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pardon my choice of words, but English isn't my native language, so I might use the wrong terms at some places.

Beside this, I do have a hypothetical situation which might be a bit off reality (because its for a fictional universe): Imagine an installation in deep space that is supposed to watch over its surroundings using radar. Beside this being impractical due to the limit of c, I noticed that its also impractical because of the nature of that radar. Its nothing like a laser, so it widens notably after short distances. Of whatever amount of power is being outputted at the senders side, only a small fraction would actually reach the target. That loss is not noticeable at, say, a distance of 400km, but...

if you insert values like a light-second of distance, 450kW sending power (like the Arecibo message), an... opening (?) of 0.0001° from a 1m² sending surface, and a hypothetical 100m² radar cross-section (with no loss due to scattering and 100% reflection) target, said target will receive 0.04 W/m² beam power.

I made a small application to find out more about this, using a cone with h of a light-second as a model for that beam. One can calculate its bottom area using that openings tangens * h * 2 if I recall correctly. I got the surface the beam (well, the cone actually) would be covering when its traveled one light second and scaled down my initial sending-power to that surface size. And received said 0.04 W/m²

So why do I create a question out of this? Because I started wondering one thing: If even the most coherent radio transmission would suffer such losses, how exactly do they keep communicating with... all the equipment spread over and around Mars? Or more extreme, the Voyager probes.

I used up to 10GW of sending power in my little experiments, and still something as "close" as 1 light-minute would receive so few Watts per square-meter, no radar receiving equipment would have a chance to pic up its reflection. This being a "do-not-know-where-to-look-at" action would simple render the target invisible, but... an aimed transmission would also suffer from these effects, and I failed to get anything in a useful Watt-range even with said 10GW sender and the 0.0001° opening beyond a light-second of distance. And I guess the opening I used might be even more coherent than one from an actual parabolic antenna. And Voyager is way more out there, and certainly does not carry around a GW antenna.

And because stuff like that seems to work in reality, my understanding (that "inner picture") of the whole physics behind that must be wrong. And my math too, by the way. Still, I do have a hard time finding the correct phrases to throw into Google, so I have to ask such a probably stupid question in here.

Actual question (tldr): So how do they keep on sending and receiving radio transmissions over astronomical distances, even when it is maybe dropping below the background-noise level once it passed one light minute of distance? And how can I calculate that by myself?

EDiT: So the first comment pointed to the DSN page and made clear, that they do receive as few as 8,5 *10-23kW from one of the Voyager probes. That... wow. Does this mean, I just underestimated the ability to receive signals with low transmission power by some orders of magnitute?

Btw, no matter how hard you try, flooding a sphere with a diameter of 1 ls with radar seems to take hours or even days when you want to see any reflections (because of the narrow opening), so radar feel like being total useless for that task... but that's another problem.

PS: Yeah, there is similar question, but I cannot answer mine with the content in it: Does the power weakening of an electromagnetic transmission over distance depends on the beam's width?

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    $\begingroup$ space.stackexchange.com/questions/9824/… This question on Space.SE has some good information. For example, apparently the signal received from the deep spacecraft is on the order of $10^{-23} W$ for some of the furthest probes. That's basically no power at all, yet they can still process the data. $\endgroup$ – JMac Apr 10 at 18:05
  • $\begingroup$ @Jmac, thank you for that link. So, they can receive transmission with 0,0000000000000000000858W signal power... I should tell my notebook about this, when it fails to connect to a hotspot in plain, unobscured view again (because of low signal power).... but does this mean, that my numbers are not as wrong as I thought they are? $\endgroup$ – Confused Merlin Apr 11 at 5:22
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There are a number of reasons for this working: The biggest components obviously to a system like this would be the transmitting antenna and the receiver at the receiver end.

The transmitting antenna needs to output both relatively high power and to be very directional. Transmitting high power is easy, just up the power in the transmitted circuit. An antenna radiates in a pattern called a beam pattern which consists of either one big beam or several beams. Usually, the boresight beam, the strongest beam, is the one in the pointing direction of the antenna. The width of this beam (the size so to speak) depends on the physical configuration of the antenna and the wavelength and several other factors as well. Depending on the design of your system, this main-beam can be very narrow. Array antennas are antennas consisting of several radiating elements which have different relative phases to one another, which causes constructive and destructive intereferences which results in a narrower beam of so desired.

The receiver on the other end needs to be sensitive enough to be able to detect the signal transmitted. The sensitivity is limited by the nosie floor in the receiver. The receiver circuit will always suffer from electrical and random noise, and if the signal is below this, detecting it will be unlikely. This noise is generally limited by the electrical quantum noise and radiation noise.

The power per unit area is inversely related to the square of the distance $\frac{1}{R^2}$ and then the receiving antenna and the wavelength determine the "receiving loss" at the receiving site (help me out, blackout on the word for this loss) which is proportional to $\lambda^2$ and there are as well losses related to atmospheric absorption unless the transmitted is in orbit around our planet.

But basically with some math, and knowledge of the receiver system, a strong enough signal can be sent without problem. What helps is the fact that there aren't any notable absorption in space to attenuate the signal much more than the $\frac{1}{R^2}$-loss.

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