## 3 Singular Perturbation Decomposition and Control Law Design
### The Slow Subsystem
Then the link's rotational angle error \(\overline{\mathbf{x}}_{1}\) is:  
\[\overline{\mathbf{x}}_{1}=\mathbf{x}_{1}-\hat{\mathbf{x}}_{1} \tag{16}\]  
And the link's rotational angular velocity error \(\overline{\mathbf{x}}_{2}\) is:  
\[\overline{\mathbf{x}}_{2}=\mathbf{x}_{2}-\hat{\mathbf{x}}_{2} \tag{17}\]  
A velocity filter is designed for the slow subsystem:  
\[\left\{ \begin{array}{l}\dot{\hat{\mathbf{x}}}_{1}=\hat{\mathbf{x}}_{2}+\mathbf{K}_{\text {p}}\overline{\mathbf{x}}_{1}+\Lambda_{1}\text{sgn}(\overline{\mathbf{x}})1\\ \dot{\hat{\mathbf{x}}}_{2}=\mathbf{D}_{0\text{s}}^{-1}(x_{0},\mathbf{x}_{1})[\mathbf{\rho}\mathbf{ \tau}_{\text{s}}-\mathbf{H}_{0\text{s}}(x_{0},\mathbf{x}_{1},\hat{\mathbf{x}}_{2})\hat\mathbf{{x}}_{2}]+\mathbf{K}_{\text{v}}\overline{\mathbf{x}}_{1}+\Lambda_{2}\text{sgn}( \overline{\mathbf{x}})1\end{array} \right. \tag{18}\]  
where, the matrix \(\mathbf{H}_{0\text{s}}(x_{0},\mathbf{x}_{1},\hat{\mathbf{x}}_{2})\) is a new matrix gotten by letting \(\mathbf{x}_{2}=\hat{\mathbf{x}}_{2}\) in \(\mathbf{H}_{0\text{s}}(x_{0},\mathbf{x}_{1},\mathbf{x}_{2})\). \(\mathbf{K}_{\text{p}}\) and \(\mathbf{K}_{\text{v}}\) are positive definite diagonal matrices. \(\Lambda_{1}=\text{diag}(\lambda_{1},\cdots,\lambda_{i},\cdots,\lambda_{n})\), \(\lambda_{i}\) is positive.\(\Lambda_{2}\)'s definition will be given below. Subtracting Equations (15) and (18), we have:  
\[\left\{ \begin{array}{l}\dot{\bar{\mathbf{x}}}_{1}=\dot{\mathbf{x}}_{1}-\dot{\hat{\mathbf{x }}}_{1}=\overline{\mathbf{x}}_{2}-\mathbf{K}_{\text{p}}\overline{\mathbf{x}}_{1}-\Lambda_ {1}\text{Sgn}(\overline{\mathbf{x}})1\\ \dot{\overline{\mathbf{x}}}_{2}=\dot{\hat{\mathbf{x}}}_{2}-\dot{\hat{\mathbf{x}}}_{2}=\mathbf{D}_{0 \text{s}}^{-1}(x_{0},\mathbf{x}_{1})[\mathbf{H}_{0\text{s}}(x_{0},\mathbf{x}_{1},\hat{\mathbf{x} }_{2})\hat{\mathbf{x}}_{2}-\mathbf{H}_{0\text{s}}(x_{0},\mathbf{x}_{1},\mathbf{x}_{2}) \mathbf{x}_{2}+\mathbf{\tau}_{\text{d}}]-\mathbf{K}_{\text{v}}\overline{\mathbf{x}}_{1}- \Lambda_{2}\text{Sgn}(\overline{\mathbf{x}})1\end{array} \right. \tag{19}\]Defining a sliding surface:  
\[S=\overline{x}_{1} \tag{20}\]  
When the velocity filter's the sliding mode occurs, that means \(\overline{x}_{1}=0\) and \(\dot{\overline{x}}_{1}=0\). Then from Equations (17) and (19), we have:  
\[ \begin{array}{c}\dot{\overline{x}}_{2}=-D_{0s}^{-1}(x_{0},\mathbf{x}_{1})[-H_{0s} (x_{0},\mathbf{x}_{1},\hat{\mathbf{x}}_{2})\hat{\mathbf{x}}_{2}+H_{0s}(x_{0},\mathbf{x}_{1},\bm {x}_{2})\mathbf{x}_{2}-\tau_{\mathrm{d}}]-\Lambda_{2}\Lambda_{1}^{-1}\overline{x}_ {2}\\ =-\Lambda_{2}\Lambda_{1}^{-1}\overline{x}_{2}-D_{0s}^{-1}(x_{0},\mathbf{x}_{1})[H_ {0s}(x_{0},\mathbf{x}_{1},\hat{\mathbf{x}}_{2})\overline{x}_{2}+H_{0s}(x_{0},\mathbf{x}_{1 },\mathbf{x}_{2})\mathbf{x}_{2}-H_{0s}(x_{0},\mathbf{x}_{1},\hat{\mathbf{x}}_{2})\mathbf{x}_{2}- \tau_{\mathrm{d}}]\end{array}  \tag{21}\]  
According to property 3, there is:  
\[H_{0s}(x_{0},\mathbf{x}_{1},\hat{\mathbf{x}}_{2})\mathbf{x}_{2}=H_{0s}(x_{0},\mathbf{x}_{1}, \mathbf{x}_{2})\hat{\mathbf{x}}_{2} \tag{22}\]  
Thus, Equation (21) can be rewritten into the following form:  
\[ \begin{array}{c}\dot{\overline{x}}_{2}=-\Lambda_{2}\Lambda_{1}^{-1}\overline {x}_{2}-D_{0s}^{-1}(x_{0},\mathbf{x}_{1})[H_{0s}(x_{0},\mathbf{x}_{1},\hat{\mathbf{x}}_{2}) \overline{x}_{2}+H_{0s}(x_{0},\mathbf{x}_{1},\mathbf{x}_{2})\mathbf{x}_{2}-H_{0s}(x_{0}, \mathbf{x}_{1},\mathbf{x}_{2})\hat{\mathbf{x}}_{2}-\tau_{\mathrm{d}}]\\ =-\Lambda_{2}\Lambda_{1}^{-1}\overline{x}_{2}-D_{0s}^{-1}(x_{0},\mathbf{x}_{1})[H_ {0s}(x_{0},\mathbf{x}_{1},\hat{\mathbf{x}}_{2})\overline{x}_{2}+H_{0s}(x_{0},\mathbf{x}_{ 1},\mathbf{x}_{2})\overline{x}_{2}-\tau_{\mathrm{d}}]\end{array}  \tag{23}\]  
Then, the stability of the system needs to be proved.