## 3 Performance Analysis
### Transmit Power of \(\mathrm{SS}_{1}\), \(\mathrm{SS}_{2}\) and \(\mathrm{SR}_{b}\)  
As proposed in [7; 8; 9], let \(\varepsilon_{\mathrm{OP}}\) denote the target QoS of the primary network, i.e., \(\mathrm{OP_{P1}}\leq\varepsilon_{\mathrm{OP}}\) and \(\mathrm{OP_{P2}}\leq\varepsilon_{\mathrm{OP}}\). To find the transmit power of \(\mathrm{SS}_{1}\), \(\mathrm{SS}_{2}\), and \(\mathrm{SR}_{b}\), we have to solve the following equations: \(\mathrm{OP_{P1}}=\varepsilon_{\mathrm{OP}}\) and \(\mathrm{OP_{P2}}=\varepsilon_{\mathrm{OP}}\). Then, using (25) to solve \(\mathrm{OP_{P2}}=\varepsilon_{\mathrm{OP}}\), we have  
\[\rho_{3}=(\Delta-1)\frac{\Omega_{\mathrm{SRPR}}}{\Omega_{\mathrm{TPTR}}} \Rightarrow P_{\mathrm{SR}_{b}}=\frac{\Delta-1}{\theta_{\mathrm{Ph}}}\frac{ \Omega_{\mathrm{SRPR}}}{\Omega_{\mathrm{TPTR}}}P_{\mathrm{PT}}, \tag{28}\]  
where  
\[\Delta=\frac{1}{1-(\varepsilon_{\mathrm{OP}})^{\frac{1}{N_{\mathrm{T}}N_{ \mathrm{R}}}}}\exp(-\Omega_{\mathrm{TPTR}}\rho_{0}). \tag{29}\]  
Because \(P_{\mathrm{SR}_{b}}\) is not negative, Equation (28) is re-written as follows:  
\[P_{\mathrm{SR}_{b}}=\biggl{[}\frac{\Delta-1}{\theta_{\mathrm{Ph}}}\frac{ \Omega_{\mathrm{SRPR}}}{\Omega_{\mathrm{TPTR}}}P_{\mathrm{PT}}\biggr{]}^{+}, \tag{30}\]  
where \([x]^{+}=\max(0,x)\). From (27)-(30), it is straightforward that \(P_{\mathrm{SR}_{b}}>0\) if  
\[\Delta-1>0 \Leftrightarrow[1-\exp(-\Omega_{\mathrm{TPTR}}\rho_{0})]^{N_{ \mathrm{T}}N_{\mathrm{R}}}<\varepsilon_{\mathrm{OP}}\] \[\Leftrightarrow\mathrm{OP_{P0}}<\varepsilon_{\mathrm{OP}}. \tag{31}\]  
**Remark 6**. : _Equation (31) implies that the primary network only shares the licensed band with the secondary network if the OP of the primary network without the interference from the secondary network \((\mathrm{OP_{P0}})\) is less than the required QoS. Otherwise, if \(\mathrm{OP_{P0}}\geq\varepsilon_{\mathrm{OP}}\), then \(SR_{b}\) is not allowed to access the licensed band, i.e., \(P_{SR_{b}}=0\). It is also noted from (30) that the transmit power of the secondary relays is the same._  
Then, we consider \(P_{\mathrm{SS}_{1}}\) and \(P_{\mathrm{SS}_{2}}\). Similarly to the power allocation method proposed in [7; 8], we have the following formula:  
\[\frac{P_{\mathrm{SS}_{1}}}{\Omega_{\mathrm{SS}_{1}\mathrm{PR}}}=\frac{P_{ \mathrm{SS}_{2}}}{\Omega_{\mathrm{SS}_{2}\mathrm{PR}}}=\delta. \tag{32}\]  
Equation (32) presents that the average interference power at PR (due to the data transmission of \(\mathrm{SS}_{1}\) and \(\mathrm{SS}_{2}\)) is the same or \(P_{\mathrm{SS}_{1}}\bigl{(}d_{\mathrm{SS}_{1}\mathrm{PR}}\bigr{)}^{-\mu}=P_{ \mathrm{SS}_{2}}\bigl{(}d_{\mathrm{SS}_{2}\mathrm{PR}}\bigr{)}^{-\mu}\). Therefore, if \(d_{\mathrm{SS}_{1}\mathrm{PR}}\geq d_{\mathrm{SS}_{2}\mathrm{PR}}\), then \(P_{\mathrm{SS}_{1}}\geq P_{\mathrm{SS}_{2}}\) and vice versa. Now, combining (24) and (32), we obtain  
\[\mathrm{OP_{P1}}=\biggl{[}1-\biggl{(}\frac{P_{\mathrm{PT}}}{P_{\mathrm{PT}}+ \Omega_{\mathrm{TPTR}}\theta_{\mathrm{Ph}}\delta}\biggr{)}^{2}\exp(-\Omega_{ \mathrm{TPTR}}\rho_{0})\biggr{]}^{N_{\mathrm{T}}N_{\mathrm{R}}}. \tag{33}\]Then, solving \(\text{OP}_{\text{P1}}=\epsilon_{\text{OP}}\), we can find \(P_{\text{SS}_{1}}\) and \(P_{\text{SS}_{2}}\), respectively, as  
\[P_{\text{SS}_{1}}=\Bigg{[}\frac{\sqrt{\Delta}-1}{\theta_{\text{Pth}}}\frac{ \Omega_{\text{SS}_{1}\text{PR}}}{\Omega_{\text{PTPR}}}P_{\text{PT}}\Bigg{]}^{+ },P_{\text{SS}_{2}}=\Bigg{[}\frac{\sqrt{\Delta}-1}{\theta_{\text{Pth}}}\frac{ \Omega_{\text{SS}_{2}\text{PR}}}{\Omega_{\text{PTPR}}}P_{\text{PT}}\Bigg{]}^{+ }. \tag{34}\]  
**Remark 7**. : _Similarly, the condition for \(P_{\text{SS}_{1}},P_{\text{SS}_{2}}>0\) is \(OP_{\text{P0}}<\epsilon_{\text{OP}}\).