# Implementation of Fault-Tolerant Control for
## 4 Fault Tolerant Control with Synchronous Sliding Mode Control
\begin{array}{ll}\dot{V}&=s^{T}\dot{s}+\sum\limits_{i=1}^{n}\dot{\varepsilon }_{i}\Lambda_{\varepsilon}\varepsilon+\sum\limits_{i=1}^{n}\kappa_{ \varepsilon}\Lambda_{\varepsilon}(\varepsilon_{i}-\varepsilon_{i-1})\int( \varepsilon_{i}-\varepsilon_{i-1})dt\\ &=A_{1}+\sum\limits_{i=1}^{n}\dot{\varepsilon}_{i}\Lambda_{\varepsilon} \varepsilon+\sum\limits_{i=1}^{n}\kappa_{\varepsilon}\Lambda_{\varepsilon} (\varepsilon_{i}-\varepsilon_{i-1})\int(\varepsilon_{i}-\varepsilon_{i-1})dt \end{array}  \tag{27}\]  
where \(A_{1}=s^{T}\dot{s}\). \(\Lambda_{\varepsilon}\) and \(\kappa_{\varepsilon}\) are positive gain. Differentiation of Equation (24) gives  
\[\dot{s}=\ddot{e}+\dot{c}\dot{e}+\alpha\dot{\xi}+\kappa\xi\,\,=\ddot{q}_{d}- \ddot{q}+\dot{c}\dot{e}+\alpha\dot{\xi}+\kappa\xi \tag{28}\]  
Substituting Equation (8) into Equation (28) yields  
\[\dot{s}=\ddot{q}_{d}+\dot{c}\dot{e}+\alpha\dot{\xi}+\kappa\xi-\mathbf{M}^{-1}( q)(\tau-\mathbf{H}(q,\dot{q}))-\mathbf{M}^{-1}(q)\zeta(q,\dot{q},\tau,t) \tag{29}\]  
Substituting Equation (25) into Equation (29), we have  
\[ \begin{array}{ll}\dot{s}&=\ddot{q}_{d}+\dot{c}\dot{e}+\alpha\dot{\xi}+ \kappa\xi-\mathbf{M}^{-1}(q)\left\{\mathbf{M}(q)\ddot{q}_{d}+\mathbf{H}(q,\dot {q})+\mathbf{M}(q)(\dot{c}\dot{e}+\alpha\dot{\xi})+k_{1}sgn(s)+k_{2}s+k_{3}\xi \\ &-\mathbf{M}(q)\widehat{\phi}-H(q,\dot{q})\right\}+\mathbf{M}^{-1}(q)\zeta(q, \dot{q},\ddot{q},\tau,t)\\ &=-\mathbf{M}^{-1}(q)k_{1}sgn(s)-\mathbf{M}^{-1}(q)k_{2}s-(\kappa+\mathbf{M}^ {-1}(q)k_{3})\xi+\widehat{\phi}+\mathbf{M}^{-1}(q)\zeta(q,\dot{q},\ddot{q}, \tau,t)\end{array}  \tag{30}\]  
It can be seen that when \(\widehat{\phi}\to\phi\) at \(t\to\infty\,\widehat{\phi}=-\mathbf{M}^{-1}(q)\zeta(q,\dot{q},\tau,t)\) then Equation (30) becomes  
\[\dot{s}=\mathbf{M}^{-1}(q)(-k_{1}sgn(s)-k_{2}s-(k_{3}-\kappa)\xi) \tag{31}\]  
From Properties, we have  
\[\dot{s}\leq\frac{1}{\mu_{2}}I(-k_{1}sgn(s)-k_{2}s-(k_{3}-\kappa)\xi) \tag{32}\]  
and  
\[ \begin{array}{ll}A_{1}=s^{T}\dot{s}&\leq s^{T}\frac{1}{\mu_{2}}I(-k_{1}sgn(s )-k_{2}s-(k_{3}-\kappa)\xi)\\ &=-\frac{1}{\mu_{2}}Ik_{1}|s|-\frac{1}{\mu_{2}}Ik_{2}|s|\|^{2}-s^{T}\frac{1}{ \mu_{2}}I(k_{3}-\kappa)\xi\\ &=-\frac{1}{\mu_{2}}Ik_{1}|s|-\frac{1}{\mu_{2}}Ik_{2}|s|\|^{2}-A_{2}\end{array}  \tag{33}\]  
In Equation (33), we have  
\[A_{2}=s^{T}\Lambda\xi,\]  
Substituting Equation (39) into Equation (33) and then substituting Equation (33) into Equation (27) gives  
\[\dot{V}\leq-\frac{1}{\mu_{2}}I_{k1}|s|-\frac{1}{\mu_{2}}I_{k2}||s||^{2}-\sum_{i=1 }^{n}c_{i+1}^{i}\Lambda_{\varepsilon}\varepsilon_{i}^{2}-\sum_{i=1}^{n}\alpha_{ \varepsilon}\Lambda_{\varepsilon}(\varepsilon_{i}-\varepsilon_{i+1})^{2}\leq 0 \tag{40}\]  
Assume that \(s_{i},\varepsilon_{i},\dot{s}_{i}\) and \(\dot{\varepsilon}_{i}\) are bounded. Therefore, \(s_{i}\) and \(\varepsilon_{i}\) are uniformly continuous. Then, from [PERSON]'s lemma, it is concluded that \(s_{i}\to 0\) and \(\varepsilon_{i}\to 0\) as time \(t\to\infty\). Now, we prove \(e_{i}=0\) when \(s_{i}\to 0\) and \(\varepsilon_{i}\to 0\). From Equations (23) and (24), \(s_{i}\to 0\) implies \(E_{i}\to 0\). Combining all equations in Equation (22) with indices \(1\) to n, one obtains  
\[e_{1}+e_{2}+\ldots+e_{n}=\sum_{i=1}^{n}E_{i}=0 \tag{41}\]  
We also have \(\varepsilon_{i}=0\) when \(t\to\infty\), which means \(e_{1}=e_{2}=e_{3}=\ldots=e_{n}\). Substituting this part into Equation (41) we have  
\[e_{1}=e_{2}=e_{3}=\ldots=e_{n}=0 \tag{42}\]  
Therefore, Theorem 2 is proven.