Wireless Power Transfer Under Secure Communication with Multiple Antennas and Eavesdroppers

Abstract

In this paper, we analyze the physical layer secrecy performance of a 5G radio frequency energy harvesting (RF-EH) network in the presence of multiple passive eavesdroppers. In this system, the source is considered as an energy-limited node, hence it harvests energy from RF signals generated by a power transfer station to use for information transmission. Additionally, in order to enhance the energy harvesting and system performance, the source is equipped with multiple antennas and employs maximal ratio combining (MRC) and transmit antenna selection (TAS) techniques to exploit the benefits of spatial diversity. Given these settings, the exact close-form expressions of existence probability of secrecy capacity and secrecy outage probability are derived. Furthermore, the obtained results indicate that multiple antennas technique applied at the source not only facilitates energy harvesting but also improves secrecy performance of the investigated network. Finally, Monte-Carlo simulation is provided to confirm our analytical results.

Appendices

Appendix A: Proof of Lemma 1

According to the probability definition, the joint CDF of X and Y is given by

$$\begin{aligned} {\begin{matrix} F_{X,Y}^{(k)}(x,y) &{}= \Pr \left( {X< x,Y< y} \right) \\ &{}= \Pr \left( {{\gamma _D}< \frac{x}{{a{\gamma _S}}},{\gamma _{E,k}} < \frac{y}{{a{\gamma _S}}}} \right) \\ &{}= \int \limits _0^\infty {{F_{{\gamma _D}}}\left( {\frac{x}{{az}}} \right) {F_{{\gamma _{E,k}}}}\left( {\frac{y}{{az}}} \right) {f_{{\gamma _S}}}(z)dz}. \end{matrix}} \end{aligned}$$

(A.1)

Substituting (2), (7), and (8) into (A.1) and then using ([16], Eq. (3.471.9)), the final result is obtained as shown in (10).

Appendix B: Proof of Proposition 1

By employing ([16], Eq. (6.561.16)) and ([16], Eq. (8.486.14)), (11) is expanded as

$$\begin{aligned} {\begin{matrix} {P_{CS,k}} &{}= \sum \limits _{p = 1}^\infty {\left( {\begin{array}{*{20}{c}} {{N_S}}\\ p \end{array}} \right) \frac{{{{\left( { - 1} \right) }^p}2p}}{{\varGamma \left( {{N_S}} \right) {2^{{N_S} - 1}}a{\lambda _S}{\lambda _D}}}\left\{ {\int \limits _0^\infty {v_x^{{N_S} - 1}{\mathcal{K}_{{N_S} - 1}}\left[ {{v_x}} \right] dx} } \right. } \\ &{}\qquad \qquad \qquad \qquad \qquad \qquad \qquad \left. { - \int \limits _0^\infty {{{\left( u \right) }^{{N_S} - 1}}{\mathcal{K}_{{N_S} - 1}}\left( u \right) dx} } \right\} \\ &{}= \sum \limits _{p = 1}^\infty {\left( {\begin{array}{*{20}{c}} {{N_S}} \\ p \end{array}} \right) \frac{{{{\left( { - 1} \right) }^p}}}{{\varGamma \left( {{N_S}} \right) {2^{{N_S} - 1}}}}\left\{ {\frac{{p{\lambda _{E,k}}}}{{p{\lambda _{E,k}} + {\lambda _D}}}\int \limits _0^\infty {{t^{{N_S}}}{\mathcal{K}_{{N_S} - 1}}\left( t \right) dx} } \right. } \qquad \\ &{}\qquad \qquad \qquad \qquad \qquad \qquad \qquad \left. { - \int \limits _0^\infty {{t^{{N_S}}}{\mathcal{K}_{{N_S} - 1}}\left( t \right) dx} } \right\} \\ &{}= \sum \limits _{p = 1}^\infty {\left( {\begin{array}{*{20}{c}} {{N_S}}\\ p \end{array}} \right) \frac{{{{\left( { - 1} \right) }^{p + 1}}{\lambda _D}}}{{p{\lambda _{E,k}} + {\lambda _D}}}}, \end{matrix}} \end{aligned}$$

(B.1)

where, ${v_x} = 2\sqrt{\frac{{\left( {p{\lambda _{E,k}} + {\lambda _D}} \right) x}}{{a{\lambda _S}{\lambda _D}{\lambda _{E,k}}}}}$ and $u = {2\sqrt{\frac{{px}}{{a{\lambda _S}{\lambda _D}}}}}$.

