[time-nuts] GPS receiver vs local oscillator

Jim Lux jimlux at earthlink.net
Thu Feb 2 15:29:10 UTC 2012


On 2/2/12 1:49 AM, Hal Murray wrote:
>
> If I want to decode a GPS signal, do I have to know (or figure out) the
> frequency of the local oscillator?  Or does it drop out of the calculations,
> somehow?
>
>
>
Yes, it comes out in the calculations.

That's why you need 4 satellites.  You solve for x,y,z and local clock 
offset. (and, in reality you also have to solve for xdot,ydot,zdot, and 
fdot)

If you have a stationary receiver, or a known position, then you have an 
additional constraint or two you can fold into the solution, so you 
don't need as many observables.



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