## 3 System Dynamics Analysis
### System Dynamics Modeling  
After the flexible space robot captures the satellite, it will form a system, and the speed of the space robot end effector and the satellite are equal, that is, \(\mathbf{V}_{\mathrm{P}}=\mathbf{V}_{\mathrm{P^{\prime}}}\). Taking the differentiation of that equation, we have:  
\[\ddot{\mathbf{q}}_{3}=\mathbf{I}_{3}^{-1}\dot{\mathbf{q}}_{z}+\mathbf{J}_{3}^{-1}\left(\dot{ \mathbf{J}}-\dot{\mathbf{J}}_{3}\mathbf{J}_{3}^{-1}\mathbf{J}\right)\dot{\mathbf{q}}_{z} \tag{11}\]  
Substituting Equation (11) into the second equation of Equation (6), we have \(\mathbf{F}_{\mathrm{P^{\prime}}}\). Substituting \(\mathbf{F}_{\mathrm{P}}\) into first equation of Equation (6), we have:  
\[\mathbf{D}_{z}\ddot{\mathbf{q}}_{z}+\mathbf{H}_{z}\dot{\mathbf{q}}_{z}+\left[ \begin{array}{c} \mathbf{0}_{2\times 1}\\ k_{\mathrm{b}}x_{\mathrm{b}}\\ 0\\ -\mathbf{K}_{\mathrm{m}}\mathbf{\sigma}\\ K_{\mathrm{S}}\mathbf{\delta}\\ K_{\mathrm{m}}\mathbf{\sigma}\\ \end{array} \right]=\left[ \begin{array}{c}\mathbf{0}_{3\times 1}\\ \tau_{0}\\ \mathbf{0}_{6\times 1}\\ \tau_{\mathrm{m}}\\ \end{array} \right] \tag{12}\]where \(\mathbf{D}_{z}=\mathbf{D}_{\rm N}+\mathbf{\mathsf{J}}^{\rm T}\big{(}\mathbf{\mathsf{J}}_{3}^{\rm T} \big{)}^{-1}\mathbf{D}_{3}\mathbf{\mathsf{J}}_{3}^{-1}\mathbf{J}\) and \(\mathbf{H}_{z}=\mathbf{H}_{\rm N}+\mathbf{\mathsf{J}}^{\rm T}\big{(}\mathbf{\mathsf{J}}_{3}^{\rm T }\big{)}^{-1}\mathbf{D}_{3}\mathbf{\mathsf{J}}_{3}^{-1}\Big{(}\mathbf{J}-\mathbf{\mathsf{J}}_{3 }\mathbf{\mathsf{J}}_{3}^{-1}\mathbf{J}\Big{)}\). For the convenience of later analysis, \(\mathbf{D}_{z}\) and \(\mathbf{H}_{z}\) are written in the form of a block matrix, as follows:  
\[\mathbf{D}_{z}=\left[ \begin{array}{cc}\mathbf{D}_{z11}&\mathbf{D}_{z12}&\mathbf{0}_{z\times 2 }\\ \mathbf{D}_{z21}&\mathbf{D}_{z22}&\mathbf{0}_{z\times 2}\\ \mathbf{0}_{2\times 2}&\mathbf{0}_{2\times 8}&\mathbf{J}_{\rm m}\end{array} \right],\mathbf{H}_{z}= \left[ \begin{array}{cc}\mathbf{0}_{2\times 2}&\mathbf{H}_{z12}&\mathbf{0}_{2\times 2}\\ \mathbf{0}_{8\times 2}&\mathbf{H}_{z22}&\mathbf{0}_{8\times 2}\\ \mathbf{0}_{2\times 2}&\mathbf{0}_{2\times 8}&\mathbf{0}_{2\times 2}\end{array} \right].\]  
Based on controller design needs, Equation (12) is rewritten as:  
\[\left[ \begin{array}{cc}\mathbf{D}_{z11}&\mathbf{D}_{z12}\\ \mathbf{D}_{z21}&\mathbf{D}_{z22}\end{array} \right]\left[ \begin{array}{c}\ddot{ \mathbf{q}}_{\rm b}\\ \ddot{\mathbf{x}}_{\rm b}\\ \ddot{\mathbf{q}}\\ \ddot{\mathbf{\delta}}\end{array} \right]+\left[ \begin{array}{cc}\mathbf{0}_{2\times 2 }&\mathbf{H}_{z12}\\ \mathbf{0}_{8\times 2}&\mathbf{H}_{z22}\end{array} \right]\left[ \begin{array}{c}\dot\mathbf{{q}}_{\rm b}\\ \dot{\mathbf{x}}_{\rm b}\\ \dot{\mathbf{q}}\\ \dot{\mathbf{\delta}}\end{array} \right]+\left[ \begin{array}{c}\mathbf{0}_{2\times 1 }\\ k_{\rm b}\mathbf{x}_{\rm b}\\ 0\\ -\mathbf{\tau}\\ \mathbf{K}_{\rm\delta}\end{array} \right]=\left[ \begin{array}{c}\mathbf{0}_{3\times 1 }\\ \tau_{0}\\ \mathbf{0}_{6\times 1}\end{array} \right] \tag{13}\]  
\[\mathbf{J}_{\rm m}\ddot{\mathbf{q}}_{\rm m}+\mathbf{K}_{\rm m}\mathbf{\sigma}=\mathbf{\tau}_{\rm m} \tag{14}\]  
\[\mathbf{\tau}=\mathbf{K}_{\rm m}\mathbf{\sigma} \tag{15}\]  
After calculating \(\ddot{\mathbf{q}}_{\rm b}\) from the first formula of Equation (13) and substituting it into the second formula, we have:  
\[\mathbf{D}_{\rm M}\left[ \begin{array}{c}\ddot{\mathbf{x}}_{\rm b}\\ \ddot{\mathbf{q}}\\ \ddot{\mathbf{\delta}}\end{array} \right]+\mathbf{H}_{\rm M}\left[ \begin{array}{c} \dot{\mathbf{x}}_{\rm b}\\ \dot{\mathbf{q}}\\ \dot{\mathbf{\delta}}\end{array} \right]+\left[ \begin{array}{c}k_{\rm b}x_{\rm b }\\ \mathbf{0}_{3\times 1}\\ \mathbf{K}_{\rm\delta}\end{array} \right]=\left[ \begin{array}{c}0\\ \tau_{0}\\ \mathbf{\tau}\\ \mathbf{0}_{4\times 1}\end{array}