In the presence of K eavesdroppers, the existence probability of secrecy capacity is given by

$$\begin{aligned} {\begin{matrix} {P_{CS}} &{}= \Pr \left( {{C_{S,1}}> 0,\mathrm{{ }}...,\mathrm{{ }}{C_{S,K}}> 0} \right) \\ &{}= \prod \limits _{k = 1}^K {\Pr \left( {{C_{S,k}} > 0} \right) }. \end{matrix}} \end{aligned}$$

(B.2)

The final result of $P_{CS}$ in (12) is obtained by substituting (B.1) into (B.2).

Appendix C: Proof of Proposition 2

Similar to the process of calculating (11) in Appendix B, the integral in (13) is solved by the help of ([16], Eq. (6.561.16)) and ([16], Eq. (8.486.14)) and the result is indicated in (C.1) as follows:

$$\begin{aligned} {\begin{matrix} {P_{Out,k}} &{}= \frac{2}{{\varGamma \left( {{N_S}} \right) {2^{{N_S} - 1}}a{\lambda _S}{\lambda _{SE,k}}}}\left\{ {\int \limits _0^\infty {{{\left( t \right) }^{{N_S} - 1}}{\mathcal{K}_{{N_S} - 1}}\left( t \right) dy} + \sum \limits _{p = 1}^{{N_S}} {\left( {\begin{array}{*{20}{c}} {{N_S}}\\ p \end{array}} \right) {{\left( { - 1} \right) }^p}} } \right. \\ &{}\qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \quad \left. { \times \int \limits _0^\infty {{{\left[ w \right] }^{{N_S} - 1}}{\mathcal{K}_{{N_S} - 1}}\left[ w \right] dy} } \right\} \\ &{}= \frac{1}{{\varGamma \left( {{N_S}} \right) {2^{{N_S} - 1}}}}\left\{ {\int \limits _0^\infty {{t^{{N_S}}}{\mathcal{K}_{{N_S} - 1}}\left( t \right) dt} } \right. \qquad \qquad \qquad \qquad \qquad \qquad \quad \\ &{}\qquad \qquad \qquad \qquad \left. { + \sum \limits _{p = 1}^{{N_S}} {\left( {\begin{array}{*{20}{c}} {{N_S}}\\ p \end{array}} \right) \frac{{{{\left( { - 1} \right) }^p}{\lambda _D}}}{{p{2^R}{\lambda _{E,k}} + {\lambda _D}}}\int \limits _u^\infty {{t^{{N_S}}}{\mathcal{K}_{{N_S} - 1}}\left( t \right) dt} } } \right\} \\ &{}= 1 - \sum \limits _{p = 1}^{{N_S}} {\left( {\begin{array}{*{20}{c}} {{N_S}}\\ p \end{array}} \right) \frac{{{{\left( { - 1} \right) }^{p + 1}}{\lambda _D}}}{{\varGamma \left( {{N_S}} \right) {2^{{N_S} - 1}}\left( {p{2^R}{\lambda _{E,k}} + {\lambda _D}} \right) }}} \\ &{}\qquad \qquad \qquad \times {\left[ {2\sqrt{\frac{{p\left( {{2^R} - 1} \right) }}{{a{\lambda _S}{\lambda _D}}}} } \right] ^{{N_S}}}{\mathcal{K}_{{N_S}}}\left[ {2\sqrt{\frac{{p\left( {{2^R} - 1} \right) }}{{a{\lambda _S}{\lambda _D}}}} } \right] , \end{matrix}} \end{aligned}$$

(C.1)

where, $t = 2\sqrt{\frac{y}{{a{\lambda _S}{\lambda _{E,k}}}}}$ and $w = 2\sqrt{\frac{{\left( {p{2^R}{\lambda _{E,k}} + {\lambda _D}} \right) y + p{\lambda _{E,k}}\left( {{2^R} - 1} \right) }}{{a{\lambda _S}{\lambda _D}{\lambda _{E,k}}}}}$.

For K eavesdroppers, the secrecy outage probability is computed as

$$\begin{aligned} {\begin{matrix} {P_{Out}} &{}= 1 - \Pr \left( {{C_{S,1}} \ge R,\mathrm{{ }}...,\mathrm{{ }}{C_{S,K}} \ge R} \right) \\ &{}= 1 - \prod \limits _{k = 1}^K {\Pr \left( {{C_{S,k}} \ge R} \right) } \\ &{}= 1 - \prod \limits _{k = 1}^K {\left[ {1 - \Pr \left( {{C_{S,k}} < R} \right) } \right] }. \end{matrix}} \end{aligned}$$

(C.2)

Substituting (C.1) into (C.2), the final result of $P_{Out}$ in (14) is derived.