# A Predefined-Time Control for the Laser Acquisition in Space Gravitational Wave Detection Mission
## 3 Mathematical Model
### Mathematic Model for the Satellite Platform  
According to the vector relationship, we can obtain the line velocity and angle velocity of the \(i\)-th body as:  
\[ \begin{bmatrix}{}^{l}\mathbf{v}_{i}\\ {}^{l}\mathbf{\omega}_{i}\end{bmatrix} = \begin{bmatrix}\mathbf{E}_{3}&({}^{l}\mathbf{v}_{0 }^{\times})^{T}\\ \mathbf{O}_{3\times 3}&\mathbf{E}_{3}\end{bmatrix}\begin{bmatrix}{}^{l}\mathbf{v}_{0}\\ {}^{l}\mathbf{\omega}_{0}\end{bmatrix} + \begin{bmatrix}\mathbf{J}_{Ti}\\ \mathbf{J}_{Ri}\end{bmatrix} \dot{\mathbf{\theta}} \tag{5}\]  
where \({}^{l}\mathbf{v}_{0i}={}^{l}\mathbf{r}_{i}-{}^{l}\mathbf{r}_{0i}\)\({}^{l}\mathbf{r}_{i}\) denotes the position vector of the centroid of body \(i\) in \(\sum_{l}\), \({}^{l}\mathbf{r}_{g}\) denotes the position vector of the centroid of the spacecraft system in \(\sum_{l}\), \(\mathbf{\theta}\in\mathbb{R}^{4\times 1}\) denotes the angle velocity of the four DOFs of two telescopes, \(\mathbf{J}_{Ti}\) and \(\mathbf{J}_{Ri}\) denote the Jacobian matrixes which can be expressed as:  
Figure 8: Schematic diagram of the topology of the satellite platform. The black numbers represent the body number and the green numbers represent the joint number. Figure 9: Vector relationship. \(C_{ij}\) denotes the link vector for the centroid of body \(i\) body to joint \(j\), \(I_{ij}\) denotes the link vector from joint \(i\) to joint \(j\), \(I_{ij}=e_{ij}-e_{ii}\), \(\mathbf{O}_{cm}^{i}\) denotes the centroid of body \(i\), \(O_{Ri}\) denotes the rotation center of telescope 1, \(O_{R_{2}}\) denotes the rotation center of telescope 2. \[\left\{ \begin{array}{l}\mathbf{I}_{T1}=\left[{}^{I}\mathbf{k}_{1}\times\left({}^{I}\mathbf{r} _{1}-{}^{I}\mathbf{R}_{1}\right),0,0,0\right]\\ \mathbf{I}_{T2}=\left[{}^{I}\mathbf{k}_{1}\times\left({}^{I}\mathbf{r}_{2}-{}^{I}\mathbf{R}_{1} \right),{}^{I}\mathbf{k}_{2}\times\left({}^{I}\mathbf{r}_{2}-{}^{I}\mathbf{R}_{2}\right),0, 0\right]\\ \mathbf{I}_{T3}=\left[0,0,{}^{I}\mathbf{k}_{3}\times\left({}^{I}\mathbf{r}_{3}-{}^{I}\mathbf{R }_{3}\right),0\right]\\ \mathbf{I}_{T4}=\left[0,0,{}^{I}\mathbf{k}_{3}\times\left({}^{I}\mathbf{r}_{4}-{}^{I}\mathbf{R }_{3}\right),{}^{I}\mathbf{k}_{4}\times\left({}^{I}\mathbf{r}_{4}-{}^{I}\mathbf{R}_{4} \right)\right]\end{array} \right. \tag{6}\]  
where \({}^{I}\mathbf{R}_{i}\) denotes the position vector of the joint \(i\) in \(\sum_{l}\):  
\[\left\{ \begin{array}{l}\mathbf{J}_{R1}=\left[{}^{I}\mathbf{k}_{1},0,0,0\right]\\ \mathbf{J}_{R2}=\left[{}^{I}\mathbf{k}_{1},{}^{I}\mathbf{k}_{2},0,0\right]\\ \mathbf{J}_{R3}=\left[0,0,{}^{I}\mathbf{k}_{3},0\right]\\ \mathbf{J}_{R4}=\left[0,0,{}^{I}\mathbf{k}_{3},{}^{I}\mathbf{k}_{4}\right]\end{array} \right. \tag{7}\]  
So, we can obtain the kinetic energy of \(i\)-th body:  
\[T_{i}=\frac{1}{2}{m_{i}}^{I}{v_{i}^{T}}\mathbf{v}_{i}+\frac{1}{2}{\left({}^{I} \mathbf{\omega}_{i}\right)}^{T}{I_{i}}^{I}\mathbf{\omega}_{i}=\frac{1}{2}{\left[{}^{I }\mathbf{v}_{i}\right]}^{T}{\left[{}^{m_{i}}\mathbf{E}_{3}\quad 0\right]}{\left[{}^{I} \mathbf{v}_{i}\atop{}^{I}\mathbf{\omega}_{i}\right]} \tag{8}\]  
and the whole kinetic energy of the satellite platform is:  
\[T=\sum_{i=0}^{4}T_{i}=\frac{1}{2}{\left[{}^{I}\mathbf{x}_{b}\atop{}^{\dot{\mathbf{ \theta}}}\right]}^{T}{\left[{}^{H}\mathbf{I}_{b}\quad{}^{H}\mathbf{I}_{b}\atop{}^{T} \quad{}^{H}\mathbf{I}_{t}\right]}^{\left[{}^{I}\mathbf{x}_{b}\atop{}^{\dot{\mathbf{\theta} }}\right]} \tag{9}\]  
where \(\mathbf{H}_{b}\in\mathbb{R}^{6\times 6}\) denotes the inertia matrix of satellite body, \(\mathbf{H}_{t}\in\mathbb{R}^{4\times 4}\) denotes the inertia matrix of telescopes, \(\mathbf{H}_{bt}\in\mathbb{R}^{6\times 4}\) denotes the coupling matrix between satellite body and telescopes.