New Look on Device to Device NOMA Systems: with and Without Wireless Power Transfer Modes

Abstract

In this paper, device to device (D2D) network is studied to support transmission in close distance among group of two users. Such two users benefit from new technique of multiple access, namely non-orthogonal multiple access. Two modes of D2D are considered, such as direct and relay links. Energy harvesting and design of multiple antennas have main impacts on system performance. We derive the closed-form expressions of outage probability for two devices in many scenarios. The Decode and Forward relaying scheme is adopted in this study. To ensure the quality of service (QoS) for the devices, suitable mode can be decided based on specific demand. We compare system performance by varying main parameters such as power allocation factors or transmit signal to noise ratio. Numerical results are performed to verify the effectiveness of the proposed D2D transmission strategies.

Appendix

Proof of Proposition 2

Based on (26), the outage probability of $U_3$ can be expressed as

$$\begin{aligned} \begin{array}{l} {O_{{U_3}}} = 1 - \Pr \left[ {\gamma _{BS}^{\left( {{z_2}} \right) } \ge {\mu _2},\gamma _{{U_3}}^{\left( {{z_1}} \right) } \ge {\mu _2},\gamma _{{U_3}}^{\left( {{z_2}} \right) } \ge {\mu _2}} \right] \\ \quad = 1 - \underbrace{\Pr \left[ {\gamma _{BS}^{\left( {{z_2}} \right) } \ge {\mu _2}} \right] }_{{\varTheta _1}}\underbrace{\Pr \left[ {\gamma _{{U_3}}^{\left( {{z_1}} \right) } \ge {\mu _2},\gamma _{{U_3}}^{\left( {{z_2}} \right) } \ge {\mu _2}} \right] }_{{\varTheta _2}}. \end{array} \end{aligned}$$

(30)

Furthermore, ${{\varTheta _1}}$ can be calculated as

$$\begin{aligned} {\varTheta _1} = \Pr \left[ {{{\left| {{l_0}} \right| }^2} \ge \frac{{{\mu _2}}}{{{\delta _2}\beta }}} \right] = \exp \left( { - \frac{{{\mu _2}}}{{{\delta _2}\beta {\lambda _{{l_0}}}}}} \right) . \end{aligned}$$

(31)

Next, ${{\varTheta _2}}$ is computed by

$$\begin{aligned} \begin{array}{l} {\varTheta _2} = \Pr \left[ {\gamma _{{D_3}}^{\left( {{z_1}} \right) } \ge {\mu _2},\gamma _{{D_3}}^{\left( {{z_2}} \right) } \ge {\mu _2}} \right] \\ \quad \quad = \Pr \left[ {\frac{{{\delta _3}\beta {{\left| {{r_{k*}}} \right| }^2}}}{{{\delta _4}\beta {{\left| {{r_{k*}}} \right| }^2} + 1}} \ge {\mu _2},{\delta _4}\beta {{\left| {{r_{k*}}} \right| }^2} \ge {\mu _2}} \right] \\ \quad \quad = \Pr \left[ {{{\left| {{r_{k*}}} \right| }^2} \ge \frac{{{\mu _2}}}{{\beta \left( {{\delta _3} - {\mu _2}{\delta _4}} \right) }},{{\left| {{r_{k*}}} \right| }^2} \ge \frac{{{\mu _2}}}{{{\delta _4}\beta }}} \right] . \end{array} \end{aligned}$$

(32)

Based on (32), let $\sigma = \max \left( {\frac{{{\mu _2}}}{{\beta \left( {{\delta _3} - {\mu _2}{\delta _4}} \right) }},\frac{{{\mu _2}}}{{{\delta _4}\beta }}} \right)$, ${{\varTheta _2}}$ can be calculated as

$$\begin{aligned} {\varTheta _2} = \Pr \left[ {{{\left| {{r_{k*}}} \right| }^2} \ge \sigma } \right] = \sum \limits _{k = 1}^K {\left( \begin{array}{l} K\\ k \end{array} \right) {{\left( { - 1} \right) }^{k - 1}}\exp \left( { - \frac{{k\sigma }}{{{\lambda _{{r_k}}}}}} \right) }. \end{aligned}$$

(33)

Plugging above values of (31) and (33) into (30) we obtain final formula.

