## 4 Mathematical Formulation of the Optimization Problem
### Transmission Power Requirements for the T-S Link  
In this subsection, we try to solve the transmission power requirements of the terminal for the T-S direct link. We consider the BER expression in Equation (3). Obviously, an explicit expression of the SNR cannot be obtained due to the fact that Equation (3) is a composite function of a polynomial and exponential. However, the BER \(\rho_{1}\) of the T-S link is a strictly decreasing function in the domain of the transmission power \(P_{s,1}\). Therefore, for the BER to maintain a minimum level \(\epsilon\) of service quality, \(P_{s,1}\) of the terminal can be iteratively computed using the gradient descent method for some \(\xi_{1}>0\) and \(\rho_{e}=\epsilon\), shown as:  
\[\hat{P}_{s,1,n+1}=\hat{P}_{s,1,n}-\xi_{1}\
abla(\hat{P}_{s,1,n}) \tag{5}\]  
where \(\
abla(\hat{P}_{s,1,n})\) is the gradient of the transmission power \(P_{s,1}\), \(\hat{P}_{s,1}\) is the desired minimum transmission power of the terminal for the T-S direct link and \(n\) is the iteration number.  
The gradient \(\
abla(\hat{P}_{s,1,n})\) can be derived as follows:  
If the transmission power of the terminal is \(P_{s,1}\), the mean received SNR for the satellite can be represented as:  
\[\gamma_{1}=\frac{\alpha_{1}E_{b}}{N_{0}}=\frac{\alpha_{1}P_{s,1}G_{s,1}G_{r,1} }{R_{b,1}L_{m}(d_{1})N_{0}BO_{o}} \tag{6}\]  
Substituting Equation (6) into Equation (3), the BER for the T-S direct transmission is given by:  
\[\begin{array}{l}\rho_{e,1}=\frac{(1+\kappa_{1})}{2(1+\kappa_{1}+\gamma_{1}) }\exp\left(\frac{-\kappa_{1}\gamma_{1}}{1+\kappa_{1}+\gamma_{1}}\right)\\ =\mathrm{B}\exp\left(\frac{-\kappa_{1}\alpha_{1}P_{s,1}G_{s,1}G_{r,1}}{R_{b,1} L_{m}(d_{1})N_{0}BO_{o}(1+\kappa_{1}+\frac{1}{R_{b,1}L_{m}(d_{1})N_{0}BO_{o}})} \right)\end{array} \tag{7}\]  
where \(\mathrm{B}=\frac{1+\kappa_{1}}{2(1+\kappa_{1}+\gamma_{1})}\).  
Then, the gradient of \(P_{1,n}\) can be obtained as:  
\[\begin{array}{l}\
abla(\hat{P}_{s,1,n})=\frac{d(\epsilon-\rho_{e}(P_{s,1}))} {dP_{s,1}}\\ \hskip 14.226378 pt=-\frac{d\mathrm{B}}{dP_{s,1}}\exp\left(\frac{-\kappa_{1} \gamma_{1}}{1+\kappa_{1}+\gamma_{1}}\right)-\mathrm{B}\frac{d\exp\left(\frac{- \kappa_{1}\gamma_{1}}{1+\kappa_{1}+\gamma_{1}}\right)}{dP_{s,1}}\\ \hskip 14.226378 pt=\frac{(1+\kappa_{1})}{2(1+\kappa_{1}+\gamma_{1})^{2}}\frac{ \alpha_{1}G_{s,1}G_{r,1}}{R_{b,1}L_{m}(d_{1})N_{0}BO_{o}}\exp\left(\frac{- \kappa_{1}\gamma_{1}}{1+\kappa_{1}+\gamma_{1}}\right)+\mathrm{B}\exp\left( \frac{-\kappa_{1}\gamma_{1}}{1+\kappa_{1}+\gamma_{1}}\right)\times\frac{ \kappa_{1}(1+\kappa_{1})}{(1+\kappa_{1}+\gamma_{1})^{2}}\times\frac{\alpha_{1} G_{s,1}G_{r,1}}{R_{b,1}L_{m}(d_{1})N_{0}BO_{o}}\\ \hskip 14.226378 pt=\frac{\alpha_{1}G_{s,1}G_{r,1}}{R_{b,1}L_{m}(d_{1})N_{0}BO_{o }}\frac{\mathrm{B}}{(1+\kappa_{1}+\kappa_{1})}+\frac{\kappa_{1}(1+\kappa_{1} )\mathrm{B}}{(1+\kappa_{1}+\gamma_{1})^{2}}\exp\left(\frac{-\kappa_{1}\gamma_{ 1}}{1+\kappa_{1}+\gamma_{1}}\right)\\ \hskip 14.226378 pt=\frac{\alpha_{1}G_{s,1}G_{r,1}\mathrm{B}(2\mathrm{B}\kappa_{ 1}+1)}{L_{1,m}(d_{1})N_{0}BO_{o}R_{b,1}(1+\kappa_{1}+\gamma_{1})}\exp\left( \frac{-\kappa_{1}\gamma_{1}}{1+\kappa_{1}+\gamma_{1}}\right)\end{array} \tag{8}\]  
Therefore, the gradient \(\
abla(\hat{P}_{1,n})\) can be represented as:  
\[\
abla(\hat{P}_{s,1,n})=\frac{\alpha G_{s,1}G_{r,1}\mathrm{B}(2\mathrm{B}\kappa _{1}+1)}{L_{1,m}(d_{1})N_{0}BO_{o}R_{b,1}(1+\kappa_{1}+\gamma_{1})}\times\exp \left(\frac{-\kappa_{1}\gamma_{1}}{1+\kappa_{1}+\gamma_{1}}\right) \tag{9}\]  
Therefore, the minimum transmission power of the terminal for the T-S direct link \(\hat{P}_{s,1}\) can be calculated by performing a sufficient number of iterations \(n\) in Equation (5).