A new look at AF two-way relaying networks: energy harvesting architecture and impact of co-channel interference

Abstract

In this paper, the performance of dual-hop amplify-and-forward (AF) two-way relaying systems is considered, where the terminals and relay are interfered by a finite number of co-channel interferers. In addition, the derived expressions are evaluated in terms of outage probability and throughput in delay-limited transmission mode. To make the analysis mathematically tractable, the unique expressions of outage probability are adopted to deal with energy harvesting protocols related to time switching and power splitting coefficients and expression of the throughput is also calculated. Based on the analytic results, this paper investigates the impact of system parameters such as energy harvesting time/power fractions, number of interferers and signal-to-noise ratio (SNR) on throughput performance. Monte Carlo simulation results are presented to prove the tightness of the proposed energy harvesting two-way relaying system.

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Notes

  1. 1.

    In practice, at the beginning of each block time, the relay node needs a certain amount of energy for operating. In this scenario, the relay node needs a certain amount of energy of previous blocks to initially operate in the next adjacent time slot. However, for simplicity, we assume that initializing energy at the beginning of each frame is not considered.

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Correspondence to Dinh-Thuan Do.

Appendix: Proof of proposition 1

Appendix: Proof of proposition 1

This appendix derives the CDF of γ u ,u = {X,Y} in {(22), (23)}, respectively. Thus, the CDF of the end-to-end SINR of γ u is expressed as

$$ \begin{array}{c} F_{\gamma_{u} } \left( \gamma \right) = \Pr \left\{ {\gamma_{u} < \gamma } \right\} \\ \!= 1 \,-\, \int\limits_{0}^{\infty} {\int\limits_{0}^{\infty} {\int\limits_{0}^{\infty} {\exp \left\{ { - \frac{\gamma }{{\lambda_{v} }}\left( {z + 1 + \frac{{y + 1}}{{\mathcal{C}_{u} x}}} \right)} \right\}f_{Z_{R} } \left( z \right)f_{Z_{u} } \left( y \right)f_{u} \left( x \right)dxdydz} } } \end{array} $$
(31)

Note that, in order to achieve (31), we applied the cumulative distribution function of X and Y which can be given as \( F_{u} \left (x \right ) = 1 - \exp \left ({ - \frac {x}{{\lambda _{u} }}} \right )\). Subsequently, the triple integral can be divided separately as follow

$$ \begin{array}{c} F_{\gamma_{u} } \left( \gamma \right) = 1 - \underbrace {\int\limits_{0}^{\infty} {\exp \left\{ { - \frac{\gamma }{{\lambda_{v} }}\left( {z + 1} \right)} \right\}f_{Z_{R} } \left( z \right)dz} }_{I_{1} } \\ {} \times \underbrace {\int\limits_{0}^{\infty} {\int\limits_{0}^{\infty} {\exp \left\{ { - \frac{\gamma }{{\lambda_{v} }}\left( {\frac{{y + 1}}{{\mathcal{C}_{u} x}}} \right)} \right\}f_{Z_{u} } \left( y \right)f_{u} \left( x \right)dxdy} } }_{I_{2} } \end{array} $$
(32)

As can be seen, in order to derive \( F_{\gamma _{u} } \left (\gamma \right )\), we have to derive I 1 and I 2 first. After substituting the PDF of Z R given in Eq. 25 into (32), I 1 can be rewritten as

$$\begin{array}{@{}rcl@{}} I_{1} &=& \exp \left( { - \frac{\gamma }{{\lambda_{Y} }}} \right)\sum\limits_{j_{1} = 1}^{\upsilon \left( \mathcal{B} \right)} {\sum\limits_{j_{2} = 1}^{\tau_{j_{1} } \left( \mathcal{B} \right)} {\chi_{j_{1} ,j_{2} } \left( \mathcal{B} \right)\frac{{\mu_{R\left\langle {j_{1} } \right\rangle }^{- j_{2} } }}{{\left( {j_{2} - 1} \right)!}}} } \int\limits_{0}^{\infty} z^{j_{2} - 1}\\ &&\times \exp \left[ { - \left( {\frac{\gamma }{{\lambda_{v} }} + \frac{1}{{\mu_{R\left\langle {j_{1} } \right\rangle } }}} \right)z} \right]dz \end{array} $$
(33)

With the help of identities [21, Eq. 3.351.3], the above integration is calculated as follow

$$ \int\limits_{0}^{\infty} {z^{j_{2} - 1} \exp \left[ { - \left( {\frac{\gamma }{{\lambda_{v} }} + \frac{1}{{\mu_{R\left\langle {j_{1} } \right\rangle } }}} \right)z} \right]dz} = \left( {j_{2} - 1} \right)!\left( {\frac{\gamma }{{\lambda_{v} }} + \frac{1}{{\mu_{R\left\langle {j_{1} } \right\rangle } }}} \right)^{- j_{2} } $$
(34)

