Secrecy Outage Analysis of an Untrusted Relaying Energy Harvesting System with Multiple Eavesdroppers

Abstract

In this paper, we study the secrecy outage probability of an untrusted relaying energy-harvesting system in the presence of an eavesdropper network, which chooses the best eavesdropper to overhear much of the sources confidential signal. To degrade the overhearing capacity and give additional energy at the untrusted relay, the destination sends artificial noise signals during the communication. For secrecy outage evaluation, we derive the analytical expression for the secrecy outage probability (SOP) and a high-power approximation for the SOP. The accuracy of the analytical results is verified by Monte Carlo simulations.

Appendices

Appendix 1: Proof of Proposition 1

The CDF of $\gamma _e$ can be computed as

$$\begin{aligned} {F_{{\gamma _e}}}\left( y \right)&= \Pr \left( {\frac{{{\rho _s}{Y_1}}}{{{\rho _d}{Y_2} + 1}} < y} \right) \nonumber \\&= \int \limits _0^{ + \infty } {{F_{{Y_1}}}\left( {\frac{{y\left( {{\rho _d}{y_2} + 1} \right) }}{{{\rho _s}}}} \right) } {f_{{Y_2}}}\left( {{y_2}} \right) d{y_2}. \end{aligned}$$

(25)

The PDF of $\gamma _e$ can be computed as

$$\begin{aligned} \frac{{d{F_{{\gamma _e}}}\left( y \right) }}{{dy}} = \frac{{{\rho _d}{y_2} + 1}}{{{\rho _s}}}\int \limits _0^{ + \infty } {{f_{{Y_1}}}\left( {\frac{{y\left( {{\rho _d}{y_2} + 1} \right) }}{{{\rho _s}}}} \right) } {f_{{Y_2}}}\left( {{y_2}} \right) d{y_2}. \end{aligned}$$

(26)

Using the PDF of $Y_1$ given by (4) and PDF of $Y_2$ given by (1), (26) can be expressed as

$$\begin{aligned} {f_{{\gamma _e}}}\left( y \right)&= \frac{{{\lambda _{{Y_1}}}{\lambda _{{y_2}}}}}{{{\rho _s}}}{e^{ - \frac{{k{\lambda _{{Y_1}}}}}{{{\rho _s}}}y}}\sum \limits _{k = 1}^K {k \left( {\begin{array}{c}K\\ k\end{array}}\right) {{\left( { - 1} \right) }^{k + 1}}} \nonumber \\&\quad \times \int \limits _0^{ + \infty } {\left( {{\rho _d}{y_2} + 1} \right) {e^{ - \left( {\frac{{k{\lambda _{{Y_1}}}{\rho _d}y}}{{{\rho _s}}} + {\lambda _{{y_2}}}} \right) {y_2}}}} d{y_2}. \end{aligned}$$

(27)

Using integration by parts, (27) can be rewritten as (10).

Appendix 2: Proof of Proposition 2

Using the PDF of $X_2$ given by (1) and the CDF of $X_1$ given by (2), (16) can be calculated as

$$\begin{aligned} {P_\text {sec,1}} = \int \limits _0^{{{{{\bar{x}}}}_1}} {{\lambda _{{X_2}}}{e^{ - {\lambda _{{X_2}}}{x_2}}}} d{x_2} + \int \limits _{{{{{\bar{x}}}}_1}}^{ + \infty } {\left( {1 - {e^{\frac{{{\lambda _{{X_1}}}\left( {\beta - 1} \right) }}{{\left( {1 - \theta } \right) {\rho _s}\varXi \left( {{X_2}} \right) }}}}} \right) } {\lambda _{{X_2}}}{e^{ - {\lambda _{{X_2}}}{x_2}}}d{x_2}. \end{aligned}$$

(28)

After some manipulations, $P_\text {sec,1}$ can be expressed as in Proposition 2.

