Low Complexity Capacity Maximizing Transmit Antenna Selection Schemes for Massive MIMO Wireless Communications

Abstract

Recently massive MIMO systems have attracted a lot of attention. A serious limitation in implementing massive MIMO is the number of RF chains required, which is due to its high implementation cost and power consumption. Antenna selection is a wise remedy, that not only reduces the number of required RF chains, but also improves the system’s performance. In this work, two low complexity capacity-maximizing transmit antenna selection (TAS) schemes, one for low SNR region and one for medium to high SNR region, are proposed. It is shown that the proposed antenna selection schemes enhance the MIMO link’s capacity at low SNRs, and select only one antenna in more than $90\%$ of channel realizations, which in terms of numbers of required RF chains is very promising. The performance of the proposed schemes are compared to an existing high performance greedy TAS technique in terms of average capacity and complexity. Finally, a close analytical approximation for average capacity of the proposed capacity-based subset selection scheme is presented.

Appendix

Proof of Proposition 2:

$\mathbf {B}_{i-1}$ has rank $i-1$, while $\mathbf {A}_i=\mathbf {b}_i\mathbf {b}_i^H$ has rank 1,

$$\begin{aligned} \begin{aligned} \lambda _{1}^{(\mathbf {A}_i)}&=\Vert \mathbf {b}_i\Vert _F^2,\\ \lambda _{k}^{(\mathbf {A}_i)}&=0, \,\,\,\,\,\,2 \le k \le N. \end{aligned} \end{aligned}$$

(18)

Assuming sorted eigenvalues (in descending order), i.e., $\lambda _{i-1}^{\mathbf {B}_{i-1}}$ being the smallest non-zero eigenvalue, and in order to maximize the lower bound of Proposition 1, $\det {(\frac{\rho }{i}\mathbf {B}_i)}$, it is straight forward to show that the only eigenvalue of $\mathbf {A}_i$, $\lambda ^{(\mathbf {A}_i)}_1$, should be added to the minimum eigenvalue of $\mathbf {B}_{i-1}$, i.e., without loss of generality assuming sorted $\lambda ^{(\mathbf {B}_i)}_k$s is descending order,

$$\begin{aligned} \begin{aligned}&\max _{p}{\left[ \left( \lambda _1^{(\mathbf {B}_{i-1})}+\lambda _{p_1}^{(\mathbf {A}_i)}\right) \left( \lambda _2^{(\mathbf {B}_{i-1})}+\lambda _{p_2}^{(\mathbf {A}_i)}\right) \cdots \left( \lambda _{i-1}^{(\mathbf {B}_{i-1})}+\lambda _{p_{i-1}}^{\mathbf {A}_i}\right) \right] }\\&\quad = \lambda _1^{\mathbf {B}_{i-1}}\lambda _2^{\mathbf {B}_{i-1}}\cdots \lambda _{i-2}^{\mathbf {B}_{i-1}}\left( \lambda _{i-1}^{\mathbf {B}_{i-1}}+\lambda _1^{(\mathbf {A}_i)}\right) . \end{aligned} \end{aligned}$$

(19)

Hence, using Proposition 1, the following inequality is obtained

$$\begin{aligned} \begin{aligned} \frac{\rho }{i^{i-1}}\left[ \lambda _1^{\mathbf {B}_{i-1}}\lambda _2^{\mathbf {B}_{i-1}} \dots \left( \lambda _{i-1}^{\mathbf {B}_{i-1}}+\lambda _{1}^{\mathbf {A}_i}\right) \right]&\le \det {\left( \frac{\rho }{i}\left( \mathbf {B}_{i-1}+\mathbf {A}_i\right) \right) }\\&=\det {\left( \frac{\rho }{i}\mathbf {B}_{i}\right) }. \end{aligned} \end{aligned}$$

(20)

In order for $\det {\left( \frac{\rho }{i}\mathbf {B}_i\right) }$ be and increasing function of i, its lower bound, obtained in (20) should be greater than $\det {\left( \frac{\rho }{i-1}\mathbf {B}_{i-1}\right) }$, which results in

$$\begin{aligned} \begin{aligned}&\det {\left( \frac{\rho }{i-1}\mathbf {B}_{i-1}\right) } = \frac{\rho ^{i-1}}{(i-1)^{i-1}}\prod _{k=1}^{i-1}{\lambda _k^{(\mathbf {B}_{i-1})}} \\&\quad < (\frac{\rho }{i})^{i-1}\left[ \lambda _1^{\mathbf {B}_{i-1}}\lambda _2^{\mathbf {B}_{i-1}} \dots \lambda _{i-2}^{\mathbf {B}_{i-1}}\left( \lambda _{i-1}^{\mathbf {B}_{i-1}}+\lambda _{1}^{\mathbf {A}_i}\right) \right] , \end{aligned} \end{aligned}$$

(21)

which is further simplified to

$$\begin{aligned} \Vert \mathbf {b}_i\Vert _F^2>\left[ \frac{i^i}{(i-1)^i}-1\right] \lambda _{min}^{\mathbf {B}_{i-1}} \end{aligned}$$

(22)

where $\lambda _{i-1}^{\mathbf {B}_{i-1}}$ and $\lambda _{1}^{\mathbf {A}_i}$ have been replaced with $\lambda _{min}^{\mathbf {B}_{i-1}}$ and $\Vert \mathbf {b}_i\Vert _F^2$, respectively.