Angle of Arrivals Estimation for Multiuser mmWave Systems

Abstract

Angle of arrival (AoA) estimation with localized hybrid arrays is challenging in mmWave communication systems. Most existing solutions quantize AoAs into limited values with relatively low accuracy. This chapter proposes a multi-AoA estimation scheme which is capable of estimating multiple AoAs from multiple users with low complexity. Section 5.1 presents the motivation of investigating the AoA estimation with localized hybrid arrays. Section 5.2 introduces system and channel models. Section 5.3 presents the proposed estimation scheme referring to a uniform linear hybrid array. Section 5.4 extends the scheme to wideband systems, consisting of the estimation of both AoAs and time delay. Section 5.5 introduces the new performance metrics and derives a lower bound of mean square error (MSE) for the estimates. Section 5.6 provides extensive simulation results and validates the effectiveness of the proposed multi-AoA estimation scheme. Section 5.7 concludes the chapter.

Appendix

5.1.1 Proof of $\frac{\pi }{N}\le B\le \frac{2\pi }{N}, N\ge 2$

In (5.11), B is defined as $2\omega _0$ with $\frac{\sin (N\omega _0)}{\sin (\omega _0)}=\frac{N}{2}$. For the right inequality, we use the inequality of $\frac{\sin (N\omega _0)}{\sin (\omega _0)}=\frac{N}{2}\ge \frac{\sin (\pi )}{\sin (\frac{\pi }{N})}=0$. Noting that $\frac{\sin (N\omega )}{\sin (\omega )}$ is a decreasing function with $\omega \in (0,\pi /N)$, we have $\omega _0\le \pi /N$, i.e., $B=2\omega _0\le \frac{2\pi }{N}$.

For the left inequality, we use the inequality of ${\sin \left( \frac{\pi }{2N}\right) }\le \frac{\pi }{2N} \le \frac{2}{N}$, where the former inequality is due to $\sin x\le x, x\ge 0$ and the latter one is valid when $N\ge 2$. Then we have $\frac{1}{\sin \left( \frac{\pi }{2N}\right) }= \frac{\sin (N\frac{\pi }{2N})}{\sin \frac{\pi }{2N}}\ge \frac{N}{2}=\frac{\sin (N\omega _0)}{\sin (\omega _0)}$. Because $\frac{\sin (N\omega )}{\sin (\omega )}$ is a decreasing function with $\omega \in (0,\pi /N)$, we have $\omega _0\ge \frac{\pi }{2N}$, i.e., $B=2\omega _0\ge \frac{\pi }{N}, N\ge 2$.

Therefore, $\frac{\pi }{N}\le B\le \frac{2\pi }{N}, N\ge 2.$

5.1.2 Proof of Theorem 5.1

Since the magnified path satisfies that $|c_{m_l}^{k,t}-\Omega _l|<\frac{2\pi }{KT}$, we have

$$\begin{aligned} \left| \omega _{l,m_l}^{k,t} \right| =\left| \frac{c_{m_l}^{k,t}-\Omega _l}{2} \right| \le \frac{\pi }{KT}\le \frac{\pi }{N}. \end{aligned}$$

(5.40)

Due to $1\le {m_l} \le M$, the gap between $c_{m_l}^{k,t}$ and $c_{m }^{k,t}$ is maximized when $m_l$ equals 1 or M. By letting $m_l=1$, we have $c_m^{k,t}=c_{m_l}^{k,t}+\frac{2\pi (m-1)}{KTM}$. Hence,

$$\begin{aligned} \left| \omega _{l,m}^{k,t} \right|&=\left| \frac{c_{m_l}^{k,t}+\frac{2\pi \left( m-1 \right) }{KTM}-\Omega _l}{2} \right| \le \left| \frac{c_{m_l}^{k,t}-\Omega _l}{2} \right| +\frac{\pi \left( m-1 \right) }{KTM} \le \frac{2\pi }{KT}\le \frac{2\pi }{N}. \end{aligned}$$

