Question

A GPS satellite is flying at a distance of 20,000 km from the observer. The phase of the L1 carrier (1575.42 MHz) in degrees as received by the observer is ............ (Rounded off to 2 decimal places).
Assume that the signal did not experience any refraction, reflection or other errors and the speed of light to be $c=3\times 10^8$ m/s.

Hint:

Received carrier phase (in degrees) $=360^\circ \!\times\! (fR/c \bmod 1)$. If $fR/c$ is an integer, the phase is exactly $0^\circ$.

Answer 1

Propagation delay for range $R=20{,}000$ km: \[ \tau=\frac{R}{c}=\frac{2\times 10^{7}}{3\times 10^{8}}=\frac{1}{15}\ \text{s}=0.066666\ldots\ \text{s}. \] Number of carrier cycles at $f=1575.42$ MHz: \[ N=f\tau=1.57542\times 10^9\times \frac{1}{15}=105{,}028{,}000.000\ \text{cycles}. \] Since this is an integer number of cycles, the residual phase at reception is \[ \phi=360^\circ. (N-\lfloor N\rfloor)=0^\circ. \] Rounded to two decimals: \(\boxed{0.00^\circ}\).