{A_{0}}+ \underbrace{\left[\begin{array}{cc}\frac{ar_{C}}{L}&\frac{\alpha}{L}\\ -\frac{\alpha}{\alpha}&0\end{array}\right]}_{A_{u_{1}}}u_{1}(t)\right]x(t)+ \underbrace{\left[\begin{array}{c}\frac{1}{L}\\ 0\end{array}\right]}_{B_{0}}V_{in}\\ y(t)=\left[\underbrace{\left[\begin{array}{cc}\alpha r_{C}&\alpha \end{array}\right]}_{C_{0}}+\underbrace{\left[\begin{array}{cc}-\alpha r_{C }&0\end{array}\right]}_{C_{u_{1}}}u_{1}(t)\right]x(t)\end{array} \right. \tag{76}\]  
with \(\alpha=R/(R+r_{C})\). We can verify that \(A_{0}\) is Hurwitz and determine the different matrices \(A_{i},\ B_{i}\) and \(C_{i}\) of the model (12). \[A_{1}=0\times A_{u_{1}}=0\hskip 28.452756 ptA_{2}=1\times A_{u_{1}}=\left[ \begin{array}{cc}\frac{ar_{C}}{L}&\frac{\alpha}{L}\\ -\frac{\alpha}{\alpha}&0\end{array} \right] \tag{77}\]  
\[B_{1}=B_{2}=B_{0}=\left[ \begin{array}{c}\frac{1}{L}\\ 0\end{array} \right],\ \ \ C_{1}=0\times C_{u_{1}}=0,\ \ C_{2}=1\times C_{u_{1}}= \left[ \begin{array}{cc}-\alpha r_{C}&0\end{array} \right],\ C_{0}=\left[ \begin{array}{cc}\alpha r_{C}&\alpha\end{array} \right]\] \begin{table}
\begin{tabular}{c c c} \hline \hline
**Parameter** & **Value** & **Unit** \\ \hline \(V_{in}\) & 150 & V \\ \(L\) & 100 & \(\mu\)H \\ \(C\) & 2 & \(\mu\)F \\ \(R\) & 100 & \(\Omega\) \\ \(r_{L}\) & 2 & \(\Omega\) \\ \(r_{C}\) & 0.2 & \(\Omega\) \\ \hline \hline \end{tabular}
\end{table} 
Table 1: Parameters of the boost converter. Then,  
\[B_{0}(x)=\left[ \begin{array}{cc}A_{1}x+B_{1}V_{in}&A_{2}x+B_{2}V_{in}\end{array}  \right]\text{ and }D_{0}(x)=\left[ \begin{array}{cc}C_{1}x&C_{2}x\end{array} \right] \tag{78}\]  
The equilibrium points are given by  
\[\left[ \begin{array}{cc}-\frac{r_{L}+\alpha r_{C}}{L}&-\frac{\alpha}{L}\\ \frac{\alpha}{C}&-\frac{\alpha}{RC}\end{array} \right]x_{e}+\left[ \begin{array}[ ]{cc}\frac{1}{L}\\ 0\end{array} \right]V_{in}(\lambda_{e_{1}}+\lambda_{e_{2}})+\left[ \begin{array} []{cc}\frac{\alpha r_{C}}{L}&\frac{\alpha}{L}\\ -\frac{\alpha}{C}&0\end{array} \right]x_{e}\lambda_{e_{2}}=0 \tag{79}\]  
with \(\lambda_{e}=\left[ \begin{array}{cc}\lambda_{e_{1}}&\lambda_{e_{2}}\end{array}  \right]^{T}\in\Lambda_{R}\). Then, because \(\lambda_{e_{1}}+\lambda_{e_{2}}=1\), we have  
\[x_{e}=-\Bigg{\{}\left[ \begin{array}{cc}-\frac{r_{L}+\alpha r_{C}}{L}&-\frac {\alpha}{L}\\ \frac{\alpha}{C}&-\frac{\alpha}{RC}\end{array} \right]+\lambda_{e_{2}}\Bigg{[} \begin{array}{cc}\frac{\alpha r_{C}}{L}&\frac{\alpha}{L}\\ -\frac{\alpha}{C}&0\end{array} \right]\Bigg{\}}^{-1}\left[ \begin{array}{cc} \frac{1}{L}\\ 0\end{array} \right]V_{in} \tag{80}\]  
The set of equilibrium states is defined by  
\[X_{eq}=\Bigg{\{}x_{e}=\left[ \begin{array}{cc}\frac{V_{in}}{r_{L}+\alpha r_{C }(1-\lambda_{e_{2}})+\alpha R(1-\lambda_{e_{2}})^{2}}&\frac{(1-\lambda_{e_{2} })RV_{in}}{r_{L}+\alpha r_{C}(1-\lambda_{e_{2}})+\alpha R(1-\lambda_{e_{2}})^{ 2}}\end{array} \right]^{T},\;\lambda_{e_{2}}\in[0,1]\Bigg{\}} \tag{81}\]  
At the equilibrium, the output voltage writes  
\[y_{e}=C_{0}x_{e}+D_{0}(x_{e})\lambda_{e}=\left[ \begin{array}{cc}\alpha(1- \lambda_{e_{2}})r_{C}&\alpha\end{array}