3.1. FBFLFJ space robot model singular perturbation decomposition

To facilitate the control analysis of the FBFLFJ space robot, the model is firstly decomposed and Eq. (5) is rewritten as

(6) \begin{equation} \boldsymbol{{D}}\!\left(\boldsymbol{{q}}_{{boj\unicode{x03B4} }}\right)\ddot{\boldsymbol{{q}}}_{{boj\unicode{x03B4} }}+\boldsymbol{{H}}\!\left(\boldsymbol{{q}}_{{boj\unicode{x03B4} }},\dot{\boldsymbol{{q}}}_{{boj\unicode{x03B4} }}\right)\dot{\boldsymbol{{q}}}_{{boj\unicode{x03B4} }}+\boldsymbol{\xi }_{{boj\unicode{x03B4} }}=\boldsymbol{\tau }_{{boj\unicode{x03B4} }} ,\end{equation}

(7) \begin{equation} \boldsymbol{{J}}_{{m}}\ddot{\boldsymbol{{q}}}_{{m}}+\boldsymbol{\tau }=\boldsymbol{\tau }_{{m}} ,\end{equation}

(8) \begin{equation} \boldsymbol{\tau }=\boldsymbol{{K}}_{{m}}\boldsymbol{\sigma } .\end{equation}

Rewrite the matrices $\boldsymbol{{D}}\!\left(\boldsymbol{{q}}_{{boj\unicode{x03B4} }}\right)$ and $\boldsymbol{{H}}\!\left(\boldsymbol{{q}}_{{boj\unicode{x03B4} }},\dot{\boldsymbol{{q}}}_{{boj\unicode{x03B4} }}\right)$ from Eq. (6) as $\left[\begin{array}{l@{\quad}l} \boldsymbol{{D}}_{{aa}} & \boldsymbol{{D}}_{{ab}}\\[4pt] \boldsymbol{{D}}_{{ba}} & \boldsymbol{{D}}_{{bb}} \end{array}\right]$ and $\left[\begin{array}{l@{\quad}l} \boldsymbol{{H}}_{{aa}} & \boldsymbol{{H}}_{{ab}}\\[4pt] \boldsymbol{{H}}_{{ba}} & \boldsymbol{{H}}_{{bb}} \end{array}\right]$ , respectively, in which $\boldsymbol{{D}}_{{aa}},\boldsymbol{{H}}_{{aa}}\in \mathrm{\mathbb{R}}^{1\times 1}, \boldsymbol{{D}}_{{ab}},\boldsymbol{{H}}_{{ab}}\in \mathrm{\mathbb{R}}^{1\times 7}, \boldsymbol{{D}}_{{ba}},\boldsymbol{{H}}_{{ba}}\in \mathrm{\mathbb{R}}^{7\times 1}, \boldsymbol{{D}}_{{bb}},\boldsymbol{{H}}_{{bb}}\in \mathrm{\mathbb{R}}^{7\times 7}$ . $\boldsymbol{{D}}\!\left(\boldsymbol{{q}}_{{boj\unicode{x03B4} }}\right)$ is invertible and can be defined by

(9) \begin{equation} \left[\begin{array}{l@{\quad}l} \boldsymbol{{N}}_{11} & \boldsymbol{{N}}_{12}\\[4pt] \boldsymbol{{N}}_{21} & \boldsymbol{{N}}_{22} \end{array}\right]=\left[\begin{array}{l@{\quad}l} \boldsymbol{{D}}_{{aa}} & \boldsymbol{{D}}_{{ab}}\\[4pt] \boldsymbol{{D}}_{{ba}} & \boldsymbol{{D}}_{{bb}} \end{array}\right]^{-1} ,\end{equation}

where $\boldsymbol{{N}}_{11}\in \mathrm{\mathbb{R}}^{1\times 1}, \boldsymbol{{N}}_{12}\in \mathrm{\mathbb{R}}^{1\times 7}, \boldsymbol{{N}}_{21}\in \mathrm{\mathbb{R}}^{7\times 1}$ and $\boldsymbol{{N}}_{22}\in \mathrm{\mathbb{R}}^{7\times 7}$ .

By substituting Eq. (9) into Eq. (6), we obtain

(10) \begin{equation} \ddot{q}_{{b}}=-\boldsymbol{{N}}_{11}\!\left(\boldsymbol{{H}}_{{aa}}\dot{q}_{{b}}+\boldsymbol{{H}}_{{ab}}\dot{\boldsymbol{{q}}}_{{oj\unicode{x03B4} }}+k_{{b}}q_{{b}}\right)-\boldsymbol{{N}}_{12}\!\left(\boldsymbol{{H}}_{{ba}}\dot{q}_{{b}}+\boldsymbol{{H}}_{{bb}}\dot{\boldsymbol{{q}}}_{{oj\unicode{x03B4} }}\right)+\boldsymbol{{N}}_{12}\!\left[\begin{array}{l} \tau _{0}\\[4pt] \boldsymbol{\tau }\\[4pt] -\boldsymbol{{K}}_{\unicode{x03B4} }\boldsymbol{\delta } \end{array}\right] ,\end{equation}

(11) \begin{equation} \ddot{\boldsymbol{{q}}}_{{oj\unicode{x03B4} }}=-\boldsymbol{{N}}_{21}\!\left(\boldsymbol{{H}}_{{aa}}\dot{q}_{{b}}+\boldsymbol{{H}}_{{ab}}\dot{\boldsymbol{{q}}}_{{oj\unicode{x03B4} }}+k_{{b}}q_{{b}}\right)-\boldsymbol{{N}}_{22}\!\left(\boldsymbol{{H}}_{{ba}}\dot{q}_{{b}}+\boldsymbol{{H}}_{{bb}}\dot{\boldsymbol{{q}}}_{{oj\unicode{x03B4} }}\right)+\boldsymbol{{N}}_{22}\!\left[\begin{array}{l} \tau _{0}\\[4pt] \boldsymbol{\tau }\\[4pt] -\boldsymbol{{K}}_{\unicode{x03B4} }\boldsymbol{\delta } \end{array}\right] .\end{equation}

Now the singular perturbation decomposition is performed for the system: treat $\boldsymbol{{q}}_{{oj\unicode{x03B4} }}$ as slow sub-variable, and treat $q_{{b}}$ and $\boldsymbol{\sigma }$ as fast sub-variables. $\mu =1/\min \!\left(k_{{b}},k_{{m}1},k_{{m}2}\right)$ is defined as the singular perturbation factor, and $t_{{f}}=t/\sqrt{\mu }$ is the fast time scale, where the equivalent stiffness coefficients of the base and the joints at this time scale are $k_{{bf}}$ and $\boldsymbol{{K}}_{{mf}}$ , respectively, and the flexible deformations are $q_{{bf}}$ and $\boldsymbol{\sigma }_{{f}}$ , respectively, where $k_{{bf}}=\mu k_{{b}}, \boldsymbol{{K}}_{{mf}}=\mu \boldsymbol{{K}}_{{m}}, q_{{bf}}=q_{{b}}/\mu$ , and $\boldsymbol{\sigma }_{\mathrm{f}}=\boldsymbol{\sigma }/\mu$ . The following relation exists

(12) \begin{equation} k_{{b}}q_{{b}}=\frac{k_{{bf}}}{\mu }q_{{bf}}\mu =k_{{bf}}q_{{bf}} ,\end{equation}

(13) \begin{equation} \boldsymbol{{K}}_{{m}}\boldsymbol{\sigma }=\frac{\boldsymbol{{K}}_{{mf}}}{\mu }\boldsymbol{\sigma }_{{f}}\mu =\boldsymbol{{K}}_{{mf}}\boldsymbol{\sigma }_{{f}} .\end{equation}

By substituting $q_{{bf}}=q_{{b}}/\mu$ , Eqs. (12) and (13) into Eqs. (10) and (11), we get

