GPS — pseudorange, trilateration, dilution of precision

1. WHY does GPS need "pseudo"-ranges?

WHY the $c\,b$ term is the same for every satellite: the receiver has one clock. Whatever it is fast/slow by, it corrupts all measurements identically. That single shared unknown is the whole trick — it makes $b$ solvable.

2. Trilateration — deriving the position from scratch

The unknowns

We have 4 unknowns: $x, y, z$ (position) and $b$ (clock bias). So we need 4 equations → 4 satellites.

\rho_i = \sqrt{(x-x_i)^2+(y-y_i)^2+(z-z_i)^2} + c\,b, \qquad i=1,2,3,4

HOW to solve: linearize about a guess (this is what real receivers do)

Formula Deriving the linearized GPS equation

WHAT: turn nasty square-root equations into a linear system by Taylor-expanding around a guess $\mathbf{x}_0=(x_0,y_0,z_0,b_0)$ .

Step 1 — true range as a function. Write $r_i(\mathbf{x})=\sqrt{(x-x_i)^2+(y-y_i)^2+(z-z_i)^2}$ . Why? We need its derivatives to linearize.

Step 2 — the gradient. Differentiate: $\frac{\partial r_i}{\partial x}=\frac{x-x_i}{r_i}\equiv e_{ix}$ and similarly $e_{iy},e_{iz}$ . Why? $(e_{ix},e_{iy},e_{iz})$ is exactly the unit vector from satellite $i$ to the receiver — the "line-of-sight direction."

Step 3 — first-order Taylor. Near the guess, $\rho_i \approx \rho_i^{(0)} + e_{ix}\,\Delta x + e_{iy}\,\Delta y + e_{iz}\,\Delta z + c\,\Delta b$ where $\rho_i^{(0)}$ is the pseudorange predicted by the guess. Why the $c\,\Delta b$ ? Because $\partial \rho_i/\partial b = c$ for every satellite.

Step 4 — the residual. Let $\Delta\rho_i = \rho_i^{\text{measured}}-\rho_i^{(0)}$ . Stack all satellites: $\boxed{\;\Delta\boldsymbol\rho = G\,\Delta\mathbf{x}\;}$ with the geometry matrix $G=\begin{bmatrix} e_{1x}&e_{1y}&e_{1z}&1\\ e_{2x}&e_{2y}&e_{2z}&1\\ \vdots&&&\vdots\\ e_{nx}&e_{ny}&e_{nz}&1\end{bmatrix}.$ Why the column of 1's? That's $\partial\rho/\partial(cb)$ — the clock direction.

Step 5 — solve & iterate. With $n\ge 4$ satellites, least squares gives $\Delta\mathbf{x}=(G^\top G)^{-1}G^\top\,\Delta\boldsymbol\rho,$ update $\mathbf{x}_0\!\leftarrow\!\mathbf{x}_0+\Delta\mathbf{x}$ , and repeat until $\Delta\mathbf{x}\to 0$ . Why iterate? The problem is nonlinear; each iteration re-linearizes at a better point.

Figure — GPS — pseudorange, trilateration, dilution of precision

3. Dilution of Precision (DOP) — how geometry amplifies errors

Formula Deriving DOP

WHAT: relate the output position error to the input pseudorange measurement error.

Assume each pseudorange has independent noise of standard deviation $\sigma_{\text{UERE}}$ (User-Equivalent Range Error), so covariance of $\Delta\boldsymbol\rho$ is $\sigma_{\text{UERE}}^2 I$ .

The least-squares solution is linear: $\Delta\mathbf{x}=(G^\top G)^{-1}G^\top\Delta\boldsymbol\rho$ . Propagate covariance ( $\mathrm{Cov}(A\mathbf y)=A\,\mathrm{Cov}(\mathbf y)\,A^\top$ ): $\mathrm{Cov}(\Delta\mathbf{x}) = (G^\top G)^{-1}G^\top\,(\sigma_{\text{UERE}}^2 I)\,G(G^\top G)^{-1} = \sigma_{\text{UERE}}^2\,(G^\top G)^{-1}.$ Why does $G^\top G(G^\top G)^{-1}$ collapse? Because $(G^\top G)^{-1}G^\top G (G^\top G)^{-1}=(G^\top G)^{-1}$ .

