Loop-Interference Based Full-Duplex Relay Selection

Abstract

This study proposes and analyses the loop-interference based selection policy for the full-duplex relaying networks. A multiple full-duplex relay-assisted network model is considered for the investigation. Computer simulations are utilized to verify the analytical and asymptotic derivations. In reference to obtained results, the loop-interference based selection policy achieves cooperative diversity order, the total number of possible paths between source-destination terminals, in high signal-to-noise ratios. Proposed selection strategy’s performance efficiency is measured by dint of outage probability, error probability, ergodic rate, and throughput. So as to provide more insight of the performance analysis, the asymptotic analysis is also provided.

Appendices

Appendix 1: Proof of Proposition 1

Since all channels have Rayleigh distribution, the $j^{th}$ relay terminal’s LI PDF expression can be written as: $f\gamma _{a_{j}}\left( \gamma \right) =\frac{1}{P _{r}\varOmega _{a_{j}}}e^{-\frac{\gamma }{P _{r}\varOmega _{a_{j}}}}$ [42]. By invoking (8) and utilizing the probability, following result can be obtained.

$$\begin{aligned} F_{{\gamma _{{a}_{j}}}}\left( \gamma _{\mathrm{th}}\right)&=P _{r}\left( \gamma _{{a}_{j}}\le \underbrace{R}_{\gamma }\right) \nonumber \\&=\int _{0}^{\gamma }f\gamma _{{a}_{j}}\left( \gamma \right) d\gamma \nonumber \\&=1-e^{-\frac{\gamma }{P _{r}\varOmega _{{a}_{j}}}} \end{aligned}$$

(15)

The final CDF can be obtained as in (2), by invoking the order statistics [43].

Appendix 2: Proof of Proposition 2

The identity product representation, which is presented as in the following expression [44, Eq. (9)], is utilized for the error probability performance analysis.

$$\begin{aligned}&\prod \limits _{k=1}^L\left( 1-x_{k}\right) =\sum \limits _{k=0}^{L}\frac{\left( -1\right) ^{k}}{k!}{ \underbrace{\sum \limits _{{n_1} = 1}^L {...\sum \limits _{{n_k} = 1}^L {} } }_{{n_1} \ne {n_2}\ne ... \ne {n_k}}}\prod \limits _{p=1}^k\left( x_{n_p}\right) \end{aligned}$$

(16)

Utilizing (16), (2) can be written as

$$\begin{aligned}&\prod \limits _{j=1}^N\left( 1-x_{j}\right) =\sum \limits _{j=0}^{N}\frac{\left( -1\right) ^{j}}{j!}{ \underbrace{\sum \limits _{{n_1} = 1}^N {...\sum \limits _{{n_j} = 1}^N {} } }_{{n_1} \ne {n_2}\ne ... \ne {n_j}}}\left( {\mathrm{exp}}\left( -x\sum _{p=1}^{j}\left( x_{n_{p}}\right) \right) \right) \end{aligned}$$

(17)

where $x_{n_{p}}=\frac{1}{P_{r}\varOmega _{a_{n_{p}}}}$. By plugging the obtained result into (3) and considering the BPSK modulation, following result can be obtained.

$$\begin{aligned} {\bar{P_{e}}}&= \frac{1}{2\sqrt{\pi }}\sum \limits _{j=0}^{N}\frac{\left( -1\right) ^{j}}{j!}{ \underbrace{\sum \limits _{{n_1} = 1}^N {...\sum \limits _{{n_j} = 1}^N {} } }_{{n_1} \ne {n_2}\ne ... \ne {n_j}}}\nonumber \\&\quad \times \int _0^\infty x^{-\frac{1}{2}}\left( {\mathrm{exp}}\left( -x\sum _{p=1}^{j}\left( \frac{1}{P_{r}\varOmega _{{a}_{{n}_{p}}}}+1\right) \right) \right) dx \end{aligned}$$

(18)

Utilizing [38, Eq. ($3.361.2^{8}$)] in (18), final result is as in (4).

Appendix 3: Proof of Proposition 3

By setting $j=0$ in the left-hand-side of (17), following expression can be obtained.

$$\begin{aligned}&\prod \limits _{j=0}^N\left( 1-x_{j}\right) =1+\sum \limits _{j=1}^{N}\frac{\left( -1\right) ^{j}}{j!}{ \underbrace{\sum \limits _{{n_1} = 1}^N {...\sum \limits _{{n_j} = 1}^N {} } }_{{n_1} \ne {n_2}\ne ... \ne {n_j}}}\nonumber \\&\times \left( {\mathrm{exp}}\left( -x\sum _{p=1}^{j}\left( x_{n_{p}}\right) \right) \right) \end{aligned}$$

(19)

Substituting (19) into (5), following result can be obtained.

$$\begin{aligned} {\overline{ER}}\left( \gamma \right)&=\frac{1}{\mathrm{ln2}}\sum \limits _{j=1}^{N}\frac{\left( -1\right) ^{j}}{j!}{ \underbrace{\sum \limits _{{n_1} = 1}^N {...\sum \limits _{{n_j} = 1}^N {} } }_{{n_{1}} \ne {n_2}\ne ... \ne {n_j}}}\nonumber \\&\quad \times {\int _{0}^{\infty }} \frac{\left( {\mathrm{exp}}\left( -x\sum _{p=1}^{j}\left( x_{{n}_{p}}\right) \right) \right) }{1+\gamma }d_{\gamma } \end{aligned}$$

(20)

Utilizing [38, Eq. (3.352.4)] in (20), the final result is obtained as in (5).