Secure communication via an untrusted relay with unreliable backhaul connections

Abstract

This paper studies the secrecy performance of a wireless energy harvesting system in which a source connected to wireless backhaul links sends information to a destination via an untrusted relay that not only helps the overall commutation but also overhears the sources confidential information. The secrecy capacity is created by using destination-assisted jamming signals. The jamming provides additional energy to the relay. To analyze the secrecy performance of the proposed system, we derived analytical expressions for the secrecy outage probability (SOP) and the average secrecy capacity, and the high-power approximations for the SOP. The accuracy of the calculations is verified by Monte Carlo simulations. Numerical results provide useful insight into the effects of the system parameters, such as the failure probability of unreliable backhaul links, the transmit powers, the power-splitting ratio, and the locations of the relay, on the secrecy performance.

Appendix

1.1 Appendix 1

The value of the function $\varXi (y)$ can be described as

$$\begin{aligned} \varXi ( y) {\left\{ \begin{array}{ll} \ge 0;&{}\quad \text {if }\;\; {\bar{y}}_p \le y< \infty \\< 0;&{}\quad \text {if}\;\; 0 \le y < {\bar{y}}_p \end{array}\right. } \end{aligned}$$

(49)

Using (49), we can rewrite the SOP in (19) as

$$\begin{aligned} P_{\text {out,low}} = 1 - \Pr \left( {X> \frac{{{\gamma _{{\text {th}}}} - 1}}{{{{\mathcal {A}}_1}\varXi \left( Y \right) }} | Y > {{\bar{y}}_p}} \right) \end{aligned}$$

(50)

Substituting the CDF of X given by (16) and PDF of Y given by $f(m_2,\lambda _2;t)$, we can derive the analytical expression for the SOP as in (20).

1.2 Appendix 2

We first study the trend of the element g(Y) in ${\mathcal {J}}_2$. It can be seen that $\mathop {\lim }\nolimits _{\begin{array}{c} (P_T,P_D) \rightarrow (\infty ,\infty ) \\ {\bar{y}}_p< Y < \infty \end{array}} g\left( Y \right) \rightarrow 0$, hence, we can use the approximating function of the CDF of X, $F_X^ \approx ( t )$ instead of using the CDF of X to simplify the calculation of ${\mathcal {J}}_2$. Using the fact $e^t=\sum _{n=0}^{+\infty }\frac{t^n}{n!}$ and (14), $F_X ( t )$ as $t \rightarrow 0$ can be approximated by $F_X^ \approx ( t )$ as follows.

$$\begin{aligned} F_X^ \approx \left( t \right)&= {\left( {1 - p + \frac{{{{\left( {{\lambda _1}t} \right) }^{{m_1}}}}}{{{m_1}!}}} \right) ^K} \nonumber \\&= {\left\{ \begin{array}{ll} \left( {{m_1}!} \right) ^{-K}\left( {{\lambda _1}t} \right) ^{Km_1}; &{}\quad \text {if}\;\; p=1\\ {\left( {1 - p} \right) }^K; &{}\quad \text {if}\;\; p<1 \end{array}\right. } \end{aligned}$$

(51)

1.2.1 The expression for ${\mathcal {J}}_2$ in the case of $p=1$

Substituting the PDF of Y given by $f(m_2,\lambda _2;t)$ and $F_X^\approx (t)$ given by (51), ${{\mathcal {J}}}_2^{p = 1}$ can be calculated as

$$\begin{aligned} {\mathcal {J}}_2^{p = 1} = \int \limits _{{{\bar{y}}_p}}^{ + \infty } {{f_Y}\left( t \right) \frac{1}{{{{\left( {{m_1}!} \right) }^K}}}{{\left( {\frac{{{\lambda _1}\left( {{\gamma _{{\text {th}}}} - 1} \right) }}{{{{\mathcal {A}}_1}}}\left( {1 + \frac{{{{\mathcal {B}}_2}}}{t}} \right) } \right) }^{K{m_1}}}} dt \end{aligned}$$

(52)

With the help of [16, Eqs. (3.351.3) and (3.351.4)], ${\mathcal {J}}_2^{p = 1}$ can be expressed as in (24).

