Let's call "radio" anything electromagnetic below 300 GHz (1 millimeter), unless there is a compelling reason to do otherwise.

The as yet unanswered question How well can Voyager 1 separate Earth signals from Solar noise these days? gives some perspective, but there must be several other sources of radio noise and interference that limit either data rates or availability of data links.

A short explanation (1-2 sentences) describing under what conditions each major source becomes important would be great!

  • One source is the cosmic microwave background en.wikipedia.org/wiki/Cosmic_microwave_background but the are other astronomical radio sources like the sun, stars and galaxies en.wikipedia.org/wiki/Astronomical_radio_source – Uwe Apr 26 '17 at 8:52
  • @Uwe are you sure the CMB qualifies as a source "that limit(s) deep space data rates and availability?" I'm asking only for sources that do that, and for a short explanation describing under what conditions each becomes important. Thanks! – uhoh Apr 26 '17 at 9:05
  • Are you looking only for sources of noise and interference, or are you looking for an explanation of why data rates are limited? If the latter, then the Shannon–Hartley theorem is very relevant, but that's not an answer to the question you are posing. FWIW, our current modulations are pretty close to the theoretical limits, so unless Shannon and Hartley were wrong, there are few rabbits left to pull out of the hats. (The modulation used by the Voyagers probably isn't ideal, though.) – a CVn Apr 26 '17 at 9:35
  • @MichaelKjörling I'm right with you there. This related question might be an excellent place to discuss it. My question is meant to be complementary to it. Shannon–Hartley doesn't extend well to strongly-non-statistical noise, like planets drifting into your line-of-sight, or perhaps some strong dispersion or scintillation by some plasma getting in the way somewhere for example; thus the inclusion of "interference" in addition to "noise". – uhoh Apr 26 '17 at 10:44
  • I think there's a more fundamental question: is the data rate of a deep space radio link limited by cosmic radio noise getting into the receiver? Or is it limited by other causes? – Hobbes Apr 26 '17 at 18:35

As mentioned in the comments, the Cosmic Microwave Background creates noise. More specifically, cosmic background creates blackbody radiation, which adds kTB noise to any receiver looking at it. It's called kTB noise, or thermal noise, because its intensity is the product of the Boltzmann constant k, the temperature of the blackbody T (around 4K for the cosmic background) and the Bandwidth B of the receiver.

A good explanation on kTB noise and receiver sensitivity:


Edit: This (https://descanso.jpl.nasa.gov/DPSummary/Descanso4--Voyager_ed.pdf) NASA pdf on Voyager telecommunications gives the following sources of noise for the downlink (Voyager to Earth) (from page 26):

Noise power of each source: Design value/Favorable Tolerance/Adverse Tolerance

Total Noise Spectral Density, dBm/Hz: –185.35/–0.97/0.80

Total System Noise Temperature, K: 21.12/–4.24/4.24

Receiver Temperature, K: 13.20/–3.00/3.00

Ground Contribution, K: 2.88/–3.00/3.00

Galactic Contribution, K: 2.68/0.00/0.00

Atmospheric Contribution, K: 2.36/0.00/0.00

Hot Body Noise, K: 0.00/0.00/0.00

Elev Angle = 58.01 deg

So the major source of noise is the receiver, followed by the blackbody radiation from the ground, from the Milky way and from the atmosphere. The CMB seems to be negligible compared to that.

  • Can you add some information that explains under what conditions CMB becomes a significant source of "radio noise and interference that limit(s) deep space data rates and availability?" Thanks! – uhoh Apr 26 '17 at 14:36
  • I can think of a few factors that might damage the link budget: 1. Pointing accuracy is low due to the uncertainty of the spacecraft position. Together with the limited antenna size that might also reduce the link. 2. Path loss became a major issue in such long distance. 3. Using high data rate also reduces the link quality. – Eviatar.E Nov 8 at 19:31

Your Answer


By clicking "Post Your Answer", you acknowledge that you have read our updated terms of service, privacy policy and cookie policy, and that your continued use of the website is subject to these policies.

Not the answer you're looking for? Browse other questions tagged or ask your own question.