Define $Q=(G^\top G)^{-1}$ with diagonal $(q_{xx},q_{yy},q_{zz},q_{tt})$ . Then: $\text{GDOP}=\sqrt{q_{xx}+q_{yy}+q_{zz}+q_{tt}},\quad \text{PDOP}=\sqrt{q_{xx}+q_{yy}+q_{zz}}$ and the position error is $\boxed{\;\sigma_{\text{pos}} = \text{PDOP}\times\sigma_{\text{UERE}}\;}$

4. Worked examples

Worked example Example A — pseudorange to range (removing the clock)

Two satellites give pseudoranges $\rho_1 = 20{,}200$ km and $\rho_2 = 21{,}000$ km. The receiver clock is later found to be fast by $1\,\mu\text{s}$ ( $b=+10^{-6}$ s). Find the true ranges.

Clock error in metres: $c\,b = (3\times10^8)(10^{-6}) = 300$ m. Why? $b$ is a time; multiply by $c$ to get distance.
$r_1 = \rho_1 - cb = 20{,}200{,}000 - 300 = 20{,}199{,}700$ m. Why subtract? A fast clock makes $t_{rx}$ read too large ⇒ overestimated distance ⇒ remove the surplus.
$r_2 = 21{,}000{,}000 - 300 = 20{,}999{,}700$ m. Note: the same 300 m is removed from both — the shared-bias insight in action.

Worked example Example C — DOP from a simple geometry

Suppose (in a flat 2D toy) two satellites at unit line-of-sight directions $90^\circ$ apart give $G=\begin{bmatrix}1&0\\0&1\end{bmatrix},\quad G^\top G=I,\quad Q=I,\quad \text{DOP}=\sqrt{1+1}=\sqrt2\approx1.41.$ Now bring them to $10^\circ$ apart: directions $(1,0)$ and $(\cos10^\circ,\sin10^\circ)$ . $G^\top G=\begin{bmatrix}1+\cos^2 10^\circ & \cos10^\circ\sin10^\circ\\ \cdot & \sin^2 10^\circ\end{bmatrix}.$ $\det(G^\top G)=\sin^2 10^\circ\approx0.030$ ⇒ $Q$ entries blow up ⇒ DOP ≫ 1.41. Why this step? A near-zero determinant means the two line-of-sight vectors are almost parallel — the geometry is nearly singular, so errors are hugely amplified. Wide angular spread = small DOP = good.

5. Common mistakes (steel-manned)

6. Flashcards

What is a pseudorange?

The measured satellite-to-receiver distance

c(t_{rx}-t_{tx})

, equal to the true range plus

c\,b

where

b

is the unknown receiver clock bias.

Why does GPS need a minimum of 4 satellites (not 3)?

There are 4 unknowns — 3 position coordinates and 1 clock bias — so 4 equations (satellites) are needed.

Why is the receiver clock bias the same in every pseudorange?

There is a single receiver clock, so its offset

b

contaminates all measurements identically — making it solvable as one shared unknown.

What is the geometry matrix

G

made of?

Rows are the line-of-sight unit vectors from each satellite to the receiver, with a final column of 1's for the clock (

cb

) term.

State the covariance of the position solution.

\mathrm{Cov}(\Delta\mathbf x)=\sigma_{\text{UERE}}^2 (G^\top G)^{-1}

Define PDOP and the position-error relation.

PDOP

=\sqrt{q_{xx}+q_{yy}+q_{zz}}

from

Q=(G^\top G)^{-1}

; position error

\sigma_{pos}=\text{PDOP}\times\sigma_{\text{UERE}}

Does DOP depend on the measurements?

No — only on satellite geometry (

G

). Lower DOP = better geometry.

