## 3 Compound Controller Design
### Stability Analysis
 \tag{39}\]  
Based on the Equations (39) and (37), Equation (35) can be further calculation as  
\[ \begin{split}\Delta V(k)=&-[s(k)-s(k+1)][s(k)+s(k+1 )]\\ =&-\left[\kappa_{1}\xi_{1}s(k)+\kappa_{1}\xi_{2} \lceil s(k)\rfloor^{\beta}-\kappa_{1}\mathbf{C}\left[\mathbf{D}(k)-\hat{ \mathbf{D}}(k)\right]\right]\\ &\times\left[2s(k)-\kappa_{1}\xi_{1}s(k)-\kappa_{1}\xi_{2} \lceil s(k)\rfloor^{\beta}+\kappa_{1}\mathbf{C}\left[\mathbf{D}(k)-\hat{ \mathbf{D}}(k)\right]\right]\\ \leq&-\mu^{2}\end{split}  \tag{40}\]  
Hence, if \(s(k)>0\), \(\Delta V(k)\leq-\mu^{2}\) holds. _Case 2_: Consider \(s(k)<-\sigma=-\psi(\beta)\cdot\Phi<0\) with Assumption 3 and Proposition 1, the proof procedure of \(\Delta V(k)\leq-\mu^{2}\) is similar to that of the _Case 1_. According to the analysis results of _Case 1_ and _Case 2_ when \(s(k)\
otin\sigma\), it can be derived from Equation (40) that \(s^{2}(k)\leq s^{2}(0)-k\mu^{2}\). We can directly solve \(s^{2}(0)-k\mu^{2}=\sigma^{2}\) to obtain the solution \(k^{*}\) defined in Equation (33). Hence, existing \(K^{*}=\lfloor k^{*}\rfloor+1\in N\) results in \(|s(K^{*})|\leq\sigma\). (2) Next, we will divide two cases to prove that once \(s(k)\in\Delta\) the sliding mode state \(s(k+1)\in\Delta\). Define \(0\leq\varsigma\leq 1\), we can obtain \(|s(k)|=\varsigma\sigma=\varsigma\psi(\beta)\Phi\). By Equation (31), it can be obtained that  
\[ \begin{split} s(k+1)&=s(k)-\kappa_{1}\xi_{1}s(k)- \kappa_{1}\xi_{2}\lceil s(k)\rfloor^{\beta}+\kappa_{1}\mathbf{C}\left[ \mathbf{D}(k)-\hat{\mathbf{D}}(k)\right]\\ &=(1-\kappa_{1}\xi_{1})s(k)-\kappa_{1}\xi_{2}|s(k)|^{\beta}sgn(s( k))+\kappa_{1}\mathbf{C}\left[\mathbf{D}(k)-\hat{\mathbf{D}}(k)\right]\end{split}  \tag{41}\]  
_Case 1_: The proof of \(-\sigma\leq s(k+1)\leq\sigma\) with \(s(k)\geq 0\). When \(s(k)\geq 0\), Equation (41) becomes  
\[s(k+1)=(1-\kappa_{1}\xi_{1})\varsigma\psi(\beta)\Phi-\kappa_{1}\xi_{2}[ \varsigma\psi(\beta)\Phi]^{\beta}sgn(\varsigma\psi(\beta)\Phi)+\kappa_{1} \mathbf{C}\left[\mathbf{D}(k)-\hat{\mathbf{D}}(k)\right] \tag{42}\]  
If \(0\leq s(k)<1\), by Proposition 1 and Assumption 3, it can be derived from Equation (42) as  
\[ \begin{split} s(k+1)&\leq(1-\kappa_{1}\xi_{1}) \varsigma\psi(\beta)\Phi-[\varsigma\psi(\beta)]^{\beta}\epsilon+\epsilon\\ &\leq(1-\kappa_{1}\xi_{1})\varsigma\psi(\beta)\Phi+\left[1-[ \varsigma\psi(\beta)]^{\beta}\right](1-\kappa_{1}\xi_{1})\Phi\\ &\leq(1-\kappa_{1}\xi_{1})\Phi\left[\varsigma\psi(\beta)-[ \varsigma\psi(\beta)]^{\beta}+1\right]\\ &\leq(1-\kappa_{1}\xi_{1})\Phi\\ &<\psi(\beta)\Phi=\sigma\end{split}  \tag{43}\]  
Otherwise, when \(s(k)\geq 1\), utilizing Proposition 1 and Assumption 3, Equation (42) can be deduced as  
\[ \begin{split} s(k+1)&\leq(1-\kappa_{1}\xi_{1}) \varsigma\psi(\beta)\Phi+\left[1-[\varsigma\psi(\beta)]^{\beta}\right]\epsilon \\ &\leq(1-\kappa_{1}\xi_{1})\varsigma\psi(\beta)\Phi\\ &\leq\varsigma\psi(\beta)\Phi\leq\sigma\end{split}  \tag{44}\]  
Next, the proof of \(-\sigma\leq s(k+1)\) with \(s(k)\geq 0\). Similarly, if \(0\leq s(k)<1\), by Proposition 1 and Assumption 3, it can be derived from Equation (42) as  
\[ \begin{split} s(k+1)&\geq(1-\kappa_{1}\xi_{1})\varsigma \psi(\beta)\Phi-\kappa_{1}\xi_{2}[\varsigma\psi(\beta)\Phi]^{\beta}sgn(\varsigma \psi(\beta)\Phi)-\varepsilon\\ &\geq(1-\kappa_{1}\xi_{1})\varsigma\psi(\beta)\Phi-\kappa_{1} \xi_{2}[\varsigma\psi(\beta)\Phi]^{\beta}-\varepsilon\\ &\geq(1-\kappa_{1}\xi_{1})\Phi\left[\varsigma\psi(\beta)-[ \varsigma\psi(\beta)]^{\beta}-1\right]\\ &\geq-(1-\kappa_{1}\xi_{1})\psi(\beta)\Phi\geq-\sigma\end{split}  \tag{45}\]  
Otherwise, when \(s(k)\geq 1\), utilizing Proposition 1 and Assumption 3, Equation (42) can be deduced as  
\[ \begin{split} s(k+1)&\geq(1-\kappa_{1}\xi_{1}) \varsigma\psi(\beta)\Phi-\kappa_{1}\xi_{2}[\varsigma\psi(\beta)\Phi]^{\beta} sgn(\varsigma\psi(\beta)\Phi)-\varepsilon\\ &\geq(1-\kappa_{1}\xi_{1})\varsigma\psi(\beta)\Phi-\kappa_{1} \xi_{2}[\varsigma\psi(\beta)\Phi]^{\beta}-\varepsilon\\ &\geq-\varepsilon\geq-(1-\kappa_{1}\xi_{1})\Phi>-\sigma\end{split}