(14) \begin{equation} \mu \ddot{q}_{{bf}}=-\boldsymbol{{N}}_{11}\!\left(\boldsymbol{{H}}_{{aa}}\mu \dot{q}_{{bf}}+\boldsymbol{{H}}_{{ab}}\dot{\boldsymbol{{q}}}_{{oj\unicode{x03B4} }}+k_{{bf}}q_{{bf}}\right)-\boldsymbol{{N}}_{12}\!\left(\boldsymbol{{H}}_{{ba}}\mu \dot{q}_{{bf}}+\boldsymbol{{H}}_{{bb}}\dot{\boldsymbol{{q}}}_{{oj\unicode{x03B4} }}\right)+\boldsymbol{{N}}_{12}\!\left[\begin{array}{l} \tau _{0}\\[4pt] \boldsymbol{{K}}_{{mf}}\boldsymbol{\sigma }_{{f}}\\[4pt] -\boldsymbol{{K}}_{\unicode{x03B4} }\boldsymbol{\delta } \end{array}\right] ,\end{equation}

(15) \begin{equation} \ddot{\boldsymbol{{q}}}_{{oj\unicode{x03B4} }}=-\boldsymbol{{N}}_{21}\!\left(\boldsymbol{{H}}_{{aa}}\mu \dot{q}_{{bf}}+\boldsymbol{{H}}_{{ab}}\dot{\boldsymbol{{q}}}_{{oj\unicode{x03B4} }}+k_{{bf}}q_{{bf}}\right)-\boldsymbol{{N}}_{22}\!\left(\boldsymbol{{H}}_{{ba}}\mu \dot{q}_{{bf}}+\boldsymbol{{H}}_{{bb}}\dot{\boldsymbol{{q}}}_{{oj\unicode{x03B4} }}\right)+\boldsymbol{{N}}_{22}\!\left[\begin{array}{l} \tau _{0}\\[4pt] \boldsymbol{{K}}_{{mf}}\boldsymbol{\sigma }_{{f}}\\[4pt] -\boldsymbol{{K}}_{\unicode{x03B4} }\boldsymbol{\delta } \end{array}\right] .\end{equation}

Consider the case when $\mu$ tends to 0, $k_{{b}}$ and $\boldsymbol{{K}}_{{m}}$ tend to infinity and $q_{{b}}$ tends to 0, we can see that the joint rotation angle $\boldsymbol{{q}}_{\mathrm{j}}=\left[\begin{array}{l@{\quad}l} \boldsymbol{{q}}_{1} & \boldsymbol{{q}}_{2} \end{array}\right]^{{T}}$ is approximately equal to the motor rotor rotation angle $\boldsymbol{{q}}_{{m}}$ . Thus, when $\mu =0$ , the robot can be considered as a rigid system. Let $\mathbf{\Re }$ be an arbitrary variable and $\overline{\mathbf{\Re }}$ denotes the new expression for $\mathbf{\Re }$ as $\mu$ tends to 0. To find the dynamics model of the slow subsystem, let $\mu =0$ , from Eqs. (14) and (15) we have

(16) \begin{equation} 0=-\overline{\boldsymbol{{N}}}_{11}\overline{\boldsymbol{{H}}}_{{ab}}\dot{\boldsymbol{{q}}}_{{oj\unicode{x03B4} }}-\overline{\boldsymbol{{N}}}_{11}k_{{bf}}\overline{q}_{{bf}}-\overline{\boldsymbol{{N}}}_{12}\overline{\boldsymbol{{H}}}_{{bb}}\dot{\boldsymbol{{q}}}_{{oj\unicode{x03B4} }}+\overline{\boldsymbol{{N}}}_{12}\!\left[\begin{array}{l} \tau _{0}\\[4pt] \boldsymbol{{K}}_{{mf}}\overline{\boldsymbol{\sigma }}_{\mathrm{f}}\\[4pt] -\boldsymbol{{K}}_{\unicode{x03B4} }\boldsymbol{\delta } \end{array}\right] ,\end{equation}

(17) \begin{equation} \ddot{\boldsymbol{{q}}}_{{oj\unicode{x03B4} }}=-\overline{\boldsymbol{{N}}}_{21}\overline{\boldsymbol{{H}}}_{{ab}}\dot{\boldsymbol{{q}}}_{{oj\unicode{x03B4} }}-\overline{\boldsymbol{{N}}}_{21}k_{{bf}}\overline{q}_{{bf}}-\overline{\boldsymbol{{N}}}_{22}\overline{\boldsymbol{{H}}}_{{bb}}\dot{\boldsymbol{{q}}}_{{oj\unicode{x03B4} }}+\overline{\boldsymbol{{N}}}_{22}\!\left[\begin{array}{l} \tau _{0}\\[4pt] \boldsymbol{{K}}_{{mf}}\overline{\boldsymbol{\sigma }}_{\mathrm{f}}\\[4pt] -\boldsymbol{{K}}_{\unicode{x03B4} }\boldsymbol{\delta } \end{array}\right] .\end{equation}

The total controller of the drive motor with joint flexibility compensation is designed as

(18) \begin{equation} \boldsymbol{\tau }_{{m}}=\left(\boldsymbol{{I}}+\boldsymbol{{K}}_{{c}}\right)\boldsymbol{\tau }_{{n}}-\boldsymbol{{K}}_{{c}}\boldsymbol{\tau } ,\end{equation}

where $\boldsymbol{{I}}\in \mathbf{R}^{2\times 2}$ is the unit matrix, $\boldsymbol{{K}}_{{c}}\in \mathbf{R}^{2\times 2}$ is the symmetric positive definite flexible compensation matrix and $\boldsymbol{\tau }_{{n}}=\boldsymbol{\tau }_{{ns}}+\boldsymbol{\tau }_{{nf}}$ is the controller to be designed. $\boldsymbol{\tau }_{{ns}}\in \mathbf{R}^{2\times 1}$ is the slow sub-controller, and $\boldsymbol{\tau }_{{nf}}\in \mathbf{R}^{2\times 1}$ is the fast sub-controller.

By substituting controller $\boldsymbol{\tau }_{{m}}, \boldsymbol{\sigma }_{{f}}=\boldsymbol{\sigma }/\mu$ and $\mu =0$ into Eq. (7), we get

(19) \begin{equation} \boldsymbol{{K}}_{{mf}}\overline{\boldsymbol{\sigma }}_{{f}}=\boldsymbol{\tau }_{{ns}}-\left(\boldsymbol{{I}}+\boldsymbol{{K}}_{c}\right)^{-1}\boldsymbol{{J}}_{{m}}\ddot{\boldsymbol{{q}}}_{\mathrm{j}} ,\end{equation}

By substituting Eqs. (16), (18) and (19) into Eq. (17), we have the dynamics model of the slow subsystem containing $\boldsymbol{{q}}$ and $\boldsymbol{\delta }$ as

(20) \begin{equation} \boldsymbol{{R}}\!\left(\boldsymbol{{q}}_{{oj\unicode{x03B4} }}\right)\ddot{\boldsymbol{{q}}}_{{oj\unicode{x03B4} }}+\boldsymbol{{S}}\!\left(\boldsymbol{{q}}_{{oj\unicode{x03B4} }},\dot{\boldsymbol{{q}}}_{{oj\unicode{x03B4} }}\right)\dot{\boldsymbol{{q}}}_{{oj\unicode{x03B4} }}+\boldsymbol{\xi }_{{oj\unicode{x03B4} }}=\boldsymbol{\tau }_{{oj\unicode{x03B4} }} ,\end{equation}