Substituting (34) into (33), new expression is then obtained as

$$ I_{1} = \exp \left( { - \frac{\gamma }{{\lambda_{v} }}} \right)\sum\limits_{j_{1} = 1}^{\upsilon \left( \mathcal{B} \right)} {\sum\limits_{j_{2} = 1}^{\tau_{j_{1} } \left( \mathcal{B} \right)} {\chi_{j_{1} ,j_{2} } \left( \mathcal{B} \right)} } \left( {\frac{{\mu_{R\left\langle {j_{1} } \right\rangle } }}{{\lambda_{v} }}\gamma + 1} \right)^{- j_{2} } $$
(35)

Secondly, I 2 is derived by first substituting \( f_{Z_{u} } \left (y \right ) \) as given in Eq. 24 and \( f_{u} \left (y \right ) = \frac {1}{{\lambda _{u} }}\exp \left ({ - \frac {y}{{\lambda _{u} }}} \right ) \) into (32), and thus, I 2 can be rewritten as

$$ I_{2} = \sum\limits_{i_{1} = 1}^{\upsilon \left( {\mathcal{A}_{u} } \right)} {\sum\limits_{i_{2} = 1}^{\tau_{i_{1} } \left( {\mathcal{A}_{u} } \right)} {\chi_{i_{1} ,i_{2} } \left( {\mathcal{A}_{u} } \right)\frac{{\mu_{u\left\langle {i_{1} } \right\rangle }^{- i_{2} } }}{{\left( {i_{2} - 1} \right)!}} \times {\Psi} } } $$
(36)

where \( {\Psi } = \int \limits _{0}^{\infty } {y^{i_{2} - 1} \exp \left ({ - \frac {y}{{\mu _{u\left \langle {i_{1} } \right \rangle } }}} \right )\int \limits _{0}^{\infty } {\frac {1}{{\lambda _{u} }}\exp \left ({ - \frac {{\gamma \left ({y + 1} \right )}}{{\lambda _{v} \mathcal {C}_{X} }}\frac {1}{x} - \frac {x}{{\lambda _{u} }}} \right )dxdy} } \)

By applying [21, Eq. 3.478.4] to the second integration, we obtain

$$\begin{array}{@{}rcl@{}} \int\limits_{0}^{\infty} {\frac{1}{{\lambda_{u} }}\exp \left( { - \frac{{\gamma \left( {y + 1} \right)}}{{\lambda_{v} \mathcal{C}_{u} }}\frac{1}{x} - \frac{x}{{\lambda_{u} }}} \right)dx} &=& \mathcal{S}_{u} \left( \gamma \right)\sqrt {y + 1} K_{1} \\&&\times\left( {\mathcal{S}_{u} \left( \gamma \right)\sqrt {y + 1} } \right) \end{array} $$
(37)

Consequently, Ψ is rewritten as

$$ {\Psi} = \mathcal{S}_{u} \left( \gamma \right)\int\limits_{0}^{\infty} {y^{i_{2} - 1} \exp \left( { - \frac{y}{{\mu_{u\left\langle {i_{1} } \right\rangle } }}} \right)\sqrt {y + 1} K_{1} \left( {\mathcal{S}_{u} \left( \gamma \right)\sqrt {y + 1} } \right)dy} $$
(38)

After some algebraic manipulations, Ψ is obtained as

$$ {\Psi} = \mathcal{S}_{u} \left( \gamma \right)\int\limits_{1}^{\infty} {\left( {\omega - 1} \right)^{i_{2} - 1} \exp \left( { - \frac{{\omega - 1}}{{\mu_{u\left\langle {i_{1} } \right\rangle } }}} \right)\sqrt \omega K_{1} \left( {\mathcal{ S}_{u} \left( \gamma \right)\sqrt \omega } \right)d\omega } $$
(39)

Since index i 2 − 1 ≥ 0 we can apply the Newton’s binomial theorem in [21, Eq. 1.111] to finally derive the following result

$$\begin{array}{@{}rcl@{}} {\Psi} &=& \mathcal{S}_{u} \left( \gamma \right)\exp \left( {\frac{1}{{\mu_{u\left\langle {i_{1} } \right\rangle } }}} \right)\sum\limits_{n = 0}^{i_{2} - 1} {\left( {\begin{array}{*{20}c} {i_{2} - 1} \\ n \end{array}} \right)\left( { - 1} \right)^{i_{2} - n - 1} } \int\limits_{1}^{\infty} \omega^{n + \frac{1}{2}} \exp \\ &&\times\left( { - \frac{\omega }{{\mu_{u\left\langle {i_{1} } \right\rangle } }}} \right)K_{1} \left( {\mathcal{S}_{u} \left( \gamma \right)\sqrt \omega } \right)d\omega \end{array} $$
(40)

This is end the proof.

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Luan, N.T., Do, DT. A new look at AF two-way relaying networks: energy harvesting architecture and impact of co-channel interference. Ann. Telecommun. 72, 669–678 (2017). https://doi.org/10.1007/s12243-017-0590-7

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Keywords

  • Energy harvesting
  • Co-channel interference (CCI)
  • Amplify-and-forward
  • Outage capacity
  • Two-way relaying