## 2 System Model Network
### Scenario 2: S-R Link Is the Rician Fading Channel, R-D Link Is the Rayleigh Fading Channel  
Similar to scenario 1, the CDF of RV \(\varphi_{1}\) and PDF of RV \(\varphi_{2}\) can be formulated as  
\[F_{\varphi_{1}}(\varsigma)=\int\limits_{0}^{\varsigma}f_{\varphi_{1}}(x)dx=1- \frac{a}{b}\sum\limits_{l=0}^{\infty}\sum\limits_{m=0}^{l}\frac{K^{l}b^{m}}{ l!m!}\varsigma^{m}e^{-b\varsigma} \tag{26}\]  
where \(a=\frac{(K+1)e^{-K}}{\lambda_{h}},b=\frac{K+1}{\lambda_{h}}\)  
\[f_{\varphi_{2}}(x)=\lambda_{g}e^{-\lambda_{g}x} \tag{27}\]  
a. Outage Probability (OP)  
\[P_{out}=\Pr\bigg{\{}\varphi_{1}<\frac{c_{1}\varphi_{2}+c_{2}}{c_{3}\varphi_{2 }}\bigg{\}}=\int\limits_{0}^{\infty}F_{\varphi_{1}}\bigg{(}\frac{c_{1}\varphi _{2}+c_{2}}{c_{3}\varphi_{2}}\bigg{)}f_{\varphi_{2}}(\varphi_{2})d\varphi_{2} \tag{28}\]  
\[P_{out}=1-\int\limits_{0}^{\infty}\begin{array}{c}\frac{d}{b}\sum\limits_{ l=0}^{\infty}\sum\limits_{m=0}^{l}\frac{K^{l}b^{m}}{l!m!}\Big{(}\frac{c_{1} \varphi_{2}+c_{2}}{c_{3}\varphi_{2}}\Big{)}^{m}\\ e^{-\frac{b\varsigma_{1}\varphi_{2}+c_{2}}{c_{3}\varphi_{2}}}\lambda_{g}e^{- \lambda_{g}\varphi_{2}}d\varphi_{2}\end{array} \tag{29}\]  
We apply the equation \((x+y)^{m}=\sum\limits_{n=0}^{m}\left(\begin{array}{c}m\\ n\end{array}\right)^{xm-n}y^{n}\), and change the variable by setting, and Equation (27) can be reformulated as the following  
\[P_{out}=1-\lambda_{g}e^{-\frac{\lambda_{h}}{c_{3}}}\int\limits_{0}^{\infty} \int\limits_{0}^{\infty}\frac{e^{\frac{\theta}{b}\sum\limits_{l=0}^{\infty} \sum\limits_{m=0}^{l}\sum\limits_{n=0}^{m}\left(\begin{array}{c}m\\ n\end{array}\right)^{\frac{K^{l}b^{m}e^{-\theta x}-c_{2}^{m}}{l!m!m!}}}t^{m-2} \\ e^{-\frac{b\varsigma_{2}\varsigma_{2}}{c_{3}^{2}}-\frac{\varsigma_{2}^{2}}{c _{3}^{2}}}dt\end{array} \tag{30}\]\[P_{out}=1-\lambda_{g}e^{-\frac{k\chi_{1}}{\alpha}}\frac{a}{b}\sum_{l=0}^{\infty} \sum_{m=0}^{l}\sum_{n=0}^{m}\frac{K^{l}b^{m}c_{1}^{m-n}c_{2}^{n}}{l!n!(m-n)!c_{3} ^{m}}\int\limits_{0}^{\infty}t^{n-2}e^{-\frac{k\chi_{2}t}{\alpha_{3}}}e^{-\frac {\lambda_{g}}{\tau}}dt \tag{31}\]  
Using the Table of Integral Equation [3.471,9] in [24], the above equation can be given as  
\[P_{out}=1-2\lambda_{g}e^{-\frac{k\chi_{1}}{\alpha}}\frac{a}{b}\sum_{l=0}^{ \infty}\sum_{m=0}^{l}\sum_{n=0}^{m}\frac{K^{l}b^{m}c_{1}^{m-n}c_{2}^{n}}{l!n!( m-n)!c_{3}^{m}}\left(\frac{\lambda_{g}c_{3}}{bc_{2}}\right)^{\frac{n-1}{2}}K_{n-1} \left(2\sqrt{\frac{bc_{2}\lambda_{g}}{c_{3}}}\right) \tag{32}\]  
\[P_{out}=1-2 ae^{-\frac{k\chi_{1}}{\alpha_{3}}}\sum_{l=0}^{\infty}\sum_{m=0}^{l} \sum_{n=0}^{m}\frac{K^{l}b^{\frac{2m-n-3}{2}}\lambda_{g}^{\frac{n+1}{2}}c_{1} ^{m-n}c_{2}^{\frac{n+1}{2}}c_{3}^{\frac{n-1-2m}{2}}}{l!n!(m-n)!}\times K_{n-1} \left(2\sqrt{\frac{bc_{2}\lambda_{g}}{c_{3}}}\right) \tag{33}\]  
where \(K_{\
u}(\bullet)\) is the modified Bessel function of the second kind and \(\
u^{\text{th}}\) order.  
b. Achievable Throughput (AT):  
\[\tau=(1-P_{out})\frac{R}{2}=2 ae^{-\frac{k\chi_{1}}{\alpha_{3}}}\sum_{l=0}^{ \infty}\sum_{m=0}^{l}\sum_{n=0}^{m}\frac{K^{l}b^{\frac{2m-n-3}{2}}\lambda_{g}^ {\frac{n+1}{2}}c_{1}^{m-n}c_{2}^{\frac{n+1}{2}}c_{3}^{\frac{n-1-2m}{2}}}{l!n!( m-n)!}\times K_{n-1}\left(2\sqrt{\frac{bc_{2}\lambda_{g}}{c_{3}}}\right)\times \frac{R}{2} \tag{34}\]  
c. The Bit Error Rate (BER)  
Similar to scenario 1, BER can be calculated with the equation below  
\[BER=\frac{\omega\sqrt{\theta}}{2\sqrt{\pi}}\int\limits_{0}^{\infty}\frac{e^{- \theta x}}{\sqrt{x}}F_{\gamma_{\text{eff}}}(x)dx \tag{35}\]