where $\boldsymbol{{R}}\!\left(\boldsymbol{{q}}_{{oj\unicode{x03B4} }}\right)=\overline{\boldsymbol{{D}}}_{{bb}}+\left[\begin{array}{l@{\quad}l@{\quad}l} 0 & \mathbf{0} & \mathbf{0}\\[4pt] \mathbf{0} & \left(\boldsymbol{{I}}+\boldsymbol{{K}}_{{c}}\right)^{-1}\boldsymbol{{J}}_{{m}} & \mathbf{0}\\[4pt] \mathbf{0} & \mathbf{0} & \mathbf{0} \end{array}\right], \boldsymbol{{S}}\!\left(\boldsymbol{{q}}_{{oj\unicode{x03B4} }},\dot{\boldsymbol{{q}}}_{{oj\unicode{x03B4} }}\right)=\overline{\boldsymbol{{H}}}_{{bb}}, \boldsymbol{\xi }_{{oj\unicode{x03B4} }}=\left[\begin{array}{l@{\quad}l} \mathbf{0}^{{T}} & \left(\boldsymbol{{K}}_{\unicode{x03B4} }\boldsymbol{\delta }\right)^{{T}} \end{array}\right]^{{T}}, $ $ \boldsymbol{\tau }_{{oj\unicode{x03B4} }}=\left[\begin{array}{l@{\quad}l} \left(\boldsymbol{\tau }_{{oj}}\right)^{{T}} & \mathbf{0}^{{T}} \end{array}\right]^{{T}}$ , and $\boldsymbol{\tau }_{{oj}}=\left[\begin{array}{l@{\quad}l} \tau _{0} & \boldsymbol{\tau }_{{ns}}^{{T}} \end{array}\right]^{{T}}$ .

Denote the fast subsystem state variable as $\boldsymbol{{q}}_{{f}}=\left[\begin{array}{l@{\quad}l@{\quad}l@{\quad}l} q_{{f}1} & q_{{f}2} & \boldsymbol{{q}}_{{f}3}^{{T}} & \boldsymbol{{q}}_{{f}4}^{{T}} \end{array}\right]^{{T}}$ , and we have $q_{{f}1}=q_{{bf}}-\overline{q}_{{bf}}, q_{{f}2}=\sqrt{\mu }\dot{q}_{{bf}}, \boldsymbol{{q}}_{{f}3}=\boldsymbol{\sigma }_{{f}}-\overline{\boldsymbol{\sigma }}_{{f}}$ and $\boldsymbol{{q}}_{{f}4}=\sqrt{\mu }\dot{\boldsymbol{\sigma }}_{{f}}$ . $\boldsymbol{{q}}_{{f}}$ is derived with respect to $t_{{f}}$ , and the dynamics model of the fast subsystem containing $q_{{b}}$ and $\boldsymbol{\sigma }$ is obtained by combining Eq. (14), controllers $\boldsymbol{\tau }_{{m}}, \boldsymbol{\sigma }_{{f}}=\boldsymbol{\sigma }/\mu$ and Eq. (7), as

(21) \begin{equation} \mathrm{d}\boldsymbol{{q}}_{{f}}/\mathrm{d}t_{{f}}=\boldsymbol{{A}}_{{f}}\boldsymbol{{q}}_{{f}}+\boldsymbol{{B}}_{{f}}\boldsymbol{\tau }_{{nf}} ,\end{equation}

where $\boldsymbol{{A}}_{{f}}=\left[\begin{array}{l@{\quad}l@{\quad}l@{\quad}l} 0 & 1 & \mathbf{0} & \mathbf{0}\\[4pt] -\overline{\boldsymbol{{N}}}_{11}k_{{bf}} & 0 & \boldsymbol{{N}}_{12}^{\ast }\boldsymbol{{K}}_{{mf}} & \mathbf{0}\\[4pt] \mathbf{0} & \mathbf{0} & \mathbf{0} & \boldsymbol{{I}}\\[4pt] \mathbf{0} & \mathbf{0} & -\boldsymbol{{J}}_{{m}}^{-1}\!\left(\boldsymbol{{I}}+\boldsymbol{{K}}_{{c}}\right)\boldsymbol{{K}}_{{mf}} & \mathbf{0} \end{array}\right], \boldsymbol{{B}}_{{f}}=\left[\begin{array}{c} \mathbf{0}\\[4pt] \mathbf{0}\\[4pt] \mathbf{0}\\[4pt] \boldsymbol{{J}}_{{m}}^{-1}\!\left(\boldsymbol{{I}}+\boldsymbol{{K}}_{{c}}\right) \end{array}\right], \boldsymbol{{N}}_{12}^{\ast }$ is the row vector consisting of the elements of the second and third terms of $\overline{\boldsymbol{{N}}}_{12}$ .

3.2. Dynamical model properties

From Eq. (20), it can be seen that the motion of the base and joints of the slow subsystem and the links vibration are coupled with each other. In order to control the base attitude and the joints, this section decouples the slow subsystem into a fully driven rigid subsystem and designs the control algorithm. Among them, $\boldsymbol{{R}}\!\left(\boldsymbol{{q}}_{{oj\unicode{x03B4} }}\right)$ and $\boldsymbol{{S}}\!\left(\boldsymbol{{q}}_{{oj\unicode{x03B4} }},\dot{\boldsymbol{{q}}}_{{oj\unicode{x03B4} }}\right)$ are written as $\left[\begin{array}{l@{\quad}l} \boldsymbol{{R}}_{11} & \boldsymbol{{R}}_{12}\\[4pt] \boldsymbol{{R}}_{21} & \boldsymbol{{R}}_{22} \end{array}\right]$ and $\left[\begin{array}{l@{\quad}l} \boldsymbol{{S}}_{11} & \boldsymbol{{S}}_{12}\\[4pt] \boldsymbol{{S}}_{21} & \boldsymbol{{S}}_{22} \end{array}\right]$ , respectively. $\boldsymbol{{R}}_{11}, \boldsymbol{{S}}_{11}\in \mathrm{\mathbb{R}}^{3\times 3}, \boldsymbol{{R}}_{12}, \boldsymbol{{S}}_{12}\in \mathrm{\mathbb{R}}^{3\times 4}, \boldsymbol{{R}}_{21}, \boldsymbol{{S}}_{21}\in \mathrm{\mathbb{R}}^{4\times 3}$ , and $\boldsymbol{{R}}_{22}, \boldsymbol{{S}}_{22}\in \mathrm{\mathbb{R}}^{4\times 4}$ . The fully driven rigid subsystem dynamic model is obtained from Eq. (20), as

(22) \begin{equation} \boldsymbol{{M}}\!\left(\boldsymbol{\delta },\boldsymbol{{q}}\right)\ddot{\boldsymbol{{q}}}+\boldsymbol{{C}}\!\left(\boldsymbol{\delta },\dot{\boldsymbol{\delta }},\boldsymbol{{q}}, \dot{\boldsymbol{{q}}}\right)\dot{\boldsymbol{{q}}}+\boldsymbol{\kappa }\!\left(\boldsymbol{\delta },\dot{\boldsymbol{\delta }},\boldsymbol{{q}},\dot{\boldsymbol{{q}}}\right)=\boldsymbol{\tau }_{{oj}} ,\end{equation}

where $\boldsymbol{{M}}=\boldsymbol{{R}}_{11}-\boldsymbol{{R}}_{12}\boldsymbol{{R}}_{22}^{-1}\boldsymbol{{R}}_{21}, \boldsymbol{{C}}=\boldsymbol{{S}}_{11}-\boldsymbol{{R}}_{12}\boldsymbol{{R}}_{22}^{-1}\boldsymbol{{S}}_{21}$ , and $\boldsymbol{\kappa }=\left(\boldsymbol{{S}}_{12}-\boldsymbol{{R}}_{12}\boldsymbol{{R}}_{22}^{-1}\boldsymbol{{S}}_{22}\right)\dot{\boldsymbol{\delta }}-\boldsymbol{{R}}_{12}\boldsymbol{{R}}_{22}^{-1}\boldsymbol{{K}}_{\unicode{x03B4} }\boldsymbol{\delta }$ is the system dynamics nonlinear term.

