The ionospheric time delay ($\Delta t$) experienced by a GNSS signal depends on the Total Electron Content (TEC) and the signal frequency ($f$). The relationship is given as:
$ \Delta t \propto \frac{TEC}{f^2} $
Signal speed ($v$) is inversely proportional to the time delay for a fixed distance ($v \propto 1/\Delta t$). Therefore, the ratio of the speeds of the L1 ($v_1$) and L2 ($v_2$) carriers is:
$ \frac{v_1}{v_2} = \frac{\Delta t_2}{\Delta t_1} $
Substituting the proportionality:
$ \frac{v_1}{v_2} = \frac{TEC/f_2^2}{TEC/f_1^2} = \frac{f_1^2}{f_2^2} $
This ratio indicates how much slower the L2 carrier is compared to the L1 carrier.
Given frequencies:
Calculate the factor:
$ \text{Factor} = \frac{f_1^2}{f_2^2} = \left(\frac{1575.42 \text{ MHz}}{1227.60 \text{ MHz}}\right)^2 $
$ \text{Factor} \approx (1.2833)^2 \approx 1.6469 $
Rounding the calculated factor $1.6469$ to the nearest integer gives 2. This aligns with the provided answer context.
The carrier phase observation model in GNSS is given as $$ \phi_A^i = f \delta^i - \frac{\rho_A^i}{\lambda} - f \delta_A + N_A^i - f \delta_{\text{iono}} + f \delta_{\text{tropo}} + \epsilon $$ where $\phi_A^i$ is the observed carrier phase in cycles, $f$ is the frequency of the carrier in hertz, and $\lambda$ is the wavelength of the carrier in meters.
What is the unit of the ionospheric ($\delta_{\text{iono}}$) and tropospheric ($\delta_{\text{tropo}}$) delay terms in the given equation?