Beamforming Only Based on Specular Component for Downlink Massive MIMO Systems

Abstract

In this paper, we are concerned with a multiuser downlink massive MIMO system operating over Ricean fading channels where each user can be equipped with one or multiple antennas. We present a beamforming (BF) transmission scheme that is only based on the specular component, which can reduce efficiently the heavy overhead for channel estimation. For the individual achievable rates, we particularly analyze their power-scaling law when the transmit power per user is scaled down by the number of antennas at the base station (BS). It is found that the individual rates of the proposed BF can be higher than the corresponding rates of the BF based on the scattered and specular components if the specular component is relatively strong. In addition, it is shown that each user can asymptotically achieve the same rate as in the single-user case when both the numbers of BS antennas and the user terminals go to infinity with a fixed ratio. Thus we conclude that the massive MIMO systems may be more suitable to be deployed in Ricean fading rather than Rayleigh fading.

Appendix

1.1 Proof of Theorem 1

Proof

By using Jensen’s Inequality, we obtain the following lower bound on the ergodic achievable rate of the k-th user

$$\begin{aligned} R_k(K) \ge \log _2\left( 1+\frac{1}{\mathbb {E}(1/\gamma _k(K))}\right) . \end{aligned}$$

(21)

Now define $\bar{\gamma }_k^{\mathtt {L}}(K)=\frac{1}{\mathbb {E}(1/ \gamma _k(K))}$. Then from (10) it follows that

$$\begin{aligned} \bar{\gamma }_k^{\mathtt {L}}(K)=\frac{\frac{p_b\beta _k\bar{\vartheta }_k}{K}MN}{\frac{p_b\beta _k\tilde{\vartheta }_k}{K} \mathbb {E}\left| \frac{\mathbf {r}^\mathtt {H}_k\tilde{\mathbf {H}}_k\mathbf {t}_k}{\sqrt{NM}}\right| ^2+\frac{p_b\beta _k}{K}\sum _{i=1,i\ne k}^K \mathbb {E}\left| \frac{\mathbf {r}^\mathtt {H}_k\mathbf {H}_k\mathbf {t}_i}{\sqrt{NM}}\right| ^2+1}. \end{aligned}$$

(22)

Since each entry $\tilde{h}_{nm}^{(k)}$ in $\widetilde{\mathbf {H}}_k$ follows $\text{ CN }(0,1)$, and all of them are i.i.d, we can have from the knowledge of statistics that

$$\begin{aligned} \frac{\mathbf {r}^\mathtt {H}_k\tilde{\mathbf {H}}_k\mathbf {t}_k}{\sqrt{NM}} \sim \text{ CN }(0,1), \;\; \frac{\mathbf {r}^\mathtt {H}_k\tilde{\mathbf {H}}_k\mathbf {t}_i}{\sqrt{NM}} \sim \text{ CN }(0,1). \end{aligned}$$

(23)

Furthermore, we can obtain the following expression

$$\begin{aligned} \mathbb {E}\left| \frac{\mathbf {r}^\mathtt {H}_k\tilde{\mathbf {H}}_k\mathbf {t}_k}{\sqrt{NM}}\right| ^2=1,\;\; \mathbb {E}\left| \frac{\mathbf {r}^\mathtt {H}_k\tilde{\mathbf {H}}_k\mathbf {t}_i}{\sqrt{NM}}\right| ^2=1. \end{aligned}$$

(24)

In addition, we also have that

$$\begin{aligned} \frac{\mathbf {r}^\mathtt {H}_k\bar{\mathbf {H}}_k\mathbf {t}_i}{\sqrt{NM}}=\frac{\sqrt{N}\mathbf {t}^\mathtt {H}_k\mathbf {t}_i}{\sqrt{M}} =\sqrt{\frac{N}{M}}\cdot \frac{1-e^{jM\varphi _{ki}}}{1-e^{j\varphi _{ki}}} \end{aligned}$$

(25)

where $\varphi _{ki}=2\pi d (\sin (\phi _k)-\sin (\phi _i))$. If let $\varrho _{ki}=\frac{1-e^{jM\varphi _{ki}}}{1-e^{j\varphi _{ki}}}$, then

$$\begin{aligned} \left| \frac{\mathbf {r}^\mathtt {H}_k\bar{\mathbf {H}}_k\mathbf {t}_i}{\sqrt{NM}}\right| ^2=\frac{N}{M}|\varrho _{ki}|^2. \end{aligned}$$

