Performance Analysis and Enhancement for Opportunistic Analog Network Coding with Imperfect CSI

Abstract

Imperfect channel state information (CSI) is among the main factors that affect system performance in wireless networks. In this paper, we investigate the impact of imperfect CSI on the performance of analog network coding (ANC) for a two-way relaying system based on opportunistic relay selection (ORS). An exact and generalized closed-form expression for system outage probability is presented in a Rayleigh flat-fading environment. To provide more insights, the closed-form asymptotic expression is then obtained. It is shown that the presence of channel estimation error causes outage probability maintain a fixed level even when a noiseless channel is adopted. Therefore, to mitigate the negative impact of imperfect CSI, we deduce the power allocation to minimize the system outage probability based on the knowledge of instantaneous channel information. Numerical results validate the accuracy of the derived expressions and highlight the effect of proposed power allocation algorithm compared with conventional uniform power allocation.

Appendices

Appendix A

$$\begin{aligned} \gamma _{ak} - \gamma _{bk}&= \frac{P_r P_b |\hat{h}_{ak} |^2 |\hat{h}_{bk} |^2 }{\alpha _{1k} |\hat{h}_{ak} |^2 + \alpha _{2k} |\hat{h}_{bk} |^2 } - \frac{P_r P_a |\hat{h}_{ak} |^2 |\hat{h}_{bk} |^2 }{\beta _{1k} |\hat{h}_{ak} |^2 + \beta _{2k} |\hat{h}_{bk} |^2} \nonumber \\&= \Phi \left( {(P_b \beta _{1k} - P_a \alpha _{1k} )|\hat{h}_{ak} |^2 + (P_b \beta _{2k} - P_a \alpha _{2k} )|\hat{h}_{bk} |^2 } \right) \end{aligned}$$

(20)

where $\Phi = \frac{{P_r |\hat{h}_{ak} |^2 |\hat{h}_{bk} |^2 }}{{\left( {\alpha _{1k} |\hat{h}_{ak} |^2 + \alpha _{2k} |\hat{h}_{bk} |^2 } \right) \left( {\beta _{1k} |\hat{h}_{ak} |^2 + \beta _{2k} |\hat{h}_{bk} |^2 } \right) }} \ge 0$.

a) Clearly, when $P_b \beta _{1k} - P_a \alpha _{1k} \ge 0$, i.e., $\beta _{1k} /\alpha _{1k} \ge P_a /P_b,\, P_b \beta _{2k} - P_a \alpha _{2k} \ge 0$. From (20), we can obtain $\gamma _{ak} \ge \gamma _{bk}$. Thus we have

$$\begin{aligned} \mathbf{Pr} \left[ {\min \left( {\gamma _{ak} ,\;\gamma _{bk} } \right) \le \gamma _{th} } \right] = \mathbf{Pr} \left[ {\gamma _{bk} \le \gamma _{th} } \right] \end{aligned}$$

(21)

b) Similarly, when $P_b \beta _{2k} - P_a \alpha _{2k} \le 0$, i.e., $\beta _{2k} /\alpha _{2k} \le P_a /P_b,\, P_b \beta _{1k} - P_a \alpha _{1k} \le 0$. From (20), we can obtain $\gamma _{ak} \le \gamma _{bk}$. Thus we have

$$\begin{aligned} \mathbf{Pr} \left[ {\min \left( {\gamma _{ak} ,\;\gamma _{bk} } \right) \le \gamma _{th} } \right] = \mathbf{Pr} \left[ {\gamma _{ak} \le \gamma _{th} } \right] \end{aligned}$$

(22)

According to the above discussions, we can arrive at the criterion in (7).

Appendix B

$$\begin{aligned} \mathcal P _{out,k} = 1 - {\mathbf{Pr}}\left[ {\gamma _{ak} > \gamma _{th} ,\;\gamma _{bk} > \gamma _{th} } \right] \end{aligned}$$

(23)

As $\gamma _{ak}$ and $\gamma _{bk}$ are not independent, it is hard to evaluate (23) straightforward. By using the fact that $\beta _{1k} /\alpha _{1k} < P_a /P_b$ and some manipulations, we can obtain

$$\begin{aligned} \mathcal P _{out,k} = 1 - {\mathbf{Pr}}\left[ {X > \max \left( {\vartheta _1 ,\;\vartheta _2 } \right) ,Y > \frac{{\alpha _{1k} \gamma _{th} }}{{P_r P_b }}} \right] \end{aligned}$$

(24)

where $X = |\hat{h}_{ak} |^2$, $Y = |\hat{h}_{bk} |^2$, $\vartheta _1 = \frac{{\alpha _{2k} \gamma _{th} Y}}{{P_r P_b Y - \alpha _{1k} \gamma _{th} }}$ and $\vartheta _2 = \frac{{\beta _{2k} \gamma _{th} Y}}{{P_r P_a Y - \beta _{1k} \gamma _{th} }}$. Furthermore, by using $\beta _{2k} /\alpha _{2k} > P_a /P_b$, (24) can be rewritten as

$$\begin{aligned} \begin{aligned} \mathcal{P }_{out}&= 1 - {\underbrace{{\mathbf{Pr}} \left[ {X > \vartheta _2 ,\;Y > \omega } \right] }}_{\xi _{ak} } \\&\quad - {\underbrace{{\mathbf{Pr}}\left[ {X > \vartheta _1 ,\;\frac{{\alpha _{1k} \gamma _{th} }}{{P_r P_b }} < Y \leqslant \omega } \right] }}_{\xi _{bk} } \\ \end{aligned} \end{aligned}$$

(25)

As $X$ and $Y$ are independent, we can evaluate $\xi _{ak}$ as

$$\begin{aligned} \xi _{ak}&= \int _\omega ^\infty {f_Y (y)\int _{\vartheta _2 }^\infty {f_X (x)dxdy} } \nonumber \\&= \int _\omega ^\infty {f_Y (y)e^{ - \frac{1}{{\Omega _{\hat{h},ak} }}\frac{{\beta _{2k} \gamma _{th} y}}{{P_r P_a y - \beta _{1k} \gamma _{th} }}} } dy \end{aligned}$$

(26)

where $f_X(x)$ and $f_Y(y)$ denote the probability density function (PDF) of $X$ and $Y$. Making the change of variable $t = P_r P_a y - \beta _{1k} \gamma _{th}$ within the integral and applying the Taylor series expansion for the exponential term and with the help of [12], we can obtain the final form.

Similarly, $\xi _{bk}$ can be evaluated as

$$\begin{aligned} \xi _{bk}&= \int _{\frac{{\alpha _{1k} \gamma _{th} }}{{P_r P_b }}}^\omega {f_Y (y)\int _{\vartheta _1 }^\infty {f_X (x)dxdy} } \nonumber \\&= \int _{\frac{{\alpha _{1k} \gamma _{th} }}{{P_r P_b }}}^\infty {f_Y (y)\int _{\vartheta _1 }^\infty {f_X (x)dxdy} }\nonumber \\&- \int _\omega ^\infty {f_Y (y)\int _{\vartheta _1 }^\infty {f_X (x)dxdy} } \end{aligned}$$

(27)

The first term of (27) can be evaluated straightforward and the second term can be deduced by following the same approach as carried out to obtain $\xi _{ak}$. Substituting $\xi _{ak}$ and $\xi _{bk}$ into (25) yields (10).