(5.41)

The interval that keeps the sign of $\frac{\sin Nx}{\sin x}$ unchanged is given by $\Delta x=\frac{2\pi }{N}$. From (5.41), it is clear that there is only one zero point, denoted as $x_0$, in $P_m^{k,t}$, such that $P_m^{k,t}>0$ when $\omega _{l,m}^{k,t}>x_0$ and $P_m^{k,t}<0$ when $\omega _{l,m}^{k,t}<x_0$. Therefore, there is at most one m, such that $P_m^{k,t}P_{m+1}^{k,t}<0$, and this m is close to $x_0$. Hence, if it exists, it has the least absolute value.

5.1.3 Proof of Theorem 5.2

We note that the efficient signal power of $y_m^{k,t}$ equals $|\alpha _u s_u(t)|^2|P_m^{k,t}|^2$. Let the variance of $I_{m}^{k,t}+n_m^{k,t}$ be $\delta$. The SINR of $y_m^{k,t}$ is given by

$$\begin{aligned} \frac{\left| \alpha _us_u\left( t \right) \right| ^2\left| P_{m}^{k,t} \right| ^2}{\delta }\triangleq \frac{\bar{P}}{\delta }. \end{aligned}$$

(5.42)

$\tilde{z}_m^{k,t}$ in (5.16) is given by

$$\begin{aligned} \tilde{z}_{m}^{k,t}=\left| I_{m}^{k,t}+z_{m}^{k,t} \right| ^2+A_{m}^{k,t*}\left( I_{m}^{k,t}+z_{m}^{k,t} \right) +A_{m+1}^{k,t}\left( I_{m}^{k,t}+z_{m}^{k,t} \right) ^*, \end{aligned}$$

(5.43)

hence the variance of $\tilde{z}_m^{k,t}$ equals $2\bar{P}\delta +\delta ^2$ and the SINR of $\rho _{m}^{k,t}$ is given by

$$\begin{aligned} \frac{\left| \alpha _us_u\left( t \right) \right| ^4\left| P_{m}^{k,t} \right| ^2\left| P_{m+1}^{k,t} \right| ^2}{2\bar{P}\delta +\delta ^2}\approx \frac{\bar{P}^2}{2\bar{P}\delta +\delta ^2}. \end{aligned}$$

(5.44)

Therefore, the SINR of $\rho _{m }^{k,t}$ is smaller than that of $y_m^{k,t}$.

5.1.4 Proof of (5.36)

Given u uniformly distributed in $(-\pi ,\pi )$ and $P_m^{k,t}=\frac{\sin (N\omega _{d,m}^{k,t})}{\sin (\omega _{d,m}^{k,t})}$, we have $\mathbb E\left( P_m^{k,t}\right) ^2=\frac{1}{2\pi }\int _{-\pi }^{\pi }$ $\left| \frac{\sin (N\omega _{d,m}^{k,t})}{\sin (\omega _{d,m}^{k,t})}\right| ^2\mathrm{d}u$. We calculate the integral by using the property of Fourier transform. We note that discrete-time Fourier transform (DTFT) of the window function is given by $F(u)=\sum _{n=0}^{N-1}1\cdot e^{-jun}=e^{j\frac{N-1}{2}u}\frac{\sin (Nu/2)}{\sin (u/2)}$. Using the frequency-shift property of DTFT, $P_m^{k,t}$ equals $\sum _{n=0}^{N-1}1\cdot e^{-ju(n+\frac{N-1}{2} )}$. As a result, we have $E\left( P_m^{k,t}\right) ^2=\frac{1}{2\pi }\int _{-\pi }^{\pi }\left| \sum _{n=0}^{N-1}1\cdot e^{-ju(n+\frac{N-1}{2} )} \right| ^2\mathrm{d}u=N$.