# Hybrid Active-Passive Reconfigurable Intelligent Surface for Cooperative Transmission Systems
## 3 Performance Analysis
### OP of the Full Active RIS System  
For \(h_{S}\[PERSON](m_{S},\Omega_{S})\), \(h_{D}\[PERSON](m_{D},\Omega_{D})\), and according to \(Y=X^{2}\) and \(F_{Y}(y)=F_{X}(\sqrt{y})\), we obtain the probability density function (PDF) and CDF of \(h_{S}^{2}\) and \(h_{D}^{2}\), respectively, as follows,  
\[f_{h_{S}^{2}}(x)=\frac{\chi_{S}^{m_{S}}}{\Gamma(m_{S})}x^{m_{S}-1}e^{-\mathrm{ Y}_{S}x} \tag{13}\]  
\[f_{h_{D}^{2}}(x)=\frac{\chi_{D}^{m_{D}}}{\Gamma(m_{D})}x^{m_{D}-1}e^{-\mathrm{ Y}_{D}x} \tag{14}\]  
\[F_{h_{S}^{2}}(x)=1-\frac{\Gamma(m_{S},\mathrm{Y}_{S})}{\Gamma(m_{S})} \tag{15}\]  
\[F_{h_{D}^{2}}(x)=1-\frac{\Gamma(m_{D},\mathrm{Y}_{D})}{\Gamma(m_{D})} \tag{16}\]  
where \(\mathrm{Y}_{S}=\frac{m_{S}}{\Omega_{S}}\) and \(\mathrm{Y}_{D}=\frac{m_{D}}{\Omega_{D}}\). For the convenience of the subsequent calculation, we take \(m_{S}\) and \(m_{D}\) as the integer values here. Now, to make it easier to compute OP, we simplify (10) to (17),  
\[\gamma_{a}=\frac{N^{2}P_{S}P_{A}h_{S}^{2}h_{D}^{2}}{P_{S}\sigma_{D}^{2}h_{S}^{ 2}+NP_{A}\sigma_{R}^{2}h_{D}^{2}}=\frac{\gamma_{A}\gamma_{B}}{\gamma_{A}+ \gamma_{B}} \tag{17}\]  
where \(\gamma_{A}=\frac{NP_{S}}{\sigma_{R}^{2}}h_{S}^{2}\) and \(\gamma_{B}=\frac{N^{2}P_{A}}{\sigma_{D}^{2}}h_{D}^{2}\). In the next step, using (17), and making \(\gamma_{A}=x\) and \(\gamma_{B}=y\), the CDF of \(\gamma_{a}\) can be indicated as  
\[F_{\gamma_{a}}(z)=\text{Pr}\bigg{(}\frac{\gamma_{A}\gamma_{B}}{\gamma_{A}+ \gamma_{B}}\leq Z\bigg{)}=1-\int_{z}^{\infty}\text{Pr}\bigg{(}\gamma_{A}\geq \frac{\gamma_{B}Z}{\gamma_{B}-Z}\bigg{)}f_{\gamma_{B}}(y)dy \tag{18}\]  
The above equation can be further expressed as  
\[F_{\gamma_{a}}(z)=1-\int_{z}^{\infty}\bigg{[}1-F_{\gamma_{A}}\bigg{(}\frac{ yz}{y-z}\bigg{)}\bigg{]}f_{\gamma_{B}}(y)dy \tag{19}\]  
By replacing \(y-z\) with the integral variable \(t\) and bringing in the above integral, CDF of \(\gamma_{a}\) can be written as  
\[F_{\gamma_{a}}(z)=1-\int_{0}^{\infty}[1-F_{\gamma_{A}}(\frac{(t+z)z}{t})]f_{ \gamma_{B}}(t+z)dt \tag{20}\]  
Next, we solve \(F_{\gamma_{A}}(x)\) and \(f_{\gamma_{B}}(y)\). These can be obtained by combining (14) and (15) with \(Y=\rho x\), \(f_{Y}(y)=\frac{1}{\rho}f_{X}\bigg{(}\frac{y}{\rho}\bigg{)}\), and \(F_{Y}(y)=F_{X}\bigg{(}\frac{y}{\rho}\bigg{)}\), respectively. \[F_{\gamma_{A}}(x)=1-\frac{\Gamma\bigg{(}m_{S},\frac{\mathrm{Y}_{S}\sigma_{R}^ {2}}{\mathrm{N}P_{S}}x\bigg{)}}{\Gamma(m_{S})} \tag{21}\]  
\[f_{\gamma_{B}}(y)=\frac{\mathrm{Y}_{D}^{m_{D}}}{\Gamma(m_{D})}\Bigg{(}\frac{ \sigma_{D}^{2}}{NP_{A}}\Bigg{)}^{2}y^{m_{D}-1}e^{-\frac{\mathrm{Y}_{D}c_{D}^ {2}}{N^{2}P_{A}}y} \tag{22}\]  
Insert (21) and (22) into (20) to obtain (23),\[ \begin{split}& F_{\gamma_{th}}(z)=1-\left(\frac{\mathrm{Y_{D}}\sigma_{D}^{2 }}{N^{2}P_{A}}\right)^{m_{D}}\frac{e^{-\frac{\mathrm{Y_{D}}\sigma_{D}^{2}}{N^{2 }P_{A}}z}}{\Gamma(m_{S})\Gamma(m_{D})}\\ &\times\int_{0}^{\infty}\Gamma\left(m_{S},\frac{\mathrm{Y_{S}} \sigma_{R}^{2}}{NP_{S}}\left(z+\frac{z^{2}}{t}\right)\right)(t+z)^{m_{D}-1}e^ {-\frac{\mathrm{Y_{D}}\sigma_{D}^{2}}{N^{2}P_{A}}t}dt\end{split}  \tag{23}\]  
Making use of ([28], (8.353.4)),  
\[\Gamma\left(m_{S},\frac{\mathrm{Y_{S}}\sigma_{R}^{2}}{NP_{S}}\left(z+\frac{z^{ 2}}{t}\right)\right)=(m_{S}-1)te^{-\frac{\mathrm{Y_{S}}\sigma_{R}^{2}}{NP_{S}} \left(z+\frac{z^{2}}{t}\right)}\sum_{k=0}^{m_{S}-1}\frac{\left(\frac{\mathrm{ Y_{S}}\sigma_{R}^{2}}{NP_{S}}\left(z+\frac{z^{2}}{t}\right)\right)^{k}}{k!} \tag{24}\]  
Bringing (24) into (23) and then by using ([28], (1.111) (3.478.4)), we can solve \(F_{\gamma_{a}}\), as shown in (25),  
\[