Visual walkthrough — GNSS — GPS, GLONASS, Galileo, BeiDou

3.5.19 · D2 · Physics › Guidance, Navigation & Control (GNC) › GNSS — GPS, GLONASS, Galileo, BeiDou

Yeh parent topic ka visual companion hai. Algebra wahan hai; yahan pictures derivation carry karti hain.

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Step 1 — Ek signal jo ek timestamp carry karta hai

KYA HAI. Ek satellite space mein ek chhota radio station hai. Woh exactly do kaam karta hai jo hamare liye matter karte hain: use pata hai woh kahan hai, aur woh ek message broadcast karta hai jo kehta hai "meri clock $t_{tx}$ read kar rahi hai abhi." Letter $t_{tx}$ ka matlab sirf hai time of transmit — woh instant jab signal ruwana hua.

KYUN. Ek radio message ko distance mein convert karne ke liye, humein ek stopwatch chahiye. Light (aur radio waves bhi light hain) hamesha same speed se travel karti hai, likha jaata hai $c\approx 3\times 10^8$ metres per second — yeh roughly dharti ke equator ki doori hai, saadhe saat baar, ek second mein. Kyunki $c$ kabhi nahi badlta, time-of-flight ek ruler hai: agar hum jaante hain signal kitni der air mein tha, toh hum jaante hain usne kitni door udaan bhari.

PICTURE. Upar wala satellite apna clock time ek wavefront par stamp karta hai aur usse uda deta hai. Green ring signal hai jo speed $c$ se bahar failti ja rahi hai.

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Step 2 — Receiver ka stopwatch: pseudorange

KYA HAI. Tum (receiver, unknown position par) signal pakad lete ho. Tumhari apni clock us waqt $t_{rx}$ read karti hai — time of receive. Elapsed time ko $c$ se multiply karo aur result ko ${\rho }_i$ kaho (Greek letter rho, satellite $i$ tak is measured distance ke liye humara naam):

$\underset{\text{measured range}}{\underbrace{{\rho }_i}}=\underset{\text{speed of light}}{\underbrace{c}}(\underset{\text{arrival par meri clock}}{\underbrace{t_{rx}}}-\underset{\text{launch par sat clock}}{\underbrace{t_{tx}}})$

KYUN. Distance = speed × time. Speed pata hai ($c$), time do stamps ke beech ka gap hai, toh distance nikal aati hai. Simple — siwaaye ek jhooth ke jo $t_{rx}$ mein chhupa hua hai: tumhari clock ek sasta quartz crystal hai, atomic clock nahi. Woh kisi unknown amount se galat hai. Toh ${\rho }_i$ true distance nahi hai; hum imaandaari se ise pseudorange kehte hain — "pseudo" = kuch jhootha.

PICTURE. Do clocks, ek orbit mein (atomic, sahi) aur ek tumhari pocket mein (drifting). Measured travel time timeline par do stamps ke beech ka horizontal gap hai.

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Step 3 — Ek measurement = ek sphere

KYA HAI. Ek second ke liye suppose karo ki tumhari clock perfect thi. Toh ${\rho }_i$ satellite $i$ tak ki true distance $r_i$ hai. Woh saare points jo ek fixed satellite se exactly $r_i$ door hain woh ek sphere banate hain $r_i$ radius ki, us satellite ko centre maan ke. Tum us sphere par kahin ho.

Radius plain 3D distance formula se aati hai (Pythagoras teen dimensions mein stretched):

$r_i=\sqrt{\underset{\text{east–west gap}}{\underbrace{(x-x_i{)}^2}}+\underset{\text{north–south gap}}{\underbrace{(y-y_i{)}^2}}+\underset{\text{up–down gap}}{\underbrace{(z-z_i{)}^2}}}$

Yahan $(x,y,z)$ ho tum (unknown) aur $(x_i,y_i,z_i)$ satellite hai (known, uske broadcast se).

KYUN. Ek number ($r_i$) tumhe teen numbers ($x,y,z$) nahi de sakta. Woh sirf possibilities ko "universe mein kahin bhi" se "is ek sphere par kahin bhi" tak shrink karta hai. Yeh sacchi progress hai par answer nahi. Humein aur spheres chahiye.

PICTURE. Ek satellite, ek shaded sphere. Tum shell par kahin bhi ho sakte ho.

