## 3 Singular Perturbation Decomposition and Control Law Design
### The Slow Subsystem  
From Equation (8) we can see that when \(\varepsilon\to 0\), the flexible-joint's stiffness coefficient matrix \(\mathbf{K}_{\text{0}}\rightarrow\infty\). At this time, the connection between the motor rotor and the manipulator can be regarded as a rigid connection. \(\mathbf{\theta}=\mathbf{q}_{\theta}\), \(\hat{\mathbf{\theta}}=\hat{\mathbf{q}}_{\theta}\), \(\hat{\mathbf{\theta}}=\tilde{\mathbf{q}}_{\theta}\), \(\mathbf{z}_{\theta}=0\). Let \(\varepsilon=0\) in Equations (9) and (10), and the slow subsystem's dynamic equation can be obtained:  
\[\mathbf{D}_{0\text{s}}(\mathbf{q})\tilde{\mathbf{q}}_{\text{0}}+\mathbf{C}_{0\text{s}}(\mathbf{q},\dot{\mathbf{q}}_{\text{0}})=\mathbf{\rho}\mathbf{\tau}_{\text{s}}+\mathbf{\tau}_{\text{d}} \tag{12}\]  
where, \(\mathbf{\tau}_{\text{s}}\in\mathbf{R}^{n\times 1}\) is the slow subsystem's controller. \(\mathbf{\rho}=\text{diag}(\rho_{1}\ \cdots\ \ \rho_{i}\ \cdots\ \ \rho_{n})\), \(\rho_{i}\in(0,1]\) is actuator's efficiency factor. \(\mathbf{D}_{0\text{s}}(\mathbf{q})=\mathbf{D}_{0}(\mathbf{q})+\mathbf{I}_{\text{0}}\), \(\mathbf{C}_{0\text{s}}(\mathbf{q},\dot{\mathbf{q}}_{\text{0}})\) is a new matrix by letting \(\hat{\mathbf{\theta}}=\hat{\mathbf{q}}_{\text{0}}\) in \(\mathbf{C}_{0}(\mathbf{q},\dot{\mathbf{q}}_{\text{0}},\dot{\mathbf{\theta}})\). Firstly, we rewrite Equation (12) into the following form:  
\[\mathbf{D}_{0\text{s}}(\mathbf{q})\tilde{\mathbf{q}}_{\text{0}}+\mathbf{H}_{0\text{s}}(\mathbf{q},\dot{\mathbf{q}}_{\text{0}})\dot{\mathbf{q}}_{\text{0}}=\mathbf{\rho}\mathbf{\tau}_{\text{s} }+\mathbf{\tau}_{\text{d}} \tag{13}\]  
The matrix in Equation (13) has the following properties:  
**Property 1**: The inertial matrix \(\mathbf{D}_{0\text{s}}(\mathbf{q})\) is symmetric positive definite and bounded. so \(0<\xi_{1}\|\mathbf{y}_{1}\|^{2}\leq\mathbf{y}_{1}^{\intercal}\mathbf{D}_{0\text{s}}(\mathbf{q })\mathbf{y}_{1}\leq\xi_{2}\|\mathbf{y}_{1}\|^{2}\), \(\forall\mathbf{y}_{1}\in\mathbf{R}^{n\times 1}\). Where, \(\xi_{1}\) is the minimum eigenvalues of \(\mathbf{D}_{0\text{s}}(\mathbf{q})\), \(\xi_{2}\) is the maximum eigenvalues of \(\mathbf{D}_{0\text{s}}(\mathbf{q})\). **Property 2**: [41]: \(\mathbf{H}_{0\text{s}}(\mathbf{q},\dot{\mathbf{q}}_{\text{0}})\in\mathbf{R}^{n\times n}\) can satisfy that:\(\mathbf{x}^{\intercal}[\mathbf{D}_{0\text{s}}(\mathbf{q})-2\mathbf{H}_{0\text{s}}(\mathbf{q}, \dot{\mathbf{q}}_{\text{0}})]\mathbf{x}=0\), \(\forall\mathbf{x}\in\mathbf{R}^{n\times 1}\). Also \(\mathbf{H}_{0\text{s}}(\mathbf{q},\dot{\mathbf{q}}_{\text{0}})\) is bounded. **Property 3**: [42]: for \(\forall\mathbf{y}_{2},\mathbf{y}_{3},\mathbf{y}_{4}\in\mathbf{R}^{n\times 1}\), there are: \(\|\mathbf{H}_{0\text{s}}(\mathbf{y}_{2},\mathbf{y}_{3})\|\leq\xi_{\text{h}}\|\mathbf{y}_{3}\|,H_{0\text{s}}(\mathbf{y}_{2},\mathbf{y}_{3})\mathbf{y}_{4}=\mathbf{H}_{0\text{s}}(\mathbf{y}_{2}, \mathbf{y}_{4})\mathbf{y}_{3}\). where, \(\xi_{\text{h}}\) is a positive constant. **Property 4**: \(\mathbf{\tau}_{\text{d}}\) is bounded, and \(\|\mathbf{\tau}_{\text{d}}\|\leq\xi_{\text{d}}\), \(\xi_{\text{d}}>0\). Defining state variables:  
\[x_{0}=q_{0},\,\mathbf{x}_{1}=\mathbf{q}_{\text{0}},\,\mathbf{x}_{2}=\dot{\mathbf{q}}_{\text{0}} \tag{14}\]  
Thus, Equation (13)'s state-space form is:  
\[\left\{ \begin{array}{l}\dot{\mathbf{x}}_{1}=\mathbf{x}_{2}\\ \dot{\mathbf{x}}_{2}=\mathbf{D}_{0\text{s}}^{-1}(x_{0},\mathbf{x}_{1})[\mathbf{\rho}\mathbf{\tau}_ {\text{s}}+\mathbf{\tau}_{\text{d}}-\mathbf{H}_{0\text{s}}(x_{0},\mathbf{x}_{1},\mathbf{x}_{2} )\mathbf{x}_{2}]\end{array} \right. \tag{15}\]  
Let \(\hat{\mathbf{x}}_{1}\) be \(\mathbf{x}_{1}\)'s estimated value, \(\hat{\mathbf{x}}_{2}\) be \(\mathbf{x}_{2}\)'s estimated value.