Hi all,
I need to understand how to compute basic pseudo-distance.
I found a post (03-09-2005) about this.
Here's an extract of the discussion
-------------------------------------------------
Question :
>I still have difficulties to understand how to determine the
> pseudo-distance.
> I read in a paper about C/A code:
>
> "At this rate of bit transitions, the full sequence of 1023 bits is
> transmitted in 1 ms. Therefore, the sequence repeats 1000 times per
> second. The chip length (distance between bit transitions) is 293 m.
> Therefore, the sequence repeats every 300 km. " and
> "The "chip length", or physical distance between binary transitions
> (between digits +1 and -1), is 293 metres. The basic information that
> the C/A code contains is the time according to the satellite clock
> when the signal was transmitted (with an ambiguity of 1 ms, which is
> easily resolved, since this corresponds to 293 km). Each satellite
has
> a different C/A code, so that they can be uniquely identified."
>
> I fully agree with that ... but I can't understand how to resolve the
> ambiguity of 1ms. In fact I don't understand how to determine
> APPROXIMATELY the pseudo distance between the satellite and the
> receiver just with the C/A code. T= N*1ms + dt that leads to
> PR=N*293km + d. The C/A code can give the "d". How can we get N ?
>
> Best regards,
>
> Clément
------------------------------------------------------------
Answer :
The 50 bit per second navigation message is on top of and synchronized
with the pseudorandom gold code. The data within the navigation
message gives you the time the next 6 second subframe begins. The
timing mark where the subframe begins is the beginning of the first bit
of the first chip of the subframe. The 1 ms chip is your coarse time
ruler. The carrier cycles are your fine time ruler.
Tom
---------------------------------------------------------------
I still don't understand how the resolve the integer ambiguity, N,
in this relation
PR=N*293km + d
I recall that PR is the pseudo-distance and "d" is resolved by the shift measurement (in time chips as each time chip correspond to 293 m) where the C/A sequence received from the satellite and the replica C/A code at the receiver side are identical.
Thanks for your help.
Regards
Michel
I need to understand how to compute basic pseudo-distance.
I found a post (03-09-2005) about this.
Here's an extract of the discussion
-------------------------------------------------
Question :
>I still have difficulties to understand how to determine the
> pseudo-distance.
> I read in a paper about C/A code:
>
> "At this rate of bit transitions, the full sequence of 1023 bits is
> transmitted in 1 ms. Therefore, the sequence repeats 1000 times per
> second. The chip length (distance between bit transitions) is 293 m.
> Therefore, the sequence repeats every 300 km. " and
> "The "chip length", or physical distance between binary transitions
> (between digits +1 and -1), is 293 metres. The basic information that
> the C/A code contains is the time according to the satellite clock
> when the signal was transmitted (with an ambiguity of 1 ms, which is
> easily resolved, since this corresponds to 293 km). Each satellite
has
> a different C/A code, so that they can be uniquely identified."
>
> I fully agree with that ... but I can't understand how to resolve the
> ambiguity of 1ms. In fact I don't understand how to determine
> APPROXIMATELY the pseudo distance between the satellite and the
> receiver just with the C/A code. T= N*1ms + dt that leads to
> PR=N*293km + d. The C/A code can give the "d". How can we get N ?
>
> Best regards,
>
> Clément
------------------------------------------------------------
Answer :
The 50 bit per second navigation message is on top of and synchronized
with the pseudorandom gold code. The data within the navigation
message gives you the time the next 6 second subframe begins. The
timing mark where the subframe begins is the beginning of the first bit
of the first chip of the subframe. The 1 ms chip is your coarse time
ruler. The carrier cycles are your fine time ruler.
Tom
---------------------------------------------------------------
I still don't understand how the resolve the integer ambiguity, N,
in this relation
PR=N*293km + d
I recall that PR is the pseudo-distance and "d" is resolved by the shift measurement (in time chips as each time chip correspond to 293 m) where the C/A sequence received from the satellite and the replica C/A code at the receiver side are identical.
Thanks for your help.
Regards
Michel