According to Kelly et al. [Reference Kelly, Santibanez and Loria30], Eq. (22) has several fundamental properties as follows:

Property 1. The inertia matrix $\boldsymbol{{M}}\!\left(\boldsymbol{{x}}\right)$ is symmetric positive definite and satisfies the following inequalities:

(23) \begin{equation} \lambda _{{m}}\!\left(\boldsymbol{{M}}\right)\left\| \boldsymbol{{y}}\right\| ^{2}\leq \boldsymbol{{y}}^{T}\boldsymbol{{M}}\!\left(\boldsymbol{{x}}\right)\boldsymbol{{y}}\leq \lambda _{{M}}\!\left(\boldsymbol{{M}}\right)\left\| \boldsymbol{{y}}\right\| ^{2}\forall \boldsymbol{{y}}\in \mathrm{\mathbb{R}}^{3\times 1} ,\end{equation}

where $\lambda _{{m}}\!\left(\cdot \right)$ and $\lambda _{{M}}\!\left(\cdot \right)$ represent the minimum and maximum eigenvalues of the matrix, respectively, and $\left\| \cdot \right\|$ represents the 2-norm.

Property 2. The centrifugal-Coriolis matrix $\boldsymbol{{C}}\in \mathrm{\mathbb{R}}^{3\times 3}$ satisfies the following relationship:

(24) \begin{equation} \boldsymbol{{C}}\!\left(\boldsymbol{{x}},\boldsymbol{{y}}\right)\boldsymbol{{z}}=\boldsymbol{{C}}\!\left(\boldsymbol{{x}},\boldsymbol{{z}}\right)\boldsymbol{{y}}\forall \boldsymbol{{y}},\boldsymbol{{z}}\in \mathrm{\mathbb{R}}^{3\times 1} ,\end{equation}

(25) \begin{equation} C_{{m}}\left\| \boldsymbol{{y}}\right\| \leq \left\| \boldsymbol{{C}}\!\left(\boldsymbol{{x}},\boldsymbol{{y}}\right)\right\| \leq C_{{M}}\left\| \boldsymbol{{y}}\right\| \forall \boldsymbol{{y}}\in \mathrm{\mathbb{R}}^{3\times 1} ,\end{equation}

where $C_{{m}}$ and $C_{{M}}$ are known positive constants.

Property 3. The matrix $\dot{\boldsymbol{{M}}}\!\left(\boldsymbol{{x}}\right)-2\boldsymbol{{C}}\!\left(\boldsymbol{{x}},\dot{\boldsymbol{{x}}}\right)$ is skew-symmetric, which implies that

(26) \begin{equation} \boldsymbol{{y}}^{{T}}\!\left[\dot{\boldsymbol{{M}}}\!\left(\boldsymbol{{x}}\right)-2\boldsymbol{{C}}\!\left(\boldsymbol{{x}},\dot{\boldsymbol{{x}}}\right)\right]\boldsymbol{{y}}=\mathbf{0}\forall \boldsymbol{{y}}\in \mathrm{\mathbb{R}}^{3\times 1} .\end{equation}

Property 4. There must exist known positive constants $k_{{M}}, k_{{C}1}, k_{{C}2}, k_{{C3}}$ and $k_{{C4}}$ for all $\boldsymbol{{x}},\boldsymbol{{y}},\boldsymbol{{z}},\boldsymbol{{x}}_{1},\boldsymbol{{y}}_{1},\boldsymbol{{x}}_{2},\boldsymbol{{y}}_{2}\in \mathrm{\mathbb{R}}^{3\times 1}$ fulfilling

(27) \begin{equation} \left\| \boldsymbol{{M}}\!\left(\boldsymbol{{x}}\right)\boldsymbol{{z}}-\boldsymbol{{M}}\!\left(\boldsymbol{{y}}\right)\boldsymbol{{z}}\right\| \leq k_{{M}}\left\| \textbf{Tanh}\!\left(\boldsymbol{{x}}-\boldsymbol{{y}}\right)\right\| \left\| \boldsymbol{{z}}\right\| ,\end{equation}

(28) \begin{equation} \left\| \boldsymbol{{C}}\!\left(\boldsymbol{{x}}_{1},\boldsymbol{{y}}_{1}\right)\boldsymbol{{z}}-\boldsymbol{{C}}\!\left(\boldsymbol{{x}}_{2},\boldsymbol{{y}}_{2}\right)\boldsymbol{{z}}\right\| \leq k_{{C}1}\left\| \textbf{Tanh}\!\left(\boldsymbol{{y}}_{1}-\boldsymbol{{y}}_{2}\right)\right\| \left\| \boldsymbol{{z}}\right\| +k_{{C}2}\left\| \boldsymbol{{y}}_{1}\right\| \left\| \textbf{Tanh}\!\left(\boldsymbol{{x}}_{1}-\boldsymbol{{x}}_{2}\right)\right\| \left\| \boldsymbol{{z}}\right\| ,\end{equation}

(29) \begin{equation} \left\| \boldsymbol{\kappa }\!\left(\boldsymbol{{x}}_{1},\boldsymbol{{y}}_{1}\right)-\boldsymbol{\kappa }\!\left(\boldsymbol{{x}}_{2},\boldsymbol{{y}}_{2}\right)\right\| \leq k_{{C3}}\left\| \boldsymbol{{y}}_{1}-\boldsymbol{{y}}_{2}\right\| +k_{{C4}}\left\| \boldsymbol{{y}}_{1}\right\| \left\| \textbf{Tanh}\!\left(\boldsymbol{{x}}_{1}-\boldsymbol{{x}}_{2}\right)\right\| ,\end{equation}

where $k_{{M}}, k_{{C}1}, k_{{C}2}, k_{{C3}}$ and $k_{\mathrm{C4}}$ are obtained according to Kelly. $\boldsymbol{\xi }=\left[\begin{array}{l@{\quad}l@{\quad}l} \xi _{1} & \xi _{2} & \xi _{3} \end{array}\right]^{{T}}\in \mathrm{\mathbb{R}}^{3\times 1}$ , and $\textbf{Tanh}\!\left(\boldsymbol{\xi }\right)=\left[\begin{array}{l@{\quad}l@{\quad}l} \tanh \!\left(\xi _{1}\right) & \tanh \!\left(\xi _{2}\right) & \tanh \!\left(\xi _{3}\right) \end{array}\right]^{{T}}$ , where $\boldsymbol{\xi }$ is an arbitrary vector.

4.1. SMC-IOR base on desired trajectory

The continuous differentiable periodic expectation trajectory satisfies the bounded condition

(30) \begin{equation} \left\| \dot{\boldsymbol{{q}}}_{{d}}\right\| \leq V_{{M}},\quad \textrm{and}\quad \left\| \ddot{\boldsymbol{{q}}}_{{d}}\right\| \leq A_{{M}} ,\end{equation}

where $V_{{M}}$ and $A_{{M}}$ are, respectively, the upper bound of the expected velocity and the acceleration norm.

For the convenience of analysis, define $\boldsymbol{{N}}\in \mathbf{R}^{3\times 1}$ as

(31) \begin{equation} \boldsymbol{{N}}\!\left(\boldsymbol{{q}},\dot{\boldsymbol{{q}}}\right)=\boldsymbol{{M}}\!\left(\boldsymbol{\delta },\boldsymbol{{q}}\right)\ddot{\boldsymbol{{q}}}_{{d}}+\boldsymbol{{C}}\!\left(\boldsymbol{\delta },\dot{\boldsymbol{\delta }},\boldsymbol{{q}}, \dot{\boldsymbol{{q}}}\right)\dot{\boldsymbol{{q}}}+\boldsymbol{\kappa }\!\left(\boldsymbol{\delta },\dot{\boldsymbol{\delta }},\boldsymbol{{q}},\dot{\boldsymbol{{q}}}\right) .\end{equation}

Let $\boldsymbol{{N}}_{{d}}=\boldsymbol{{N}}\!\left(\boldsymbol{{q}}_{{d}},\dot{\boldsymbol{{q}}}_{{d}}\right)$ , where $\boldsymbol{{N}}_{{d}}$ is a periodic function with period $T$ . Expanded $\boldsymbol{{N}}_{{d}}$ in Fourier series as

(32) \begin{equation} \boldsymbol{{N}}_{{d}}=\overline{\boldsymbol{{a}}}_{0}+\sum _{k=1}^{\infty }\!\left[\overline{\boldsymbol{{a}}}_{{k}}\cos \!\left(k\omega t\right)+\overline{\boldsymbol{{b}}}_{{k}}\sin \!\left(k\omega t\right)\right]\enspace ,\end{equation}

where $\omega =2\pi /T$ is the angular frequency. $\overline{\boldsymbol{{a}}}_{0}, \overline{\boldsymbol{{a}}}_{{k}}$ and $\overline{\boldsymbol{{b}}}_{{k}}$ are the unknown constant vectors.