Appendix 3: Proof of Proposition 3

The expression in (18) can be rewritten as

$$\begin{aligned} {P_\text {sec,2}}&= \int \limits _{{{{{\bar{x}}}}_1}}^{ + \infty } {{f_{{{{X}}_2}}}\left( z \right) } \int \limits _{{\varXi _2}\left( {{X_2}} \right) }^{ + \infty } {{f_{{\gamma _e}}}\left( t \right) } \left( {F_{{X_1}}}\left( {\frac{{\beta t + \beta - 1}}{{\left( {1 - \theta } \right) {\rho _s}}}\left( {1 + \frac{\kappa }{z}} \right) } \right) \right. \nonumber \\&\quad \left. - {F_{{X_1}}}\left( {\frac{{\beta - 1}}{{\left( {1 - \theta } \right) {\rho _s}{\varXi _1}\left( z \right) }}} \right) \right) dyd{x_2} \nonumber \\&= \underbrace{\int \limits _{{{{{\bar{x}}}}_1}}^{ + \infty } {{f_{{X_2}}}\left( {{x_2}} \right) } {e^{ - \frac{{{\lambda _{{X_1}}}\left( {\beta - 1} \right) }}{{\left( {1 - \theta } \right) {\rho _s}{\varXi _1}\left( {{x_2}} \right) }}}}d{x_2}\int \limits _{{\varXi _2}\left( {{x_2}} \right) }^{ + \infty } {{f_{{\gamma _e}}}\left( y \right) } dy}_{{P_\text {sec,2,1}}} \nonumber \\&\quad - \underbrace{\int \limits _{{{{{\bar{x}}}}_1}}^{ + \infty } {{f_{{X_2}}}\left( {{x_2}} \right) } \int \limits _{{\varXi _2}\left( {{x_2}} \right) }^{ + \infty } {{f_{{\gamma _e}}}\left( y \right) } {e^{ - {\lambda _{{X_1}}}\frac{{\beta \left( {y + 1} \right) - 1}}{{\left( {1 - \theta } \right) {\rho _s}}}\left( {1 + \frac{\kappa }{{{x_2}}}} \right) }}dyd{x_2}}_{{P_\text {sec,2,2}}}. \end{aligned}$$

(29)

We focus on deriving the expression of $P_\text {sec,2,2}$. Substituting the PDF of $X_2$ given by (1) and the PDF of $\gamma _e$ given by (10) in (29), and letting $u = y - {\varXi _2}\left( {{x_2}} \right)$, $P_\text {sec2,2}$ can be calculated as

$$\begin{aligned} {P_\text {sec,2,2}}&= {\lambda _{{X_2}}}\sum \limits _{k = 1}^K { \left( {\begin{array}{c}K\\ k\end{array}}\right) {{\left( { - 1} \right) }^{k + 1}}} \int \limits _{{{{{\bar{x}}}}_1}}^{ + \infty } {{e^{ - {\lambda _{{X_2}}}{x_2}}}} {e^{ - \frac{{{\lambda _{{X_1}}}}}{{\left( {1 - \theta } \right) {\rho _s}}}\left( {1 + \frac{\kappa }{{{x_2}}}} \right) \left( {\beta {\varXi _2}\left( {{x_2}} \right) + \beta - 1} \right) - \frac{{k{\lambda _{{Y_1}}}}}{{{\rho _s}}}{\varXi _2}\left( {{x_2}} \right) }} \nonumber \\&\quad \times \left( {\underbrace{\int \limits _0^{ + \infty } {\frac{{{\lambda _{{y_2}}}{e^{ - \left( {\frac{{k{\lambda _{{Y_1}}}}}{{{\rho _s}}} + \frac{{\beta {\lambda _{{X_1}}}}}{{\left( {1 - \theta } \right) {\rho _s}}}\left( {1 + \frac{\kappa }{{{x_2}}}} \right) } \right) u}}}}{{{\rho _d}\left( {u + {\varXi _2}\left( {{X_2}} \right) + {{{\beta _Y}} / k}} \right) }}} du}_{{I_1}} + \underbrace{\int \limits _0^{ + \infty } {\frac{{{\beta _Y}{e^{ - \left( {\frac{{k{\lambda _{{Y_1}}}}}{{{\rho _s}}} + \frac{{\beta {\lambda _{{X_1}}}}}{{\left( {1 - \theta } \right) {\rho _s}}}\left( {1 + \frac{\kappa }{{{x_2}}}} \right) } \right) u}}}}{{k{{\left( {u + {\varXi _2}\left( {{X_2}} \right) + {{{\beta _Y}}/ k}} \right) }^2}}}} du}_{{I_2}}} \right) d{x_2}. \end{aligned}$$

(30)

