# Hybrid Active-Passive Reconfigurable Intelligent Surface for Cooperative Transmission Systems
## Appendix A The CDF Derivation of SNR for HAPR Systems  
Let \(h_{S}^{2}\) be the random variable \(x\) and \(h_{D}^{2}\) be the random variable \(y\), \(F_{\gamma}(\gamma_{th})\) can be further written as  
\[ \begin{split} F_{\gamma}(\gamma_{th})&=\Pr\bigl{(} \max\bigl{(}\gamma_{a},\gamma_{p}\bigr{)}\leq\gamma_{th}\bigr{)}=\Pr\bigl{(} \gamma_{a}\leq\gamma_{th},\gamma_{p}\leq\gamma_{th}\bigr{)}\\ &=\iint\limits_{D=D_{1}+D_{2}+D_{3}}f(x,y)dxdy\end{split}  \tag{1}\]  
where \(D\) is the common region of curves \(\gamma_{a}<\gamma_{th}\) and \(\gamma_{p}\geq\gamma_{th}\), the specific divisions are shown in Figure 11.  
Introducing (10) and (12) into them, can be expressed as  
\[\gamma_{a}<\gamma_{th}\Leftrightarrow y<\frac{x}{ax-b} \tag{2}\]  
\[\gamma_{p}\leq\gamma_{th}\Leftrightarrow y\leq\frac{c}{x} \tag{3}\]  
\[D=\left\{(x,y)\middle|\left(y<\frac{x}{ax-b}\right)\bigcap\left(y\leq\frac{c}{ x}\right)\right\} \tag{4}\]  
It can be found that coordinates \((x_{1},y_{1})\) and \((x_{2},y_{2})\) at the intersection of two curves are  
\[ \begin{split}(x_{1},y_{1})&=\left(\frac{ac-\sqrt{a^ {2}c^{2}-4 bc}}{2},\frac{2c}{ac-\sqrt{a^{2}c^{2}-4 bc}}\right)\\ (x_{2},y_{2})&=\left(\frac{ac+\sqrt{a^{2}c^{2}-4 bc} }{2},\frac{2c}{ac+\sqrt{a^{2}c^{2}-4 bc}}\right)\end{split}  \tag{5}\]  
The integral region in (4) can be divided into three, and their double integrals are denoted, respectively, as follows,  
\[I_{1}=\iint\limits_{D_{1}}f(x,y)dxdy=\int_{y_{1}}^{\infty}\int_{0}^{\frac{c}{ 5}}f(x,y)dxdy \tag{6}\]  
Figure 11: OP versus \(P_{S}\) for HAPR with a different number of \(N\). \[I_{2}=\iint\limits_{D_{2}}f(x,y)dxdy=\int_{0}^{x_{2}}\int_{0}^{\frac{x}{\pi-8}}f(x,y)dydx\] (A7)  
\[I_{3}=\iint\limits_{D_{3}}f(x,y)dxdy=\int_{x_{2}}^{\infty}\int_{0}^{\frac{c}{x}}f (x,y)dydx\] (A8)  
where \(f(x,y)\) is the joint probability density of \(x\) and \(y\), and as \(x\) and \(y\) are independent of each other, \(f(x,y)\) can be obtained directly from (13) and (14), as shown below  
\[f(x,y)=f_{h_{3}^{2}}(x)\times f_{h_{3}^{2}}(y)=\frac{\mathrm{Y}_{S}^{m_{S}} \mathrm{Y}_{D}^{m_{D}}}{\Gamma(m_{S})\Gamma(m_{D})}x^{m_{S}-1}y^{m_{D}-1}e^{- \mathrm{Y}_{S}x-\mathrm{Y}_{D}y}\] (A9)  
Firstly, we calculate \(I_{1}\) by introducing (A9) into (A6) as follows,  
\[I_{1}=\int_{y_{1}}^{\infty}\int_{0}^{\frac{c}{y}}f(x,y)dydx=\frac{\mathrm{Y}_{ S}^{m_{S}}\mathrm{Y}_{D}^{m_{D}}}{\Gamma(m_{S})\Gamma(m_{D})}\int_{y_{1}}^{ \infty}y^{m_{D}-1}e^{-\mathrm{Y}_{D}y}dy\int_{0}^{\frac{c}{y}}x^{m_{S}-1}e^{- \mathrm{Y}_{S}x}dx\] (A10)  
The integral with respect to \(x\) can be obtained by using Equations (3.381.1) and (8.353.6) in [28],  
\[\int_{0}^{\frac{c}{y}}x^{m_{S}-1}e^{-\mathrm{Y}_{S}x}dx=\mathrm{Y}_{S}^{-m_{S} }(m_{S}-1)!\left[1-e^{-\frac{\mathrm{Y}_{S}c}{y}}\sum_{k=0}^{m_{S}-1}\frac{ \left(\frac{\mathrm{Y}_{S}c}{y}\right)^{k}}{k!}\right]\] (A11)  
Introduce (A11) into (A10) and rewrite it as follows,  
\[ \begin{split}& I_{1}=\frac{\mathrm{Y}_{D}^{m_{D}}}{\Gamma(m_{D})} \int_{y_{1}}^{\infty}y^{m_{D}-1}e^{-\mathrm{Y}_{D}y}\left(1-e^{-\frac{\mathrm{ Y}_{S}c}{y}}\sum_{k=0}^{m_{S}-1}\frac{\left(\mathrm{Y}_{S}c\right)^{k}y^{-k}}{k! }\right)dy\\ &=\frac{\mathrm{Y}_{D}^{m_{D}}}{\Gamma(m_{D})}\left(\int_{y_{1} }^{\infty}y^{m_{D}-1}e^{-\mathrm{Y}_{D}y}dy-\sum_{k=0}^{m_{S}-1}\frac{\left( \mathrm{Y}_{S}c\right)^{k}}{k!}\int_{y_{1}}^{\infty}y^{m_{D}-k-1}e^{-\mathrm{ Y}_{D}y-\frac{\mathrm{Y}_{S}c}{y}}dy\right)\end{split}