# How to compute footprint(size) of beam of sat.system?

I need to compute the footprint (ground area that its transponders offer coverage) of beam diameter. Given different frequencies, distance between a satellite and ground station.

I have found the same calculation:

it was presented by mynaric

Could someone explain how the beam sizes were computed?

In the far field, the size of a simply-shaped beam will increase proportional to distance:

$$w \approx \text{factor} \ \times L$$

where $$w$$ is the width and $$L$$ is the distance. Let's see if we can find the factor that explains all that data.

There are two common simple shapes we use for the initial beam. One is a circular disk of uniform brightness (like a mask with a circular hole in it, lit from behind) and the other is a Gaussian-shaped beam.

In the far field, the shape of the beam will be the Fourier transform of those. For the uniform circular aperture it's the Airy disk and for the Gaussian it's another Gaussian beam.

For an Airy pattern in the far field a common formula for the size is

$$w_{Airy} = 1.22 \frac{\lambda}{d} L$$

where $$\lambda$$ is the wavelength and $$d$$ is the diameter of the initial circular uniform beam.

For a Gaussian beam with an initial profile $$\exp(-2 r^2/w_0^2)$$ where $$w_0 = \text{FWHM} / \sqrt{2 \ln(2)}$$ we have

$$w_{Gaussian} = \frac{2 \lambda}{\pi w_0} L = \frac{2 \lambda \sqrt{2 \ln(2)}}{\pi \ \text{FWHM}} L = 0.75 \frac{\lambda}{d} L$$

where $$d$$ is the initial beam diameter measured as the full width at half-maximum (FWHM).

It looks exactly like the other equation except the factor in front is a little smaller.

Now let's to look at that table.

In my other answer I get the following wavelengths for laser, Ka-band and X-band and include their diameters $$d$$, and the factor $$\lambda/d$$

type          λ (m)         d(m)        λ/d
--------      ----------    ----      -------
laser         1.55E-06      0.08      1.94-05
Ka-band       9.375E-03     0.3       3.13E-02
X-band        3.75E-02      0.6       6.25E-02


Using just $$\lambda / d$$ times the distance $$L$$ to get width, we get the following.

L (km)       laser     Ka-band     X-band
-------     -------    -------     -------
50          1          1563        3125
200          4          6250      12,500
1,400         27        43,750      87,500
4,000         78       125,000     250,000


which matches pretty closely their table.

Could someone explain how the beam sizes were computed?

Mostly it's just $$\lambda L / d$$ but they may have used some pre-factor based on the beam shape. The laser would have been near-Gaussian because they will be using single mode fiber to guide the laser beam to the focus of their 8 cm telescope, but the Ka-band and X-band antennas will have had some other shape. There's no way to tell exactly, but every antenna will be somewhat different.

• When computing angular beamwidth a lot more than $\lambda$/d goes into it. The illumination pattern on the primary reflector (or intensity distribution pattern on a phased array) modify the beam shape as compared to a uniformly illuminated, filled aperture. Occultations of a primary reflector by such as a subreflector also affect it. Illuminating the edges of a primary less than the center (such as would happen with the typical RF illumination source) widens the beam, and the degree depends on the degree of illumination decrease as you move from the center toward the edges. ... Sep 29, 2021 at 18:46
• Occulting the center of a primary actually narrows the beam! Sep 29, 2021 at 18:47
• @TomSpilker 1) for the purposes of understanding the numbers in the table in the question based on available information, $\lambda / d$ is sufficient. They don't even describe the antennas. 2) In this answer I've done a full-blown calculation for a specific telescope using the correct central obstruction and illumination filling pattern. But that simply can't be done here. 3) Along with central obstruction that filling pattern is important. For dish antennas the feed horn radiation pattern is important as well as the secondary mirror shape.
– uhoh
Sep 30, 2021 at 2:21
• @TomSpilker from what I recall the Voyagers have a secondary mirror that is shaped and positioned slightly differently than a proper point-to-parallel Cassegrain design in order to lower the central peak and make the radiation pattern a little flatter over a certain angular range so that gain changes less for small pointing errors over a limited range. I think that that may be helping the Voyagers now, they just keep an attitude such that the dish points towards the Sun and the Earth wiggles around it circa 0.5°?
– uhoh
Sep 30, 2021 at 2:26

Below the table is the note:

Laser aperture size: 80mm, Ka-band antenna size: 300mm, X-band antenna size: 600mm. Assuming physical limits for lowest possible beam size.

