# Performance Analysis of Dual-Hop AF Relaying with Non-Linear/Linear Energy Harvesting
## Appendix A Calculation of \(A(w)\)  
In (16), substituting (14), \(A(w)\) is re-expressed as  
\[A(w)=\frac{a}{2}\int_{0}^{P_{th}}\left[1-\frac{1}{\sqrt{C+\frac{B}{Z}}}\right] f_{Z}(z)\mathrm{d}z, \tag{17}\]  
where \(C=(1+wg)/wg\), \(B=1/\bar{x}\) and \(P_{th}=P_{th_{s}}=P_{th_{r}}\) and \(f_{Z}(z)\) is given in (21). Substituting (21), (17) is written as  
\[A(w)=\frac{a}{2}(E_{1}-F_{1}(w)), \tag{18}\]  
where  
\[E_{1}=\int_{0}^{P_{th}}f_{Z}(z)\mathrm{d}z=D\gamma(m_{z},P_{th}\phi_{z})/\phi_{ z}^{m_{z}} \tag{19}\]  
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and  
\[N_{2}(w)=\beta(j)(\frac{w\overline{y}}{1+w\overline{y}})^{j+0.5}.\] (A12)  
In (A10) and (A12), we have  
\[\Lambda(k,n)=\sum_{k=0}^{\mathrm{Y}}\sum_{n=0}^{\mathrm{Y}}\frac{(0.5)_{k}(-1)^ {k}(-1)^{n}\phi_{z}^{n}}{k!n!(m_{z}+k+n+0.5)}B^{n+m_{z}+0.5}\] (A13)  
and  
\[\beta(j)=\sum_{j=0}^{\mathrm{Y}}(0.5)_{j}\frac{(-1)^{j}B^{j+0.5}\Gamma(m_{z}-j,\phi_{z}P_{th})}{j!\phi_{z}^{(m_{z}-j)}},\] (A14)  
where \(k,n\) and \(j\) are the indices of infinite summations defined in \(\Lambda(k,n)\) and \(\beta(j)\), respectively. Using ([52], 06.06.26.0005.01) and ([50], eq. 9.931-5), simplifying and substituting in (A11), we have  
\[N_{1}(w)=\alpha(j)G\,_{12}^{\,20}\left(\phi_{z}B\frac{w\overline{y}}{1+w \overline{y}}\left| \begin{matrix}j+1.5\\ j+0.5,m_{z}+0.5\end{matrix} \right.\right),\] (A15)  
where  
\[\alpha(j)=(\frac{1}{\phi_{z}})^{m_{z}+0.5}\sum_{j=0}^{\mathrm{Y}}(0.5)_{j} \frac{(-1)^{j}}{j!}.\] (A16)  
Finally, the simplified version of \(F_{1}(w)\) in (A9) is calculated by replacing (A10), (A15) and (A12). Then, substituting (A9) and (A3) in (A2), and, after some mathematical simplifications, \(P_{s1}\) in (16) is given as  
\[P_{s1}=\int_{0}^{P_{th}}A(w)f_{W}(w)\mathrm{d}w=\frac{a}{2}[E_{1}E_{2}-L],\] (A17)  
where  
\[E_{1}E_{2}=DI\gamma(m_{z},P_{th}\phi_{z})\gamma(m_{w},P_{th}\phi_{w})/\phi_{w }^{m_{w}}\phi_{z}^{m_{z}}\] (A18)  
and  
\[L=\frac{D}{\sqrt{B}}[L_{1}+L_{2}-L_{3}].\] (A19)  
and  
\[L_{3}=\int_{0}^{P_{th}}N_{2}(w)f_{W}(w)\mathrm{d}w=l\beta(j)\Delta_{2}. \tag{101}\]  
Please note that (100) is calculated numerically since a closed-form solution is not tractable. In (101) and (101), we have  
\[\Delta_{i}=\int_{0}^{P_{th}}w^{m_{w}-1}\exp(-w\phi_{w})(\frac{w \bar{g}}{1+w\bar{g}})^{u_{i}}\mathrm{d}w, \tag{102}\]  
where for \(i=\{1,2\}\), we define \(u_{1}=n+m_{z}+0.5\) and \(u_{2}=j+0.5\), respectively, for (101) and (102). Taking \(x=w\bar{g}\) and simplifying (102), (101) and (101) are rewritten as  
\[L_{1}=I\Lambda(k,n)(\frac{1}{\bar{y}})^{m_{w}}(\lambda_{1}+ \varpi_{1}) \tag{103}\]  
and  
\[L_{3}=I\beta(j)(\frac{1}{\bar{y}})^{m_{w}}(\lambda_{2}+\varpi_{ 2}), \tag{104}\]  
respectively, where  
\[\lambda_{v} =\int_{0}^{1}x^{m_{w}+u_{v}-1}(1+x)^{-u_{v}}\exp(-x\phi_{w}/\bar {y})\mathrm{d}x\] \[=\sum_{\delta=0}^{\mathrm{Y}}(u_{v})_{\delta}\frac{(-1)^{\delta} }{\delta!}\int_{0}^{1}x^{m_{w}+u_{v}+\delta-1}\exp(-x\phi_{w}/\bar{y})\mathrm{ d}x\] \[P_{s1}=\int_{0}^{P_{th}}A(w)f_{W}(w)\mathrm{d}w=\frac{a}{2}[E_{1}E_{2}-L],\]0  
(105) is calculated and simplified using ((56), eq. 3-a), ((51), eq. 13.2.1) and considering ((50), eq. 8.338-1) and ((50), eq. 8.331-1). Please note that in (105), \(v=1\) and \(v=2\) for (103) and (104), respectively. Moreover, in (103) and (104),  
\[P_{s1}=\int_{0}^{P_{th}}A(w)f_{W}(w)\mathrm{d}w=\frac{a}{2}[E_{1}E_{2}-L],\]1 \[P_{s1}=\int_{0}^{P_{th}}A(w)f_{W}(w)\mathrm{d}w=\frac{a}{2}[E_{1}E_{2}-L],\]2  
(106) is calculated using ((56), eq. 3-b) and ((50), eq. 3.351-2). (103) and (104) are calculated by substituting (105) and (106), respectively, for \(v=1\) and \(v=2\). (107) is calculated by substituting (103), (100) and (104). Finally, (108) is calculated by substituting (107) and (109). As seen from (108), \(P_{s1}\) is dependent on \(L\), \(E_{1}\) and \(E_{2}\).