Define the filtered tracking error $\boldsymbol{\eta }=\dot{\boldsymbol{{e}}}+\textbf{Tanh}\!\left(\boldsymbol{{e}}\right)+\textbf{Tanh}\!\left(\boldsymbol{{v}}\right)$ , where $\boldsymbol{{e}}=\boldsymbol{{q}}_{{d}}-\boldsymbol{{q}}$ is the trajectory tracking error and $\boldsymbol{{v}}\in \mathbf{R}^{3\times 1}$ is the filter vector. For the rigid subsystem Eq. (22), design the algorithm of sliding mode control with input restriction, output feedback and repetitive learning based on the desired trajectory (SMC-IOR-D) as

(33) \begin{equation} \boldsymbol{\tau }_{{oj}}=\boldsymbol{{K}}_{0}\textbf{Sgn}\!\left(\textbf{Tanh}\!\left(\boldsymbol{{e}}\right)+\alpha \boldsymbol{{e}}_{{f}}\right)-\left(k^{\prime}+1\right)\textbf{Tanh}\!\left(\boldsymbol{{v}}\right)+\textbf{Tanh}\!\left(\boldsymbol{{e}}\right)+\textbf{Sat}\!\left(\boldsymbol{{Q}}_{0}\boldsymbol{{z}}_{0}+\sum _{k=1}^{N}\boldsymbol{{Q}}_{k}\dot{\boldsymbol{{z}}}_{k}\right) ,\end{equation}

(34) \begin{equation} \left\{\begin{array}{l} \textbf{Tanh}\!\left(\boldsymbol{{v}}\right)=\boldsymbol{{p}}_{1}-\left(k^{\prime}+1\right)\boldsymbol{{e}}\\[4pt] \dot{\boldsymbol{{p}}}_{1}=-k^{\prime}\textbf{Tanh}\!\left(\boldsymbol{{e}}\right)-\left(k^{\prime}+2\right)\textbf{Tanh}\!\left(\boldsymbol{{v}}\right)-\alpha ^{-1}\boldsymbol{{e}}_{{f}}\\[4pt] \boldsymbol{{p}}_{1}\!\left(0\right)=\left(k^{\prime}+1\right)\boldsymbol{{e}}\!\left(0\right) \end{array}\right. ,\end{equation}

(35) \begin{equation} \left\{\begin{array}{l} \boldsymbol{{e}}_{{f}}=\boldsymbol{{p}}_{2}+\alpha ^{-1}\!\left(\boldsymbol{{e}}-\textbf{Tanh}\!\left(\boldsymbol{{e}}\right)\right)\\[4pt] \dot{\boldsymbol{{p}}}_{2}=\alpha ^{-1}\textbf{Tanh}\!\left(\boldsymbol{{v}}\right)-\boldsymbol{{e}}_{{f}}\\[4pt] \boldsymbol{{p}}_{2}\!\left(\mathrm{0}\right)=\alpha ^{-1}\!\left(\textbf{Tanh}\!\left(\boldsymbol{{e}}\!\left(0\right)\right)-\boldsymbol{{e}}\!\left(0\right)\right) \end{array}\right. ,\end{equation}

(36) \begin{equation} \left\{\begin{array}{l} \boldsymbol{{z}}_{0}=\boldsymbol{{p}}_{0}+\boldsymbol{{Q}}_{0}\boldsymbol{{e}}\\[4pt] \dot{\boldsymbol{{p}}}_{0}=\boldsymbol{{Q}}_{0}\!\left(\textbf{Tanh}\!\left(\boldsymbol{{e}}\right)+\textbf{Tanh}\!\left(\boldsymbol{{v}}\right)\right) \end{array}\right. ,\end{equation}

(37) \begin{equation} \left\{\begin{array}{l} \dot{\boldsymbol{{z}}}_{k}=\boldsymbol{{p}}_{k}+\boldsymbol{{Q}}_{k}\boldsymbol{{e}}\\[4pt] \dot{\boldsymbol{{p}}}_{k}=\boldsymbol{{Q}}_{k}\!\left(\textbf{Tanh}\!\left(\boldsymbol{{e}}\right)+\textbf{Tanh}\!\left(\boldsymbol{{v}}\right)\right)-k^{2}\omega ^{2}\boldsymbol{{z}}_{k}\enspace \end{array}\right.k=1,\cdots N ,\end{equation}

where $\boldsymbol{{K}}_{0}\in \mathrm{\mathbb{R}}^{3\times 3}$ is a constant diagonal matrix, $\boldsymbol{{e}}_{{f}}\in \mathrm{\mathbb{R}}^{3\times 1}$ is a filter vector and $\alpha$ and $k^{\prime}$ are positive constants. $\boldsymbol{{Q}}_{0},\boldsymbol{{Q}}_{k}\in \mathrm{\mathbb{R}}^{3\times 3}$ are constant positive definite diagonal matrices, $\boldsymbol{{Q}}_{0}\boldsymbol{{z}}_{0}+\sum _{k=1}^{N}\boldsymbol{{Q}}_{k}\dot{\boldsymbol{{z}}}_{k}\enspace \enspace$ is the repetitive learning controller (RC), and $N$ is the number of harmonic oscillator terms. $\textbf{Sgn}\!\left(\boldsymbol{\xi }\right)=\left[\begin{array}{l@{\quad}l@{\quad}l} \mathrm{sgn}\!\left(\xi _{1}\right) & \mathrm{sgn}\!\left(\xi _{2}\right) & \mathrm{sgn}\!\left(\xi _{3}\right) \end{array}\right]^{{T}}$ , $\textbf{Sat}\!\left(\boldsymbol{\xi }\right)=\left[\begin{array}{l@{\quad}l@{\quad}l} \mathrm{sat}\!\left(\xi _{1}\right) & \mathrm{sat}\!\left(\xi _{2}\right) & \mathrm{sat}\!\left(\xi _{3}\right) \end{array}\right]^{{T}}$ , $\boldsymbol{\xi }\in \mathrm{\mathbb{R}}^{3\times 1}$ is an arbitrary vector. $\textbf{Sat}\!\left(\cdot \right)$ is a saturation function whose elements are as follows:

(38) \begin{equation} \mathrm{sat}\!\left(\xi _{i^{\prime}}\right)=\left\{\begin{array}{l} \xi _{i^{\prime}},\quad\qquad \left| \xi _{i^{\prime}}\right| \leq \beta _{i^{\prime}}\\[5pt] \beta _{i^{\prime}}\mathrm{sgn}\!\left(\xi _{i^{\prime}}\right),\left| \xi _{i^{\prime}}\right| \gt \beta _{i^{\prime}} \end{array}\right. \, \forall \xi _{i^{\prime}}\in \mathrm{\mathbb{R}},i^{\prime}=1,2,3 ,\end{equation}

where $\beta _{i^{\prime}}$ is the maximum threshold value of the saturation function.

