Training Power Allocation Based on MSE-Minimization for Multi-Relay Amplify-and-Forward Cooperation in Wireless Sensor Networks

Abstract

In this paper, a training scheme for multi-relay clustered wireless sensor networks is proposed for the destination cluster head node to estimate the individual source-relay and relay-destination channels. Compared with an existing training scheme, the proposed one has shown improved efficiency with reduced power and computation time in the training of the source relay channels. Furthermore, the training power allocation problem is analyzed. Based on minimizing the total mean-square-error (MSE) of all channel estimates, two approximate solutions for the power allocation among all network nodes and all training links are derived. The first solution is adaptive to the instantaneous relay-destination channel estimates, thus requires feedback from the destination during the training process. The other solution depends on channel variances only and has lower complexity in implementation. Simulation results demonstrate that the proposed training scheme and power allocation solutions obtain lower total MSE and higher network throughput than existing schemes.

Appendices

Appendix 1: Proof of Lemma 1

Define

$$\begin{aligned} {\tilde{\gamma }}^* \triangleq \left( {\frac{{\sum \limits _{i = 1}^N {{\tilde{c}}_i^{ - 1} \sqrt{a_i {\tilde{c}}_i - {\tilde{b}}_i } } }}{{1 + \sum \limits _{i = 1}^N {{\tilde{c}}_i^{ - 1} } }}} \right) ^2 \end{aligned}$$

(35)

and

$$\begin{aligned} {\tilde{\theta }} _i^* \triangleq \frac{1}{{{\tilde{c}}_i }}\left( \sqrt{\frac{a_i {\tilde{c}}_i - {\tilde{b}}_i }{{\tilde{\gamma }}^*}}- 1 \right) ,\ i = 1, \ldots ,N. \end{aligned}$$

(36)

We will show that the defined ${\tilde{\gamma }}^*$ and $\theta _i^*$’s are the solutions to (22). This is equivalent to showing that $\tilde{\theta }_i^*\in (0,1)$ and $\sum _{i=1}^N\tilde{\theta }_i^*=1$. From the definitions of $\tilde{\theta }_i^*$ and $\tilde{\gamma }^*$ in (36) and (35), it is obvious that $\sum _{i=1}^N\tilde{\theta }_i^*=1$. Next, we show that for $i=1,\ldots ,N$, $\tilde{\theta }_i^*\in (0,1)$. Define the following functions:

$$\begin{aligned} \phi _i(x) \triangleq \frac{a_i {\tilde{c}}_i - \tilde{b}_i}{{\left( {1+{\tilde{c}}_i x}\right) }^2}, \ i = 1, \ldots ,N. \end{aligned}$$

(37)

Obviously, $\phi _i$ is a monotonically decreasing function of x. For $x\in [0,1]$, by evaluating $\phi _i(x)$ at $x=1$ and $x=0$, the minimum and maximum values of $\phi _i(x)$ are achieved as follows

$$\begin{aligned} \phi _{i,\min }\triangleq & {} \frac{{a_i {\tilde{c}}_i - {\tilde{b}}_i }}{{\left( {1 + {\tilde{c}}_i } \right) ^2 }} = \frac{{P_s P_f \sigma _{f_i}^2 \sigma _{g_i}^2 }}{{\left( {1 + P_f \sigma _{g_i }^2 } \right) ^2 }} \approx \frac{P_s}{P_f }\frac{{ \sigma _{f_i}^2 }}{{\sigma _{g_i }^2 }}, \end{aligned}$$

(38)

$$\begin{aligned} \phi _{i,\max }\triangleq & {} a_i {\tilde{c}}_i - {\tilde{b}}_i = P_s P_f \sigma _{f_i}^2 \sigma _{g_i}^2. \end{aligned}$$

(39)

From the definition in (36), we have ${\tilde{\gamma }}^*=\phi _i({\tilde{\theta }}_i^*)$. Since $\phi _i(x)$ is a continuous function, it is sufficient to show that ${\tilde{\theta }}_i^*\in (0,1)$ by showing that ${\tilde{\gamma }}^*\in (\phi _{i,\min },\phi _{i,\max })$ for $i=1,\ldots ,N$. By using (14) and (20) in (35),

$$\begin{aligned} {\tilde{\gamma }}^* = \left( \frac{\sqrt{\frac{P_s}{P_f}}\sum _{i = 1}^N \frac{\sigma _{f_i}}{\sigma _{g_i}}}{1 + P_f^{-1}\sum _{i= 1}^N {\sigma _{g_i}^{-2} }} \right) ^2. \end{aligned}$$

(40)

With the high SNR assumption, we have $P_{f_i}\sigma _{g_i}^2\gg 1$. Thus $P_{f_i}\gg \sigma _{g_i}^{-2}$ and

$$\begin{aligned} P_f=\sum _{i=1}^N P_{f_i}\gg \sum _{i=1}^N \sigma _{g_i}^{-2}, \end{aligned}$$

(41)

from which we can obtain that $P_f^{-1}\sum _{i= 1}^N \sigma _{g_i}^{-2} \ll 1$. Thus from (40),

$$\begin{aligned} {\tilde{\gamma }}^* \approx \frac{P_s}{P_f}\left( \sum _{i = 1}^N \frac{\sigma _{f_i}}{\sigma _{g_i}}\right) ^2. \end{aligned}$$

(42)

