## 4 Simulation Results
### Rejection of Measured Perturbation  
In this paragraph, we suppose that the load is varying, the state measurable and the load current is considered as a measurable perturbation. For the boost converter depicted in Figure 5, the basic equations are  
\[\left\{ \begin{array}{l}L\frac{di_{L}(t)}{dt}=V_{in}-(1-u_{1}(t))y(t)-r_{L}i_{L} (t)\\ C\frac{dv_{\mathbb{C}}(t)}{dt}=(1-u_{1}(t))i_{L}(t)-i_{0}(t)=\frac{y(t)-v_{ \mathbb{C}}(t)}{r_{\mathbb{C}}}\end{array} \right. \tag{86}\]  
Taking the state vector, \(x(t)=\left[ \begin{array}{cc}i_{L}(t)&v_{\mathbb{C}}(t)\end{array} \right]\) and \(w(t)=i_{0}(t)\) we have  
\[\left\{ \begin{array}{l}\frac{dx(t)}{dt}=\left[\underbrace{\left[\begin{array} []{cc}-\frac{x_{1}+v_{\mathbb{C}}}{L}&-\frac{t}{L}\\ \frac{1}{L}&0\end{array}\right]}_{A_{0}}+\underbrace{\left[\begin{array}{ cc}\frac{x_{1}}{L}&\frac{1}{L}\\ -\frac{1}{L}&0\end{array}\right]}_{A_{0}}u_{1}(t)\right]x(t)+\underbrace{\left[ \begin{array}{cc}\frac{x_{1}}{L}\\ 0\end{array}\right]}_{B_{0}}V_{in}+\left[\underbrace{\left[\begin{array}{ cc}\frac{x_{1}}{L}\\ -\frac{1}{L}\end{array}\right]}_{B_{0}}+\underbrace{\left[\begin{array}{ cc}-\frac{x_{1}}{L}\\ 0\end{array}\right]}_{B_{0}}\right]_{in}w(t)\\ y(t)=\left[\underbrace{\left[\begin{array}{cc}r_{\mathbb{C}}&1\end{array} \right]}_{C_{0}}+\underbrace{\left[\begin{array}{cc}-r_{\mathbb{C}}&0\end{array} \right]}_{C_{0}}u_{1}(t)\right]x(t)+\underbrace{-r_{\mathbb{C}}w(t)}_{D_{0 \infty}}\end{array} \right. \tag{87}\]  
We can note that \(A_{0}\) is Hurwitz. The main matrices associated with the model (60) are given by  
\[A_{1}=0\times A_{u_{1}}=0\ \ \ A_{2}=1\times A_{u_{1}}=\left[ \begin{array}{ cc}\frac{r_{\mathbb{C}}}{L}&\frac{1}{L}\\ -\frac{1}{L}&0\end{array} \right]\]  
\[B_{1}=B_{2}=B_{0}=\left[ \begin{array}{cc}\frac{1}{L}\\ 0\end{array} \right],\ \ C_{1}=0\times C_{u_{1}}=0,\ \ C_{2}=1\times C_{u_{1}}=\left[ \begin{array} []{cc}-r_{\mathbb{C}}&0\end{array} \right],\ C_{0}=\left[ \begin{array}{cc}r_{ \mathbb{C}}&1\end{array} \right]\]  
\[B_{w_{1}}=B_{w0}+0\times B_{wu_{1}}=\left[ \begin{array}{cc}\frac{r_{\mathbb{C }}}{L}\\ -\frac{1}{L}\end{array} \right],\ B_{w_{2}}=B_{w0}+1\times B_{wu_{1}}=\left[ \begin{array}{cc}0\\ -\frac{1}{L}\end{array} \right] \tag{88}\]  
\[B_{0}(x,w)=\left[ \begin{array}{cc}A_{1}x+B_{w_{1}}w+B_{1}V_{in}\ \ \ A_{2}x+B_{w_{2}}w+B_{2}V_{in}\end{array} \right] \tag{89}\]  
\[D_{0}(x,w)=\left[ \begin{array}{cc}C_{1}x+D_{w_{1}}w&C_{2}x+D_{w_{2}}w\end{array} \right] \text{and}\,\ D_{w_{1}}=D_{w_{2}}=D_{w0}=-r_{\mathbb{C}} \tag{90}\]  
The set \(\mathcal{X}_{e}\) is the set of the function vectors defined by  
Figure 5: Boost Converter with a variation of load represented by \(i_{0}\). \[x_{e}(t) =-\Bigg{\{}\Bigg{[} \begin{array}{cc}-\frac{r_{L}+r_{C}}{L}&-\frac{1}{L}\\ \frac{1}{C}&0\end{array} \Bigg{]}+\lambda_{e_{2}}(t)\Bigg{[} \begin{array}{cc} \frac{r_{C}}{L}&\frac{1}{L}\\ -\frac{1}{C}&0\end{array} \Bigg{]}\Bigg{\}}^{-1}\Bigg{\{}\Bigg{[} \begin{array}{ cc}\frac{1}{L}\\ 0\end{array} \Bigg{]}V_{in}+\Bigg{[} \begin{array}{cc}\lambda_{e1}\frac{r_{C}}{L} \\ -\frac{1}{C}\end{array} \Bigg{]}w(t)\Bigg{\}}\] (91) \[=\Bigg{[} \begin{array}{cc}\frac{w(t)}{1-\lambda_{e_{2}}(t)}& \frac{V_{in}}{1-\lambda_{e_{2}}(t)}-\frac{r_{C}\lambda_{e_{2}(t)}w(t)}{1- \lambda_{e_{2}}(t)}-\frac{r_{L}w(t)}{\left(1-\lambda_{e_{2}}(t)\right)^{2}} \end{array} \Bigg{]}^{T}\] with \[\lambda_{e}\,:\mathbb{R}\rightarrow\Lambda_{R}\] At equilibrium, the output voltage is written as \[y_{e}(t) = C_{0}x_{e}(t)+D_{0}(x_{e}(t),w(t))\lambda_{e}(t)=\big{[} \begin{array} []{cc}\lambda_{e_{2}}(t)r_{C}&1\end{array} \big{]}x_{e}(t)-r_{C}w(t)\] \[= \frac{V_{in}}{1-\lambda_{e_{2}}(t)}-\frac{r_{L}w(t)}{\left(1- \lambda_{e_{2}}(t)\right)^{2}}-r_{C}w(t)\] Case 1: \[i_{0}\] variable and constant output voltage.