The ionosphere slows radio signals by an amount that depends on frequency, so comparing two frequencies lets a receiver measure and remove that error.
Navigation signals come down through the ionosphere, a layer of electrically charged gas that stretches from about 60 km to more than 2,000 km above the ground. Free electrons there slow the signal down. If a receiver assumes the signal travelled at full speed, it thinks the satellite is farther away than it is.
ESA's Navipedia says the uncorrected error for single-frequency users can reach several tens of metres.
The useful trick is that the delay depends on frequency: it falls as one over the frequency squared. A lower frequency is slowed more than a higher one.
So if a satellite sends the same timing on two frequencies, the receiver sees two slightly different distances. The difference tells it how much ionosphere the signals passed through. It can then calculate what the distance would be with no ionosphere at all. Navipedia says this removes more than 99.9% of the effect.
GPS.gov says combining two civil frequencies lets receivers correct the distortion caused by Earth's atmosphere and improves accuracy.
GPS's main civil signal is on L1 at 1575.42 MHz. GPS added L2C and L5 (1176.45 MHz). L5 sits in a band protected for aviation and is designed for safety-of-life uses.
India's NavIC has used two frequencies from the start: L5 and the S-band. Its newer NVS satellites add L1. Over India's equatorial region the ionosphere is especially variable, which is why India's GAGAN system needed its own ionosphere model.
Pick any NavIC satellite: its signal to you must cross the whole ionosphere, from about 60 km to beyond 2,000 km up. See the NavIC fleet
Navipedia gives the first-order ionospheric delay as:
I = 40.3 × STEC / f²
with STEC the slant total electron content in electrons per square metre, f in hertz and I in metres. Using TEC units (1 TECU = 10¹⁶ electrons/m²), the factor is 40.3 × 10¹⁶ / f² metres per TECU.
Worked example with STEC = 50 TECU:
L1: f² = (1575.42 × 10⁶)² = 2.482 × 10¹⁸, so 40.3 × 10¹⁶ / 2.482 × 10¹⁸ = 0.162 m/TECU, and I₁ = 0.162 × 50 ≈ 8.1 m.
L5: f² = (1176.45 × 10⁶)² = 1.384 × 10¹⁸, so 0.291 m/TECU, and I₅ = 0.291 × 50 ≈ 14.6 m.
The pseudoranges are P₁ = ρ + I₁ and P₅ = ρ + I₅. Because I₁ × f₁² = I₅ × f₅², the ionosphere-free combination is:
P_IF = (f₁² × P₁ − f₅² × P₅) / (f₁² − f₅²)
The ratio f₁² / f₅² = 2.482 / 1.384 ≈ 1.793. Plugging in, the ionosphere terms cancel and P_IF = ρ. The price is noise: the combination amplifies each signal's random error, so receivers smooth it over time.
The example also shows why lower frequencies feel the ionosphere more: L5's delay here is 14.6 / 8.1 ≈ 1.8 times L1's, exactly the ratio of the frequencies squared.
Free electrons slow the radio signal, so it takes longer to arrive, and the receiver turns the extra time into extra distance.
The delay varies as 1/f², so the two frequencies are delayed differently; comparing them lets the receiver calculate and remove the delay.
L5, because it has the lower frequency and the delay is proportional to 1/f².