Completely off-topic comment while on the subject: I can't see how randomly selected 1ms of data can contain unambiguously detectable start of C/A code. If NAV data bit phase flip occurred within this 1ms of data then correlation search will most probably fail. Yes, the probability of this is only 5% (assuming equal number of 1s and 0s in NAV message) but still not zero. Perhaps the quote should have added "carefully chosen 1ms of data" At the minimum, one would typically select 2ms of sampled data when they are doing a correlation search and then they are guaranteed a full PRN sequence somewhere in there. Leo > From: Tom Van Baak Thu, 30 Nov 2017 14:43:02 -0800 > In order to find the beginning of a C/A code in the received signal only a > very limited data record is needed such as 1 ms. If there is no Doppler effect > on the received signal, then one millisecond of data contains all the 1,023 > chips.

On 11/30/17 3:05 PM, Leo Bodnar wrote: > Completely off-topic comment while on the subject: > > I can't see how randomly selected 1ms of data can contain unambiguously detectable start of C/A code. > If NAV data bit phase flip occurred within this 1ms of data then correlation search will most probably fail. > Yes, the probability of this is only 5% (assuming equal number of 1s and 0s in NAV message) but still not zero. I believe that Nav bits only flip at code epoch boundaries (actually, every 20 code epochs), it can't change in the middle of a 1023 chip sequence. You use the epoch sync signal to run your symbol integrator. > Perhaps the quote should have added "carefully chosen 1ms of data" > At the minimum, one would typically select 2ms of sampled data when they are doing a correlation search and then they are guaranteed a full PRN sequence somewhere in there. > > Leo > >> From: Tom Van Baak Thu, 30 Nov 2017 14:43:02 -0800 >> In order to find the beginning of a C/A code in the received signal only a >> very limited data record is needed such as 1 ms. If there is no Doppler effect >> on the received signal, then one millisecond of data contains all the 1,023 >> chips. Doppler does make it harder (ask the folks who did the Huygens probe) > _______________________________________________ > time-nuts mailing list -- time-nuts@febo.com > To unsubscribe, go to https://www.febo.com/cgi-bin/mailman/listinfo/time-nuts > and follow the instructions there. >