When the system beam outage probability is analyzed above, the transmission timeslot in uplink is evenly divided to each sensor for convenience of calculation. However, in practical applications, uplink sub-timeslots can be allocated more reasonably according to different application requirements. In order to further study how to improve the system performance, we aim to maximize the energy efficiency (EE) of the system, and realize the efficient green communications by optimizing the allocations of WET sub-timeslot in downlink and WIT sub-timeslots based on TDMA operation. Therefore, we aim to maximize the EE of the considered system by optimizing the allocation of WIT sub-intervals for sensors in the service sector. The EE is defined as
$${\text{EE}} = \frac{{R_{{{\text{sum}}}} }}{{E_{T} }},$$
(17)
where
\(R_{{{\text{sum}}}}\) and
\(E_{T}\) denote the system throughput with
K sensors and total energy consumption, respectively, which are given by
$$R_{{{\text{sum}}}} { = }\sum\limits_{k = 1}^{K} {\alpha_{k} \log_{2} \left( {1 + \frac{{\alpha_{0} }}{{\alpha_{k} }}B_{k} } \right)} ,$$
(18)
$$E_{T} { = }\sum\limits_{k = 1}^{K} {\alpha_{0} C_{k} } ,$$
(19)
with
\(B_{k} { = }\frac{{\eta P_{{{\text{HAP}}}} \left| {g_{k} } \right|^{4} }}{{\sigma_{0}^{2} \left( {1 + d_{k}^{\beta } } \right)^{2} }}F_{M}^{2} \left( {\delta_{k} } \right)\) and
\(C_{k} { = }\frac{{\eta P_{{{\text{HAP}}}} \left| {g_{k} } \right|^{2} }}{{1 + d_{k}^{\beta } }}F_{M} \left( {\delta_{k} } \right)\). Note that in the TDMA mode, the HAP can rely on channel reciprocity to acquire full CSI of uplink when the uplink and downlink share the same band [
43]. According to the channel reciprocity, the result of effective complex gain for uplink
\(\left| {g_{k}^{\prime } } \right|^{2}\) in (9) is equal to
\(\left| {g_{k} } \right|^{2}\).
In (20),
\(\alpha_{0}\) and
\(\left\{ {\alpha_{k} } \right\}\) represent the downlink energy transmission duration and uplink data transmission duration for each sensor, respectively. Therefore, the constraint C1 in (20) indicates that the sum of
\(\alpha_{0}\) and
\(\left\{ {\alpha_{k} } \right\}\) is less than or equal to a whole transmission interval. In the constraint C2, we set
\(\alpha_{0} > 0\) and
\(\alpha_{k} \ge 0\), where
\(\alpha_{k} = 0\) occurs when the
k-th sensor does not harvest RF energy. To obtain the optimal pair of
\(\left( {\alpha_{0} ,\left\{ {\alpha_{k} } \right\}} \right)\), we first prove that the objective function in (20) is a concave function. Since the constraint C1
\(\, \alpha_{0} + \sum\nolimits_{k = 1}^{K} {\alpha_{k} } \le T\) illustrating the parameters between the
\(\alpha_{0}\) and
\(\left\{ {\alpha_{k} } \right\}\) are coupled, we thus need to the concavity of the subject function when
\(\, \alpha_{0}\) is fixed. Correspondingly, the second derivative of the subjective function can be equivalently expressed as
$$\begin{gathered} f^{\prime \prime } \left( {\alpha_{k} } \right) = \left( {\alpha_{k} \log_{2} \left( {1 + \frac{{\alpha_{0} }}{{\alpha_{k} }}B_{k} } \right)} \right)^{\prime \prime } \hfill \\ = - \frac{{\alpha_{0}^{2} B_{k}^{2} }}{{\alpha_{k}^{3} + 2\alpha_{0} B_{k} \alpha_{k}^{2} + \alpha_{0}^{2} B_{k}^{2} \alpha_{k} }} < 0, \hfill \\ \end{gathered}$$
