Visual walkthrough — GPS — pseudorange, trilateration, dilution of precision

3.5.18 · D2 · Physics › Guidance, Navigation & Control (GNC) › GPS — pseudorange, trilateration, dilution of precision

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Step 1 — Ek signal jo timestamp carry karta hai

PICTURE. Neeche diye figure mein, upar ka black dot satellite hai. Red line woh radio signal hai jo receiver (bottom pe black dot) tak travel kar raha hai. Do clocks draw ki gayi hain: satellite ki clock send-time read karti hai, aapki clock catch-time read karti hai.

KYUN. Distance speed times time hota hai. Agar hum exactly jaante ki trip kitni der mein hui, toh hum distance instantly jaante:

$$\underset{\text{metres}}{\underbrace{\text{range}}}=\underset{\text{speed of light}}{\underbrace{c}}\times \underset{\text{travel time}}{\underbrace{(t_{rx}-t_{tx})}}$$

• $t_{tx}$ — transmit time, satellite ki atomic clock se stamped (extremely accurate).
• $t_{rx}$ — receive time, aapki clock se padha hua.
• $c$ — speed of light, $\approx 3\times 10^8$ m/s, "seconds of travel" aur "metres of distance" ke beech fixed conversion.

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Step 2 — Aapki clock jhooth bolti hai, isliye range ek "pseudo" range hai

PICTURE. Red bar satellite tak ki true distance dikhata hai. Upar se laga hua extra black segment woh fake distance $cb$ hai jo purely clock error ki wajah se inject hui hai. Aap actually jo measure karte ho woh poora bar hai — true plus fake.

KYUN. Kyunki receiver ke paas ek clock hai, wahi same error har measurement ko corrupt karta hai. Ise likho:

$$\underset{\text{measured}}{\underbrace{{\rho }_i}}=\underset{\text{true distance}}{\underbrace{r_i}}+\underset{\text{clock error in metres}}{\underbrace{cb}}$$

• ${\rho }_i$ — satellite $i$ ka pseudorange: jo hum actually observe kar sakte hain.
• $r_i$ — true geometric distance jo hum chahte hain.
• $cb$ — clock bias $b$ (seconds) jo $c$ se metres mein convert kiya gaya. Har $i$ ke liye same.

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Step 3 — Ek range ek sphere hai

PICTURE. Red circle possible positions ka sphere hai (2D mein circle ki tarah draw kiya gaya). Satellite uske centre pe hai; tum rim pe kahin ho.

KYUN. 3D mein distance formula hi sphere ki equation hai:

$$r_i=\sqrt{(x-x_i{)}^2+(y-y_i{)}^2+(z-z_i{)}^2}$$

• $(x,y,z)$ — tumhari unknown position (jo hum chahte hain).
• $(x_i,y_i,z_i)$ — known satellite position, message mein broadcast kiya gaya, ECEF coordinates mein expressed.
• Square root teen coordinate differences ko ek straight-line distance mein convert karta hai (3D mein Pythagoras).

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Step 4 — Spheres stack karna: humein kitni chahiye?

PICTURE. Left: do circles do points pe cross kar rahi hain. Right: teen wali circle ek single point select kar rahi hai. Red dot located receiver hai.

KYUN count matter karta hai. Har sphere ek equation hai. Pure geometry mein 3 unknowns $x,y,z$ hain, toh 3 spheres kaafi hain. Lekin hum clock bhool gaye:

$${\rho }_i=\sqrt{(x-x_i{)}^2+(y-y_i{)}^2+(z-z_i{)}^2}+cb$$

Ab unknowns honestly gino: $x,y,z,$ aur $b$ — yeh 4 unknowns hain. Chaar unknowns ko 4 equations → 4 satellites chahiye. Chauthe satellite ka poora kaam clock error $b$ pin down karna hai.

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Step 5 — Square roots ugly hain, toh hum ek guess ke around linearize karte hain

PICTURE. Red curve true sphere hai. Straight red-dashed line guess point ${\mathbf{x}}_0$ pe uska tangent hai. ${\mathbf{x}}_0$ ke paas tangent aur curve almost same hain — yeh linearizing ka poora idea hai.

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Step 6 — Line-of-sight unit vector, draw kiya gaya

PICTURE. Red arrow unit line-of-sight vector ${\mathbf{e}}_i=(e_{ix},e_{iy},e_{iz})$ hai. Uske components axes pe uske shadows hain. Uski length exactly 1 hai.

KYUN length 1? Kyunki humne $r_i$ se divide kiya. Pythagoras se check:

$$e_{ix}^2+e_{iy}^2+e_{iz}^2=\frac{(x-x_i{)}^2+(y-y_i{)}^2+(z-z_i{)}^2}{r_i^2}=\frac{r_i^2}{r_i^2}=1$$

Toh ${\mathbf{e}}_i$ sirf direction carry karta hai, kabhi magnitude nahi. Yeh satellite $i$ ki steering hai.

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Step 7 — Har satellite ke liye ek linear equation

$$\underset{\text{measured}-\text{predicted}}{\underbrace{\Delta {\rho }_i}}\approx \underset{\text{position correction along line of sight}}{\underbrace{e_{ix}\Delta x+e_{iy}\Delta y+e_{iz}\Delta z}}+\underset{\text{clock correction}}{\underbrace{c\Delta b}}$$

• $\Delta {\rho }_i={\rho }_i^{\text{measured}}-{\rho }_i^{(0)}$ — residual: satellite $i$ ke liye hamara current guess kitna galat hai.
• $\Delta x,\Delta y,\Delta z$ — chhota position fix jo hum solve kar rahe hain.
• $\Delta b$ — chhota clock fix. Uska coefficient har satellite ke liye $c$ hai (shared-clock insight phir se).

