# Energy Harvesting over Rician Fading Channel:
## Appendix A Proof of Theorem 1  
In this section, the detailed proof of Theorem 1 is provided. Without loss of generality, we can consider the node \(S_{1}\) only. From Equation (14), we can fix \(X_{1}\) at some value \(x_{1}\) and compute the probability in Equation (14) in terms of \(x_{1}\). Then, we take the average of the obtained function with respect to the distribution of \(X_{1}\) to get the outage probability. Using this approach, Equation (14) can be rewritten as  
\[P_{out\_1}=\int\limits_{0}^{\infty}F_{X_{2}}\left(\frac{\gamma_{th}x_{1}^{2}a+ \gamma_{th}x_{1}b+\gamma_{th}c}{x_{1}-\gamma_{th}x_{1}a}\right)f_{X_{1}}(x_{1}) dx_{1}. \tag{16}\]  
Let us denote \(\tilde{\gamma}_{th}\triangleq\frac{\gamma_{th}}{1-a\gamma_{th}}\). Then, we have \(\frac{\gamma_{th}x_{1}^{2}a+\gamma_{th}x_{1}b+\gamma_{th}c}{x_{1}-\gamma_{th}x _{1}a}=\tilde{\gamma}_{th}\left(x_{1}a+\frac{c}{x_{1}}+b\right)\). Now, Equation (16) can be rewritten as  
\[\begin{split}& P_{out\_1}=\\ &=1-\int\limits_{0}^{\infty}\sum\limits_{l=0}^{\infty}\sum \limits_{m=0}^{l}\frac{e^{-K}k^{l}(\zeta_{2}\tilde{\gamma}_{th})^{m}}{l!m!} \left(x_{1}a+\frac{c}{x_{1}}+b_{1}\right)^{m}\!e^{-\zeta_{2}\tilde{\gamma  
\[P_{out\_1} =1-\xi_{12}\sum_{0}^{\infty}\sum_{l=0}^{\infty}\sum_{k=0}^{\infty} \sum_{m=0}^{l}\sum_{p=0}^{n}\frac{K^{l+k}\xi_{12}^{k}\xi_{12}^{m}\xi_{lh}^{m}a^{m -n}p^{c}a^{m-p}}{l!p!(n-p)!(m-n)!(k!)^{2}}x_{1}^{m-2n+k+p}e^{-\xi_{2}\tilde{\gamma }_{ln}\left(x_{1}+\frac{\varepsilon}{\varepsilon_{1}}\right)}e^{-\xi_{1}x_{1}}dx _{1} \tag{10}\] \[=1-\xi_{12}\sum_{0}^{\infty}\sum_{l=0}^{\infty}\sum_{k=0}^{\infty} \sum_{m=0}^{l}\sum_{n=0}^{m}\frac{K^{l+k}\xi_{12}^{k}\xi_{12}^{m}\xi_{lh}^{m}a^ {m-n}}{l!p!(n-n)!(k!)^{2}}x_{1}^{m-n+k}\sum_{p=0}^{n}\binom{n}{p}\left(\frac{x _{1}}{x_{1}}\right)^{n-p}b^{p}e^{-\xi_{2}\tilde{\gamma}_{ln}\left(x_{1}+\frac {\varepsilon}{\varepsilon_{1}}\right)}e^{-\xi_{1}x_{1}}dx_{1}\] \[=1-\xi_{12}\sum_{0}^{\infty}\sum_{l=0}^{\infty}\sum_{k=0}^{\infty} \sum_{m=0}^{l}\sum_{n=0}^{m}\frac{K^{l+k}\xi_{12}^{k}\xi_{12}^{m}\xi_{lh}^{m}a ^{m-n}p^{c}a^{m-p}}{l!p!(n-p)!(m-n)!(k!)^{2}}x_{1}^{m-2n+k+p}e^{-\xi_{2}\tilde{ \gamma}_{ln}\left(x_{1}+\frac{\varepsilon}{\varepsilon_{1}}\right)}e^{-\xi_{1} x_{1}}dx_{1}\] \[=1-\xi_{12}\sum_{l=0}^{\infty}\sum_{k=0}^{\infty}\sum_{m=0}^{l} \sum_{n=0}^{m}\sum_{p=0}^{n}\frac{K^{l+k}\xi_{12}^{k}\xi_{12}^{m}\xi_{lh}^{m}a ^{m-n}p^{c}a^{m-p}}{l!p!(n-n)!(m-n)!(k!)^{2}}\int_{0}^{\infty}x_{1}^{m-2n+k+p} e^{-(\xi_{1}+4\xi_{2}\tilde{\gamma}_{ln})x_{1}-\zeta_{2}\tilde{\gamma}_{ln} \frac{\varepsilon}{\varepsilon_{1}}}dx_{1}.\]  
Now, by letting \(\xi_{12}\triangleq\zeta_{1}+a\xi_{2}\tilde{\gamma}_{th}\) and using Equation (3.471.9) in [32], we obtain the outage probability:  
\[P_{out\_1} =1-\xi_{12}\!\!\sum_{l=0}^{\infty}\sum_{k=0}^{\infty}\sum_{m=0}^ {l}\sum_{n=0}^{m}\sum_{p=0}^{n}\frac{K^{l+k}\xi_{12}^{k}\xi_{12}^{m}\xi_{12}^{ m}a^{m-n}p^{c}a^{m-p}}{l!p!(n-p)!(n-n)!(k!)^{2}}\left(\frac{\xi_{2}\tilde{ \gamma}_{th}c}{\xi_{12}}\right)^{\frac{-m_{2}+p+k+1}{2}}K_{m-2n+p+k+1}\left(2 \sqrt{\xi_{12}\xi_{2}\tilde{\gamma}_{th}c}\right) \tag{11}\] \[=1-2\xi_{12}\sum_{l=0}^{\infty}\sum_{k=0}^{\infty}\sum_{m=0}^{l} \sum_{n=0}^{m}\sum_{p=0}^{n}\frac{K^{l+k}\xi_{1}^{k}(\xi_{2}\tilde{\gamma}_{ln })^{\frac{3m-2n+p+k+1}{2}}a^{m-n}p^{c}\frac{-p+k+1}{2}}{l!p!(n-p)!(m-n)!(k!)^{2} \xi_{12}^{\frac{-3m+p+k+1}{2}}}K_{m-2n+p+k+1}\left(2\sqrt{\xi_{12}\xi_{2} \tilde{\gamma}_{ln}c}\right),\]  
where \(K_{v}(\cdot)\) is the \(v^{th}\)-order modified Bessel function of the second kind.