Millimeter-Wave Communications with Beamforming for UAV-Assisted Railway Monitoring System

Abstract

Railway is an important means of transportation for passenger and freight, and the maintenance and repair work is essential. Unmanned aerial vehicle (UAV) can be deployed along the railway track for monitoring, reconnaissance and ensuring the safe operation of the railway because of its flexible mobility and low labor cost. In this paper, a railway monitoring system based on aerial platform is formed, which takes UAV monitoring as the main body. In order to solve the problem of long-distance and wide bandwidth backhaul communication, this paper adopts mmWave array communication scheme, and designs the large antenna array beamforming on the base station receiver with zero-forcing (ZF) and minimum mean square error (MMSE). Finally, the simulation results show that the minimum mean square error method is effective, and the simulation results also provide a reference for the UAV deployment along the railway.

This work was supported in part by the National Key Research and Development Program (Grant Nos. 2016YFB1200100), and the National Natural Science Foundation of China (NSFC) (Grant Nos. 61827901 and 91738301).

Appendix

Equation (14) derives the $\mathbf{{\tilde{D}}}$ matrix as follows. First, convert Eq. (14) to expression:

$$\begin{aligned} f(\mathbf{{\tilde{D}}}) = Tr\{ E[({\mathbf{{\tilde{D}}}^H}{\mathbf{{H}}_{eff}}{} \mathbf{{Ps}} + {\mathbf{{\tilde{D}}}^H}{\mathbf{{A}}^H}{} \mathbf{{z}} - \mathbf{{s}})({\mathbf{{s}}^H}{} \mathbf{{P}}{\mathbf{{H}}_{eff}}^H\mathbf{{\tilde{D}}} + {\mathbf{{z}}^H}{} \mathbf{{A\tilde{D}}} - {\mathbf{{s}}^H})]\} \end{aligned}$$

(15)

Then expand the above formula and simplify:

$$\begin{aligned} f(\mathbf{{\tilde{D}}}) = Tr(P{\mathbf{{\tilde{D}}}^H}{\mathbf{{H}}_{eff}}{\mathbf{{H}}_{eff}}^H\mathbf{{\tilde{D}}} - {\mathbf{{\tilde{D}}}^H}{\mathbf{{H}}_{eff}}{} \mathbf{{P}} + {\delta ^2}{\mathbf{{\tilde{D}}}^H}{\mathbf{{A}}^H}{} \mathbf{{A\tilde{D}}} - \mathbf{{P}}{\mathbf{{H}}_{eff}}^H\mathbf{{\tilde{D}}} + \mathbf{{I}}) \end{aligned}$$

(16)

Derivative the above formula to $\mathbf{{\tilde{D}}}$ and make it equal to zero

$$\begin{aligned} \frac{{df}}{{d({{\mathbf{{\tilde{D}}}}^H})}} = P{\mathbf{{\tilde{D}}}^T}{\mathbf{{H}}_{eff}}^*{\mathbf{{H}}_{eff}}^T - \sqrt{P} {\mathbf{{H}}_{eff}}^T + {\delta ^2}{\mathbf{{\tilde{D}}}^T}{\mathbf{{A}}^T}{\mathbf{{A}}^*} = 0 \end{aligned}$$

(17)

$$\begin{aligned} \Rightarrow {\mathbf{{\tilde{D}}}^T}(P{\mathbf{{H}}_{eff}}^*{\mathbf{{H}}_{eff}}^T + {\delta ^2}{\mathbf{{A}}^T}{\mathbf{{A}}^*}) - \sqrt{P} {\mathbf{{H}}_{eff}}^T\mathrm{{ = }}\,0 \end{aligned}$$

(18)

$$\begin{aligned} \begin{aligned} \therefore \mathrm{{ }}{} \mathbf{{\tilde{D}}}\,\mathrm{{ = }}\sqrt{P} {\mathbf{{J}}^{ - 1}}{\mathbf{{H}}_{eff}}~~~~~~~~~~~~\\where, \mathbf{{J}}\,\mathrm{{ = (}}P{\mathbf{{H}}_{eff}}{\mathbf{{H}}_{eff}}^H + {\delta ^2}{\mathbf{{A}}^H}{} \mathbf{{A}}) \end{aligned} \end{aligned}$$

(19)