Curved Earth Correction of Elevation and Azimuth angles between earth stations (BSs and UEs) from IMT networks and satellites used in large areas simulation in SHARC

[judsonbraga]

Objectives

• Find values ​​for the Elevation ($\theta$) and Azimuth ($\varphi$) angles, as well as the distance (sland range (D)) between satellites and IMT ground stations (BSs and UEs) considering a curved Earth with a mean radius of 6371 km.

• The objective is not to modify the flat system in which the IMT network and the projected satellite itself are located. Since the satellite position is projected onto this plane with parameters relative to the center of the IMT system, and the positions (x, y) of the BSs and UEs are also relative to the center of the IMT system, the simple use of the corrected values ​​of $\theta$, $\varphi$ and D sphereizes the system without needing to modify any parameter of the satellite or the IMT network.

Algorithm summary

• IMT System Emulated by 1 or 7 Cluster Topology;
• The IMT system considered is flat with 57 or 399 subregions of equal size and hexagonal shape;
• The satellite with altitude h is assumed to point to the center of the IMT system;
• Lat/Long of the satellite ($𝐿𝑎𝑡_{sat}$, $𝐿𝑜𝑛𝑔_{sat}$) and the center of the IMT system are known;
• IMT base stations (BS) and user equipments (UE) are distributed within each subregion following specific models;
• The Cartesian positions (x, y) of each BS and UE can be calculated relative to the center of the IMT system which is located at ($𝐿𝑎𝑡_{𝑐𝑒𝑛𝑡}$, $𝐿𝑜𝑛𝑔_{𝑐𝑒𝑛𝑡}$)
• Lat/Long of each earth station (BS or UE) is computed as ($𝐿𝑎𝑡_{es}$, $𝐿𝑜𝑛𝑔_{es}$) using positions (x, y) with respect to IMT center and $𝐿𝑎𝑡_{𝑐𝑒𝑛𝑡}$, $𝐿𝑜𝑛𝑔_{𝑐𝑒𝑛𝑡}$.
• Using $𝐿𝑎𝑡_{es}$, $𝐿𝑜𝑛𝑔_{es}$, $𝐿𝑎𝑡_{sat}$, $𝐿𝑜𝑛𝑔_{sat}$, h, it is possible to calculate $\theta$, $\varphi$ and D between each IMT station and the satellite.
• These values ​​can be used directly in the propagation model and to obtain the antenna gains from the satellite and the IMT station (BS or EU).

Problem geometry

Given that:

• $\theta = 180\degree (1 - \sigma/\pi)$ is the elevation angle of the BS (or UE) with respect to Zenith;
• 𝜑(𝐴) is the azimuth with respect to the North axis (North Pole direction);
• x and y are the coordinates of the BS (or UE) with respect to the center of the IMT system;
• $𝐿𝑎𝑡_{𝑐𝑒𝑛𝑡}$ and $𝐿𝑜𝑛𝑔_{𝑐𝑒𝑛𝑡}$ are the Lat/Long of the center of the IMT system;
• $𝐿𝑎𝑡_{s𝑎𝑡}$ and $𝐿𝑜𝑛𝑔_{s𝑎𝑡}$ are the Lat/Long of the satellite;
• 𝐵 = $𝐿𝑜𝑛𝑔_{es}^{rad}$ − $𝐿𝑜𝑛𝑔_{s𝑎𝑡}^{rad}$;
• 𝛾 = $𝐿𝑎𝑡_{es}^{rad}$ − $𝐿𝑎𝑡_{s𝑎𝑡}^{rad}$;
• R = 6371 is the average radius of the Earth;
• h is the altitude of the satellite relative to the mean altitude of the terrain;
• D is the distance from the BS (or UE) to the satellite (slant range);
• $\sin⁡\sigma = \sin⁡(180−\theta^{rad}) = \sin\theta^{rad}$ (𝜎 is supplementary to the elevation angle 𝜃);
• $𝑐 = (\pi/2 − 𝐿𝑎𝑡_{𝐵𝑆}^{rad}) $;
• $𝑎 = (\pi/2 − 𝐿𝑎𝑡_{s𝑎𝑡}^{rad})$;
• $x^{rad}$ indicates angle $x$ in radians.

And considering the values of x and y available from the IMT plane structure, the earth station Lat/Long can be computed as:

$𝐿𝑎𝑡_{es}^{rad} = 𝐿𝑎𝑡_{𝑐𝑒𝑛𝑡}^{rad} + \frac{𝑦}{𝑅} $

$𝐿𝑜𝑛𝑔_{es}^{rad} = 𝐿𝑜𝑛𝑔_{𝑐𝑒𝑛𝑡}^{rad} + \frac{𝑥}{𝑅 \overline{\cos (𝐿𝑎𝑡_{c𝑒𝑛𝑡−es}^{rad})}} $

where $\overline{\cos (𝐿𝑎𝑡_{c𝑒𝑛𝑡−es}^{rad})}$ is the mean value of $cos(Lat^{rad})$ between ES and IMT centre. Then

$𝐿𝑜𝑛𝑔_{es}^{rad} = 𝐿𝑜𝑛𝑔_{𝑐𝑒𝑛𝑡}^{rad} + \frac{𝑥}{𝑅 \frac{\sin(𝐿𝑎𝑡_{es}^{rad}) − \sin⁡(𝐿𝑎𝑡_{c𝑒𝑛𝑡}^{rad})}{𝐿𝑎𝑡_{es}^{rad} − 𝐿𝑎𝑡_{c𝑒𝑛𝑡}^{rad}} } $

From law of sines we have:

$\frac{𝐷}{\sin(𝑏)} = \frac{(𝑅+ℎ)}{\sin(𝜎)} → \theta^{rad} = \sin^{-1} {\frac{(𝑅+ℎ)\sin (b)}{𝐷} }$

𝑏 is computed through trigonometric spherical identity:

$\cos(𝑏) = \cos(𝑐) \cos(𝑎) + \sin(𝑐) \sin(𝑎) \cos(𝐵)$

and 𝐷 through From law of cosines:

$𝐷 = \sqrt{ {(𝑅+ℎ)}^2 + {𝑅}^2 − 2𝑅(𝑅+ℎ) \cos(𝑏)}$

From the law of sines we find the azimuth angle 𝜑(𝐴):

$\frac{\sin(𝐴)}{\sin(𝑎)} = \frac{\sin(|𝐵|)}{\sin(𝑏)} → 𝐴 = \sin^{-1} \frac{ \sin(|𝐵|) \sin(𝑎) }{ \sin(𝑏) }$

To compute $\varphi$, the A parameter must be associated with positions of ES and S, as follows:

Remembering that

$𝛾 = 𝐿𝑎𝑡_{es}−𝐿𝑎𝑡_{s𝑎𝑡}$
$𝐵 = 𝐿𝑜𝑛𝑔_{es}−𝐿𝑜𝑛𝑔_{s𝑎𝑡}$

$\varphi$ can be computed using:

Finally, azimuth and elevatoin angles can be computed by:

$\varphi^{rad} = \frac{1 + \mathbf{sign}(\gamma)}{2} \pi + \mathbf{sign}(\gamma) \mathbf{sign}(B) A$

$𝜃^{rad} = \sin^{-1} \frac{(𝑅+ℎ)}{𝐷 \sin(𝑏) }$