What geometry gives a bad (high) DOP?

Satellites clustered together / nearly collinear line-of-sight vectors, making

G^\top G

near-singular (small determinant).

Convert a

1\,\mu s

clock bias to a range error.

c\,b = 3\times10^8\times10^{-6}=300

Recall Feynman: explain it to a 12-year-old

Imagine you're blindfolded and friends far away shout "NOW!" at the same instant. You hear each shout a bit later because sound takes time — the farther the friend, the longer the delay. If you knew exactly how long each shout took, you could figure out where you're standing. GPS does this with radio "NOW!"s from satellites. One problem: your stopwatch is broken and always off by the same little amount. Luckily, if you listen to four friends instead of three, the math can figure out both where you are and how broken your stopwatch is at the same time. And if all four friends are bunched together in the same corner of the sky, your guess is fuzzy; if they're spread all around you, your guess is sharp. That spreading-out quality is called DOP.

Connections

Trilateration and Multilateration — geometric root of the method.
Least Squares Estimation — solves the over-determined $\Delta\boldsymbol\rho=G\Delta\mathbf x$ .
Kalman Filter in GNC — recursive successor that fuses pseudoranges over time.
Clock Bias and Atomic Clocks — why the 4th unknown exists.
Covariance Propagation — the machinery behind the DOP derivation.
Reference Frames — ECEF and WGS84 — coordinate system for $(x_i,y_i,z_i)$ .
Time of Flight and Ranging — the $c\Delta t$ physics.

Concept Map

Hinglish (regional understanding)

Intuition Hinglish mein samjho

Dekho, GPS ka basic funda simple hai: har satellite apne signal pe exact transmit time stamp kar deta hai (uske paas atomic clock hai). Receiver dekhta hai ki signal kab pahuncha, aur time difference ko $c$ se multiply karke distance nikaalta hai. Problem ye hai ki receiver ki clock sasti hai — thodi fast ya slow hoti hai. Is wajah se jo distance milta hai wo pura sach nahi hota, usme ek extra $c\,b$ term ghusa hota hai (yaha $b$ = clock bias). Isiliye ise pseudorange kehte hain, seedha "range" nahi.

Ab trilateration: ek satellite bolta hai "tum meri sphere pe kahin ho," teen satellites milkar ek point pin kar dete hain. Par yaha ek chhupa hua unknown hai — clock bias $b$ . Total unknowns 4 ho gaye: $x, y, z$ aur $b$ . Isiliye kam se kam 4 satellites chahiye. Chautha satellite ka kaam hi yahi hai ki clock error bhi solve ho jaaye. Bas 3 se kaam nahi chalega, chahe geometry class me kuch bhi padha ho — kyunki wahan clock ka concept nahi tha.

DOP (Dilution of Precision) ye batata hai ki satellite ki geometry tumhare error ko kitna badha degi. Socho agar saare satellites aasman ke ek hi kone me hain — unki spheres bahut halke angle pe cut karti hain, aur chhoti si measurement galti bhi position ko bahut faila deti hai. Agar satellites chaaro taraf spread hain, toh spheres tez angle pe cut karti hain aur point sharp banta hai. Formula: position error $=$ PDOP $\times \sigma_{\text{UERE}}$ . Yaad rakho — DOP sirf geometry pe depend karta hai, measurement quality pe nahi. Isiliye kam DOP hamesha better. Mantra: "4 se fix, spread se slick."

GPS — pseudorange, trilateration, dilution of precision

1. WHY does GPS need "pseudo"-ranges?

2. Trilateration — deriving the position from scratch

The unknowns

HOW to solve: linearize about a guess (this is what real receivers do)

3. Dilution of Precision (DOP) — how geometry amplifies errors

4. Worked examples

5. Common mistakes (steel-manned)

6. Flashcards

Connections

Concept Map

Hinglish (regional understanding)

Go deeper — visual, from zero

Test yourself — Guidance, Navigation & Control (GNC)

Connections