Beam Radius Calculations
The half-power point occurs when the power drops to half of its maximum value. Mathematically, this is when
\[\begin{equation} G(\theta_{HPBW})= 4\times \left|\frac{J_{1}(ka\sin(\theta_{HPBW}))}{ka\cdot\sin(\theta_{HPBW})}\right|^{2}=\frac{1}{2} \end{equation}\]
Taking square root
\[\begin{equation} \left|\frac{2\cdot J_{1}(ka\sin(\theta_{HPBW}))}{ka\cdot\sin(\theta_{HPBW})}\right|=\frac{1}{\sqrt{2}} \end{equation}\]
Solving numerically for \(J_{1}(x)=\frac{x}{2\sqrt{2}}\), we get
\[\begin{equation} ka\sin(\theta_{HPBW})\approx 1.616 \end{equation}\]
Substituting \(k=2\pi /\lambda\) and \(a=\frac{D}{2}\)
\[\begin{equation} \frac{2\pi }{\lambda }\cdot\frac{D}{2}\sin(\theta_{HPBW})\approx 1.616 \end{equation}\]
Using the fact that for small angles \(\sin(\theta )\approx\theta\), and on converting to degrees, we get
\[\begin{equation} \theta_{HPBW}\;[\text{degrees}]\approx\frac{70\lambda }{D} \end{equation}\]
For aperture, \(a\) and wavelength \(\lambda\), since \(2a=D\) and substituting \(\lambda =\frac{c}{f}\) we get
\[\begin{equation} \theta_{HPBW}\;[\text{degrees}]=\frac{35\times c}{a\times f} \end{equation}\]
| Parameters | LEO 1200 | MEO 10000 | GEO 35786 |
|---|---|---|---|
| Altitude (km) | 1200 | 10000 | 35786 |
| Band | S band | S band | S band |
| Antenna Aperture Radius (m) | 1 | 5 | 11 |
| Channel Frequency (GHz) | 2 | 2 | 2 |
| \(\theta_{3dB}\) (or \(\theta_{HPBW}\)) (Calculation provided below) |
5.25 | 1.05 | 0.476 |
| Beam Radius (km) | 110.26 | 183.28 | 297.31 |
\[\begin{equation} \theta_{HPBW}=\frac{35\times \lambda }{a}= \frac{35\times c}{a\times f} \end{equation}\]
\[\begin{equation} \text{LEO 1200: } \theta_{HPBW}=\frac{35\times 3 \times 10^{8}}{1\times 2\times 10^{9}}=5.25^{\circ} \end{equation}\]
\[\begin{equation} \text{MEO 10000: } \theta_{HPBW}=\frac{35\times 3 \times 10^{8}}{5\times 2\times 10^{9}}=1.05^{\circ} \end{equation}\]
\[\begin{equation} \text{GEO 35786: } \theta_{HPBW}=\frac{35\times 3 \times 10^{8}}{11\times 2\times 10^{9}}=0.476^{\circ} \end{equation}\]
| Parameters | LEO 1200 | MEO 10000 | GEO 35786 |
|---|---|---|---|
| Altitude (km) | 1200 | 10000 | 35786 |
| Band | K band | K band | K band |
| Antenna Aperture Radius (m) | 0.25 | 1 | 2.5 |
| Channel Frequency (GHz) | 20 | 20 | 20 |
| \(\theta_{3dB}\) (or \(\theta_{HPBW}\)) (Calculation provided below) |
2.1 | 0.525 | 0.21 |
| Beam Radius (km) | 44 | 91.63 | 131.16 |
\[\begin{equation} \text{LEO 1200: } \theta_{HPBW}=\frac{35\times 3 \times 10^{8}}{0.25\times 20\times 10^{9}}=2.1^{\circ} \end{equation}\]
\[\begin{equation} \text{MEO 10000: } \theta_{HPBW}=\frac{35\times 3 \times 10^{8}}{1\times 20\times 10^{9}}=0.525^{\circ} \end{equation}\]
\[\begin{equation} \text{GEO 35786: } \theta_{HPBW}=\frac{35\times 3 \times 10^{8}}{2.5\times 20\times 10^{9}}=0.21^{\circ} \end{equation}\]
Hexagonal tessellation based on beam radius¶
In a hexagonal tessellation:
Each beam is represented by a hexagon.
The beams are arranged in a honeycomb pattern to maximize coverage and minimize overlap.
The beam radius, \(R\) is the distance from the centre of the beam to its edge, where the power drops to a specified level (e.g., half-power or \(-\)3 dB).
In a hexagonal grid, the distance between beam centers (beam spacing) is \(\sqrt{3}\cdot R\)