PICTURE. Red residual bar woh gap hai jo hamara guess predict karta hai aur jo actually measure kiya gaya hai ke beech. Derivation ka kaam har aise bar ko zero karna hai.

KYUN. Yeh corrections mein linear equation hai — square root gone. Har satellite ek aisi line contribute karta hai. Inhe stack karo.

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Step 8 — Geometry matrix $G$ aur least-squares solve

$$\underset{\text{residuals}}{\underbrace{\Delta \mathbf{\rho }}}=\underset{\text{geometry}}{\underbrace{G}}\underset{\text{correction}}{\underbrace{\Delta \mathbf{x}}},G=\left[\begin{array}{cccc}{e}_{1x}& e_{1y}& e_{1z}& 1\\ e_{2x}& e_{2y}& e_{2z}& 1\\ ⋮& & & ⋮\\ e_{nx}& e_{ny}& e_{nz}& 1\end{array}\right]$$

• Har row = ek satellite = ek line-of-sight arrow + clock column of $1$.
• $1$'s ka last column $\partial \rho /\partial (cb)$ hai — "clock direction," sab ke liye identical.
• $\Delta \mathbf{x}=(\Delta x,\Delta y,\Delta z,c\Delta b{)}^{\top }$ — chaar corrections.

PICTURE. $G$ ki har row ek red arrow (ek satellite direction) ki tarah draw ki gayi hai jo apne $1$ ke saath baitha hai. Dur spread arrows matrix ko independent rows se bharte hain; bunched arrows near-duplicate rows banate hain.

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Step 9 — Degenerate case: satellites ek saath bunched

PICTURE. Left: arrows wide spread, spheres sharply cross karte hain — intersection ek crisp red point hai. Right: arrows bunched, spheres shallow angle pe graze karte hain — intersection ek fat red blob of uncertainty mein smear ho jaata hai.

KYUN. Parent note ke flat 2D toy mein, $\theta$ angle apart do directions $\det (G^{\top }G)={\sin }^2\theta$ deti hain. Jab $\theta \to 0$ (parallel), determinant $\to 0$ aur $(G^{\top }G{)}^{-1}$ explode karta hai. Woh amplification factor hi dilution of precision hai:

$${\sigma }_{\text{pos}}=\text{PDOP}\times {\sigma }_{\text{UERE}}$$

• Wide angular spread ⇒ small determinant avoid hota hai ⇒ low DOP ⇒ good fix.
• Bunched satellites ⇒ near-singular $G$ ⇒ high DOP ⇒ poor fix — even with perfect measurements.

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Ek-picture summary

Yeh final drawing poora walk compress karta hai: satellites timestamped signals broadcast karte hain (Step 1); ek jhoothi clock range ko pseudorange banati hai (Step 2); har range ek sphere hai (Step 3); chaar spheres position aur clock pin karte hain (Step 4); hum curves ko derivatives ke through tangents se replace karte hain (Steps 5–6), unhe $\Delta \mathbf{\rho }=G\Delta \mathbf{x}$ mein stack karte hain (Steps 7–8), least-squares-solve karte hain, aur geometry se precision read off karte hain (Step 9).

Recall Feynman retelling — plain words mein wapas bolo

Ek GPS satellite chillata hai "time hai abhi aur main hoon yahan." Tum use ek pal baad suno, toh delay se tum satellite tak apni distance jaanoge — except tumhari watch sasti hai aur har satellite ke liye same amount se galat hai. Ek satellite se fixed distance pe hona matlab tum uske around ek badi invisible ball pe kahin ho. Do balls ek ring mein cross karti hain, teen ek point mein — lekin galat watch ek chautha mystery number add karti hai, toh use bhi solve karne ke liye chautha ball chahiye. Distances ugly square roots ki tarah aate hain, toh tum guess karo ki tum kahan ho, pretend karo ki har ball wahan flat hai (yahi derivative tumhe deta hai — har satellite se tumhari taraf point karta ek arrow), aur har satellite tumhe ek easy straight-line equation deta hai. Un arrows ko ek table $G$ mein bundle karo, aur thodi least-squares algebra se pata chalta hai apna guess kaise nudge karna hai. Do baar repeat karo aur tum move karna band kar dete ho — yeh hai tumhara fix. Last trick: agar satellites sab sky ke ek kone mein huddle hain, unke arrows almost overlap karte hain, balls sirf ek doosre ko graze karte hain, aur ek tiny timing error tumhari position ko ek wide blur pe smear kar deta hai. Unhe spread karo aur balls cleanly cut karte hain. Woh spread-vs-huddle amplification dilution of precision hai.

Recall Quick self-test

Exactly 4 satellites kyun chahiye, 3 nahi? ::: Kyunki 4 unknowns hain — position $x,y,z$ aur clock bias $b$ — aur har satellite ek equation deta hai. Unit vector ${\mathbf{e}}_i$ kis direction mein point karta hai? ::: Satellite $i$ se seedha receiver ki taraf (line of sight); uski length 1 hai. Residual $\Delta {\rho }_i$ kya hai? ::: Measured pseudorange minus current guess se predicted pseudorange. $G$ mein 1's ka column kyun appear hota hai? ::: Yeh $\partial \rho /\partial (cb)$ hai, clock direction, har satellite ke liye identical. DOP bada kya banata hai? ::: Satellites bunched together ⇒ near-parallel line-of-sight arrows ⇒ near-singular $G^{\top }G$ ⇒ amplified error.