After taking a basic signals & systems class and learning about the frequency domain, I started wondering:

How the heck do scientists still communicate with Voyagers I and II??

Do they send/receive signals in a frequency that is otherwise unusually silent? If not, then how do they send and receive faint signals from something 10 billion miles away, when the spacecraft has so little energy and where it seems practically impossible to aim everything 100% correctly (since things are so distant and moving at the same time)?

(I wasn't sure what to tag this, so please retag as appropriate, thanks!)

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    $\begingroup$ At least for the aim: It's not that voyager sends a divergence-less beam. But it is quite an impressive feat, I agree on that. One first hint is the Deep Space Network $\endgroup$
    – Lagerbaer
    Aug 6, 2011 at 5:09
  • $\begingroup$ @Lagerbaer: Yeah I realize that... even lasers diverge a lot over distance, and they probably didn't even use those. I'm almost tempted to post this on Skeptics.SE... $\endgroup$
    – user541686
    Aug 6, 2011 at 5:11
  • $\begingroup$ @downvoter: Care to comment? $\endgroup$
    – user541686
    Aug 6, 2011 at 16:51
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    $\begingroup$ Some data on antenna power/gain of Voyager 2 can be found here. $\endgroup$
    – Leonidas
    Aug 7, 2011 at 0:07
  • $\begingroup$ @Leonidas: Yeah exactly -- how in the world do you detect something so minuscule? (For some reason, "by having REALLY large dishes" simply doesn't seem to quite convince me.) $\endgroup$
    – user541686
    Aug 7, 2011 at 0:18

3 Answers 3


We know exactly where the spacecraft is, and it knows pretty well where we are. Distance does not aggravate the accuracy of aim problem, indeed the further apart the less relative motion, so aim gets easier. The problem is signal attenuation by dispersal. i.e. at twice the distance, the signal will be a quarter of the strength.

The solution, for Voyager, has been that our earth-based recievers and transmitters have increased in both accuracy and sensitivity several times faster than the signal from Voyager has been weakening due to distance. Indeed, if the power source would have lasted, we already have the ability to recieve a signal such a voyager's from thousands of times the distance. The bigger your antenna, the tighter you can focus its view. Once we get around to building the square-kilometer-array (2025 or so) we would have ability to hear a similar signal from several light years range.

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    $\begingroup$ Created an account just to upvote this awesome positive answer. $\endgroup$
    – Aditya M P
    Jun 29, 2013 at 23:31
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    $\begingroup$ I think exactly is rather strong... Most likely the spacecraft position is inaccurate to at least several hundred kilometers. However, that inaccuracy is negligible compared to the distance from Earth, so as you said it is not an issue. $\endgroup$ Apr 15, 2016 at 13:53
  • $\begingroup$ From both Voyager 1 and 2's out-of-plane-and-very-far-away perspectives, these days earth moves in a 0.5 x 0.3, or 0.4 x 0.2 degree ellipse around the sun. At S-band (circa 2GHz uplink) it can aim at the sun because the beam is so broad. At X-band (circa 8GHz downlink) it should offset a bit to point at earh, based on the time of the year since the beam is much narrower at higher frequency. $\endgroup$
    – uhoh
    Jul 23, 2016 at 2:23

As always, a communication via electromagnetic radiation depends on both ends. Uplink from earth can be done with a lot of power and big dishes, of course. Downlink is limited to the power of the nuclear battery on board but has a rather impressive 2.7 meters dish!. On top of that they use a rather slow bitrate, I think with a lot of redundancy. All this is commonplace information technology when dealing with bad connections with signal not much above noise.
Details are given in Wikipedia:

Uplink communications is via S band (16-bit/s command rate) while an X band transmitter provides downlink telemetry at 160 bit/s normally and 1.4 kbit/s for playback of high-rate plasma wave data. All data is transmitted from and received at the spacecraft via the 3.7-meter high-gain antenna.

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    $\begingroup$ +1 a partial explanation on how they correct errors, but it doesn't explain at all how they detect the signals in the first place. (A 2.7-meter dish doesn't seem to quite explain to me how it solves the problem of communicating over billions of miles.) $\endgroup$
    – user541686
    Aug 6, 2011 at 9:28
  • $\begingroup$ The problem is, that You "imagine" problems. Detecting is done by so called "receivers". Here on earth one would use a parametric receiver, maybe cooled down cryoscopically. That is common technolagy for radioastronomy and such X-band in general. I gave You hints to wikipedia, and some buzzwords to google for. $\endgroup$
    – Georg
    Aug 6, 2011 at 9:36
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    $\begingroup$ I'm just imagining problems? So you're saying it's easy to communicate across 10 billion miles? Pardon me if it's an unreasonable question, but what exactly is the range of the signals emitted by these tools? You seem to be implying that it's unlimited, which is pretty impossible given that signals drop in strength in proportional to the square of distance... $\endgroup$
    – user541686
    Aug 6, 2011 at 11:16
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    $\begingroup$ @Lagerbaer: (1) Explaining how DSN can extract such sensitive data definitely answers part of my question... (2) Sure, but how how does Voyager send data so accurately/strongly in the first place? It seems like any signal sent would be lost in the cosmos... $\endgroup$
    – user541686
    Aug 6, 2011 at 15:57
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    $\begingroup$ I deleted a few inappropriate comments. Georg, you really need to keep your comments under control. There have been quite a few complaints recently. I'll ask both you and Mehrdad not to continue this, at least not here. $\endgroup$
    – David Z
    Aug 7, 2011 at 20:56

The question has already been answered pretty well, but I think more explanation of why big antennas help would be worthwhile ...

Antennas have a property called "gain" which means that they work best in one direction (or several directions, depending on the type of antenna) and less well in others -- so a transmitting antenna with high gain sends its signal in one precise beam, and a receiving antenna with high gain receives best from one precise direction and mostly ignores signals from other directions.

Well, the gain of an antenna depends on the type of antenna it is and the ratio of its size to the wavelength of the signal. What this means for us is that if you want a really high gain antenna ... it has to be large compared to the wavelength of the signal.

The size of the antenna on the probe is fixed, but it's quite large, so it's got a high gain -- it sends its signal directly at where it knows the Earth is. On the Earth, we keep using bigger and bigger antennas to talk to the probe, so we get higher and higher gains, and a high gain antenna aimed properly picks up more of the signal and less noise.

(There are other factors involved, of course, but this is why the big antennas matter.)

Note that the higher the gain of your antenna, the more important it is that it is precisely aimed. But when we're talking to probes like this, we're aiming the antenna extremely precisely.

But for something like the WiFi in your house ... high gain antennas tend to be a bad thing rather than a good thing, because it means that the range is good (better than a lower gain antenna) in the given direction(s) but bad (worse than a lower gain antenna) in others.

  • $\begingroup$ I like to think that engineers give antennas the property of gain. Antennas have mass, angular momentum, charge etc, but the concept of assigning gain to antennas was invented to make dB math easier for them. Just like they call I ~ 1/r^2 "attenuation" which it really isn't. $\endgroup$
    – uhoh
    Jul 6, 2016 at 1:33
  • $\begingroup$ Are you really quibbling with me over what "property" means here? $\endgroup$
    – dougmc
    Dec 8, 2016 at 0:41
  • $\begingroup$ I liked your answer and had upvoted it. The comment isn't a personal message directed at you, so nope, I'm not quibbling with you. $\endgroup$
    – uhoh
    Dec 8, 2016 at 1:26

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