Since $\sigma _{f_i},\sigma _{g_i}>0$, we have $\tilde{\gamma }^*>\phi _{i,\min }$. Next, we show that $\tilde{\gamma }^*<\phi _{i,\max }$. From (1), we have

$$\begin{aligned} \sigma _{f_i}^2\sigma _{g_i}^2=\frac{1}{\mu _i^2(1-\mu _i)^2} \overset{(a)}{\ge }16, \quad \frac{\sigma _{f_i}^2}{\sigma _{g_i}^2}=\frac{(1-\mu _i)^2}{\mu _i^2}, \end{aligned}$$

where the inequality (a) is because $\mu _i\in (0,1)$. Thus, when $P_f>N(1-\mu _a)/(4\mu _a)$, we have

$$\begin{aligned}&{\tilde{\gamma }}^* \approx \frac{P_s}{P_f}\left( \sum _{i = 1}^N \frac{\sigma _{f_i}}{\sigma _{g_i}}\right) ^2 =\frac{P_sP_f}{P_f^2}\left( \sum _{i = 1}^N \frac{\sigma _{f_i}}{\sigma _{g_i}}\right) ^2\\&\quad <P_sP_f\frac{16\mu _a^2}{N^2(1-\mu _a)^2}\left( \sum _{i = 1}^N \frac{1-\mu _i}{\mu _i}\right) ^2\\&\quad \overset{(b)}{\le } P_sP_f\frac{16\mu _a^2}{N^2(1-\mu _a)^2}\left( \sum _{i = 1}^N \frac{1-\mu _a}{\mu _a}\right) ^2=16P_sP_f \\&\quad \overset{(c)}{\le } P_sP_f\sigma _{f_i}^2\sigma _{g_i}^2=\phi _{i,\max }. \end{aligned}$$

In these derivations, the inequality (b) is because $0<\mu _a \le \mu _i<1$ and the inequality (c) is due to the inequality (a). We have shown ${\tilde{\gamma }}^*\in (\phi _{i,\min },\phi _{i,\max })$ for $i=1,\ldots ,N$. Finally, (24) is shown for the proof by using (20), (21), and (42) in (22).

Appendix 2: Proof of Lemma 2

For homogeneous relay networks, we have $\sigma _{f_i}^2=\sigma _f^2$ and $\sigma _{g_i}^2=\sigma _g^2$, $i=1, \ldots , N$. By using (1), after some tedious but straightforward calculations, the threshold of ${\tilde{P}}_t$ in (33) can be replaced with

$$\begin{aligned} {\tilde{P}}_t(\mu )= & {} {\left( 1-\mu \right) }{\left( \mu \sqrt{N}+N \sqrt{\mu ^2+(1-\mu )^2}\right) }. \end{aligned}$$

(43)

where the function ${\tilde{P}}_t(\mu )$ is decreasing for $\mu$ in the interval $\left[ \mu _a,\mu _b \right]$. To prove this, let $\alpha = \sqrt{\mu ^2+(1-\mu )^2}$. since $0\le \mu _a\le \mu \le \mu _b\le 1$, we have

$$\begin{aligned} 1/2\le \alpha ^2\le 1\ \Longrightarrow 0\le 2\alpha ^2-1\le 1. \end{aligned}$$

(44)

Thus, the following two solutions to $\mu$ can be obtained by

$$\begin{aligned} \mu _1 = \frac{1}{2}\sqrt{2\alpha ^2-1}+\frac{1}{2}, \quad \mu _2 = -\frac{1}{2}\sqrt{2\alpha ^2-1}+\frac{1}{2}. \end{aligned}$$

(45)

On substituting $\mu _1$ and $\mu _2$ into (43) and calculating the first-order derivatives with respect to $\alpha$, respectively, we have

$$\begin{aligned} \frac{{\partial {{\tilde{P}}_t}(\mu ) }}{\partial \alpha }|_{\mu =\mu _1}& {}= - \sqrt{N}\alpha -\frac{N\sqrt{2\alpha ^2-1}}{2}-\frac{N\left( {2\alpha ^2-\sqrt{2\alpha ^2-1}}\right) }{2\sqrt{2\alpha ^2-1}},\\ \frac{{\partial {{\tilde{P}}_t}(\mu ) }}{\partial \alpha }|_{\mu =\mu _2}& {}= \frac{N}{2}\left( {1+\sqrt{2\alpha ^2-1}}\right) +\frac{\sqrt{N}\alpha \left( {\sqrt{N\alpha ^2}-\sqrt{2\alpha ^2-1}}\right) }{\sqrt{2\alpha ^2-1}}. \end{aligned}$$

From (44), we have $2\alpha ^2\ge \sqrt{2\alpha ^2-1}$ and ${{\partial {\tilde{P}_t}(\mu )|_{\mu =\mu _1} } /{\partial \alpha }} <0$. For $N\ge 2$, ${{\partial {{\tilde{P}}_t}(\mu )|_{\mu =\mu _2} } / {\partial \alpha }} >0$. Thus, when $1/2\le \alpha ^2\le 1$, ${\tilde{P}_t}(\mu )|_{\mu =\mu _1}$ is a decreasing function of $\alpha$ while ${{\tilde{P}}_t}(\mu )|_{\mu =\mu _2}$ is an increasing function for $\alpha$. Besides, we know from (45) that the $\mu _1$-value increases while the $\mu _2$-value decreases as the $\alpha$-value increases. Therefore, ${\tilde{P}}_t(\mu )$ is always a decreasing function for any $\mu \in \left[ \mu _a,\mu _b\right]$.

Thus, when all the relays are closer to the destination, the less the threshold of total training power is. In other words, for $\mu _i$’s$=\mu _a$, the total training power threshold of making $P_f^*>0$ is maximum, while for $\mu _i$’s $=\mu _b$, the total training power threshold of making $P_f^*>0$ is minimum. Then, we have

$$\begin{aligned} {\tilde{P}}_t^{LB}={\tilde{P}}_t(\mu _b)\le {\tilde{P}}_t(\mu ) \le {\tilde{P}}_t(\mu _a)={\tilde{P}}_t^{UB}. \end{aligned}$$

Lemma 2 is proved.