Question

According to the first order ionospheric delay term, the time delay experienced by the GNSS signal is directly proportional to the Total Electron Content (TEC) in the ionosphere, and inversely proportional to the square of the frequency of the carrier wave. Based on this, the GPS L2 (1227.60 MHz) carrier is slower than the GPS L1 (1575.42 MHz) carrier by a factor of ________ for a given TEC (Rounded off to the nearest integer).

Hint:

Ionospheric delay is inversely proportional to the square of the carrier frequency, so compare (f_L1 divided by f_L2) squared for a fixed TEC.

Answer 1

Correct Answer: 2

Step 1: Write the first-order ionospheric delay relation.
The ionospheric time delay experienced by a GNSS signal follows \[ \Delta t \propto \frac{TEC}{f^2} \] for a fixed Total Electron Content (TEC), so the delay on any carrier is inversely proportional to the square of its frequency.

Step 2: Set up the ratio of delays for L2 and L1.
Since TEC is the same for both signals (they pass through essentially the same ionospheric path), \[ \frac{\Delta t_{L2}}{\Delta t_{L1}} = \frac{1/f_{L2}^2}{1/f_{L1}^2} = \left(\frac{f_{L1}}{f_{L2}}\right)^2 \]

Step 3: Substitute the carrier frequencies.
\(f_{L1} = 1575.42\ MHz\) and \(f_{L2} = 1227.60\ MHz\): \[ \frac{f_{L1}}{f_{L2}} = \frac{1575.42}{1227.60} = 1.28333 \]

Step 4: Square the frequency ratio.
\[ \left(\frac{f_{L1}}{f_{L2}}\right)^2 = (1.28333)^2 = 1.64694 \]

Step 5: Round off to the nearest integer.
\[ \frac{\Delta t_{L2}}{\Delta t_{L1}} \approx 1.647 \approx 2 \] So the L2 carrier, being at the lower frequency, is delayed (slowed) by a factor of about \(2\) relative to L1 for the same TEC, matching the accepted answer.

\[ \boxed{\text{Factor} \approx 2} \]