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Step 4 — Teen spheres ek point pin karte hain (perfect clocks ke saath)

KYA HAI. Doosra satellite add karo. Do spheres ek circle mein overlap karti hain. Teesri add karo: teesri sphere us circle ko (generally) do points par slice karti hai, aur unme se ek bakwaas hai — woh space mein bahar hai ya galat direction mein ghoom raha hai — toh hum sensible wala rakhte hain. Teen spheres → ek point → tumhara $(x,y,z)$.

KYUN. Mystery count karo: teen unknowns $(x,y,z)$, toh teen independent facts (teen spheres) chahiye. Yeh perfect-clock answer hai — aur yeh classic galti ka source hai "3 satellites kaafi hain." Hote, agar tumhari clock atomic hoti. Nahi hai.

PICTURE. Sphere 1 ∩ Sphere 2 ek circle banati hai (teal). Sphere 3 us circle ko do orange dots par chhedti hai; real receiver filled wala hai.

Recall

Perfect clocks ke saath, 3D position ke liye kitne satellites chahiye? ::: Teen — ek per unknown $(x,y,z)$.

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Step 5 — Chhupa hua fourth unknown: clock bias $b$

KYA HAI. Tumhari clock off hai. Offset ko $b$ kaho (seconds): "meri clock true time se $b$ aage hai." Kyunki tumne travel time ko $b$ se zyada count kiya, har pseudorange $cb$ metres zyada lamba hai. Toh honest relationship hai:

$\boxed{{\rho }_i=\underset{\text{true distance}}{\underbrace{r_i}}+\underset{\text{har sphere par same error}}{\underbrace{cb}}}$

Note karo $cb$ saare satellites ke liye identical hai — ek clock, ek bias. Yahi woh key hai jo hume ise solve karne deti hai.

KYUN. Yeh error kitna bada hai? Ek tiny $1\mu \text{s}$ (ek millionth of a second) clock error $cb=(3\times 10^8)(10^{-6})=300$ metres ban jaata hai — poora ek football stadium. Hum $b$ ko ignore nahi kar sakte; hume ise fourth unknown ki tarah solve karna hoga. Chaar unknowns $(x,y,z,b)$ ko chaar facts chahiye → ek fourth satellite.

PICTURE. Teeno spheres same amount $cb$ se inflate hain; woh ek point par nahi milti balki ek fuzzy triangle par milti hain. Fourth sphere hume batati hai ki har sphere ko kitna deflate karna hai jab tak sab agree na kar lein — woh shared shrink $b$ reveal karta hai.

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Step 6 — Poori observation equation (chaar satellites)

KYA HAI. Steps 2–5 ko saath rakhho, ek baar per satellite $i=1,2,3,4$:

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

Chaar aisi equations, chaar unknowns $(x,y,z,b)$. Solve karo → position milti hai aur true time muft mein.

KYUN. Har satellite ek equation contribute karta hai aur chaar unknowns hain, toh chaar satellites honest minimum hai. Yeh parent note ka central result hai — ab iska har piece stopwatch aur Pythagoras se build kiya ja chuka hai.

PICTURE. Aasman mein chaar satellites, chaar line-of-sight arrows receiver par converge karte hue, unme se ek par poori equation labelled hai.

Recall

Teen ki jagah chaar satellites kyun? ::: Receiver ki clock ek fourth unknown $b$ add karti hai; chaar unknowns ko chaar equations chahiye.

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Step 7 — Ise solve karna: nishana lagao, miss measure karo, correct karo, repeat karo

KYA HAI. Square roots equations ko nonlinear banate hain — tum directly high-school elimination nahi kar sakte. Toh hum wahi karte hain jo ek dart player karta hai: andaaza lagao, kitna off hai dekho, thoda nudge karo, repeat karo.

1. Ek starting position ${\mathbf{x}}_0$ guess karo (Earth ka center bhi kaam karta hai).
2. Har satellite ke liye, tumhare guess se us satellite ki taraf ki direction ek unit vector ${\hat{\mathbf{u}}}_i$ hai — ek arrow length 1 ka jo satellite ki taraf point karta hai. Iske ingredients distance formula ke slope se aate hain: $\frac{\partial r_i}{\partial x}=\frac{x-x_i}{r_i}=-{\hat{u}}_{ix}$ Yeh kehta hai: "agar main ek metre east step karun, toh satellite $i$ tak meri distance kitni change hogi?" — aur answer pointing arrow ka east-component hai. Isliye ek derivative aata hai: yeh ek curved sphere ko ek flat local slope mein convert karta hai jise hum actually solve kar sakte hain.
3. Yeh arrows (plus shared clock term ke liye 1's ka ek column) ek table $\mathbf{H}$ mein stack karo, geometry matrix, aur ab-linear correction $\Delta \mathbf{x}=({\mathbf{H}}^{\mathsf{T}}\mathbf{H}{)}^{-1}{\mathbf{H}}^{\mathsf{T}}\Delta \mathbf{\rho }$ solve karo.
4. Correction apply karo, re-guess karo, repeat karo do ya teen baar jab tak woh rukna na band kar de.

