Can a multi-hop link relying on untrusted amplify-and-forward relays render security?

Abstract

Cooperative relaying is utilized as an efficient method for data communication in wireless sensor networks and the Internet of Things. However, sometimes due to the necessity of multi-hop relaying in such communication networks, it is challenging to guarantee the secrecy of cooperative transmissions when the relays may themselves be eavesdroppers, i.e., we may face with the untrusted relaying scenario where the relays are both necessary helpers and potential adversary. To obviate this issue, a new cooperative jamming scheme is proposed in this paper, in which the data can be confidentially communicated from the source to the destination through multiple untrusted relays. In our proposed secure transmission scheme, all the legitimate nodes contribute to providing secure communication by intelligently injecting artificial noises to the network in different communication phases. For the sake of analysis, we consider a multi-hop untrusted relaying network with two successive intermediate nodes, i.e, a three-hop communications network. Given this system model, a new closed-form expression is presented in the high signal-to-noise ratio (SNR) regime for the Ergodic secrecy rate (ESR). Furthermore, we evaluate the high-SNR slope and power offset of the ESR to gain an insightful comparison of the proposed secure transmission scheme and the state-of-arts. Our numerical results highlight that the proposed secure transmission scheme provides better secrecy rate performance compared with the two-hop untrusted relaying as well as the direct transmission schemes.

Appendices

A Proof of Lemma 2

The CDF of $Z=\frac{X}{Y}$ has been derived in [14]. To obtain the CDF of $W=\frac{XY}{X+Y}$, we start from the definition of CDF as

$$\begin{aligned} F_W(\omega )&=\Pr \Big \{\frac{XY}{X+Y}<\omega \Big \}\nonumber \\ &=\Pr \Big \{XY-\omega (X+Y)<0\Big \}\nonumber \\ &=\Pr \Big \{X<\frac{\omega Y}{Y-\omega }|Y-\omega \ge 0\Big \}\Pr\{Y-\omega \ge 0\}\nonumber \\ &\quad +{{\Pr \Big \{X\ge \frac{\omega Y}{Y-\omega }|Y-\omega<0\Big \}}}\Pr \{Y-\omega <0\}\nonumber \\ &=\int _{\omega }^{\infty }F_X\left(\frac{\omega y}{y-\omega }\right) f_Y(y){\mathrm{d}}y+\int _{0}^{\omega }f_Y(y){\mathrm{d}}y\nonumber\\ &=\int _{\omega }^{\infty }\left[ 1-\exp \left(-\frac{\omega y}{m_x(y-\omega)}\right)\right] f_Y(y){\mathrm{d}}y +\int _{0}^{\omega }f_Y(y){\mathrm{d}}y\nonumber \\ &=1-\frac{1}{m_y}\int _{\omega }^{\infty }\exp \left(-\frac{\omega y}{m_x(y-\omega )} -\frac{y}{m_y}\right) {\mathrm{d}}y\nonumber\\ &=1-\frac{1}{m_y}\exp \left( -\frac{\omega }{m_x}-\frac{\omega}{m_y}\right) \int _{0}^{\infty }\nonumber \\ &\quad \exp \left(-\frac{\omega ^2}{m_x y}-\frac{y}{m_y}\right) {\mathrm{d}}y\nonumber\\ &{\mathop {=}\limits ^{(a)}}1-\frac{2\omega }{\sqrt{m_x m_y}}\exp \left( -\frac{\omega }{m_x}-\frac{\omega }{m_y}\right) {\mathrm {K}}_1\left( \frac{2\omega }{\sqrt{m_xm_y}}\right) ,\end{aligned} $$

(24)

Finally, after calculating the integral term using [33, Eq. (3.324.1)], one can obtain the expression given in (11). $\square$

B Proof of Lemma 3

In the following, we proceed to prove Lemma 3 wherein the different exact/approximate expressions for ${\mathcal {P}}$, ${\mathcal {T}}_1$, and ${\mathcal {T}}_2$ are given.

A. Calculating ${\mathcal {P}}$ Plugging (5) into ${\mathcal {P}}=\Pr \{\gamma ^{(1)}_{R_1}>\gamma ^{(2)}_{R_2}\}$, and then defining $X=\gamma _f$, $Y=\gamma _h$ and $Z=\gamma _g$, we get

