GNSS — GPS, GLONASS, Galileo, BeiDou: Worked examples

3.5.19 · D3 · Physics › Guidance, Navigation & Control (GNC) › GNSS — GPS, GLONASS, Galileo, BeiDou

Yeh page parent GNSS note ka drill page hai. Parent ne theory build ki; yahan hum har tarah ki situation walk karte hain jo ek GNSS problem mein aa sakti hai — chote numbers, bade numbers, zero cases, degenerate geometry, ek real-world word problem, aur ek exam twist. Kuch bhi naya assume nahi kiya gaya hai: neeche use kiya gaya har symbol parent note pe define hua tha, aur jaise-jaise aata hai hum usse re-anchor karte hain.

Scenario matrix

Neeche har cell mein kam se kam ek worked example hai. Har example mein label (jaise [Cell B2]) batata hai ki yeh kaun sa case cover karta hai.

#	Case class	Tricky kya hai	Example
A1	Time → distance, normal size	seedha "distance = c·t"	Ex 1
A2	Time → distance, tiny input	nanosecond ⇒ centimetres	Ex 2
B1	Clock bias in metres	1 unknown = saikdon metre	Ex 3
B2	Clock bias, zero case	perfect clocks ⇒ 3 sats	Ex 4
C1	Unknowns / equations count karna	kitne satellites chahiye?	Ex 4
D1	Geometry / DOP, good spread	low DOP, well-conditioned	Ex 5
D2	Geometry / DOP, degenerate	satellites bunched ⇒ singular	Ex 6
E1	Relativity drift (SR + GR)	net +38 μs/day, sign matters	Ex 7
F1	Real-world word problem	urban canyon, multi-constellation	Ex 8
G1	Exam twist / limiting value	signal at horizon, extra travel	Ex 9

Prerequisite links jo aap kholi rakh sakte ho: Trilateration and Multilateration, Dilution of Precision (DOP), Least Squares Estimation, Special Relativity — Time Dilation, General Relativity — Gravitational Time Dilation.

Poore time, yaad rakho yeh constants:

$c = 3 \times 1 0^{8} m/s$ — speed of light, hamara ruler. Travel time ka ek second = $3 \times 1 0^{8}$ metres distance.
$1 μ s = 1 0^{- 6} s$ (microsecond), $1 ns = 1 0^{- 9} s$ (nanosecond), $1 ms = 1 0^{- 3} s$ (millisecond).

Ex 1 — Time to distance, normal size [Cell A1]

Worked example Ek signal 67 ms mein arrive karta hai. Satellite kitni door hai?

Forecast: pehle guess karo — kya answer 100 km, 20,000 km, ya 400,000 km ke kareeb hoga? Likh lo.

Step 1. Time ko seconds mein convert karo: $67 ms = 67 \times 1 0^{- 3} s = 0.067 s$ . Yeh step kyun? Speed $c$ metres per second mein hai, isliye units cleanly cancel hone ke liye time seconds mein hona chahiye.

Step 2. Ruler $c$ se multiply karo: $ρ = c t = (3 \times 1 0^{8} \frac{m}{s}) (0.067 s) = 2.01 \times 1 0^{7} m .$ Yeh step kyun? Distance = speed × time. Yahan $ρ$ (pseudorange) woh raw distance hai jo timing imply karta hai.

Step 3. Km mein padho: $2.01 \times 1 0^{7} m = 20, 100 km$ . Yeh step kyun? Is scale pe metres feel karna mushkil hai; 1000 se divide karke kilometres mein laane se hum known orbit altitudes se compare karke answer sanity-check kar sakte hain.

Verify: seconds cancel ho gaye: $\frac{m}{s} \cdot s = m$ ✓. Aur $20, 100$ km bilkul MEO mein baithta hai (~20,200 km GPS ke liye) — answer physical sanity check pass karta hai.

Ex 2 — Time to distance, tiny input [Cell A2]

Worked example Receiver arrival time ko 1 ns precision se measure karta hai. Woh ek nanosecond kitne metres ki position uncertainty represent karta hai?

Forecast: ek nanosecond invisible lagta hai. Guess karo: millimetres, centimetres, ya metres?

Step 1. Convert karo: $1 ns = 1 0^{- 9} s$ . Yeh step kyun? Same unit rule jaise Ex 1 — ruler seconds mein kaam karta hai.

Step 2. $Δ ρ = c \cdot Δ t = (3 \times 1 0^{8}) (1 0^{- 9}) = 0.30 m = 30 cm .$ Yeh step kyun? Distance mein error bilkul usi factor $c$ se time mein error ke saath scale hoti hai. Yeh famous "light 30 cm per nanosecond travel karta hai" fact hai.

Verify: units $\frac{m}{s} \cdot s = m$ ✓. Isliye GNSS engineers timing ke baare mein itna obsess karte hain: ek nanosecond bachao, 30 cm bachao. Yeh bhi explain karta hai ki 38 μs/day relativistic drift (Ex 7) catastrophic kyun hai — 38 μs matlab 38,000 ns hai.

Ex 3 — Clock bias in metres [Cell B1]

Shuru karne se pehle, parent ke observation equation $ρ_{i} = r_{i} + c b + ε_{i}$ se borrowed do symbols re-anchor karte hain:

$r_{i}$ = true geometric range — aapki position se satellite $i$ tak ki honest straight-line distance, metres mein. Yeh woh hai jo aap perfect clocks ke saath measure karte.
$ε_{i}$ = satellite $i$ ke liye residual measurement noise (metres mein) — atmosphere se signal bend hone, electronic noise, etc. se bachne wali leftover errors. Choti aur random, lekin kabhi exactly zero nahi.