1.2.2 The expression for ${\mathcal {J}}_2$ in the case of $p<1$

Substituting the PDF of Y given by $f(m_2,\lambda _2;t)$ and $F_X^\approx (t)$ given by (51), ${\mathcal {J}}_2^{p < 1}$ can be calculated as

$$\begin{aligned} {\mathcal {J}}_2^{p = 1} = \int \limits _{{{\bar{y}}_p}}^{ + \infty } {{f_Y}\left( t \right) {{\left( {1 - p} \right) }^K}} dt \approx {\left( {1 - p} \right) ^K} \end{aligned}$$

(53)

1.3 Appendix 3

The CDF of $\gamma _D$ can be calculated as

$$\begin{aligned} {F_{{\gamma _D}}}\left( \gamma \right)&= 1 - \Pr \left( {\left( {{{\mathcal {A}}_1}X - \gamma } \right) Y > \gamma {{\mathcal {B}}_2}} \right) \end{aligned}$$

(54)

$$\begin{aligned}&= 1 - \int \limits _0^\infty {\left( {1 - {F_Y}\left( {\frac{{\gamma {{\mathcal {B}}_2}}}{{{{\mathcal {A}}_1}t}}} \right) } \right) } {f_X}\left( {t + \frac{\gamma }{{{{\mathcal {A}}_1}}}} \right) dt \end{aligned}$$

(55)

Substituting the CDF of Y given by $F(m_2,\lambda _2;t)$ and CDF of X given by (15), we have

$$\begin{aligned}&{F_{{\gamma _D}}}\left( \gamma \right) = 1 - \frac{{Kp}}{{\varGamma \left( {{m_1}} \right) }}\sum \limits _{{{\left\| {\mathbf{u}} \right\| }_1} = K - 1} \left( {\begin{array}{c}K-1\\ {\mathbf{u}} \end{array}}\right) {\left( { - p} \right) ^{{{\left\| {\mathbf{u }_{{m_1}}} \right\| }_1}}}\lambda _1^{{m_1} + \omega \left( {\mathbf{u }_{{m_1}}} \right) } \left( {\prod \limits _{i = 0}^{{m_1} - 1} {{{\left( {i!} \right) }^{ - {u_i}}}} } \right) \nonumber \\&\quad \times \, {e^{ - {\lambda _1}\left( {{{\left\| {\mathbf{u }_{{m_1}}} \right\| }_1} + 1} \right) \frac{\gamma }{{{{\mathcal {A}}_1}}}}} \sum \limits _{i = 0}^{{m_2} - 1} {\frac{1}{{i!}}{{\left( {\frac{{{\lambda _2}{{\mathcal {B}}_2}}}{{{{\mathcal {A}}_1}}}\gamma } \right) }^i}} \sum \limits _{j = 0}^{{m_1} + \omega \left( {\mathbf{u }_{{m_1}}} \right) - 1} {{\left( {\frac{\gamma }{{{{\mathcal {A}}_1}}}} \right) }^{{m_1} + \omega \left( {\mathbf{u }_{{m_1}}} \right) - j - 1}} \nonumber \\&\quad \times \, \left( {\begin{array}{c}{m_1} + \omega \left( {\mathbf{u }_{{m_1}}} \right) - 1\\ j\end{array}}\right) \int \limits _0^\infty {{t^{j - i}}{e^{ - \frac{{{\lambda _2}{{\mathcal {B}}_2}}}{{{{\mathcal {A}}_1}t}}\gamma - {\lambda _1}\left( {{{\left\| {\mathbf{u }_{{m_1}}} \right\| }_1} + 1} \right) t}}} \end{aligned}$$

(56)

With the help of [16, Eq. (3.471.9)], we can obtain (35).