(21)
which means that optimization problem (P1) is actually a concave optimization problem [
44]. Therefore, the optimal solution can be obtained by transferring the objective function in (P1) into parametric form and then adopting the Lagrange dual multipliers method. The parametric form of objective function in optimization problem (P1) can be rewritten as
$$F\left( \theta \right) = \mathop {\max }\limits_{{\alpha_{0} ,\left\{ {\alpha_{k} } \right\}}} \, R_{{{\text{sum}}}} - \theta E_{T} ,$$
(22)
where
\(\theta\) denotes a non-negative parameter. For a given
\(\theta\), the Lagrangian function of optimization problem (P1) is given by
$$\begin{gathered} L\left( {\alpha_{0} ,\left\{ {\alpha_{k} } \right\}} \right) = \sum\limits_{k = 1}^{K} {\left[ {\alpha_{k} \log_{2} \left( {1 + \frac{{\alpha_{0} }}{{\alpha_{k} }}B_{k} } \right) - \theta \alpha_{0} C_{k} } \right]} \hfill \\ { + }\xi \left[ {T - \alpha_{0} - \sum\limits_{k = 1}^{K} {\alpha_{k} } } \right], \hfill \\ \end{gathered}$$
(23)
where
\(\xi\) is Lagrangian coefficient. Based on the complementary slackness, the optimal lagrange multipliers are zero for constrain C2. We thus omit the corresponding lagrangian multipliers in (23). According to the KKT conditions of lagrangian function (23), we have
$$\frac{\partial L}{{\partial \alpha_{0} }} = \sum\limits_{k = 1}^{K} {\left[ {\frac{{B_{k} \log_{2} \left( e \right)}}{{B_{k} {{\alpha_{0} } \mathord{\left/ {\vphantom {{\alpha_{0} } {\alpha_{k} }}} \right. \kern-0pt} {\alpha_{k} }} + 1}} - \theta C_{k} } \right] - \xi } { = }0,$$
(24)
$$\frac{\partial L}{{\partial \alpha_{k} }} = \left[ {\log_{2} \left( {1 + \frac{{\alpha_{0} B_{k} }}{{\alpha_{k} }}} \right)} \right.\left. { - \left( {\frac{{B_{k} }}{{1 + B_{k} {{\alpha_{0} } \mathord{\left/ {\vphantom {{\alpha_{0} } {\alpha_{k} }}} \right. \kern-0pt} {\alpha_{k} }}}}\frac{{\alpha_{0} }}{{\alpha_{k} }}} \right)\log_{2} \left( e \right)} \right] - \xi { = }0,$$
(25)
$$\xi \left( {T - \alpha_{0} - \sum\limits_{k = 1}^{K} {\alpha_{k} } } \right){ = }0,$$
(26)
respectively. From (25), result of
\(\xi\) can be obtained and then substituting it into (24), we have
$$\begin{gathered} \sum\limits_{k = 1}^{K} {\left[ {\frac{{B_{k} \log_{2} \left( e \right)}}{{B_{k} \Delta_{k} + 1}} - \theta C_{k} } \right]} - \log_{2} \left( {1 + B_{k} \Delta_{k} } \right) \hfill \\ { + }\left( {\frac{{B_{k} \Delta_{k} }}{{1 + B_{k} \Delta_{k} }}} \right)\log_{2} \left( e \right){ = }0, \hfill \\ \end{gathered}$$
(27)
where
\(\Delta_{k} = {{\alpha_{0} } \mathord{\left/ {\vphantom {{\alpha_{0} } {\alpha_{k} }}} \right. \kern-0pt} {\alpha_{k} }}\). Therefore, all
\(\Delta_{k} {\text{s}}\) \(\forall k \in \left\{ {1,2, \ldots ,K} \right\}\) can be obtained from (27). For (26), we have
\(T{ = }\alpha_{0} { + }\sum\nolimits_{k = 1}^{K} {\alpha_{k} }\). The optimal solution for pair of
\(\left( {\alpha_{0} ,\left\{ {\alpha_{k} } \right\}} \right)\) can be derived as follows
$$\alpha_{0} = \, \frac{T}{{1 + \sum\nolimits_{k = 1}^{K} {{1 \mathord{\left/ {\vphantom {1 {\Delta_{k} }}} \right. \kern-0pt} {\Delta_{k} }}} }}{\text{ and }}\alpha_{k} = \frac{{\alpha_{0} }}{{\Delta_{k} }}.$$
(28)