GPS L1 Carrier Phase Calculation Details

This solution details the calculation for the phase of a GPS L1 carrier signal received from a satellite, considering the signal's path length and frequency.

1. GPS L1 Signal Information

Key parameters provided are:

• Distance ($d$): 20,000 km = $2 \times 10^7$ m
• Frequency ($f$): 1575.42 MHz = $1575.42 \times 10^6$ Hz
• Speed of light ($c$): $3 \times 10^8$ m/s

2. Wavelength Calculation for GPS L1

The wavelength ($\lambda$) is the spatial period of the wave, calculated using the speed of light and frequency:

$\lambda = \frac{c}{f}$

Substituting the values:

$\lambda = \frac{3 \times 10^8 \text{ m/s}}{1575.42 \times 10^6 \text{ Hz}} = \frac{3}{1575.42} \text{ m}$

3. Number of Wavelengths in GPS Signal Path

To find the phase, we determine the total number of wavelengths ($N$) spanning the distance ($d$). This represents the number of cycles the signal completes.

$N = \frac{d}{\lambda}$

Calculation:

$N = \frac{2 \times 10^7 \text{ m}}{3 / 1575.42 \text{ m}} = \frac{2 \times 10^7 \times 1575.42}{3}$

$N = \frac{31508.4 \times 10^6}{3} = 10502.8 \times 10^6 = 10,502,800,000$

This is the total number of wavelengths covering the distance.

4. GPS Signal Phase Determination

The phase ($\phi$) in degrees corresponds to the fractional part of $N$, multiplied by $360^\circ$.

Phase $= (N \mod 1) \times 360^\circ$

Since $N = 10,502,800,000$ is an exact integer:

Fractional Part $= 10,502,800,000 \mod 1 = 0$

Phase $= 0 \times 360^\circ = 0^\circ$

Rounded to two decimal places, the phase is $0.00^\circ$.