1.4 Appendix 4

Substituting the CDF of $\gamma _D$ into (36), $C_D$ can be obtained as in (37) where ${\mathcal {I}}_1 \left( \alpha ,\beta , v,\mu \right)$ is calculated by

$$\begin{aligned} {{\mathcal {I}}_1}\left( {\alpha ,\beta ,v,\mu } \right) = \frac{1}{{\ln \left( 2 \right) }}\int \limits _0^\infty {\ln \left( {1 + \gamma } \right) } \left( {\underbrace{\frac{d}{{d\gamma }}{\gamma ^\alpha }{e^{ - \beta \gamma }}{K_v}\left( {2\sqrt{\mu \gamma } } \right) }_{{{\mathcal {I}}_2}\left( {\gamma ;\alpha ,\beta ,v,\mu } \right) }} \right) d\gamma \end{aligned}$$

(57)

Using the fact that $\frac{{d{K_v}\left( x \right) }}{{dx}} = - \frac{1}{2}\left( {{K_{v - 1}}\left( x \right) + {K_{v + 1}}\left( x \right) } \right)$, ${{\mathcal {I}}_2}\left( {\gamma ;\alpha ,\beta ,v,\mu } \right)$ can be calculated as

$$\begin{aligned} {{\mathcal {I}}_2}\left( {\gamma ;\alpha ,\beta ,v,\mu } \right) = \alpha {\gamma ^{\alpha - 1}}{e^{ - \beta \gamma }}{K_v}\left( {2\sqrt{\mu \gamma } } \right) - \beta {\gamma ^\alpha }{e^{ - \beta \gamma }}{K_v}\left( {2\sqrt{\mu \gamma } } \right) \nonumber \\ - \frac{{\sqrt{\mu }}}{2}{\gamma ^{\alpha - \frac{1}{2}}}{e^{ - \beta \gamma }}{K_{v - 1}}\left( {2\sqrt{\mu \gamma } } \right) - \frac{{\sqrt{\mu }}}{2}{\gamma ^{\alpha - \frac{1}{2}}}{e^{ - \beta \gamma }}{K_{v + 1}}\left( {2\sqrt{\mu \gamma } } \right) \end{aligned}$$

(58)

By expressing $\ln \left( {1 + \gamma } \right)$ in (58) by $H_{2,2}^{1,2}\left[ {\gamma | \begin{array}{c} (1,1),(1,1)\\ (1,1),(0,1) \end{array}} \right]$ and ${K_v}\left( {2\sqrt{\mu \gamma } } \right)$ in (58) by $\frac{1}{2}H_{2,0}^{0,2}\left( \mu \gamma |{(v/2,1),(v/2,1)} \right)$, and using [21, Eq. (2.6.2)], we can obtain (38). Then, Proposition 4 can be proved.

1.5 Appendix 5

Substituting (40) in to (39), $C_R$ can be expressed as in (43) where ${\mathcal {I}}_3 (\mu _2)$ is calculated by

$$\begin{aligned} {{\mathcal {I}}_3}\left( {{\mu _2}} \right) = \int \limits _0^\infty {\underbrace{\frac{{{\gamma ^{\omega \left( {\mathbf{u }_{{m_1}}} \right) }}}}{{\left( {\gamma + 1} \right) {{\left( {\gamma + {\mu _2}} \right) }^{\omega \left( {\mathbf{u }_{{m_1}}} \right) + {m_2}}}}}}_{{{\mathcal {I}}_4}\left( {{\mu _2};\gamma } \right) }} d\gamma \end{aligned}$$

(59)

For the case of $\mu _2 \ne 1$, we can decompose ${{\mathcal {I}}_4}\left( {{\mu _2};\gamma } \right)$ as

$$\begin{aligned} {{\mathcal {I}}_4}\left( \gamma \right) = \frac{{{{\mathcal {C}}_{1,0}}}}{{\gamma + 1}} + \sum \limits _{n = 1}^{\omega \left( {\mathbf{u }_{{m_1}}} \right) + {m_2}} {\frac{{{{\mathcal {C}}_{1,n}}}}{{{{\left( {\gamma + {\mu _2}} \right) }^n}}}} \end{aligned}$$

(60)

For the case of $\mu _2=1$, we can decompose ${{\mathcal {I}}_4}\left( {{\mu _2};\gamma } \right)$ as

$$\begin{aligned} {{\mathcal {I}}_4}\left( \gamma \right) = \sum \limits _{n = 1}^{\omega \left( {\mathbf{u }_{{m_1}}} \right) + {m_2} + 1} {\frac{{{{\mathcal {C}}_{2,n}}}}{{{{\left( {\gamma + 1} \right) }^n}}}} \end{aligned}$$

(61)

Substituting (60) and (61) into (59), and after some manipulation, we can prove Proposition 7.