Power Adaptation for Energy Efficient Bi-directional Relaying with Quality-of-Service Requirement and Individual Peak-Power Limit

Abstract

This paper addresses an optimal power adaptation (PA) problem of a two-time-slot bi-directional relaying network with a half-duplex amplify-and-forward relay. Unlike the existing studies, our goal is to develop effective PA strategies that can dynamically adjust the transmit-power levels of all the terminals to achieve energy efficiency, while satisfying the individual peak-power limit on each terminal and the quality-of-service (QoS) requirement of the network. By using the instantaneous channel state information (ICSI) and the statistical CSI (SCSI) knowledge, respectively, and with the aid of traffic information, the PA problem is analytically solved, leading to the so-called ICSI and SCSI based PA strategies with closed-form PA solutions for individual transmit-powers at the relay and the two end-terminals. Simulation results have verified the correctness of the derived expressions and confirmed the efficiency of our proposed strategies. It is shown that the proposed PA strategies can significantly reduce the total transmit-power of the network with guaranteed network QoS.

Appendices

Appendix 1

By the observation of (8), it can be found that the global minimum of (8) is achieved only when $\gamma _A = \gamma _B = \upsilon$, and the final solution of (8), i.e., $\left( {{\widehat{E}}_A ,\widehat{E}_B ,{\widehat{E}}_R } \right)$, yields the maximum $\gamma _A$ and $\gamma _B$ under the constraint $E_A + E_B + E_R = {\widehat{E}}_A + {\widehat{E}}_B + {\widehat{E}}_R$. Thus, problem (8) can be decomposed into the following two sub-problems.

$$\begin{aligned} {\text{Sub}}{\text{-}}{\text{problem}} 1: ~ \left( {E_A^{\mathrm{{I}}} ,E_B^{\mathrm{{I}}} ,E_R^{\mathrm{{I}}} } \right) = \arg \mathop {\max }\limits _{E_A ,E_B ,E_R } \gamma _A \end{aligned}$$

(35a)

$$\begin{aligned} ~{\mathrm{{subject\;to }}} \; \gamma _B = \gamma _A \end{aligned}$$

(35b)

$$\begin{aligned} E_A + E_B + E_R = E_T \end{aligned}$$

(35c)

$$\begin{aligned} E_A> 0,E_B> 0,E_R > 0 \end{aligned}$$

(35d)

$$\begin{aligned} {\text{Sub}}{\text{-}}{\text{problem}} 2: ~ \left( {E_A^{\mathrm{{II}}} ,E_B^{\mathrm{{II}}} ,E_R^{\mathrm{{II}}} } \right) = \arg \mathop {\min }\limits _{E_A^1 ,E_B^1 ,E_R^1 } \frac{{E_A^{\mathrm{{I}}} + E_B^{\mathrm{{I}}} + E_R^{\mathrm{{I}}} }}{2} \end{aligned}$$

(36a)

$$\begin{aligned} {\mathrm{{subject\;to }}}\quad \gamma _A = \upsilon \end{aligned}$$

(36b)

It’s worth-mentioning that a widely used approximation of the received SNR at the end-terminal as given by (37) is employed to facilitate the solution of (35) and (36).

$$\begin{aligned} \gamma _i \approx \frac{{E_j E_R \left| {h_{AR} } \right| ^2 \left| {h_{BR} } \right| ^2 }}{{\left( {E_i + E_R } \right) \left| {h_{iR} } \right| ^2 + E_j \left| {h_{jR} } \right| ^2 }} \end{aligned}$$

(37)

where $j = \left\{ {A,B} \right\} - i$. Then, by using some algebraic manipulations, (35) can be solved and its solution can be given by

$$\begin{aligned} \left\{ \begin{aligned}&E_A^{\mathrm{{I}}} = \frac{{\left| {g_{BR} } \right| E_T }}{{2\left( {\left| {g_{AR} } \right| + \left| {g_{BR} } \right| } \right) }} \\&E_B^{\mathrm{{I}}} = \frac{{\theta \left| {g_{AR} } \right| E_T }}{{2\left( {\left| {g_{AR} } \right| + \left| {g_{BR} } \right| } \right) }} \\&E_R^{\mathrm{{I}}} = \frac{{\left( {\left| {g_{AR} } \right| + \theta \left| {g_{BR} } \right| } \right) E_T }}{{2\left( {\left| {g_{AR} } \right| + \left| {g_{BR} } \right| } \right) }} \\ \end{aligned} \right. . \end{aligned}$$

(38)

Next, substituting (38) into (36) and performing some simple calculations, (36) can be solved straightforwardly and its solution can be given by the following expressions.

