Satellite communication generally involves three types of nodes: the ground station (also known as teleport), the satellite in geosynchronous orbit, and the satellite receiver/transceiver. I am interested in knowing how much a signal leaks into space during the ground-space-ground communication process.

The following images are two possible scenarios which I drew to help illustrate the path for a signal uplink, and I'm unsure which scenario is more accurate.

Possible Scenario One: The satellite uplink maintains it's path/signal width into geosynchronous orbit. Either little, or none of the signal leaks into space.

Possible Satellite Uplink - Scenario One

Possible Scenario Two: The signal of the satellite uplink spreads from the source of the uplink, similar to the path of a light of a flashlight. As a result, this spread results in much of the signal leaking into space.

Possible Satellite Uplink - Scenario Two

Which of these scenarios seems to be more accurate? Does the type of satellite uplink affect how much a signal spreads? (Ku-band, Ka-band, S-band?) Is a standard household satellite internet transceiver uplink more likely to leak then a teleport uplink due to the dish design?

Another note to consider, I discovered while doing some Google-fu that satellite signal uplink also consists of "lobes". The main lobe has the strength to travel into geosynchronus orbit, but I'm unsure if the side lobes are capable of leaking into space.

Satellite Lobes. Source: http://www.marinesatellitesystems.com/index.php?page_id=1086

Source: http://www.marinesatellitesystems.com/index.php?page_id=1086

The above points are things to consider with satellite uplink leaking into space, but I am also curious if satellite downlink can also leak. As a final note, the following is a diagram showing the main lobe from a satellite leaking during the downlink process. I am unsure if the diagrammed scenario happens universally with all geosynchronous satellites, or only with the one listed in the image source.

enter image description here

Source: https://www.cambridge.org/core/journals/journal-of-navigation/article/characterisation-of-gnss-space-service-volume/2EB9FAF84543144E198900B0A0271B91/core-reader


They have to spread their signal. Also, it requires a really big dish to not spread your signal at all.

Imagine that there was no spreading at all. The dish would have to be pointed exactly at the satellite, to the margin of the size of the dish. That is far beyond the technology of today! DSS has it's dishes pointed to 0.1 degrees. At the distance of Geosynchronous orbit, that is 62 km. Lasers can be pointed a bit more precisely, but still aren't that accurate.

The amount that a signal spreads is inversely proportional to the size of the antenna (In most cases). There are calculators out there, but in general the higher the frequency and the larger the dish, the smaller the beam width is. But it cannot ever be pointed only in one direction without an infinitely large antenna.

  • $\begingroup$ Dish? Who uses dishes anymore? Phased arrays are dirt cheap and have better radiation patterns anyway. $\endgroup$ – Sam Mar 9 '18 at 22:54
  • 5
    $\begingroup$ @Sam small dishes are dirtier cheaper $\endgroup$ – immibis Mar 10 '18 at 1:54
  • 1
    $\begingroup$ @Sam phased dirt cheap arrays are great for many applications, but they have their own unique problems that the nice clean optics of a Cassegrain reflector will not have. $\endgroup$ – uhoh Mar 10 '18 at 3:19
  • $\begingroup$ @PearsonArtPhoto Beautiful description about the signal being constrained to the size of the dish as improbable. Thank you for the response! $\endgroup$ – JaceVoracek Mar 12 '18 at 13:00

As an example, an antenna 10 m wide has a beam width of 0.14 degrees at 14 GHz (where I've taken the beam width as the -3 dB point).
At 36,000 km (geostationary orbit), a beam sent by this antenna will be 88 km wide.

36000 * tan(0.14 degrees) = 88

For any reasonable antenna size, you will have leakage because the main lobe will be much wider than the receiving antenna. The sidelobes will be at a fairly large angle relative to the main lobe. For a dish, the sidelobes are a lot weaker than the main lobe, I suspect.

  • $\begingroup$ Fantastic example along with the accompanied equation. I also appreciate the description regarding the lobes. Thanks for the response! $\endgroup$ – JaceVoracek Mar 12 '18 at 13:02

For the uplink, there is a theoretical lower limit of the width of the spread that is ~ (distance between receiver and transmitter)*(wavelength)/(size of transmitter). There are several ways of arriving at this, such as Heisenberg's Uncertainty Principle (the smaller the transmitter, the more you know the position of the beam, therefore the less you know about the direction). This is just an order of magnitude formula; obviously the beam doesn't have a sharp edge. Taking the numbers from @Hobbes' answer:

15 GHz -> wavelength ~ 0.0214 m, distance = 36*10^6 m and the transmitter = 10 m. 36*10^6*.0214/10 = 77*10^3. So my formula gives a rough approximation of Hobbes' result.

Note that this is the lower limit; it's physically impossible to do better than this. In the real world, any practical system is going to do worse.

For the downlink, this spread is going to be significantly smaller than the radius of the Earth as long as it's in the MHz range, so if it wouldn't violate quantum mechanics to have only a small portion of the beam miss the Earth, but again practical engineering is going to be another story.

Also, if "downlink" is understood to mean satellite to Earth, then your last picture appears to be showing satellite to satellite transmissions, not downlinks.

PS "receiver/transceiver" is redundant; "transceiver" already indicates this it is capable of being both a transmitter and receiver.

  • $\begingroup$ I'm trying to understand how this is really using the uncertainty principle. You've created similar triangles and paired the ratios of the sides, but I don't see Planck's constant anywhere. Where is ΔpΔx~ħ? Of course it is in there somewhere, somehow, but can you add some reference to how this is really using the uncertainty principle, or why it works? $\endgroup$ – uhoh Mar 11 '18 at 22:43
  • $\begingroup$ @Accumulation Great answer and further analysis regarding the limit for the width of spread. Thanks for the response! $\endgroup$ – JaceVoracek Mar 12 '18 at 13:04

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