Joint relay selection and opportunistic physical layer network coding for two-way relay channels

Abstract

Physical layer network coding can significantly increase the throughput of two-way relay networks. However, fading phenomenon usually causes great asymmetry between two multiple access channels which can drastically degrade the performance of PNC protocol. To handle this issue, we propose an alternative detection method to improve the detection of physical layer network coded signal in multi-relay networks. First, we introduce single node detection to extract a single data from superimposed signal in the multiple access phase of physical layer network coding protocol. Then, we propose a relay selection method based on single node detection (RS-SND) and compare it to conventional relay selection method based on physical network coding (RS-PNC) in terms of average bit error rate for detection of XOR-ed data of the two source nodes. Closed-form expressions are derived for average BER of the proposed scheme and its asymptotic approximation at high signal to noise ratio. It is shown that RS-SND outperforms RS-PNC in Rayleigh fading channels when more than two relays are employed. Finally, we propose a relay selection-opportunistic physical layer network coding (RS-OPNC) method by dynamically selecting between RS-SND and RS-PNC schemes based on channel state information. Simulation results verify that RS-OPNC offers considerable SNR gain over conventional RS-PNC method.

Appendix A (Proof of Lemma I)

The joint cumulative distribution function of two Rayleigh random variables $R_A$ and $R_B$ is given by [28]

$$\begin{aligned} \begin{aligned}&F_{R_A,R_B}(r_A,r_B,\mu ) \\&\quad =1-\exp {\left( -\frac{r_A^2}{\varOmega _A}\right) }Q\left( \sqrt{\frac{2}{1-\mu }}\frac{r_B}{\sqrt{\varOmega _B}} ,\sqrt{\frac{2\mu }{1-\mu }}\frac{r_A}{\sqrt{\varOmega _A}}\right) \\&\qquad -\exp {\left( -\frac{r_B^2}{\varOmega _B}\right) }\left[ 1-Q\left( \sqrt{\frac{2\mu }{1-\mu }}\frac{r_B}{\sqrt{\varOmega _B}} ,\sqrt{\frac{2}{1-\mu }}\frac{r_A}{\sqrt{\varOmega _A}}\right) \right] \end{aligned} \end{aligned}$$

(40)

where $\varOmega _A$ and $\varOmega _B$ are average powers of random variables $R_A$ and $R_B$, respectively, and $\mu$ is their power correlation coefficient, which can be computed by the method proposed in [29] as follows.

Let, $h_1=X_{c_1}+jX_{s_1}$, and $h_2=X_{c_2}+jX_{s_2}$ where $X_{c_1}$, $X_{s_1}$, $X_{c_2}$ and $X_{s_2}$ are independent normal random variables such that $X_{c_1},X_{s_1}\sim {\mathcal {N}}(0,\sigma _1^2)$, and $X_{c_2},X_{s_2}\sim {\mathcal {N}}(0,\sigma _2^2)$. Then, define random variables $R_A=|h_1|$ and $R_B=|h_1-h_2|$. Clearly, $R_A$ and $R_B$ are Rayleigh random variables with scale variables $\sigma _A=\sigma _1$ and $\sigma _B=\sqrt{\sigma _1^2+\sigma _2^2}$, respectively [30]. Let, ${\mathbf {K}}_{cc}=E[{\mathbf {X}}_c {\mathbf {X}}_c^T]$ and ${\mathbf {K}}_{cs}=E[{\mathbf {X}}_c {\mathbf {X}}_s^T]$ in which ${\mathbf {X}}_{c}=[X_{c_1}~~X_{c_1}-X_{c_2}]^T$ and ${\mathbf {X}}_{s}=[X_{s_1}~~X_{s_1}-X_{s_2}]^T$. Then, using [29, eqs. 109 and 118], we get

$$\begin{aligned} \begin{aligned} \mu =\text {corr}(R^2_A, R^2_B)&=\frac{\sigma ^2_1}{\sigma ^2_1+\sigma ^2_2} \\&=\frac{\varOmega _1}{\varOmega _1+\varOmega _2} \end{aligned} \end{aligned}$$

(41)

where $\varOmega _i=2\sigma ^2_i$ is the average power of $|h_i|$, i=1,2. Then, substituting (41) in (40) and considering the fact that $F_{R^2_A,R^2_B}(z_1,z_2,\mu )=F_{R_A,R_B}(\sqrt{z_1},\sqrt{z_2} ,\mu )$, the joint CDF of random variables $R^2_A$ and $R^2_B$ is obtained as

$$\begin{aligned} \begin{aligned}&F_{R^2_A,R^2_B}(z_1,z_2,\mu ) \\&\quad =1-\exp {\left( -\frac{z_1}{\varOmega _1}\right) }Q\left( \sqrt{\frac{2}{\varOmega _2}}\sqrt{z_2},\sqrt{\frac{2}{\varOmega _2}}\sqrt{z_1}\right) \\&\qquad -\exp {\left( -\frac{z_2}{\varOmega _1+\varOmega _2}\right) }\left[ 1-Q\left( \sqrt{\frac{2\mu }{\varOmega _2}}\sqrt{z_2},\sqrt{\frac{2}{\mu \varOmega _2}}\sqrt{z_1}\right) \right] \\ \end{aligned} \end{aligned}$$

(42)

Also, it is obvious that the CDF of exponential random variables $R^2_A$ and $R^2_B$ are given by $F_{R^2_A}(z)=1-e^{-\frac{z}{\varOmega _1}}$ and $F_{R^2_B}(z)=1-e^{-\frac{z}{\varOmega _1+\varOmega _2}}$, respectively. Therefore, employing $F_{R^2_A}(z)$, $F_{R^2_B}(z)$, and (42) in the general formula $F_{\min {(R^2_A,R^2_B)}}(z)=F_{R^2_A}(z)+F_{R^2_B}(z)-F_{R^2_A,R^2_B}(z,z)$, we obtain (13). $\square$