Limited Feedback Assisted Dynamic Decode-and-Forward Relaying for Multiple Access Relay Channel

Abstract

Feedback about the channel state information (CSI) enables the transmitter nodes to exploit channel conditions to yield large improvements in almost any performance metric. However, in practice, channel adaptive techniques based on full CSI have been deemed impractical due to the finite capacity of the feedback links. This work considers a multiple-access relay channel (MARC) where two source nodes communicate with one destination node assisted by one half-duplex dynamic decode-and forward (DDF) relay. Using the diversity-multiplexing tradeoff as a figure-of-merit, we propose a practical limited-feedback (LF) mechanism for MARC and show that a small number of information bits about the channel conditions leads to near optimal performance. With no CSI, this system incurs diversity-gain loss at higher multiplexing gains. However, by using an LF scheme where the destination sends an ACK or NACK to the relay, we show that the optimal diversity gain can be achieved across all the multiplexing gains for the considered DDF based MARC system.

Appendix

1.1 Proof of Lemma 1

The constraint region corresponding to linear programming problem defined in (26) can be characterized into three parts based on the value of f. These possibilities are graphically depicted in Fig. 4a, b and c.

• 1. $0 \le f \le \frac{1}{2}$ : Using Fig. 4a, we can conclude that the optimal values $v_1^*=v_r^*=1-\frac{r}{2}$, i.e.,

  $$\begin{aligned} \inf (v_1 + v_2 + v_r) = 2 \left( 1-\frac{r}{2} \right) = 2-r \end{aligned}$$

  (46)

• 2. $\frac{1}{2} \le f \le (1-\frac{r}{2})$ : As shown in Fig. 4b, solving (26) for $v_r^*$ at $v_{1}^*=1$ gives

  $$\begin{aligned} v_r=1-\frac{r}{2(1-f)}, \end{aligned}$$

  thus

  $$\begin{aligned} \inf (v_1 + v_2 + v_r) = 2-\frac{r}{2(1-f)} \end{aligned}$$

  (47)

• 3. $\left( 1-\frac{r}{2}\right) \le f \le 1$ : As can be seen from Fig. 4c, in this case $v_1^* = \frac{2-r}{2f}$ and $v_r^*=0$, hence

  $$\begin{aligned} \inf (v_1 + v_2 + v_r) = \frac{2-r}{2f} \end{aligned}$$

  (48)

Fig. 4

Outage regions for individual rate related error event (${\mathscr {E}}_1$) in Mode $M_2$. a For $0 \le f \le \frac{1}{2}$, b for $\frac{1}{2} \le f \le (1-\frac{r}{2})$, c for $(1-\frac{r}{2}) \le f \le 1$

Finally, by using (46), (47) and (48), we get (27) and thus completes the proof. $\square$

1.2 Proof of Lemma 2

We follow essentially similar steps as in the case of individual rate related error event. For $r \le 2/3$, the constraint region corresponding to the linear programming problem defined in (36) can be characterized into three distinct portions based on the value of f. These possibilities are graphically depicted in Fig. 5a, b and c. However for $r > 2/3$, the region depicted in Fig. 5b is subsumed in to the region defined by Fig. 5a, i.e., $\frac{1-r}{1-\frac{r}{2}} \le \frac{1}{2}$.

• 1. $0 \le f \le \frac{1}{2}$ : Using Fig. 5a, we can conclude that the optimal values $v_1^*=v_r^*=1-\frac{r}{2}$, i.e.,

  $$\begin{aligned} \inf (v_1 + v_2 + v_r) = 2 \left( 1-r \right) = \left( 1-\frac{r}{2} \right) + 2(1-r) \end{aligned}$$

  (49)

• 2. $\frac{1}{2} \le f \le \frac{1-r}{1-\frac{r}{2}}$ : As shown in Fig. 5b, solving (36) for $v_r^*$ at $v_{1}^*=1-\frac{r}{2}$ gives

  $$\begin{aligned} v_r=1-\frac{r}{2} \left( \frac{2-f}{1-f} \right) , \end{aligned}$$

  thus

  $$\begin{aligned} \inf (v_1 + v_2 + v_r) = \left( 1-\frac{r}{2} \right) + 2 - \frac{r}{2} \left( \frac{3-2f}{1-f} \right) \end{aligned}$$

  (50)

• 3. $\frac{1-r}{1-\frac{r}{2}} \le f \le 1$ : As can be seen from Fig. 5c, in this case $v_1^* = \frac{1-r}{f}$ and $v_r^*=0$, hence

  $$\begin{aligned} \inf (v_1 + v_2 + v_r) = \left( 1-\frac{r}{2} \right) + \frac{1-r}{f} \end{aligned}$$

  (51)

Fig. 5

Outage regions for sum-rate related error event (${\mathscr {E}}_2$) in Mode $M_2$. a For $0 \le f \le \frac{1}{2}$, b for $\frac{1}{2} \le f \le \frac{1-r}{1-\frac{r}{2}}$, c for $\frac{1-r}{1-\frac{r}{2}} \le f \le 1$

Finally, by using (49), (50) and (51), we get (37) and (38) and thus completes the proof. $\square$