(26)

Finally, (12) can be easily derived with the help of the following expression

$$\begin{aligned} \mathbb {E}\left| \frac{\mathbf {r}^\mathtt {H}_k\mathbf {H}_k\mathbf {t}_i}{\sqrt{NM}}\right| ^2=\bar{\vartheta }_k\left| \frac{\mathbf {r}^\mathtt {H}_k\bar{\mathbf {H}}_k\mathbf {t}_i}{\sqrt{NM}}\right| ^2+\tilde{\vartheta }_k\mathbb {E}\left| \frac{\mathbf {r}^\mathtt {H}_k\tilde{\mathbf {H}}_k\mathbf {t}_i}{\sqrt{NM}}\right| ^2. \end{aligned}$$

(27)

On the other hand, from (10) it is obvious that

$$\begin{aligned} \gamma _k(K) \le \frac{p_b\beta _k\bar{\vartheta }_k}{K}MN. \end{aligned}$$

(28)

Now define $\bar{\gamma }_k^{\mathtt {U}}(K)=\frac{p_b\beta _k\bar{\vartheta }_k}{K}MN$. Then (13) holds.

1.2 Proof of Corollaries 1 and 2

Under the condition given in Corollary 1 or Corollary 2, we only need to prove that (12) can tend to (13) as $M \rightarrow \infty$. (12) can be rewritten as

$$\begin{aligned} \bar{\gamma }_k^{\mathtt {L}}(K)=\frac{E_u\beta _k\bar{\vartheta }_k}{1+\frac{E_u}{M^2}\beta _k\bar{\vartheta }_k\sum _{i=1,i\ne k}^K|\varrho _{ki}|^2+\frac{E_uK}{M N}\tilde{\vartheta }_k\beta _k}. \end{aligned}$$

(29)

Due to the fact that $\frac{K}{M N} \rightarrow 0$, now we only need to prove that

$$\begin{aligned} \frac{1}{M^2}\sum _{i=1,i\ne k}^K|\varrho _{ki}|^2 \rightarrow 0. \end{aligned}$$

(30)

When K is finite, it is obvious that (30) holds since $|\varrho _{ki}|^2 \le \frac{4}{|1-e^{j\varphi _{ki}}|^2}$ is still finite as M grows large. Thus Corollary 1 holds. For infinite K, we define $\mathbb {E}|\varrho _{ki}|^2=\omega _k$. According to the law of large numbers, we know that

$$\begin{aligned} \lim _{K \rightarrow \infty }\frac{1}{K}\sum _{i=1,i\ne k}^K|\varrho _{ki}|^2=\omega _k. \end{aligned}$$

(31)

Furthermore, under the given condition of Corollary 2, we have that (30) holds and thus Corollary 2 holds.

1.3 Proof of Theorem 2

Suppose that $N=1$. From Corollary 1 it follows that

$$\begin{aligned} \lim _{M \rightarrow \infty } R_k(K)=\log _2(1+E_u\beta _k\bar{\vartheta }_k). \end{aligned}$$

(32)

On the other hand, by following a similar line of reasoning as in [7], we can obtain

$$\begin{aligned} \lim _{M \rightarrow \infty }\hat{R}_k (K)=\frac{T-\tau }{T}\log _2(1+E_u\beta _k(\bar{\vartheta }_k+\sigma _k^2\tilde{\vartheta }_k)). \end{aligned}$$

(33)

Based on Bernoulli’s Inequality in [14] and the given condition $\vartheta _k\ge \frac{T-\tau }{\tau }\sigma ^2_k$, we further have

$$\begin{aligned} \lim _{M \rightarrow \infty }\hat{R}_k (K) & = \log _2\left( 1+E_u\beta _k\left( \bar{\vartheta }_k+\sigma _k^2\tilde{\vartheta }_k\right) \right) ^{\frac{T-\tau }{T}} \nonumber \\ & < \log _2\left( 1+E_u\beta _k\frac{T-\tau }{T}\left( \bar{\vartheta }_k+\sigma _k^2\tilde{\vartheta }_k\right) \right) \nonumber \\\ & \le \log _2\left( 1+E_u\beta _k\bar{\vartheta }_k\right) =\lim _{M \rightarrow \infty }R_k (K). \end{aligned}$$

(34)

So we finally arrive at the desired result in (20).