Solving $I_1$ and $I_2$ with the help of [12, Eq. (3.352.4) and Eq. (3.353.3)], $P_\text {sec,2,2}$ can be calculated as

$$\begin{aligned} {P_\text {sec,2,2}}&= {P_{\sec ,2,1}} + \frac{{{\beta _Y}\beta {\lambda _{{X_1}}}{\lambda _{{X_2}}}}}{{\left( {1 - \theta } \right) {\rho _s}}}\sum \limits _{k = 1}^K {\left( {\begin{array}{c}K\\ k\end{array}}\right) \frac{{{{\left( { - 1} \right) }^{k + 1}}}}{k}} {e^{\frac{{{\lambda _{{Y_2}}}}}{{{\rho _d}}} + \frac{{{\lambda _{{X_1}}}}}{{\left( {1 - \theta } \right) {\rho _s}}}\left( {\frac{{\beta {\beta _Y}}}{k} - \beta + 1} \right) }} \nonumber \\&\quad \times \int \limits _{{{{{\bar{x}}}}_1}}^{ + \infty } {\left( {1 + \frac{\kappa }{{{x_2}}}} \right) } {e^{ - {\lambda _{{X_2}}}{x_2} + \frac{{{\lambda _{{X_1}}}\kappa }}{{\left( {1 - \theta } \right) {\rho _s}{x_2}}}\left( {\frac{{\beta {\beta _Y}}}{k} - \beta + 1} \right) }} \nonumber \\&\quad \times Ei\,\left( { - \frac{{{\varXi _2}\left( {{X_2}} \right) + {\textstyle {{{\beta _Y}} \over k}}}}{{{\rho _s}}}\left( {k{\lambda _{{Y_1}}} + \frac{{\beta {\lambda _{{X_1}}}}}{{\left( {1 - \theta } \right) }}\left( {1 + \frac{\kappa }{{{x_2}}}} \right) } \right) } \right) . \end{aligned}$$

(31)

By substituting (31) into (29), $P_\text {sec,2}$ can be expressed as in (19).

Appendix 4: Proof of Proposition 5

Substituting the PDF of $X_2$ given by (1), the PDF of $\gamma _e$ given by (10), and the CDF of $X_1$ given in (2), $P_\text {sec,2}^\infty$ can be calculated as

$$\begin{aligned} P_{{\text {sec,2}}}^\infty&= \int \limits _{{{{{\bar{x}}}}_0}}^{ + \infty } {{\lambda _{{X_2}}}{e^{ - {\lambda _{{X_2}}}{x_2}}}} {e^{ - {\lambda _{{X_1}}}\frac{{\beta - 1}}{{\left( {1 - \theta } \right) {\rho _s}}}\left( {1 + \frac{\kappa }{{{x_2}}}} \right) }} \nonumber \\&\quad \times \left( {1 - \sum \limits _{k = 1}^K { \left( {\begin{array}{c}K\\ k\end{array}}\right) {{\left( { - 1} \right) }^{k + 1}}} \int \limits _0^{ + \infty } {\left( {\frac{{{\lambda _{{y_2}}}{e^{ - {{{{\mathcal {J}}}}}\left( {{x_2}} \right) y}}}}{{{\rho _d}\left( {y + {\textstyle {{{\beta _Y}} \over k}}} \right) }} + \frac{{{\beta _Y}{e^{ - {{{{\mathcal {J}}}}}\left( {{x_2}} \right) y}}}}{{k{{\left( {y + {\textstyle {{{\beta _Y}} \over k}}} \right) }^2}}}} \right) } dy} \right) d{x_2}, \end{aligned}$$

(32)

where ${{{{\mathcal {J}}}}}\left( {{x_2}} \right) = \frac{{\beta {\lambda _{{X_1}}}}}{{\left( {1 - \theta } \right) {\rho _s}}}\left( {1 + \frac{\kappa }{{{x_2}}}} \right) + \frac{{k{\lambda _{{Y_1}}}}}{{{\rho _s}}}$.

Solving the integrals in (32) with the help of [12, Eq. (3.352.4) and (3.353.3)], after some manipulations, $P_\text {out,2}^\infty$ can be approximated as

$$\begin{aligned} P_{{\text {out,2}}}^\infty&= \frac{{{\beta _Y}\beta {\lambda _{{X_1}}}{\lambda _{{X_2}}}}}{{\left( {1 - \theta } \right) {\rho _s}}}\sum \limits _{k = 1}^K {\left( {\begin{array}{c}K\\ k\end{array}}\right) \frac{{{{\left( { - 1} \right) }^k}}}{k}} \nonumber \\&\quad \times \int \limits _{{{{{\bar{x}}}}_0}}^{ + \infty } {\left( {1 + \frac{\kappa }{{{x_2}}}} \right) {e^{ - {\lambda _{{X_2}}}{x_2}}}} Ei\left( { - {{{{\mathcal {J}}}}}\left( {{x_2}} \right) {\textstyle {{{\beta _Y}} \over k}}} \right) d{x_2}. \end{aligned}$$

