# Fusion of Land-Based and Satellite-Based Localization Using Constrained Weighted Least Squares
## 5 Proposed Method
### Proposed Method for Satellite  
#### 5.2.1 Localization Method for Satellite  
Substituting the measurements into (32) gives \begin{table}
\begin{tabular}{l c} \hline
**Step 1:** Set \(k=0\) and choose appropriate \(\delta_{a}>0\) and \(\kappa_{a}\). Let \(\mathbf{W}_{a}=\mathbf{Q}_{\mathbf{z}}^{-1}\) to obtain the initial result; **Step 2:** Set \(k=k+1\). Substituting \(\hat{\mathbf{q}}_{a}^{k-1}\) and \(\hat{\mathbf{a}}_{a}^{k-1}\) into (56) and (58) yields \(\mathbf{W}_{a}^{k-1}\), \(\mathbf{D}_{a}^{k-1}\), and \(\hat{\mathbf{a}}_{a}^{k-1}\); **Step 3:** Calculate \(\hat{\mathbf{\eta}}_{a}^{k}\) based on (61) and update the iteration result by (62); **Step 4:** If \[||\hat{\mathbf{q}}_{a}^{k}-\hat{\mathbf{\eta}}_{a}^{k-1}||_{2}/||\hat{\mathbf{q}}_{a}^{k}| |_{2}<\delta_{a}\] (63) Stop the iteration and output \(\hat{\mathbf{q}}_{a}=\hat{\mathbf{q}}_{a}^{k}\). If not, go to **Step 2**. \\ \hline \end{tabular}
\end{table} 
Table 2: Summary of land-based localization. \begin{table}
\begin{tabular}{c c} \hline
**Computational Element** & **Computational Complexity** \\ \hline \(\tilde{G}_{a}^{o}\) & \(3N_{1}+4N_{1}\) \\ \(\mathbf{y}_{a}^{o}\) & \(2N_{1}\) \\ \(\hat{\mathbf{y}}_{a}^{0}\) & \(8N_{1}^{2}+32N_{1}+O(4^{3})\) \\ \(\hat{\mathbf{W}}_{a}^{k}\) and \(\Big{(}\mathbf{W}_{a}^{k}\Big{)}^{-1}\) & \(2(2N_{1})^{3}+6N_{1}^{3}+O\Big{(}\big{(}2N_{1}\big{)}^{3}\Big{)}\) \\ \(\tilde{\mathbf{G}}_{a}^{k}\) & \(16N_{1}^{2}+32N_{1}\) \\ \(\tilde{\mathbf{y}}_{a}^{k}\) & \(8(2N_{1})^{2}\) \\ \(\tilde{\mathbf{P}}_{a}^{k}\) & \(32+O\big{(}2^{3}\big{)}\) \\ \(\tilde{\mathbf{\tau}}_{a}^{k}\) & \(280+O\big{(}2^{3}\big{)}+O\big{(}4^{3}\big{)}\) \\ \(\hat{\mathbf{\eta}}_{a}^{k}\) & \(56+O\big{(}2^{3}\big{)}\) \\ \(\mathbf{cov}(\hat{\mathbf{u}}_{a})\) & \(276+O\big{(}2^{3}\big{)}\) \\ \hline \end{tabular}
\end{table} 
Table 3: Computational complexity of land-based localization. \[\varepsilon_{t}=\mathbf{y}_{t}-\mathbf{G}_{t}\mathbf{\eta}_{t} \tag{68}\]  
Performing the first-order Taylor expansion of \(\mathbf{y}_{t}\) and \(\mathbf{G}_{t}\) gives  
\[\left\{ \begin{array}{l}\mathbf{y}_{t}\approx\mathbf{y}_{t}^{o}+\mathbf{A}_{3}\mathbf{n}_{t}+ \mathbf{A}_{4}\mathbf{n}_{s}\\ \mathbf{G}_{t}\approx\mathbf{G}_{t}^{o}+\sum_{d_{1}=1}^{N_{2}-1}<\mathbf{r}{>}_{d_{1}}\mathbf{ B}_{3,d_{1}}+\sum_{d_{2}=1}^{3N_{2}}<\mathbf{s}_{t}{>}_{d_{2}}\mathbf{B}_{4,d_{2}} \end{array} \right. \tag{69}\]  
where  
\[\left\{ \begin{array}{l}\mathbf{A}_{3}=\frac{\partial\mathbf{y}_{t}}{\partial\mathbf{r} },\mathbf{B}_{3,d_{1}}=\frac{\partial\mathbf{G}_{t}}{\partial\mathbf{r}<\mathbf{r}_{d_{1}}},1 \leq d_{1}\leq N_{2}-1\\ \mathbf{A}_{4}=\frac{\partial\mathbf{y}_{t}}{\partial\mathbf{s}_{t}^{2}}\mathbf{B}_{4,d_{2}}= \frac{\partial\mathbf{G}_{t}}{\partial\mathbf{s}_{t}{>}_{d_{2}}},1\leq d_{2}\leq 3N_{2} \end{array} \right. \tag{70}\]  
Substituting (69) into (68) gives  
\[\mathbf{C}_{3} =\left(\mathbf{A}_{3}-\left[\mathbf{B}_{3,1}\mathbf{\eta}_{t},\mathbf{B}_{3,2}\bm {\eta}_{t},...,\mathbf{B}_{3,N_{1}-1}\mathbf{\eta}_{t}\right]\right)\in\mathbb{R}^{(N _{2}-1)\times(N_{2}-1)} \tag{71}\] \[\mathbf{C}_{4} =\left(\mathbf{A}_{4}-\left[\mathbf{B}_{4,1}\mathbf{\eta}_{t},\mathbf{B}_{4,2}\bm {\eta}_{t},...,\mathbf{B}_{4,3N_{2}}\mathbf{\eta}_{t}\right]\right)\in\mathbb{R}^{(N _{2}-1)\times 3N_{2}}\]  
Therefore, the TDOA localization for the satellite can be modeled as  
\[\left\{\begin{array}{l}\min_{\mathbf{\eta}_{t}}\mathrm{J}_{t}(\mathbf{\eta}_{t})= \left(\mathbf{y}_{t}-\mathbf{G}_{t}\mathbf{\eta}_{t}\right)^{\mathrm{T}}\mathbf{W}_{t}(\mathbf{y} _{t}-\mathbf{G}_{t}\mathbf{\eta}_{t})\\ \mathrm{s.t. }\left(\mathbf{u}-\mathbf{s}_{N_{1}+1}^{o}\right)^{\mathrm{T}}\mathbf{u}-||\bm {u}-\mathbf{s}_{N_{1}+1}^{o}\mathbb{E}\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\! \!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\! \!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\! \!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\! \!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!