This is end of proof. $\square$

Proof of Proposition 3

The outage probability for $U_2$ can therefore be obtained as

$$\begin{aligned} \begin{array}{l} O_{{U_2}}^{\left( {EH} \right) } = 1 - \Pr \left[ {\gamma _{E - {U_2}}^{\left( {{z_1}} \right) } \ge {\mu _1}} \right] \\ \quad \quad \quad = 1 - \Pr \left[ {{{\left| {{l_{k*}}} \right| }^2} \ge \frac{{{\mu _1}}}{{{{\left| {{v_0}} \right| }^2}\beta \xi \psi \left( {{\delta _3} - {\mu _1}{\delta _4}} \right) }}} \right] \\ \quad \quad \quad = 1 - \int \limits _0^\infty {{F_{{{\left| {{l_{k*}}} \right| }^2}}}} \left( {\frac{{{\mu _1}}}{{x\beta \xi \psi \left( {{\delta _3} - {\mu _1}{\delta _4}} \right) }}} \right) {f_{{{\left| {{v_0}} \right| }^2}}}\left( x \right) dx. \end{array} \end{aligned}$$

(34)

With ${\delta _3} > {\mu _1}{\delta _4}$, it can be given that

$$\begin{aligned} \begin{array}{l} O_{{U_2}}^{\left( {EH} \right) } = 1 - \sum \limits _{k = 1}^K {\left( \begin{array}{l} K\\ k \end{array} \right) {{\left( { - 1} \right) }^{k - 1}}\frac{1}{{{\lambda _{{v_0}}}}}} \int \limits _0^\infty {\exp \left( { - \frac{{k{\mu _1}}}{{x{\lambda _{{l_k}}}\beta \xi \psi \left( {{\delta _3} - {\mu _1}{\delta _4}} \right) }} - \frac{x}{{{\lambda _{{v_0}}}}}} \right) } dx\\ \quad \quad \quad = 1 - \sum \limits _{k = 1}^K {\left( \begin{array}{l} K\\ k \end{array} \right) {{\left( { - 1} \right) }^{k - 1}}} \sqrt{\frac{{4k{\mu _1}}}{{{\lambda _{{l_k}}}{\lambda _{{v_0}}}\beta \xi \psi \left( {{\delta _3} - {\mu _1}{\delta _4}} \right) }}} \\ \quad \quad \quad \times {\mathrm{{K}}_{{1}}}\left( {\sqrt{\frac{{4k{\mu _1}}}{{{\lambda _{{l_k}}}{\lambda _{{v_0}}}\beta \xi \psi \left( {{\delta _3} - {\mu _1}{\delta _4}} \right) }}} } \right) . \end{array} \end{aligned}$$

(35)

By using the $\int _0^\infty {{e^{\left( { - \frac{\beta }{{4x}} - \gamma x} \right) }}dx = \sqrt{\frac{\beta }{\gamma }} {\mathrm{K}_1}\left( {\sqrt{\beta \gamma } } \right) }$ [27].

Similarly, it can be obtained the closed-form of $U_3$ such as

$$\begin{aligned} \begin{array}{l} O_{{U_3}}^{\left( {EH} \right) } = 1 - \Pr \left[ {\gamma _{E - {U_3}}^{\left( {{z_2}} \right) } \ge {\mu _2}} \right] \\ \quad \quad = 1 - \Pr \left[ {{{\left| {{l_{k*}}} \right| }^2} \ge \frac{{{\mu _2}}}{{{{\left| {{r_0}} \right| }^2}{\delta _4}\beta \xi \psi }}} \right] \\ \quad \quad = 1 - \int \limits _0^\infty {{F_{{{\left| {{l_{k*}}} \right| }^2}}}} \left( {\frac{{{\mu _2}}}{{x{\delta _4}\beta \xi \psi }}} \right) {f_{{{\left| {{r_0}} \right| }^2}}}\left( x \right) dx\\ \quad \quad = 1 - \sum \limits _{k = 1}^K {\left( \begin{array}{l} K\\ k \end{array} \right) {{\left( { - 1} \right) }^{k - 1}}\frac{1}{{{\lambda _{{r_0}}}}}} \int \limits _0^\infty {\exp \left( { - \frac{{k{\mu _2}}}{{x{\lambda _{{l_k}}}{\delta _4}\beta \xi \psi }} - \frac{x}{{{\lambda _{{r_0}}}}}} \right) } dx\\ \quad \quad = 1 - \sum \limits _{k = 1}^K {\left( \begin{array}{l} K\\ k \end{array} \right) {{\left( { - 1} \right) }^{k - 1}}} \sqrt{\frac{{4k{\mu _2}}}{{{\lambda _{{l_k}}}{\lambda _{{r_0}}}{\delta _4}\beta \xi \psi }}} {\mathrm{{K}}_{{1}}}\left( {\sqrt{\frac{{4k{\mu _2}}}{{{\lambda _{{l_k}}}{\lambda _{{r_0}}}{\delta _4}\beta \xi \psi }}} } \right) . \end{array} \end{aligned}$$

(36)

It completes Proposition 3. $\square$