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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.
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()