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tl;dr:

Question: Have deep-space spacecraft always use some form of spread-spectrum for data downlink?

note: I'm looking for some insight into why, and any possible exceptions, not just a "yes" or "no". Thanks!

I was telling someone that the use of spread-spectrum in one form or another has become almost universal in civilian wireless communication and signaling (cell phone voice and 3G/4G/5G, WiFi, Bluetooth, GPS, ZigBee, LoRa (e.g. Understanding the relationship between LoRa chips, chirps, symbols and bit) etc. and it is interesting to note that WiFi and its resistance to multi-path interference has its origins in a couple of radio astronomers from Australia's CSRIO!

In April 2009, 14 technology companies agreed to pay CSIRO \$1 billion for infringements on CSIRO patents. This led to Australia labeling Wi-Fi as an Australian invention, though this has been the subject of some controversy. CSIRO won a further \$220 million settlement for Wi-Fi patent-infringements in 2012 with global firms in the United States required to pay the CSIRO licensing rights estimated to be worth an additional \$1 billion in royalties. In 2016, the wireless local area network Test Bed was chosen as Australia's contribution to the exhibition A History of the World in 100 Objects held in the National Museum of Australia.

I believe that the Gold code used in GPS can trace part of its origins to NASA's deep-space communications as well, though I can't find a reference for that right now. Note also that the importance of the Gold code is for timing reconstruction using correlation in the same way that NASA used it for range/rate, as well as for signal/noise benefits; see this and this answer for example.

Spread spectrum has many advantages, and one of them is signal-to-noise, which can be understood in terms of the Shannon-Hartley Theorem (see Am I using Shannon-Hartley Theorem and thermal noise correctly here?), and I believe that Voyager's data downlink always employs a bandwidth wider than than its bits-per-second rate would require, but I'm not sure; see the last paragraph in this answer to "How to calculate data rate of Voyager 1?" for example.

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  • $\begingroup$ Just to note, instead of having to add the "$xyz" in a code block, you can add a back slash (\) in front of the dollar sign (\$xyz) to prevent it from rendering as Latex or whatever funky script it's rendering as. $\endgroup$
    – Edlothiad
    Mar 2, 2018 at 9:32
  • $\begingroup$ @Edlothiad excellent! So it's the same as the MathJax \$ in some other sites (e.g. Electronics) just backwards. $\endgroup$
    – uhoh
    Mar 2, 2018 at 11:30
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    $\begingroup$ I don't know MathJax, I just know the \ means I can stop things from rendering, like *\*waves** allows me to make the word "waves" italic and keep the second set of asterisks: *waves* $\endgroup$
    – Edlothiad
    Mar 2, 2018 at 11:47
  • $\begingroup$ @Edlothiad left is MathJax in this site, right is MathJax in Electronics SE. i.stack.imgur.com/GpXKs.png It's the "same thing except backwards". $\endgroup$
    – uhoh
    Mar 3, 2018 at 5:43
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    $\begingroup$ @uhoh: Oops, link works for me (shortened Amazon URL). The title is "Deep Space Communications" / Jim Taylor(editor), Wiley 2016. Covers selected NASA missions. $\endgroup$
    – Andreas
    Mar 12, 2018 at 22:28

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Pseudo-random codes are used for ranging, i.e. to calculate the round-trip time of a signal. That takes up some bandwidth, so I would consider that to be spread-spectrum. That has been in use since, I guess, the 60's.

I am not aware of any applications of spread-spectrum techniques in deep space data communications (as opposed to tracking). Sure, they will use wide bandwidths, just because they can. So the modulation scheme is chosen for performance, not for minimizing spillage into adjacent bands as is often a concern on Earth. However spread spectrum means not just a wide bandwidth, but also some sort of time-dependent coding or other structure used at the transmitter and then necessarily duplicated at the receiving end in order to decode the signal through correlation. That is what I am not aware of in deep space comm.

For reference, the definition of spread spectrum as quoted in this paper:

Spread spectrum is a means of transmission in which the signal occupies a bandwidth in excess of the minimum necessary to send the information; the band spread is accomplished by means of a code which is independent of the data, and a synchronized reception with the code at the receiver is used for despreading and subsequent data recovery.

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  • $\begingroup$ If I understand correctly, Shannon-Harltey sez (and by "sez" I mean I'm paraphrasing) that if you spread the spectrum of your signal out across a bandwidth significantly larger than you need (with some constraints on how you do it) you can improve your signal to noise. Now I don't know if that alone counts as a "spread-spectrum* technique, but I think it does. The coding can be just a way to do the spreading so that the receiver will know how to decode. It doesn't have to be a "secret" code. This is just my own take though $\endgroup$
    – uhoh
    Mar 12, 2018 at 19:38
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    $\begingroup$ The key here is that in spread spectrum the 'a code which is independent of the data'. If you increase the bandwidth with a code that is dependant on the data (i.e. an error correcting code) then increasing the bandwidth helps SNR; if it is independent then it just shuffles the noise and gives no advantage. Spread spectrum is great against narrowband interference, but doesn't help against broadband noise. It is also good for tracking (hence its use for ranging & GPS). And for multiple access - but that's not currently a problem for deep space missions... $\endgroup$
    – Paul Norridge
    Mar 13, 2018 at 17:42
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    $\begingroup$ @uhoh The key point glossed over in that application note (leaving aside the questionable maths) is that the bandwidth increase only helps when the SNR is fixed. But normally an increase in bandwidth also increases the noise by the same factor, so you have no benefit. As far as I can see, LoRa just increases the SNR by effectively lowering the rate of the useful data, so increasing the energy per data bit. But that isn't dependant on spread spectrum. You could decrease the data rate on an unspread signal and see the same effect. $\endgroup$
    – Paul Norridge
    Mar 19, 2018 at 22:47
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    $\begingroup$ @uhoh On the maths: although the conclusion that C/B is proportional to S/N is correct, the explanations of how equation 2 & 3 are derived don't make sense (or, at least, they are very confused). $\endgroup$
    – Paul Norridge
    Mar 20, 2018 at 23:26
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    $\begingroup$ @uhoh More importantly, C is the potential capacity of the link, not the guaranteed performance. So, "only the transmitted signal bandwidth need be increased" is not correct; you need some way to exploit that bandwidth effectively. And, you are relying on the S/N not changing. If you look at the wikipedia page for S–H there is an example showing why C is typically independent of bandwidth in low SNR. A similar argument works for spread spectrum. $\endgroup$
    – Paul Norridge
    Mar 20, 2018 at 23:27

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