Take the derivative of $\boldsymbol{\eta }$ , left multiply both sides of the resultant equation by $\boldsymbol{{M}}\!\left(\boldsymbol{\delta },\boldsymbol{{q}}\right)$ and substituting $\boldsymbol{{e}}=\boldsymbol{{q}}_{{d}}-\boldsymbol{{q}}$ , Eqs. (22), (31) and (34) into the right side of the equation, then we have

(39) \begin{equation} \boldsymbol{{M}}\!\left(\boldsymbol{\delta },\boldsymbol{{q}}\right)\dot{\boldsymbol{\eta }}=-k^{\prime}\boldsymbol{{M}}\!\left(\boldsymbol{\delta },\boldsymbol{{q}}\right)\boldsymbol{\eta }-\boldsymbol{{M}}\!\left(\boldsymbol{\delta },\boldsymbol{{q}}\right)\left(\boldsymbol{{I}}_{3}-\textbf{Sec}\textbf{h}^{2}\!\left(\boldsymbol{{e}}\right)\right)\boldsymbol{\eta }+ {\tilde{{\boldsymbol{{N}}}}}+\boldsymbol{{N}}_{{d}}-\boldsymbol{\tau }_{{oj}}-\frac{1}{2}\dot{\boldsymbol{{M}}}\!\left(\boldsymbol{\delta },\boldsymbol{{q}}\right)\boldsymbol{\eta } ,\end{equation}

(40) \begin{align} {\tilde{{\boldsymbol{{N}}}}} & =\boldsymbol{{N}}-\boldsymbol{{N}}_{{d}}-\boldsymbol{{M}}\!\left(\boldsymbol{\delta },\boldsymbol{{q}}\right)\left[\textbf{Sec}\textbf{h}^{2}\!\left(\boldsymbol{{e}}\right)\left(\textbf{Tanh}\!\left(\boldsymbol{{e}}\right)+\right.\right.\left.\textbf{Tanh}\!\left(\boldsymbol{{v}}\right)\right)\left.+\textbf{Tanh}\!\left(\boldsymbol{{v}}\right)-\textbf{Tanh}\!\left(\boldsymbol{{e}}\right)+\alpha ^{-1}\boldsymbol{{e}}_{{f}}\right]\nonumber\\& +\frac{1}{2}\dot{\boldsymbol{{M}}}\!\left(\boldsymbol{\delta },\boldsymbol{{q}}\right)\boldsymbol{\eta } .\end{align}

Convert Eq. (40) into the normal form, and according to $\boldsymbol{{e}}=\boldsymbol{{q}}_{{d}}-\boldsymbol{{q}}$ , Properties 1−4 and Eq. (30), it is obtained from the expression of $\boldsymbol{\eta }$ that

(41) \begin{equation} \left\| {\tilde{{\boldsymbol{{N}}}}}\right\| \leq 4\lambda _{{M1}}\left\| \boldsymbol{{z}}\right\| +\lambda _{{M2}}\left\| \boldsymbol{{z}}\right\| ^{2}=\left(4\lambda _{{M1}}+\lambda _{{M2}}\left\| \boldsymbol{{z}}\right\| \right)\left\| \boldsymbol{{z}}\right\| =\rho \!\left(\left\| \boldsymbol{{z}}\right\| \right)\left\| \boldsymbol{{z}}\right\| ,\end{equation}

where $\lambda _{{M1}}=\max \!\left\{\wp _{1},\wp _{2},\wp _{3},\wp _{4}\right\}, \lambda _{{M2}}=8C_{{M}}, \boldsymbol{{z}}=\left[\begin{array}{l@{\quad}l@{\quad}l@{\quad}l} \textbf{Tan}\textbf{h}^{{T}}\!\left(\boldsymbol{{e}}\right) & \boldsymbol{{e}}_{{f}}^{{T}} & \textbf{Tan}\textbf{h}^{{T}}\!\left(\boldsymbol{{v}}\right) & \boldsymbol{\eta }^{{T}} \end{array}\right]^{{T}}$ . $\wp _{1}=k_{{M}}A_{{M}}+$ $k_{{C}2}V_{{M}}^{2}+2C_{{M}}V_{{M}}+k_{{C4}}V_{{M}}+\lambda _{{M}}\!\left(\boldsymbol{{M}}\right)+k_{{C3}}, \wp _{2}=2C_{{M}}V_{{M}}+\frac{1}{2}\lambda _{{M}}\!\left(\dot{\boldsymbol{{M}}}\right)+k_{{C3}}, \wp _{3}=\frac{\lambda _{{M}}\left(\boldsymbol{{M}}\right)}{\alpha }, \wp _{4}=2C_{{M}}V_{{M}}+$ $\lambda _{{M}}\!\left(\boldsymbol{{M}}\right)+k_{{C3}}$ . $\rho \!\left(\left\| \boldsymbol{{z}}\right\| \right)\in \mathrm{\mathbb{R}}$ is the positive definite invertible non-decreasing function.

Lemma 1. Define the auxiliary intermediate variable $L_{n}(t)\in \mathrm{\mathbb{R}}$ as shown below

(42) \begin{equation} L_{n}(t)=\boldsymbol{\eta }^{{T}}\!\left[\boldsymbol{{f}}_{n}(t)-\boldsymbol{{K}}_{0}\textbf{sgn}\!\left(\textbf{Tanh}\!\left(\boldsymbol{{e}}\right)+\alpha \boldsymbol{{e}}_{{f}}\right)\right] ,\end{equation}

If $K_{0ii}$ satisfies sufficient $K_{0ii}\gt \left\| \boldsymbol{{f}}_{n}(t)\right\| _{\infty }+\left\| \dot{\boldsymbol{{f}}}_{n}(t)\right\| _{\infty }$ , then

(43) \begin{equation} \int _{0}^{t}L_{n}\!\left(\tau \right)\mathrm{d}\tau \leq \xi _{n} ,\end{equation}

where $K_{0ii}$ denotes the i-th element on the diagonal of the diagonal matrix $\boldsymbol{{K}}_{0}$ . $\boldsymbol{{K}}_{0ii}=\left[\begin{array}{l@{\quad}l@{\quad}l} K_{011} & K_{022} & K_{033} \end{array}\right]^{{T}}, \boldsymbol{{f}}_{n}(t),\dot{\boldsymbol{{f}}}_{n}(t)\in \boldsymbol{{l}}_{\infty }$ , and $\xi _{n}=\sum _{i=1}^{n}K_{0ii}\!\left| \tanh \!\left(e_{i}\!\left(0\right)\right)\right| -\textbf{Tan}\textbf{h}^{{T}}\!\left(\boldsymbol{{e}}\!\left(0\right)\right)\boldsymbol{{f}}_{n}\!\left(0\right), n=1,2$ .

Thus, the stability theorem for SMC-IOR-D is formulated as follows:

Theorem 1. When Eqs. (33)−(37) of SMC-IOR-D are under the input constraints from Eq. (44), if the controller parameters meet the sufficient conditions shown in Eqs. (45)−(48), the asymptotic stable trajectory tracking control of the FBFLFJ space robot can be realized

(44) \begin{equation} K_{0ii}+k^{\prime}+\mathrm{2}+\beta _{i}\leq \tau _{{oj},i,\max } ,\end{equation}

(45) \begin{equation} K_{0ii}\gt \max \!\left\{\left\| \boldsymbol{{f}}_{1}\left(t\right)\right\| _{\infty }+\left\| \dot{\boldsymbol{{f}}}_{1}(t)\right\| _{\infty },\left\| \boldsymbol{{f}}_{2}(t)\right\| _{\infty }+\left\| \dot{\boldsymbol{{f}}}_{2}(t)\right\| _{\infty }\right\} ,\end{equation}

(46) \begin{equation} \alpha \gt 1.5 ,\end{equation}

(47) \begin{equation} k^{\prime}=\left(k_{{n}}+1\right)/\lambda _{{m}}\!\left(\boldsymbol{{M}}\right) ,\end{equation}

