## 3 The Full-Duplex Dual Antenna Energy Harvesting  
### The Optimal Time Allocation Scheme  
The energy efficiency optimization problem is written as  
\[\underset{\alpha}{max}\xi(\alpha)=\frac{(1-\alpha)log_{2}\bigg{(}1+min\left\{ \frac{y_{h_{1}}|^{2}(1-\alpha)}{\frac{y_{h_{1}}|^{2}(1-\alpha)}{\eta\big{|}^{ 2}(y_{h_{1}}|^{2}+y_{h_{2}}|^{2})\alpha}},\frac{\eta P_{s}\big{(}y_{h_{1}}|^{2 }+|y_{h_{2}}|^{2}\big{)}|g\big{|}^{2}\alpha}{N_{0}(1-\alpha)}\right\}\bigg{)}}{ \beta P_{s}+P_{c}}. \tag{17}\]  
When the source node is equal to the relay node and the relay node to the target node, \(\frac{|y_{h_{1}}|^{2}(1-\alpha)}{\eta\big{|}^{2}(y_{h_{1}}|^{2}+|y_{h_{2}}|^{ 2})\alpha}=\frac{\eta P_{s}\big{(}y_{h_{1}}|^{2}+|y_{h_{2}}|^{2}\big{)}|g\big{|} ^{2}\alpha}{N_{0}(1-\alpha)}\). The time switch threshold can be calculated as  
\[\alpha_{1}=\frac{1}{1+\sqrt{\frac{\eta^{2}\big{|}f\big{|}g\big{|}^{2}\big{(} y_{h_{1}}|^{2}+y_{h_{2}}|^{2}\big{)}P_{s}}{\beta P_{s}+P_{c}}}}. \tag{18}\]  
Therefore, the optimal time allocation scheme to maximize the energy efficiency can be transformed into two processes. For the first case, \(0<\alpha<\alpha_{1}\), the time allocation problem to maximize energy efficiency can be presented as  
\[\underset{\alpha}{max}\xi(\alpha)=\frac{(1-\alpha)log_{2}\Big{(}1+\frac{a_{1} \alpha}{1-\alpha}\Big{)}}{\beta P_{s}+P_{c}}, \tag{19}\]  
\[\text{s.t.}0<\alpha<\alpha_{1}, \tag{20}\]  
where \(a_{1}=\frac{\eta|g\big{|}^{2}P_{s}(y_{h_{1}}|^{2}+|y_{h_{2}}|^{2})}{N_{0}}\).  
In order to require the time allocation factor of the maximum energy efficiency, the first derivative of the solution in (19) is  
\[\frac{\partial\xi(\alpha)}{\partial\alpha}=\frac{-log_{2}\Big{(}1+\frac{a_{1} \alpha}{1-\alpha}\Big{)}+\frac{a_{1}}{(1-\alpha)\left(1+\frac{1+a}{1-\alpha} \right)ln2}}{\beta P_{s}+Pc}. \tag{21}\]  
Therefore, we can get the optimal time allocation factor in the range (\(0<\alpha<\alpha_{1}\)).  
\[\alpha^{*}=\frac{exp\Big{(}W\Big{(}\frac{a_{1}-1}{e}\Big{)}+1\Big{)}-1}{a_{1} -1+exp\Big{(}W\Big{(}\frac{a_{1}-1}{e}\Big{)}+1\Big{)}}, \tag{22}\]  
where \(a_{1}=\frac{\eta|g|^{2}P_{S}\left(|h_{1}|^{2}+|h_{2}|^{2}\right)}{N_{0}}\), \(W(x)\) is the Lambert function, and \(W(x)\) is the solution of \(Wexp(W)=x\).  
For the second case, \(\alpha_{1}<\alpha<1\), we construct the energy efficiency optimization problem  
\[\underset{\alpha}{max}\xi(\alpha)=\frac{(1-\alpha)log_{2}\Big{(}1+\frac{1- \alpha}{a_{2}\alpha}\Big{)}}{\beta P_{s}+P_{c}}, \tag{23}\]  
\[\text{s.t.}\alpha_{1}<\alpha<1\]  
where \(a_{2}=\frac{\eta|f|^{2}\left(|h_{1}|^{2}+|h_{2}|^{2}\right)}{|h_{1}|^{2}}\).  
The derivative function of the Equation (27) can be obtained and expressed as  
\[\frac{\partial\xi(\alpha)}{\partial\alpha}=-\frac{1}{\beta P_{s}+P_{c}}\Bigg{(} log_{2}\Big{(}1+\frac{1-\alpha}{a_{2}\alpha}\Big{)}+\frac{(1-\alpha)}{a_{2} \alpha^{2}}*\frac{1}{1+\frac{(1-\alpha)}{a_{2}\alpha}}\Bigg{)}. \tag{24}\]  
Since \(\frac{\partial\xi(\alpha)}{\partial\alpha}\) is always less than zero in the range \(\alpha\in(\alpha_{1},1)\), the energy efficiency is a monotonically decreasing function in the range \(\alpha\in(\alpha_{1},1)\). At \(\alpha=\alpha_{1}\), the energy efficiency \(\xi(\alpha)\) achieves the maximum value within the range \(\alpha\in(\alpha_{1},1)\). Therefore, we present both cases as follows.  
\[\alpha^{*}=\begin{cases}\frac{exp\Big{(}W\Big{(}\frac{a_{1}-1}{e}\Big{)}+1 \Big{)}-1}{a_{1}-1+exp\Big{(}W\Big{(}\frac{a_{1}-1}{e}\Big{)}+1\Big{)}},&if \Big{(}\Big{(}\frac{a_{1}-1}{e}\Big{)}+1\Big{)}<ln\frac{a_{1}}{\sqrt{a_{1}a_{ 2}}}\\ \frac{1}{1+\sqrt{a_{1}a_{2}}},&otherwise\end{cases}. \tag{25}\]