## 3 BER Performance Estimation
### Average BER Performances Due to Turbulence and SJ Influence  
By averaging the instantaneous BER expression above over \(I\), using the joint PDF, the average BER performance expression is taken:  
\[\widetilde{P}_{b,N,TP,MS}(\mathbf{I}) = \int\limits_{\mathbf{I}}P_{N,MS}(\mathbf{I})f_{TP,\mathbf{I}}( \mathbf{I})d\mathbf{I}=\int\limits_{\mathbf{I}}p_{MS}\mathcal{Q}\Bigg{(}q_{MS} \sqrt{\sum\limits_{n}^{N}\frac{\left(1+\xi_{n}^{-2}\right)^{2}\mu_{n}I_{n}^{2} }{A_{0,n}^{2}g_{n}^{2}(c_{n}+\Omega_{n})^{2}}/N}\Bigg{)}f_{TP,\mathbf{I}}( \mathbf{I})d\mathbf{I} \tag{11}\]  
Taking the Q function approach \(Q(x)\approx 0.208e^{-0.971x^{2}}+0.147e^{-0.525x^{2}}\) into account [Equation 13c of [45]], the above expression is transformed to the next equation:  
\[ \begin{array}{c}\widetilde{P}_{b,N,TP,MS}(\mu_{n})\ =\ p_{MS}\Bigg{[}0.208\prod\limits_{n=1}^{N }\int\limits_{0}^{+\infty}\exp\left(-\frac{0.971u_{MS}^{2}(1+\xi_{n}^{-2})^{2} \mu_{n}I_{n}^{2}}{NA_{0,n}^{2}g_{n}^{2}(c_{n}+\Omega_{n})^{2}}\right)f_{TP,I_{ n}}(I_{n})dI_{n}+\\ +0.147\prod\limits_{n=1}^{N}\int\limits_{0}^{+\infty}\exp\left(-\frac{0.525g_{ MS}^{2}\left(1+\xi_{n}^{-2}\right)^{2}\mu_{n}I_{n}^{2}}{NA_{0,n}^{2}g_{n}^{2}(c_{n}+ \Omega_{n})^{2}}\right)f_{TP,I_{n}}(I_{n})dI_{n}\Bigg{]}\\ \text{or}\ \ \widetilde{P}_{b,N,TP,MS}(\mu_{n})\ \ \ =\ p_{MS}\Bigg{[}0.208\prod\limits_{n=1}^{N }\Psi_{n}(\mu_{n},0.971)+0.147\prod\limits_{n=1}^{N}\Psi_{n}(\mu_{n},0.525) \Bigg{]}\ \,\ \ \text{with}\] \[\Psi_{n}(\mu_{n};\psi)\ =\ \int\limits_{0}^{+\infty}\exp\left(-\frac{\psi_{MS}^{2}(1+ \xi_{n}^{-2})^{2}\mu_{n}I_{n}^{2}}{NA_{0,n}^{2}g_{n}^{2}(c_{n}+\Omega_{n})^{2} }\right)f_{TP,I_{n}}(I_{n})dI_{n},\text{ where the corresponding solution of integral }\Psi_{n}\) is given by [33; 46; 47]:  
\[\Psi_{n}(\mu_{n};\psi)=\frac{\chi_{n}^{2}A_{n}}{64\pi}\sum\limits_{k=1}^{\omega }C_{k,n}\Bigg{(}\frac{4}{B_{n}}\Bigg{)}^{\frac{\mu_{n}+k}{2}}C_{6,3}^{16} \Bigg{(}\frac{16\psi_{MS}^{2}\mu_{n}\left(1+\xi_{n}^{-2}\right)^{2}}{B_{n}^{2} N(c_{n}+\Omega_{n})^{2}}\Bigg{|}\ \frac{1-\xi_{n}^{2}}{2},1-\frac{\xi_{n}^{2}}{2},1-\frac{a_{n}}{2},1-\frac{a_{n}} {2},1-\frac{k}{2}\over 2}\Bigg{)} \tag{13}\]  
Then, the ABER performance due to turbulence and SJ is obtained by:  
\[\begin{array}{c}\widetilde{P}_{b,N,TP,MS}(\mu_{n})\Psi=\frac{p_{MS}}{(64\pi)^ {N}}\Bigg{[}0.208\prod\limits_{n=1}^{N}\xi_{n}^{2}A_{n}\sum\limits_{k=1}^{ \omega}C_{k,n}\Big{(}\frac{4}{B_{n}}\Big{)}^{\frac{\mu_{n}+k}{2}}C_{6,3}^{16} \Bigg{(}\frac{15.536u_{MS}^{2}\mu_{n}\left(1+\xi_{n}^{-2}\right)^{2}}{B_{n}^{2} N(c_{n}+\Omega_{n})^{2}}\Bigg{|}\ \frac{1-\xi_{n}^{2}}{2},1-\frac{\xi_{n}^{2}}{2},1-\frac{ \xi_{n}}{2},1-\frac{k}{2}\over 2},1-\frac{\xi_{n}}{2},1-\frac{k}{2}\over 2},\\ 0,-\frac{\xi_{n}^{2}}{2},1-\frac{\xi_{n}^{2}}{2}\end{array}\Bigg{)}+\\ +0.147\prod\limits_{n=1}^{N}\xi_{n}^{2}A_{n}\sum\limits_{k=1}^{\omega}C_{k,n} \Big{(}\frac{4}{B_{n}}\Big{)}^{\frac{\mu_{n}+k}{2}}C_{6,3}^{16}\Bigg{(}\frac{8.4 4u_{MS}^{2}\mu_{n}\left(1+\xi_{n}^{-2}\right)^{2}}{B_{n}^{2}N(c_{n}+\Omega_{n})^ {2}}\Bigg{|}\ \frac{1-\xi_{n}^{2}}{2},1-\frac{\xi_{n}^{2}}{2},1-\frac{\xi_{n}}{2},1- \frac{k}{2}\over 2}\Bigg{)}\Bigg{]}\\ 0,-\frac{\xi_{n}^{2}}{2},\frac{1-\xi_{n}^{2}}{2}\end{array}  \tag{14}\]  
Finally, supposing that all receivers are both close to each other and in adequate distance from the transmitter, then all parameters of the system are independent of \(n\) index, so a simplified expression is derived:  
\[