(48) \begin{equation} k_{n}\gt \frac{1}{4\lambda _{3}}\max \!\left\{\rho ^{2}\!\left(\sqrt{\lambda _{12}/\lambda _{11}}\left\| \boldsymbol{{y}}_{1}\!\left(0\right)\right\| \right),\rho ^{2}\!\left(\sqrt{\lambda _{22}/\lambda _{21}}\left\| \boldsymbol{{y}}_{2}\!\left(0\right)\right\| \right),\rho ^{2}\!\left(\left\| \boldsymbol{{z}}\!\left(0\right)\right\| \right)\right\} ,\end{equation}

where $\lambda _{11}=0.5\min \!\left\{1, \lambda _{{m}}\!\left(\boldsymbol{{M}}\right)\right\}, \lambda _{12}=\max \!\left\{1, \frac{1}{2}\lambda _{{M}}\!\left(\boldsymbol{{M}}\right)\right\}, \lambda _{21}=\frac{1}{2}\min \!\left\{1,\lambda _{{m}}\!\left(\boldsymbol{{Q}}_{0}\right),\omega ^{2},\lambda _{{m}}\!\left(\boldsymbol{{M}}\right)\right\},$ $\lambda _{22}=\frac{1}{2}\max \!\left\{2,\lambda _{{M}}\!\left(\boldsymbol{{Q}}_{0}\right),N^{2}\omega ^{2},\lambda _{{M}}\!\left(\boldsymbol{{M}}\right)\right\}, \lambda _{3}=1-\frac{3}{2\alpha }, \boldsymbol{{f}}_{1}(t)\boldsymbol{=}\boldsymbol{{N}}_{{d}}-\boldsymbol{{B}}\cdot \textbf{sgn}\!\left(\boldsymbol{{Q}}_{0}\boldsymbol{{z}}_{0}+\sum _{k=1}^{N}\boldsymbol{{Q}}_{k}\dot{\boldsymbol{{z}}}_{k}\right)$ , and $\boldsymbol{{f}}_{2}(t)=\sum _{k=N+1}^{\infty }\left[\overline{\boldsymbol{{a}}}_{k}\cos \!\left(k\omega t\right)+\overline{\boldsymbol{{b}}}_{k}\sin \!\left(k\omega t\right)\right]$ .

4.2. SMC-IOR based on hybrid trajectory

The SMC-IOR-D in the previous section achieves asymptotic stable tracking of the base attitude and joint angle, but the vibration of the flexible link is not suppressed. In this section, we use the concept of virtual control force to modify the original desired trajectory and generate a hybrid trajectory $\boldsymbol{{q}}_{{h}}$ that reflects both the flexible vibration of the link and the rigid motion of the system. Based on this, the sliding mode control with input restriction, output feedback and repetitive learning based on the hybrid trajectory (SMC-IOR-H) is designed to achieve the dual functions of motion tracking and simultaneous suppression of the links vibration. The virtual force $\boldsymbol{{F}}\in \mathrm{\mathbb{R}}^{3\times 1}$ is introduced, and the error between the desired trajectory and the hybrid trajectory is $\boldsymbol{{e}}_{{h}}=\boldsymbol{{q}}_{{d}}-\boldsymbol{{q}}_{{h}}$ , which is generated by the second-order command generator $\ddot{\boldsymbol{{e}}}_{{h}}+\boldsymbol{{a}}\dot{\boldsymbol{{e}}}_{{h}}+\boldsymbol{{b}}\boldsymbol{{e}}_{{h}}=\boldsymbol{{F}}$ , where $\boldsymbol{{a}}, \boldsymbol{{b}}\in \mathrm{\mathbb{R}}^{3\times 3}$ are the constant positive definite diagonal matrix.

The hybrid error is defined as $\boldsymbol{{e}}_{{r}}=\boldsymbol{{q}}_{{h}}-\boldsymbol{{q}}$ , and the hybrid filter tracking error is defined as $\boldsymbol{\eta }_{{r}}=\dot{\boldsymbol{{e}}}_{{r}}+\textbf{Tanh}\!\left(\boldsymbol{{e}}_{{r}}\right)+\textbf{Tanh}\!\left(\boldsymbol{{v}}_{{r}}\right)$ . $\boldsymbol{{v}}_{{r}}$ and $\boldsymbol{{e}}_{{fr}}$ are the hybrid filter vectors. Then, the SMC-IOR-H is obtained as follows:

(49) \begin{equation} \left(\boldsymbol{\tau }_{{oj}}\right)_{{r}}=\boldsymbol{{K}}_{0}\textbf{Sgn}\!\left(\textbf{Tanh}\!\left(\boldsymbol{{e}}_{{r}}\right)+\alpha \boldsymbol{{e}}_{{fr}}\right)-\left(k^{\prime}+1\right)\textbf{Tanh}\!\left(\boldsymbol{{v}}_{{r}}\right)+\textbf{Tanh}\!\left(\boldsymbol{{e}}_{{r}}\right)+\textbf{Sat}\!\left(\boldsymbol{{Q}}_{0}\boldsymbol{{z}}_{0{r}}+\sum _{k=1}^{N}\boldsymbol{{Q}}_{k}\dot{\boldsymbol{{z}}}_{k{r}}\right) ,\end{equation}

(50) \begin{equation} \left\{\begin{array}{l} \textbf{Tanh}\!\left(\boldsymbol{{v}}_{{r}}\right)=\boldsymbol{{p}}_{1{r}}-\left(k^{\prime}+1\right)\boldsymbol{{e}}_{{r}}\\[4pt] \dot{\boldsymbol{{p}}}_{1{r}}=-k^{\prime}\textbf{Tanh}\!\left(\boldsymbol{{e}}_{{r}}\right)-\left(k^{\prime}+2\right)\textbf{Tanh}\!\left(\boldsymbol{{v}}_{{r}}\right)-\alpha ^{-1}\boldsymbol{{e}}_{{fr}}\\[4pt] \boldsymbol{{p}}_{1{r}}\!\left(0\right)=\left(k^{\prime}+1\right)\boldsymbol{{e}}_{{r}}\!\left(0\right) \end{array}\right. ,\end{equation}

(51) \begin{equation} \left\{\begin{array}{l} \boldsymbol{{e}}_{{fr}}=\boldsymbol{{p}}_{2{r}}+\alpha ^{-1}\!\left(\boldsymbol{{e}}_{{r}}-\textbf{Tanh}\!\left(\boldsymbol{{e}}_{{r}}\right)\right)\\[4pt] \dot{\boldsymbol{{p}}}_{2{r}}=\alpha ^{-1}\textbf{Tanh}\!\left(\boldsymbol{{v}}_{{r}}\right)-\boldsymbol{{e}}_{{fr}}\\[4pt] \boldsymbol{{p}}_{2{r}}\!\left(\mathrm{0}\right)=\alpha ^{-1}\!\left(\textbf{Tanh}\!\left(\boldsymbol{{e}}_{{r}}\!\left(0\right)\right)-\boldsymbol{{e}}_{{r}}\!\left(0\right)\right) \end{array}\right. ,\end{equation}

(52) \begin{equation} \left\{\begin{array}{l} \boldsymbol{{z}}_{0{r}}=\boldsymbol{{p}}_{0{r}}+\boldsymbol{{Q}}_{0}\boldsymbol{{e}}_{{r}}\\[4pt] \dot{\boldsymbol{{p}}}_{0{r}}=\boldsymbol{{Q}}_{0}\!\left(\textbf{Tanh}\!\left(\boldsymbol{{e}}_{{r}}\right)+\textbf{Tanh}\!\left(\boldsymbol{{v}}_{{r}}\right)\right) \end{array}\right. ,\end{equation}

(53) \begin{equation} \left\{\begin{array}{l} \dot{\boldsymbol{{z}}}_{k{r}}=\boldsymbol{{p}}_{k\mathrm{r}}+\boldsymbol{{Q}}_{k}\boldsymbol{{e}}_{{r}}\enspace \\[4pt] \dot{\boldsymbol{{p}}}_{k{r}}=\boldsymbol{{Q}}_{k}\!\left(\textbf{Tanh}\!\left(\boldsymbol{{e}}_{{r}}\right)+\textbf{Tanh}\!\left(\boldsymbol{{v}}_{{r}}\right)\right)-k^{2}\omega ^{2}\boldsymbol{{z}}_{k{r}}\enspace \end{array}\right.k=1,\cdots N .\end{equation}

