GPS Pseudorange
Measured distance to a GPS satellite from signal travel time, biased by the receiver clock error — hence “pseudo” range.
Formula
Variables
Example
Travel time 67 ms: ρ ≈ 3×10⁸×0.067 ≈ 20100 km
How it works
A GPS receiver estimates how far away a satellite is by measuring how long the radio signal took to arrive and multiplying that travel time by the speed of light. The catch is that the receiver’s clock is not perfectly synchronized with GPS time, so the measured delay includes an extra timing offset; converting that offset into distance gives the clock-bias term cδt. The result is called a pseudorange because it is a range estimate contaminated by receiver clock error, not a pure geometric distance.
When to use it
Use this relation whenever you are turning a satellite signal time-of-flight measurement into a GPS range observation, especially in navigation, timing, and positioning calculations. It is a core input to least-squares or Kalman-filter solutions that combine several satellite pseudoranges to solve for position and receiver clock bias.
Worked example, step by step
Given: t_r - t_t = 74.6 ms = 0.0746 s δt = 18 μs = 18 × 10^-6 s c = 299,792,458 m/s Step 1: Compute the raw signal travel distance c(t_r - t_t) = 299,792,458 × 0.0746 ≈ 22,360,639.81 m Step 2: Compute the clock-bias distance term cδt = 299,792,458 × 18 × 10^-6 ≈ 5,396.26 m Step 3: Add them to get the pseudorange ρ = c(t_r - t_t) + cδt ρ ≈ 22,360,639.81 + 5,396.26 ρ ≈ 22,366,036.07 m Final answer: ρ ≈ 2.2366 × 10^7 m ≈ 22,366 km
Common mistakes to avoid
- Using the receiver clock bias with the wrong sign, which shifts the pseudorange in the opposite direction
- Forgetting to convert milliseconds or microseconds into seconds before multiplying by c
- Treating pseudorange as the true satellite-receiver distance without accounting for clock error and other later corrections
Did You Know?
A GPS receiver solves for four unknowns — x, y, z and its own clock error — which is why at least four satellites are needed for a 3-D fix.
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