3 ms·
> With India, China and Russia launching their own positionning systems, I wonder if it would be possible to interpolate an even higher precision by listening t
by wolfgke 10y ago
> With India, China and Russia launching their own positionning systems, I wonder if it would be possible to interpolate an even higher precision by listening to all 5 signals at the same time.
First: Only Navstar-GPS, GLONASS and (in near future) Galileo will have worldwide coverage. Perhaps in future (but according to http://www.navipedia.net/index.php/BeiDou_Future_and_Evolutions#BeiDou_Status http://www.navipedia.net/index.php/BeiDou_Future_and_Evoluti... not before 2020) BeiDou will also have global coverage. IRNSS only has regional coverage.
So in near future we will only have three sources (not five).
Now for the simple mathematical trick: Model the position of the three systems as normal distributed random variable with same \mu, but different \sigma (\sigma_1, \sigma_2, \sigma_3). These three sigmas have to be measured. Now a convex combination of these random variables is again normal distributed, but with different \sigma (let's call it \sigma^* ) (easy exercise or look it up). Now just compute the coefficients of the convex combination that minimize \sigma^*.
- jessriedel 10y agoOr you could just say that, for N independently distributed samples, the accuracy improves by a factor of at most sqrt(N) :)
- wolfgke 10y agoThis does only hold if the standard deviation \sigma is the same for all of the normal distributed variables. This is not the case here: Galileo is supposed to be more precise than Navstar-GPS (no surprise, since it is the newer system) - though I know that the US government is working on to increase the precision for Navstar-GPS. On the other hand (source: http://www.navipedia.net/index.php/GLONASS_Performances#GLONASS_Accuracy_Comparison http://www.navipedia.net/index.php/GLONASS_Performances#GLON...) GLONASS is less precise than Navstar-GPS.
- jessriedel 10y agoWell, I said "at most", so it holds for all sigma_i. As you know, when the sigma_i are very different, then the accuracy is approximately given by the best accuracy (smallest sigma_i), so there's very little improvement from averaging.
- wolfgke 10y ago> Well, I said "at most", OK, you are right. This can indeed be shown.