GPS Trilateration
GPS Trilateration
A GPS receiver determines position by measuring the travel time of signals from at least four satellites. Each time-of-arrival measurement gives a pseudorange — distance plus a clock error. Four equations in four unknowns (latitude, longitude, altitude, clock bias) yield the fix.
The geometric principle is trilateration: with three distances to known points, you localize to the intersection of three spheres. The fourth satellite is needed to solve for the receiver's clock bias, which otherwise would scale every distance.
Accuracy depends on DOP (dilution of precision): the geometry of the visible satellites. A wide spread gives low DOP and good accuracy; satellites clustered overhead give high DOP and poor fix.
$$\rho_i = \sqrt{(x - x_i)^2 + (y - y_i)^2 + (z - z_i)^2} + c\,\delta t$$
where $c$ is speed of light and $\delta t$ is the receiver clock bias (unknown).
If measurement noise has standard deviation $\sigma_\rho$, position error $\sigma_{\text{pos}} \approx \text{PDOP} \cdot \sigma_\rho$. Typical PDOP: 1.5–3 (good), > 6 (poor).
Why do we need at least four satellites for 3-D GPS, not three?
Three would suffice if the clock were perfect: three spheres intersect in at most two points. But the receiver clock has bias that adds a common distance offset to every measurement. Treating $\delta t$ as a fourth unknown means we need a fourth equation — the fourth satellite.
Three beacons at $A=(0,0)$, $B=(10,0)$, $C=(5, 8.66)$ km. Measured ranges: $r_A=7, r_B=7, r_C=5$ km. Find position.
From $A, B$ (symmetric): $x = 5$. From $A$: $25 + y^2 = 49 \Rightarrow y = \pm\sqrt{24} = \pm 4.899$.
Check with $C$: $|C - (5, 4.899)| = |(0, 3.76)| = 3.76$ — mismatch with $r_C=5$. Try $(5, -4.899)$: $|C-P| = |(0, 13.56)| = 13.56$ — worse.
Least-squares fit gives $(5, 4.3)$ approximately; the range set is slightly inconsistent (real GPS observations always are).
A GPS receiver clock is off by $1\,\mu\text{s}$. How much range error does that create?
$\Delta\rho = c \cdot \Delta t = 3\times 10^8 \cdot 10^{-6} = 300\,\text{m}$. Tiny clock errors matter a lot — this is why GPS needs nanosecond-level timing.
Practice Problems
Show Answer Key
1. Three sphere surfaces can fix 3-D geometry if the clock is perfect; the clock bias is a fourth unknown, needing a fourth satellite.
2. $c\cdot 10\times 10^{-9} = 3\,\text{m}$.
3. Horizontal dilution of precision — the horizontal-component geometry factor.
4. Ionospheric delay, tropospheric refraction, multipath, satellite ephemeris error.
5. A reference station broadcasts corrections that cancel correlated errors (ionosphere, clock) at nearby users.
6. At most two (or one if tangent, none if disjoint).