Defining the state variable as $\boldsymbol{{q}}_{{s}}=\left[\begin{array}{l@{\quad}l@{\quad}l@{\quad}l} \boldsymbol{\delta }^{{T}} & \boldsymbol{{e}}^{{T}} & \dot{\boldsymbol{\delta }}^{{T}} & \dot{\boldsymbol{{e}}}^{{T}} \end{array}\right]^{{T}}$ , the equation of state is obtained from Eqs. (49)–(53) and Eq. (22) as follows:

(54) \begin{equation} \dot{\boldsymbol{{q}}}_{{s}}=\boldsymbol{{A}}_{{s}}\boldsymbol{{q}}_{{s}}+\boldsymbol{{B}}_{{s}}\boldsymbol{{F}}+\boldsymbol{{L}}_{{s}} ,\end{equation}

where $\boldsymbol{{A}}_{{s}}=\left[\begin{array}{l@{\quad}l@{\quad}l@{\quad}l} \mathbf{0} & \mathbf{0} & \boldsymbol{{I}} & \mathbf{0}\\[4pt] \mathbf{0} & \mathbf{0} & \mathbf{0} & \boldsymbol{{I}}\\[4pt] -\boldsymbol{{R}}_{22}^{-1}\boldsymbol{{K}}_{\unicode{x03B4} } & -\boldsymbol{{R}}_{22}^{-1}\boldsymbol{{R}}_{21}\boldsymbol{{b}} & -\boldsymbol{{R}}_{22}^{-1}\boldsymbol{{S}}_{22} & -\boldsymbol{{R}}_{22}^{-1}\boldsymbol{{R}}_{21}\boldsymbol{{a}}\\[4pt] \mathbf{0} & -\boldsymbol{{b}} & \mathbf{0} & -\boldsymbol{{a}} \end{array}\right], \boldsymbol{{B}}_{{s}}=\left[\begin{array}{l} \mathbf{0}\\[4pt] \mathbf{0}\\[4pt] \boldsymbol{{R}}_{22}\boldsymbol{{R}}_{21}\\[4pt] \boldsymbol{{I}} \end{array}\right]$ , and $\boldsymbol{{L}}_{{s}}=\left[\begin{array}{l} \mathbf{0}\\[4pt] \mathbf{0}\\[4pt] \boldsymbol{{R}}_{22}^{-1}\boldsymbol{{R}}_{21}\!\left(\boldsymbol{{G}}-\ddot{\boldsymbol{{q}}}_{{d}}\right)-\boldsymbol{{R}}_{22}^{-1}\boldsymbol{{S}}_{21}\dot{\boldsymbol{{q}}}\\[4pt] \boldsymbol{{G}} \end{array}\right]$ .

If $\boldsymbol{{L}}_{{s}}$ satisfies the following conditions:

(55) \begin{equation} \frac{\left\| \boldsymbol{{L}}_{{s}}(t)\right\| _{{E}}}{\left\| \boldsymbol{{q}}_{{s}}\right\| _{{E}}}\leq \frac{1}{2\left\| \boldsymbol{{N}}_{{s}}^{-1}\right\| _{{s}}\left\| \boldsymbol{{P}}_{{s}}\right\| _{{s}}}=\frac{\lambda _{\min }\left(\boldsymbol{{N}}_{{s}}\right)}{2\lambda _{\max }\left(\boldsymbol{{P}}_{{s}}\right)},\forall \boldsymbol{{q}}_{{s}}\in \mathbf{R}^{3\times 1} ,\end{equation}

then the virtual force $\boldsymbol{{F}}$ is

(56) \begin{equation} \boldsymbol{{F}}=-\boldsymbol{{R}}_{{s}}^{-1}\boldsymbol{{B}}_{{s}}^{{T}}\boldsymbol{{P}}_{{s}} \boldsymbol{{q}}_{{s}} .\end{equation}

In Eq. (56), $\boldsymbol{{R}}_{{s}}\gt \mathbf{0}$ , and $\boldsymbol{{P}}_{{s}}$ is the solution of Riccati equation $\boldsymbol{{P}}_{{s}}\boldsymbol{{A}}_{{s}}+\boldsymbol{{A}}_{{s}}^{{T}}\boldsymbol{{P}}_{\mathrm{s}}-\boldsymbol{{P}}_{{s}}\boldsymbol{{B}}_{\mathrm{s}}\boldsymbol{{R}}_{{s}}^{-1}\boldsymbol{{B}}_{{s}}^{{T}}\boldsymbol{{P}}_{{s}}+\boldsymbol{{Q}}_{{s}}=\mathbf{0}$ . Constructing the Lyapunov function $V\!\left(\boldsymbol{{q}}_{{s}}\right)=\boldsymbol{{q}}_{{s}}^{{T}}\boldsymbol{{P}}_{{s}} \boldsymbol{{q}}_{{s}}, V\!\left(\boldsymbol{{q}}_{{s}}\right)$ is derived and combining with Eqs. (55) and (56) yields $\dot{V}\!\left(\boldsymbol{{q}}_{{s}}\right)\leq -\left(\boldsymbol{{q}}_{{s}}^{{T}}\boldsymbol{{N}}_{{s}}\boldsymbol{{q}}_{{s}}-\lambda _{\min }\left(\boldsymbol{{N}}_{{s}}\right)\left\| \boldsymbol{{q}}_{{s}}\right\| ^{2}\right)\leq 0$ , so the system is stable.

4.3. Multiple flexible suppression control based on linear quadratic optimal method

As analyzed above, the state equation $\boldsymbol{{q}}_{{f}}$ at the fast time scale $t_{{f}}$ is shown in Eq. (21). The performance index function is constructed as

(57) \begin{equation} J_{{qf}}=0.5\int _{0}^{\infty }\left(\boldsymbol{{q}}_{{f}}^{{T}}\boldsymbol{{Q}}_{{f}}\boldsymbol{{q}}_{{f}}+\boldsymbol{\tau }_{{nf}}^{{T}}\boldsymbol{{R}}_{{f}}\boldsymbol{\tau }_{{nf}}\right)\mathrm{d}t_{{f}} .\end{equation}

In Eq. (57), symmetric matrix $\boldsymbol{{Q}}_{{f}}\geq \mathbf{0}, \boldsymbol{{R}}_{{f}}\gt \mathbf{0}$ . The optimal control of the fast subsystem is designed as

(58) \begin{equation} \boldsymbol{\tau }_{{nf}}=-\boldsymbol{{R}}_{{f}}^{-1}\boldsymbol{{B}}_{{f}}^{{T}}\boldsymbol{{P}}_{{f}}\boldsymbol{{q}}_{{f}} ,\end{equation}

where $\boldsymbol{{P}}_{{f}}$ is the solution of the Riccati equation $\boldsymbol{{P}}_{{f}}\boldsymbol{{A}}_{{f}}+\boldsymbol{{A}}_{{f}}^{{T}}\boldsymbol{{P}}_{{f}}-\boldsymbol{{P}}_{{f}}\boldsymbol{{B}}_{{f}}\boldsymbol{{R}}_{{f}}^{-1}\boldsymbol{{B}}_{{f}}^{{T}}\boldsymbol{{P}}_{{f}}+\boldsymbol{{Q}}_{{f}}=\mathbf{0}$ .

Then, ISMC-IOR is composed of SMC-IOR-H, virtual force Eq. (56), fast sub-controller Eq. (58) and motor general controller Eq. (18). The overall scheme of this controller is shown in flowchart Fig. 4, where the blue dashed area is the fast subsystem controller and the red dashed area is the slow subsystem controller. According to the above analysis, the implementation of the ISMC-IOR does not need the dynamic model information of space robot and the motion speed information of base and joint, and can meet the requirements of limited driving torque of FBFLFJ space robot.

Figure 4.

Integrated sliding mode control with input restriction, output feedback and repetitive learning.