90% of the problem is solved by looking at $$d/\lambda$$ which is the initial diameter of the beam expressed in wavelengths.

From the following:

I'll choose wavelengths for Ka-band and X-band of $$c/32$$ GHz and $$c/8$$ GHz where $$c$$ is the speed of light.

For the optical link we can go to this page https://mynaric.com/products/space/condor-mk3/ where they show they're using standard long haul single mode fiber communications wavelengths around 1550 nm which makes sense since they are pushing extremely high data rates and the fiber lasers and erbium doped fiber amplifier technology and fancy modulation/demodulation schemes are very well developed.

With the initial free-space beam diameters of 0.08, 0.3, and 0.6 meters from the table that gives $$d/\lambda$$ values of 94,000, 32 and 16.

The wider the initial beam size in wavelengths, the slower it will expand. This is a basic result of any circular diffraction calculation, or as Uwe reminds us, of Gaussian shaped beams as well.

So I predict that the light beam will be 52000/32 or about 1,600 times narrower than the Ka-band beam. It's not exact, but it's quite close; about 1,200 times smaller. So we've found the right metric.

Below I've plotted the beam size, the ratio size/distance and the size/distance ratio scaled by $$d/\lambda$$ and for the third metric we see that all three are almost the same.

The final value is ~1000 because the y axis in meters and the x axis is in kilometers. If the same units, it would be around 1.

import numpy as np
import matplotlib.pyplot as plt

distance = np.array([50, 200, 1400, 4000.]) # km
laser = np.array([1, 5, 35, 111.])
Ka_band = np.array([1600, 6500, 45000, 145000.])
X_band = np.array([3200, 13000, 90000, 290000.])
beams = laser, Ka_band, X_band
names = 'laser', 'Ka_band', 'X_band'
c = 3E+08  # m/s
f_Ka_band, f_X_band = 32E+09, 8E+09 # Hz
wavelengths = np.array([1550E-09, c/f_Ka_band, c/f_X_band])
diameters = np.array([0.08, 0.3, 0.6])
lines = '-', '-', '--'

fig, (ax1, ax2, ax3) = plt.subplots(3, 1)
for beam, name, lam, d, line in zip(beams, names, wavelengths, diameters, lines):
ax1.plot(distance, beam, line)
ax2.plot(distance, beam/distance, line)
ax3.plot(distance, (beam/distance) * (d/lam), line)
if True:
ax1.set_xscale('log')
ax2.set_xscale('log')
ax3.set_xscale('log')
ax1.set_yscale('log')
ax2.set_yscale('log')
ax3.set_yscale('log')
ax2.set_ylim(1E-03, 1E+03)
ax3.set_ylim(100, 10000)
else:
ax3.set_ylim(0, None)
ax3.set_xlabel('distance (km)')
ax1.set_title('size')
ax2.set_title('size/distance')
ax3.set_title('(size/distance) / (initial size/wavelength)')
plt.show()

• i read your last post and tried to ude the same calculation as you. But with this calculation we get different value of the beam sizes, mynaric computed it using a different math expressions, i think Sep 28, 2021 at 6:06
• @Adil.Kolenko I will add something to my answer within 24 hours; I get almost exactly the same result as they do, it's fine. The final plot shows all points for all three beams at 1000 because it's meters. Had I used kilometers for both the line would be near 1 for all distances of all beams. But I can write a clearer single equation that fits them all as soon as I get a chance.
– uhoh
Sep 28, 2021 at 7:29
• @Adil.Kolenko I found it easier to just add a second answer with equations. It's mostly just $$\lambda L / d$$
– uhoh
Sep 29, 2021 at 0:06
• it seems it isn’t active question, but I will try to ask here. What is a difference between footprint of a satellite and beam size you calculated here? Nov 7, 2022 at 12:11
• @NoelMiller Here's a quick and simple response that be enough. With the far-field radiation pattern of antennas, we usually express the beam width as an angle, and it will be of order $\lambda / d$. The size of the corresponding footprint in linear dimension would then be $\lambda L / d$ where $L$ is the distance to a surface perpendicular to the direction of the satellite. If you were in the footprint in that case, the satellite would be overhead. In reality most satellite signals are coming in at a significant angle so the footprint is stretched in one direction. And of course it's fuzzy...
– uhoh
Nov 7, 2022 at 13:06