Question

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?

• A. Second
• B. Meter
• C. Cycle
• D. Cycles/second

Hint:

Every additive term in the equation must come out in cycles like phi itself; since these delay terms are multiplied by the frequency f (in cycles per second), work out what unit delta must have on its own to cancel out to cycles.

Answer 1

The Correct Option is A

Step 1: Check the overall dimensional consistency of the equation.
The left-hand side, \(\phi_A^i\), is stated to be in cycles, so every term added or subtracted on the right-hand side must also work out to cycles.
Step 2: Check the \(\rho_A^i / \lambda\) term as a reference.
\(\rho_A^i\) is a geometric range, measured in meters, and \(\lambda\) is the wavelength, also in meters, so \(\rho_A^i/\lambda\) is a pure (dimensionless) ratio, which is exactly how a number of cycles is expressed: a range that is 3.5 wavelengths long corresponds to 3.5 cycles of phase.
Step 3: Check the \(f\delta^i\) and \(f\delta_A\) clock terms.
\(f\) is the carrier frequency, in hertz, i.e. cycles per second. The satellite and receiver clock offsets \(\delta^i\) and \(\delta_A\) are, by GNSS convention, expressed as time offsets, in seconds. So \(f \times \delta\) has units \((\text{cycles/second}) \times \text{second} = \text{cycles}\), consistent with the left-hand side.
Step 4: Apply the identical logic to \(f\delta_{\text{iono}}\) and \(f\delta_{\text{tropo}}\).
These terms have exactly the same structure, frequency multiplied by a delay symbol \(\delta\). For the product \(f \times \delta_{\text{iono}}\) to come out in cycles (matching every other term in the sum), and since \(f\) is in cycles per second, \(\delta_{\text{iono}}\) itself (before being multiplied by \(f\)) must be a time quantity, in seconds. The same reasoning applies identically to \(\delta_{\text{tropo}}\).
Step 5: Physically interpret this.
This matches the physical meaning of \(\delta_{\text{iono}}\) and \(\delta_{\text{tropo}}\): they represent the extra time it takes the signal to propagate through the ionosphere and troposphere respectively (a propagation delay), which is naturally expressed in seconds, exactly like the clock-offset terms \(\delta^i\) and \(\delta_A\) that sit right next to them in the same equation.
Step 6: Rule out the other options.
Meter (option B) would be the unit of \(\delta_{\text{iono}}\) only if it were being divided by \(\lambda\) like the range term, which it is not, it is multiplied by \(f\) instead. Cycle (option C) and Cycles/second (option D) describe the unit of the already-multiplied term \(f\delta_{\text{iono}}\) (cycles) or of \(f\) itself (cycles/second, i.e. hertz), not of \(\delta_{\text{iono}}\) in isolation.
Step 7: Conclude.
\(\delta_{\text{iono}}\) and \(\delta_{\text{tropo}}\), being propagation time delays, are expressed in seconds.
\[ \boxed{\text{Unit of } \delta_{\text{iono}}, \delta_{\text{tropo}} = \text{Second}} \]