$$\begin{aligned} {\mathcal {P}}&=\Pr \left\{ \gamma_f>\frac{\gamma ^2_h}{\gamma _g+\gamma _h}\right\} =1-\Pr \left\{ X<\frac{Y^2}{Y+Z}\right\} \nonumber\\ &=1-{\mathbb{E}}_Y\left\{ {\mathbb{E}}_Z\left\{ F_X\left(\frac{y^2}{y+z}\right) \right\} \right\} \nonumber\\ &=\frac{1}{m_ym_z}\int _{0}^{\infty }\int _{0}^{\infty }\nonumber \\ &\quad \exp\left( -\frac{y^2}{(y+z)m_x}-\frac{y}{m_y}-\frac{z}{m_z}\right){\mathrm{d}}z{\mathrm{d}}y \nonumber \\ &{\mathop {=}\limits ^{(a)}}\frac{1}{m_ym_z}\int _{0}^{\infty }\int _{0}^{\infty }\nonumber \\ &\quad \exp \left( -\frac{v^2}{um_x}-\frac{v}{m_y}-\frac{u-v}{m_z}\right) {\mathrm{d}}u{\mathrm{d}}v\nonumber \\ &{\mathop {=}\limits ^{(b)}}\sqrt{\frac{4}{m_xm_ym_z}}\int _{0}^{\infty }\nonumber\\ &\quad \exp \left( -v\frac{m_y-m_z}{m_ym_z}\right) v{{K}_1}\left( \sqrt{\frac{2}{{m_xm_z}}v}\right){\mathrm{d}}v\nonumber \\ &{\mathop {\approx }\limits ^{(c)}}\frac{\sqrt{m_{x}}{m_{z}}^{3/2}}{m_{z}-m_{y}\,\sqrt{m_{x}}\sqrt{m _{z}}}+2\,m_{z}\,m_{{ y}}\nonumber \\ &\quad \sum _{n=1}^{M}\sum_{i=1}^{n}\Lambda (1,n,i)i!\left( {\frac{2m_{z}\,m_{y}}{m_{z}-m_{y}\,\sqrt{m_{x}\,m_{z}}+2\,m_{z}\,m_{y}}} \right) ^i,\nonumber \\ &{\mathop {\approx }\limits ^{(d)}}\frac{4\sqrt{m_{x}}{m_{z}}^{5/2}m_{y}}{3\left(m_{z}-m_{y}\,\sqrt{m_{x}}\sqrt{m_{z}}+\,2\,m_{z}\,m_{y} \right) ^{2}} {\mathop {=}\limits ^{\tiny \Delta }}{\mathcal {P}}_1, \end{aligned}$$

(25)

where (a) follows from defining the auxiliary variables $u=y+z$ and $v=y$, (b) follows from using [33, Eq. (3.324.1)] and [33, Eq. (3.351.3)], (c) follows from substituting the equivalent series of modified Bessel function of the second kind and first order as presented in [34], which is a well-tight approximation with finite series, as observed later in numerical results. For $\nu >0$ and positive integer M which controls the accuracy of infinite series, we have [34]

$$\begin{aligned} { K}_{\nu }(\beta x)\approx \exp (-\beta x)\sum \limits _{n=0}^{M}\sum _{i=0}^{n}\Lambda (\nu , n, i)(\beta x)^{i-\nu }, \end{aligned}$$

Finally, (d) presents the first term of the infinite series given for $M=1$ to have a closed-form approximation. We will show in the simulation results how this simple closed-form expression works well.

B. Calculating $T_1$ Using Lemma 2, we can derive a closed-form expression for $T_{1}$, after assuming $X=\frac{\gamma _g}{\gamma _h}$, as

$$\begin{aligned} T_{1} &= {\mathbb{E}}\left\{ \ln (1+\frac{\gamma _g}{\gamma _h})\right\} \\ &= \int _{0}^{\infty }\ln (1+x)f_X(x)~{\mathrm{d}}x\nonumber \\&{\mathop {=}\limits ^{(a)}}&\int _{0}^{\infty }\frac{1-F_X(x)}{1+x}{\mathrm{d}}x, \end{aligned}$$

(26)

where (a) follows from integration by parts law. Then, computing the last integral, considering $F_X(x)$ given in Lemma 2, leads to the closed-form expression for $T_1$ given in (27).

C. Calculating $T_2$ The part $T_{2}$ can be mathematically calculated as

$$\begin{aligned} T_{2}={\mathbb{E}}\left\{ \ln \left( 1+\frac{\gamma _g\gamma _h}{\gamma _f(\gamma _g+\gamma _h)}\right) \right\} {\mathop {\approx }\limits ^{(a)}}\ln \left( 1+\frac{{\mathbb{E}}\left\{ \frac{\gamma _g\gamma _h}{\gamma _g+\gamma _h}\right\} }{{\mathbb{E}}\{\gamma _f\}}\right) , \end{aligned}$$

(27)

where (a) follows after using the approximation ${\mathbb{E}}\left\{ \log \left( 1+\frac{X}{Y}\right) \right\} \approx \log \left( 1+\frac{{\mathbb{E}}\{X\}}{{\mathbb{E}}\{Y\}}\right)$ given in [35]. Thus, after further calculation, using Lemma 2, one can obtain $T_2$ in (27). $\square$