3
\$\begingroup\$

I work in developing software that integrates with GPS and Galileo receivers. I've struggled for years with inconsistencies in documentation for the NMEA emitted by such receivers, and I've decided to put it to bed.

Is it true that, since GPS is a digitally modulated signal, signal-to-noise and carrier-to-noise are equivalent values for any given receiver? Is this the case for other GNSS systems?

I have done some reading but I am not an electrical engineer.

\$\endgroup\$
3
  • \$\begingroup\$ Interesting question and I'm not sure if they're theoretically equivalent but I've found in practice how different receivers report values is a bit over the place and often not documented, for example as per gauss.gge.unb.ca/papers.pdf/SNR.memo.pdf \$\endgroup\$
    – PeterJ
    Commented Jul 28, 2015 at 13:27
  • \$\begingroup\$ @PeterJ: Interesting PDF, and further adds to the bloomin' mess and mystery of this oft-messed up term. \$\endgroup\$
    – Lightness Races in Orbit
    Commented Jul 28, 2015 at 13:36
  • 1
    \$\begingroup\$ Anonymous editor: adding acronyms for common engineering terms and replacing "GNSS" for no apparent reason (there is more to GNSS than just the USA's GPS) are not "clarifications". \$\endgroup\$
    – Lightness Races in Orbit
    Commented Oct 17, 2015 at 13:18

2 Answers 2

Reset to default
1
\$\begingroup\$

I edited my answer to the following:

As i understand, if a transmitted signal has no (or insignificantly small amount of) unmodulated carrier and/or unmodulated signal itself, for that signal C/N = S/N, and there is no difference in this rule for which the modulation is: digital, analog or mixed/complex, such is DSSS.

(I have found no direct claims for DSSS, but found ref for DSBSC, SSB--FakeMoustache refers to---claiming that.)

Therefore, in the power (math) context, C/N and S/N are equal for DSSS.

Meanwhile, the term C/N is used to distinguish a digitally modulated signal quality from a analogue demodulated signal quality expressed with help of the term S/N.

Therefore, in that context, C/N and S/N are not equal for DSSS (which transmits digital signals only) 'cause the S/N is not applicable here at all.

How to deal with this? Be careful and remember that the terms are context-specific.

Also, it is necessary to say that both in the NMEA native GSV message and in the NMEA extending Garmin PGRMB message, the measure of signal quality is called "SNR" and expressed in dB. IMO, it's a good example of using the terms in the power context.

As for another GNSS, it depends on the modulation the GNSS employs.

Below is my previous answer, please keep it for history.

Maybe, you are not enough strict in terms?

This html and its typeset PDF version describes (including mathematically) the difference between signal-to-noise ratio (SNR) and carrier-to-noise density (C/N0) in the scope of GNSS application. Is this what you are searching for?

Also, if you are interested in a concrete GNSS --- GPS in your case --- may be it is better to refer to by its individual name (NAVSTAR) GPS, not the common name GNSS that covers any of them: GPS, GLONASS, Galileo, Beidou, QZSS, IRNSS, WAAS, EGNOS, GAGAN, MSAS, SDCM, because principal differences may exist.

In my comments below the phrase "these two" relates to SNR and C/N0 therefore there is no collision with my newer answer (after edit). Please insight. Thanks.

\$\endgroup\$
3
  • \$\begingroup\$ My title ---is--- was more accurate than the body; I am looking for a general rule for GNSS, or a statement that there are differences if any. If you have found an external resource that answers the question, could you use it to produce an answer here please? \$\endgroup\$
    – Lightness Races in Orbit
    Commented Oct 18, 2015 at 19:05
  • \$\begingroup\$ @LightnessRacesinOrbit Short answer: as stated in the reference, these two are physically different, but both significant and useful. For a long answer, please read the reference. \$\endgroup\$
    – asndre
    Commented Oct 18, 2015 at 19:15
  • \$\begingroup\$ I'm asking you to kindly write the answer in the answer! SE is a Q&A repo not a link index! \$\endgroup\$
    – Lightness Races in Orbit
    Commented Oct 18, 2015 at 19:37
0
\$\begingroup\$

I would say that in general signal-to-noise and carrier-to-noise are the same thing. However, not all modulation techniques have a signal in their spectrum that can be considered to be the "carrier", single-sideband modulation for example.

Therefore I think signal-to-noise is a better name since it says "signal" meaning the signal containing the information you're interested in and "noise" in which you're of course not interested.

Fun fact about GPS: the signal is often lower than the noise ! This is because so very little power is received from the satellites that often the antenna itself introduces more signal power because of (thermal) noise. It is still possible to receive the data because it is sent at a very slow datarate. This means the noise can be supressed by averaging so the signal can be retrieved.

\$\endgroup\$
3
  • \$\begingroup\$ Can you then explain why I keep encountering people asking whether a piece of documentation "is really S/N, or does it mean C/N?" Almost everybody I encounter seems to think there is a difference, but they can't explain what that difference is. I used to think I grokked the topic but have started to think S/N and C/N may be equivalent for typical GNSS modulations, hence the question ... but would definitely like an explanation/proof of such if it's true! \$\endgroup\$
    – Lightness Races in Orbit
    Commented Oct 17, 2015 at 13:19
  • \$\begingroup\$ OK, I've read a bit more, GNSS uses DSSS modulation so it does not have a visible (on a spectrum analyser) carrier ! The DSSS modulation makes the signal look like noise. So to anyone who talks about C/N in relation to a GNSS signal I would ask: please show me the carrier. Obviously they cannot because there is no visible carrier, it is spread out (in frequency). So therefore I think it is much better to talk about S/N ratio. \$\endgroup\$
    – Bimpelrekkie
    Commented Oct 17, 2015 at 15:26
  • \$\begingroup\$ Now we're getting somewhere :) \$\endgroup\$
    – Lightness Races in Orbit
    Commented Oct 17, 2015 at 16:31

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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