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.
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Notes
This model is an extremely special case of practical scenario and it is used for validating the proposed methods due to its yielding easier theoretical analysis. Subsequent simulations also demonstrate that the same conclusions drawn for this special case go to more practical one.
Due to the space limit, this BER performance of various training schemes over optimal PA are not given below. Besides, in the subsequent simulations, the regular schemes without considering training power allocation would be omitted for a more clear comparison.
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This work has been presented in part at the Canadian Workshop on Information Theory, St. John’s, Canada, July 2015 [1].
This work was partly supported by the project of Natural Science and Engineering Research Council, Canada.
Appendices
Appendix 1: Proof of Lemma 1
Define
and
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:
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
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),
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
from which we can obtain that \(P_f^{-1}\sum _{i= 1}^N \sigma _{g_i}^{-2} \ll 1\). Thus from (40),
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
where the inequality (a) is because \(\mu _i\in (0,1)\). Thus, when \(P_f>N(1-\mu _a)/(4\mu _a)\), we have
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
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
Thus, the following two solutions to \(\mu\) can be obtained by
On substituting \(\mu _1\) and \(\mu _2\) into (43) and calculating the first-order derivatives with respect to \(\alpha\), respectively, we have
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
Lemma 2 is proved.
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Yang, L., Wang, D. & Zhao, Q. Training Power Allocation Based on MSE-Minimization for Multi-Relay Amplify-and-Forward Cooperation in Wireless Sensor Networks. Wireless Pers Commun 106, 593–619 (2019). https://doi.org/10.1007/s11277-019-06181-9
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DOI: https://doi.org/10.1007/s11277-019-06181-9