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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