(33)

Using [12, Eq.(8.214.1)], (33) can be approximated as

$$\begin{aligned} P_{{\text {out,2}}}^\infty&= \frac{{{\beta _Y}\beta {\lambda _{{X_1}}}{\lambda _{{X_2}}}}}{{\left( {1 - \theta } \right) {\rho _s}}}\sum \limits _{k = 1}^K { \left( {\begin{array}{c}K\\ k\end{array}}\right) \frac{{{{\left( { - 1} \right) }^k}}}{k}} \int \limits _{{{{{\bar{x}}}}_0}}^{ + \infty } {\left( {1 + \frac{\kappa }{{{x_2}}}} \right) {e^{ - {\lambda _{{X_2}}}{x_2}}}} \nonumber \\&\quad \times \ln \left( {\underbrace{\frac{{{\lambda _{{Y_2}}}}}{{{\rho _d}}}\left( {1 + \frac{{\beta {\lambda _{{X_1}}}}}{{k\left( {1 - \theta } \right) {\lambda _{{Y_1}}}}}} \right) }_{{I_1}} + \underbrace{\frac{{{\beta _Y}\beta {\lambda _{{X_1}}}\kappa }}{{k\left( {1 - \theta } \right) {\rho _s}}}}_{{I_2}}\frac{1}{{{x_2}}}} \right) d{x_2}. \end{aligned}$$

(34)

Using L’Hospital’s Rule, it is easy to show that $I_2$ is dominant over $I_1$, where ${x_2} \in \left[ {{x_0},M} \right)$ and $M \ll + \infty$. Then, (34) can be rewritten as

$$\begin{aligned} P_{{\text {out,2}}}^\infty&= \frac{{{\beta _Y}\beta {\lambda _{{X_1}}}{\lambda _{{X_2}}}}}{{\left( {1 - \theta } \right) {\rho _s}}}\sum \limits _{k = 1}^K {\left( {\begin{array}{c}K\\ k\end{array}}\right) \frac{{{{\left( { - 1} \right) }^{k + 1}}}}{k}} \nonumber \\&\quad \times \underbrace{\int \limits _{{{{{\bar{x}}}}_0}}^{ + \infty } {\left( {1 + \frac{\kappa }{{{x_2}}}} \right) {e^{ - {\lambda _{{X_2}}}{x_2}}}} \ln \left( {{\rho _s}{x_2}} \right) d{x_2}}_{{I_3}}. \end{aligned}$$

(35)

With the help of [12, Eq. (4.337.1) and Eq. (3.351.5)], $I_3$ is expressed as

$$\begin{aligned} {I_3}&= \frac{{{e^{ - {\lambda _{{X_2}}}{{{{\bar{x}}}}_0}}}\ln \left( {{\rho _2}{{{{\bar{x}}}}_0}} \right) }}{{{\lambda _{{X_2}}}}} - \left( {\kappa \ln \left( {{\rho _2}} \right) + \frac{1}{{{\lambda _{{X_2}}}}}} \right) Ei\left( { - {\lambda _{{X_2}}}{{{{\bar{x}}}}_0}} \right) \nonumber \\&\quad + \kappa \int \limits _{{{{{\bar{x}}}}_0}}^{ + \infty } {\frac{{\ln \left( {{x_2}} \right) }}{{{x_2}}}{e^{ - {\lambda _{{X_2}}}{x_2}}}} d{x_2}. \end{aligned}$$

(36)

With the help of Mathematica, Eq. (36) can rewritten as

$$\begin{aligned} {I_3}&= \frac{{{e^{ - {\lambda _{{X_2}}}{{{{\bar{x}}}}_0}}}\ln \left( {{\rho _2}{{{{\bar{x}}}}_0}} \right) }}{{{\lambda _{{X_2}}}}} - \left( {\kappa \ln \left( {{\rho _2}{{{{\bar{x}}}}_0}} \right) + \frac{1}{{{\lambda _{{X_2}}}}}} \right) Ei\left( { - {\lambda _{{X_2}}}{{{{\bar{x}}}}_0}} \right) \nonumber \\&\quad + \kappa G_{2,3}^{3,0}\left( {{\lambda _{{X_2}}}{{{{\bar{x}}}}_0}|_{0,0,0}^{1,1}} \right) . \end{aligned}$$

(37)