On Secure mmWave RSMA Systems

TL;DR

This work investigates the secrecy performance of a downlink mmWave RSMA system with an external eavesdropper, modeling four eavesdropping scenarios determined by overlapped resolvable paths. It derives exact secrecy outage probability expressions for User 1 using Gaussian–Chebyshev quadrature and validates them via Monte Carlo simulations, revealing how path overlap and power allocation ($\tau_c$, $P$) influence security. By transmitting the common stream on overlapped paths and private streams on non-overlapping paths, the study shows RSMA can improve secrecy performance compared to NOMA in multiple settings, and it identifies conditions under which there exists an optimal power allocation and transmit power to minimize SOP. The results offer design insights for beamforming and resource allocation in secure mmWave RSMA systems and motivate future work on additional eavesdropping strategies and secrecy optimization.

Abstract

This work considers a multiple-input-single-output mmWave RSMA system wherein a base station serves two users in the presence of a passive eavesdropper. Different eavesdropping scenarios are considered corresponding to the overlapped resolvable paths between the main and the wiretap channels under the considered transmission schemes. The analytical expressions for the secrecy outage probability are derived respectively through the Gaussian Chebyshev quadrature method. Monte Carlo simulation results are presented to validate the correctness of the derived analytical expressions and demonstrate the effects of system parameters on the SOP of the considered mmWave RSMA systems.

Figures (8)

• Figure 1: A downlink mmWave RSMA system consists a base station ($S$), two user ($U_1$ and $U_2$), and a external passive eavesdropper ($E$).
• Figure 2: Scenarios for the different number of the overlapped overlapped paths between $U_i$ and $E$ with $L = 5$ and ${L_{c}} = 3$.
• Figure 3: SOP versus the $P$ with $\tau_c =\tau_1 =\tau_2 = \frac{1}{3}$, $r_e = 30$, $L = 8$, $L_c = 4$, and $R_{{1,c}}^{ {\mathrm{th}}} = R_{1,p}^{\mathrm{th}} = 0.1$.
• Figure 4: SOP versus the $P$ with $N_s = 50$, $\tau_c =\tau_1 =\tau_2 = \frac{1}{3}$, $r_1 =15$, $r_e = 25$, $L = 9$, and $R_{1,c}^{\mathrm{th}} = R_{1,p}^{\mathrm{th}} = 0.1$.
• Figure 5: SOP versus the $\tau_1$ and $\tau_2$ ($\tau_c = 1 - \tau_1 - \tau_2$) with $N_s = 50$, $r_1 =15$, $L = 8$, $L_c = 4$, and $R_{1,c}^{\mathrm{th}} = R_{1,p}^{\mathrm{th}} = 0.1$.