Outage Performance Analysis of Energy Harvesting Wireless Sensor Networks for NOMA Transmissions

Abstract

In this paper, we investigate radio frequency (RF) energy harvesting (EH) in wireless sensor networks (WSNs) using non-orthogonal multiple access (NOMA) uplink transmission with regard to a probable secrecy outage during the transmission between sensor nodes (SNs) and base station (BS) in the presence of eavesdroppers (EAVs). In particular, the communication protocol is divided into two phases: 1) first, the SNs harvest energy from multiple power transfer stations (PTSs), and then, 2) the cluster heads are elected to transmit information to the BS using the harvested energy. In the first phase, we derive a 2D RF energy model to harvest energy for the SNs. During the second phase, the communication faces multiple EAVs who attempt to capture the information of legitimate users; thus, we propose a strategy to select cluster heads and implement the NOMA technique in the transmission of the cluster heads to enhance the secrecy performance. For the performance evaluation, the exact closed-form expressions for the secrecy outage probability (SOP) at the cluster heads are derived. A nearly optimal EH time algorithm for the cluster head is also proposed. In addition, the impacts of system parameters, such as the EH time, the EH efficiency coefficient, the distance between the cluster heads and the BS, and the number of SNs as well as EAVs on the SOP, are investigated. Finally, Monte Carlo simulations are performed to show the accuracy of the theoretical analysis; it is also shown that the secrecy performance of NOMA in RF EH WSN can be improved using the optimal EH time.

Appendices

Appendix A: Proof of Lemma 1

Now, we prove Lemma 1 by considering the following integral:

$$ {\Gamma} \left( z \right) = \int\limits_{0}^{\infty} {\frac{{{e^{Cz}}}}{{{{\left( {z + A} \right)}^{2}}\left( {z + D} \right)}}} dz. $$

(78)

Using some simplified transformations, we can rewrite Eq. 78 as follows:

$$\begin{array}{@{}rcl@{}} {\Gamma} \left( z \right) \!&=&\! \int\limits_{0}^{\infty} {\frac{{\left( {D - A} \right)\left( {z + B} \right) - \left( {z + A} \right)\left( {z + D} \right) + {{\left( {z + A} \right)}^{2}}}}{{{{\left( {z + A} \right)}^{2}}\left( {z + D} \right){{\left( {D - A} \right)}^{2}}}}}\\ &&\times {e^{Cz}}dz\\ &=&{{\Gamma}_{1}}\left( z \right)+{{\Gamma}_{2}}\left( z \right)+{{\Gamma}_{3}}\left( z \right), \end{array} $$

(79)

where ${{\Gamma }_{1}}\left (z \right )$, ${{\Gamma }_{2}}\left (z \right )$, and ${{\Gamma }_{3}}\left (z \right )$ are defined as follows:

$$\begin{array}{@{}rcl@{}} {{\Gamma}_{1}}\left( z \right) &=& \int\limits_{0}^{\infty} {\frac{{{e^{Cz}}}}{{{{\left( {z + A} \right)}^{2}}\left( {D - A} \right)}}} dz, \end{array} $$

(80)

$$\begin{array}{@{}rcl@{}} {{\Gamma}_{2}}\left( z \right) &=& -\int\limits_{0}^{\infty} {\frac{{{e^{Cz}}}}{{\left( {z + A} \right){{\left( {D - A} \right)}^{2}}}}} dz, \end{array} $$

(81)

$$\begin{array}{@{}rcl@{}} {{\Gamma}_{3}}\left( z \right) &=& \int\limits_{0}^{\infty} {\frac{{{e^{Cz}}}}{{\left( {z + D} \right){{\left( {D - A} \right)}^{2}}}}} dz. \end{array} $$

(82)

Using (3.352.4) and (3.353.2) in [58] to solve these integrals, ${\Gamma } \left (z \right )$ is obtained as in Eq. 50. The proof is completed.

Appendix B: Proof of Lemma 2

Similar to the approach of Lemma 1, we have

$$\begin{array}{@{}rcl@{}} {\Omega} \left( z \right) &=& \int\limits_{0}^{\infty} {\frac{{{e^{Cz}}}}{{\left( {z + A} \right)\left( {z + D} \right)}}} dz\\ &=& \int\limits_{0}^{\infty} {\frac{{z + D - \left( {z + A} \right)}}{{\left( {z + A} \right)\left( {z + D} \right)\left( {D - A} \right)}}{e^{Cz}}} dz\\ &=& {{\Omega}_{1}}\left( z \right)+{{\Omega}_{2}}\left( z \right), \end{array} $$

(83)

where ${{\Omega }_{1}}\left (z \right )$ and ${{\Omega }_{2}}\left (z \right )$ are defined as

$$\begin{array}{@{}rcl@{}} {{\Omega}_{1}}\left( z \right) &=& \int\limits_{0}^{\infty} {\frac{{{e^{Cz}}}}{{\left( {z + A} \right)\left( {D - A} \right)}}} dz, \end{array} $$

(84)

$$\begin{array}{@{}rcl@{}} {{\Omega}_{2}}\left( z \right) &=& \int\limits_{0}^{\infty} {\frac{{{e^{Cz}}}}{{\left( {z + D} \right)\left( {A - D} \right)}}} dz. \end{array} $$

(85)

Using (3.352.4) and (3.353.2) in [58] to calculate these integrals, the function ${\Omega } \left (z \right )$ can be obtained as in Eq. 54. The proof is completed.