$$\begin{aligned} E_A^{\mathrm{{II}}} & = \frac{{\upsilon \left( {\left| {g_{AR} } \right| + \left| {g_{BR} } \right| } \right) }}{{\left| {g_{AR} } \right| ^2 \left| {g_{BR} } \right| }} \end{aligned}$$

(39)

$$\begin{aligned} E_B^{\mathrm{{II}}} & = \frac{{\upsilon \left( {\left| {g_{AR} } \right| + \left| {g_{BR} } \right| } \right) }}{{\left| {g_{AR} } \right| \left| {g_{BR} } \right| ^2 }} \end{aligned}$$

(40)

$$\begin{aligned} E_R^{\mathrm{{II}}} & = \frac{{\upsilon \left( {\left| {g_{AR} } \right| + \left| {g_{BR} } \right| } \right) ^2 }}{{\left| {g_{AR} } \right| ^2 \left| {g_{BR} } \right| ^2 }} \end{aligned}$$

(41)

It’s noteworthy that (39)–(41) can be considered as the final solution of (8). However, since the approximate SNRs of the two end-terminals given by (37) are used to solve the optimization problems, the end-to-end data rates of the two ends cannot be stringently satisfied. To this end, we propose a scheme to modify the solution obtained above. By the observation, one can find that the received SNRs of the two end-terminals are both monotonically increasing functions of $E_R$. Therefore, to satisfy the data rates of the two end-terminals, we modify $E_R^{\mathrm{{II}}}$ in (41). Our idea is to increase the transmit-power level of $T_R$ such that the resulting SNRs of the two end-terminals can satisfy the initial requirement. As explained, substituting $E_A^{\mathrm{{II}}}$ and $E_B^{\mathrm{{II}}}$ into (37) and letting $\gamma _A = \gamma _B = \upsilon$ yields

$$\begin{aligned} {\widehat{E}}_R = \frac{{\upsilon \left( {\left| {g_{AR} } \right| + \left| {g_{BR} } \right| } \right) ^2 + \left| {g_{AR} } \right| \left| {g_{BR} } \right| }}{{\left| {g_{AR} } \right| ^2 \left| {g_{BR} } \right| ^2 }} \end{aligned}$$

(42)

So the final solution of (8) can be given by (39), (40) and (42).

Appendix 2

First of all, it can readily be obtained that ${\widehat{E}}_R$ is the maximum of ${\widehat{E}}_A$, ${\widehat{E}}_B$ and ${\widehat{E}}_R$. Thus, the MSOP can be rewritten as

$$\begin{aligned} \Pr \left( {{\widehat{E}}_R> P_L } \right) = \Pr \left[ {\frac{{\upsilon \left( {\left| {g_{AR} } \right| + \left| {g_{BR} } \right| } \right) ^2 + \left| {g_{AR} } \right| \left| {g_{BR} } \right| }}{{\left| {g_{AR} } \right| ^2 \left| {g_{BR} } \right| ^2 }} > P_L } \right] . \end{aligned}$$

(43)

Based on ${{a \mathord{\left/ {\vphantom {a b}} \right. \kern-\nulldelimiterspace} b} + {b \mathord{\left/ {\vphantom {b a}} \right. \kern-\nulldelimiterspace} a} \ge 2} {\text{for}} a,b > 0$ for $a,b > 0$, we have

$$\begin{aligned} \frac{{\upsilon \left( {\left| {g_{AR} } \right| + \left| {g_{BR} } \right| } \right) ^2 + \left| {g_{AR} } \right| \left| {g_{BR} } \right| }}{{\left| {g_{AR} } \right| ^2 \left| {g_{BR} } \right| ^2 }} \le \left( {\upsilon + 0.25} \right) \left( {\frac{1}{{\left| {g_{AR} } \right| }} + \frac{1}{{\left| {g_{BR} } \right| }}} \right) ^2. \end{aligned}$$

(44)

Thus, the MSOP can be upper-bounded as

$$\begin{aligned} \Pr \left[ {\max \left( {{\widehat{E}}_A ,{\widehat{E}}_B ,{\widehat{E}}_R } \right) > P_L } \right] \le \Pr \left[ {\left( {\upsilon + 0.25} \right) \left( {\frac{1}{{\left| {g_{AR} } \right| }} + \frac{1}{{\left| {g_{BR} } \right| }}} \right) ^2 } \right] . \end{aligned}$$

(45)

Then, after some algebraic manipulations on (45), we can obtain the upper-bound of the MSOP as presented in Proposition 2.