KYUN. Linearizing ek unsolvable curved problem ko guess ke aas paas ek solvable straight-line problem mein turn karta hai; iterate karna us guess ko true answer ki taraf le jaata hai. 5+ satellites ke saath system overdetermined hota hai aur hum Least Squares Estimation use karte hain best fit point dhundhne ke liye — aur Kalman Filter yeh continuously time ke through karta rehta hai, Inertial Navigation System (INS) ke saath fused hokar.

PICTURE. Ek guess dot, ek correction arrow $\Delta \mathbf{x}$ true dot ki taraf point karta hua, aur residual "misses" $\Delta \rho$ har iteration ke baad shrink hote hue.

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Step 8 — Degenerate case: kharab geometry (DOP)

KYA HAI. Chaar satellites kaafi hain tabhi agar woh sky mein spread ho. Agar woh sab ek jagah huddle karein, unke pointing arrows almost parallel ho jaate hain, aur geometry matrix ${\mathbf{H}}^{\mathsf{T}}\mathbf{H}$ almost invert karna impossible ho jaata hai. Tab ek small range error ek huge position error mein blast ho jaata hai. Woh amplification Dilution of Precision hai.

${\sigma }_{\text{position}}=\text{DOP}\times {\sigma }_{\text{measurement}}$

Yahan $\sigma$ (sigma) ka matlab hai "error ka typical size." DOP ek pure multiplier hai: DOP = 1 ideal hai, DOP = 10 matlab tumhara ek metre ranging error das metre position error ban jaata hai.

KYUN. Same reason jis wajah se center mein bunched legs wali table hilti hai jabki corners par legs wali table rock-steady rehti hai. Widely separated lines-of-sight tumhare position estimate ko kai independent directions se brace karti hain. Yahi GPS + GLONASS + Galileo + BeiDou saath use karne ka asli fayda hai — zyada satellites, behtar spread, lower DOP, khaaskar urban canyons mein.

PICTURE. Left: chaar spread-out satellites, arrows fan out karte hue, tight small error box (accha). Right: chaar clustered satellites, near-parallel arrows, huge stretched error box (bura).

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Ek-picture summary

Upar sab kuch ek single frame mein compress kiya gaya: ek satellite ek time stamp karta hai (Step 1–2), woh ek sphere banta hai (Step 3), chaar inflated spheres plus ek shared clock error tumhare point par milti hain (Steps 4–6), aur satellites ka spread decide karta hai woh point kitna sharp hai (Step 8).

Recall Feynman retelling — poora walk plain words mein (click to open)

Tum ek bade andhare maidan mein kho gaye ho. Chaar doost tall, door-door towers par khade hain. Har doost exactly jaanta hai uski tower kahan hai, aur har ek perfect wristwatches use karke ek hi sahi instant par "ABHI!" chillata hai.

Tumhari apni watch thodi fast ya slow hai — tum nahi jaante kitni. Jab tum ek "ABHI!" sunte ho, tum apni watch par time note karte ho. Jo doost door hai woh lagta hai jaise usne baad mein chillaya, kyunki uski awaaz tumhak pahunchne mein zyada der lagi. Har delay ko distance mein convert karo (delay × speed of sound), aur har doost tumhe ek ring deta hai "tum mujhse itne door ho."

Teen rings ek jagah cross karti agar tumhari watch perfect hoti. Nahi hai — toh har ring thodi zyada badi draw hoti hai same amount se (tumhari watch error). Ek fourth doost tumhe fourth clue deta hai jo tumhe kahan ho aur tumhari watch exactly kitni galat hai dono ek saath figure out karne deta hai. Aur agar chaar doost sab tumhare ek taraf bunched hain, tumhara fix mushy hai; unhe charo taraf spread karo aur woh ek sharp point ban jaata hai.

"Sound" ki jagah "radio at the speed of light" aur "towers par doost" ki jagah "orbit mein satellites" rakh do, aur woh GNSS hai.

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Parent topic par wapas jaao · related: Trilateration and Multilateration, Special Relativity — Time Dilation, General Relativity — Gravitational Time Dilation, Orbital Mechanics — MEO/GEO/IGSO, Signal Modulation — CDMA & FDMA.