[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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