Physical layer security for IoT over Nakagami-m and mixed Rayleigh–Nakagami-m fading channels

Abstract

The Internet of Things (IoT) is a new para-digm for achieving ubiquitous connectivity by enormously deploying physical objects like sensors, actuators, and controllers. Wireless communication is one of the main enabling technologies that make the IoT a reality. However, due to the open nature of wireless communication and constraints on energy consumption, this raises concerns about the security of the IoT. Eavesdropping is a major threat to the wireless communications security. To counteract such an attack, physical layer security is among the promising solutions. The aim of this work is to evaluate the secrecy performance of two-hop wireless communication systems. This evaluation is done in the presence of potential eavesdropping threats over Nakagami-m and mixed Rayleigh–Nakagami-m fading channels. To enhance the security of the system, the use of cooperative jamming and opportunistic relaying has been considered. In particular, based on classical probability theory, we first analyze the secrecy outage performance for scenario 1, where all channels (S-R, R-D, R-R, S-E, R-E) experience Nakagami-m fading. Then, we examine the secrecy performance for scenario 2, where the channel coefficients for links (S-R, R-D, S-E, R-E) are subject to Nakagami-m fading and the channel coefficients for the link (R-R) follow Rayleigh fading. Finally, simulation and numerical results are presented to validate our theoretical achievements. The results suggest that as the fading parameter m increases, the probability of secrecy outages decreases, leading to an improvement in the security performance of the wireless IoT data collection process.

Appendix A: Proof of Lemma 2

From the definition of event B, we can write

$$\begin{aligned} \begin{aligned} p_{{B|J^{n}_{1}}} = P(\delta _{S,R_{c}}<\alpha |J^{n}_{1}).\\ \end{aligned} \end{aligned}$$

(27)

In this scenario 2, the channel coefficients $|h_{i,j}|^{2}$ for (S-R, R-D, S-E, R-E) are Gamma distributed, whereas the channel coefficients $|h_{i,j}|^{2}$ for (R-R) are exponentially distributed. Therefore, the cdf, i.e., $P (V_{j} \le V_{i})$ in (8), will also be exponentially distributed. By following the same procedure as for scenario 1, $P(|h_{S,R_{c}}|^{2}<z)$ in (8), becomes equal to $\int _{0}^{\infty }uP\Big (\vert h_{S,R_{i}}\vert ^{2}<z, V_{i}=t\Big ) (1-e^{-2t})^{u-1}dt$ for scenario 2. Now, by applying the total probability law, we obtain

$$\begin{aligned} \begin{aligned}&P(|h_{S,R_{c}}|^{2}<z)\\&\quad = \int _{0}^{z}u\varGamma (m,t)(2\varGamma (m,t)-\varGamma (m,z))(1-e^{-2t})^{u-1}dt,\\&\quad =2u\int _{0}^{z}\Big (e^{-t}\sum _{k=0}^{m-1}\dfrac{t^{k}}{k!}\Big )^{2}(1-e^{-2t})^{u-1}dt-u\varGamma (m,z)\\&\qquad \times \int _{0}^{z}e^{-t}\sum _{k_{3}=0}^{m-1}\dfrac{t^{k_{3}}}{k_{3}!}(1-e^{-2t})^{u-1}dt,\\&\quad =2u\sum _{r=0}^{u-1}\sum _{k_{1}=0}^{m-1}\sum _{k_{2}=0}^{m-1}\left( {\begin{array}{c}u-1\\ r\end{array}}\right) (-1)^{r}\int _{0}^{z}e^{-2t(r+1)}\dfrac{t^{k_{1}+k_{2}}}{k_{1}!k_{2}!}dt\\&\qquad -u\varGamma (m,z)\sum _{r=0}^{u-1}\sum _{k_{3}=0}^{m-1}\left( {\begin{array}{c}u-1\\ r\end{array}}\right) (-1)^{r}\int _{0}^{z}e^{-t(2r+1)}\dfrac{t^{k_{3}}}{k_{3}!}dt,\\&\quad =2u\sum _{r=0}^{u-1}\left( {\begin{array}{c}u-1\\ r\end{array}}\right) (-1)^{r}\Bigg [\sum _{k_{1}=0}^{m-1}\sum _{k_{2}=0}^{m-1}\sum _{k_{5}=0}^{k_{1}+k_{2}}\Big [\dfrac{1}{k_{1}!k_{2}!}\\&\qquad \times \dfrac{(k_{1}+k_{2})!}{(2(r+1))^{k_{1}+k_{2}+1}}-e^{-2(r+1)z}\\&\qquad \times \dfrac{(k_{1}+k_{2})!z^{k_{5}}}{{k_{1}!k_{2}!k_{5}!(2(r+1))^{k_{1}+k_{2}+1-k_{5}}}}\Big ]-\sum _{k_{3}=0}^{m-1}\sum _{k_{4}=0}^{m-1}\sum _{k_{6}=0}^{k_{3}}\\&\qquad \times \Big [\dfrac{e^{-z}z^{k_{4}}}{k_{4}!(2r+1)^{k_{3}+1}}-e^{-2(r+1)z} \dfrac{z^{k_{6}+k_{4}}}{k_{6}!k_{4}!(2r+1)^{k_{3}+1-k_{6}}}\Big ]\Bigg ].\\ \end{aligned} \end{aligned}$$

(27)

After substituting $z = \alpha \sum \limits _{j\in \mathcal {J}}|h_{R_{j},R_{c}}|^{2}|J^{n}_{1}$ in (29), rearranging and taking expectation by using the pdf of $|h_{R_{j},R_{c}}|^{2}$ for any $j\in \mathcal {J}$ given below, Lemma 2 follows.

$$\begin{aligned} \begin{aligned} f_{|h_{R_{j},R_{c}}|^{2}}(z)= {\left\{ \begin{array}{ll} \frac{z^{m-1}e^{-z}}{\varGamma m(1-e^{-\sigma })},&{} 0\le z< \sigma \\ 0,&{} z\ge \sigma . \,\,\blacksquare \end{array}\right. } \end{aligned} \end{aligned}$$

(28)