Worked example Receiver ki sasti quartz clock true time se

$b = 1 μ s$ aage hai. Yeh har pseudorange mein kitni fake distance inject karti hai? Forecast: guess karo — decimetres, metres, ya saikdon metres?

Step 1. Observation equation yaad karo: $ρ_{i} = r_{i} + c b + ε_{i}$ , jahan (upar re-anchor kiya) $r_{i}$ true range hai aur $ε_{i}$ chota noise hai. Bias term $c b$ hai. Yeh step kyun? Bias $b$ ek time error hai; $c$ se multiply karne pe yeh woh distance error ban jaata hai jise yeh fake karta hai. Yeh har satellite mein equally add hota hai — precisely yahi cheez hume isse 4th unknown ke roop mein solve karne deti hai.

Step 2. $c b = (3 \times 1 0^{8}) (1 \times 1 0^{- 6}) = 300 m .$ Yeh step kyun? Numbers seedha $c b$ mein plug karo.

Verify: units ✓. 300 m ek football-stadium-sized error hai sirf ek microsecond se. Isliye $b$ ko solve karna padta hai (4th satellite chahiye), assume zero nahi kar sakte — Ex 4 dekho. Ex 2 se compare karo: 1 μs = 1000 ns, aur $1000 \times 0.30 m = 300$ m ✓, internally consistent.

Ex 4 — Zero-bias case aur unknowns count karna [Cells B2, C1]

Worked example Maan lo aapke receiver mein ek

perfect atomic clock hai jo satellites ke clocks ke saath synchronized hai. Ab 3D fix ke liye kitne satellites chahiye — aur kyun? Forecast: 3, 4, ya phir bhi 4?

Step 1. Unknowns list karo. Generally yeh $x, y, z, b$ hain — chaar hain. Yeh step kyun? Har unknown ko pin karne ke liye ek equation (ek satellite) chahiye.

Step 2. Bias ko uski zero/known value pe set karo: $b = 0$ . Tab $c b = 0$ , aur observation equation $ρ_{i} = r_{i} + ε_{i}$ mein collapse ho jaata hai, ek pure sphere. Yeh step kyun? Ek perfect synchronized clock hidden unknown remove kar deta hai. Yeh problem ka degenerate "no-bias" corner hai.

Step 3. Jo bacha usse count karo: unknowns $= x, y, z = 3$ . Isliye 3 satellites (3 spheres ek point pe intersect karte hain). Yeh step kyun? Ek unknown remove karna ek required equation remove karta hai — pure dimension counting.

Verify: general case ke saath consistency check. Bias ke saath: $4$ unknowns ⇒ $4$ sats. Bias ke bina: $4 - 1 = 3$ unknowns ⇒ $3$ sats. Arithmetic $4 - 1 = 3$ ✓. (Reality mein koi consumer clock itna achha nahi hota, isliye hum hamesha 4 use karte hain — yeh example isolate karta hai kyun 4th ki zaroorat hai.)

Ex 5 — Achhi geometry, low DOP [Cell D1]

Figure neeche (s01): receiver origin pe slate square hai; chaar colored arrows chaar satellites (stars) ki taraf shoot karte hain jo sky mein wide spread hain — ek East ki taraf neeche, ek bilkul upar, do alag directions mein upar. Kyunki arrows genuinely alag directions mein point karte hain, yeh aapki position ko kai sides se brace karte hain (jaise ek achhe se lage tripod ki tarah). Yahi wide spread hai jo DOP ko small banata hai.

Figure — GNSS — GPS, GLONASS, Galileo, BeiDou

Worked example Do satellites unit line-of-sight directions

$\overset{u}{^}_{1} = (1, 0)$ aur $\overset{u}{^}_{2} = (0, 1)$ pe hain — matlab ek due East horizon pe, ek bilkul upar. Ise simplified 2D fix maanke (clock column ignore karke), geometry matrix banao aur position-error amplification estimate karo. Forecast: kya yeh geometry error bahut amplify karegi, thodi, ya unchanged chhod degi?

Step 1. Line-of-sight rows se geometry matrix $H$ banao. Har row ek satellite ki direction hai: $H = [1001] .$ Yeh step kyun? Parent note ne $\partial r_{i} / \partial x = (x - x_{i}) / r_{i}$ dikhaya, unit direction ke components. Perpendicular directions ek identity-jaisi matrix dete hain.

Step 2. $H^{T} H = [1001]$ compute karo, aur uska inverse same identity hai. Yeh step kyun? DOP $(H^{T} H)^{- 1}$ se banta hai — iska diagonal jitna chhota, measurement error utni kam amplify hogi.

Step 3. DOP $= tr ((H^{T} H)^{- 1}) = 1 + 1 = 2 \approx 1.41$ . Yeh step kyun? Yeh do perpendicular directions ke liye smallest possible value hai — geometry raw measurement noise ke upar almost kuch nahi add karta.

Verify: $2 \approx 1.414$ ✓. Toh $σ_{position} = 1.41 \times σ_{measurement}$ : 3 m ranging error ~4.2 m position error ban jaati hai. Ex 6 se compare karo, jahan same ranging error explode ho jaati hai.

Ex 6 — Degenerate geometry, singular matrix [Cell D2]

Figure neeche (s02): same receiver (slate square), lekin ab chaaon saare arrows ek narrow cone mein East horizon pe neeche point karte hain — satellites bunched hain. Arrows nearly parallel hain, isliye yeh sab aapki position ko ek hi side se brace karte hain aur us direction ke perpendicular almost koi leverage nahi dete (jaise ek tripod jiske teeno legs ek saath jame ho). Yahi near-parallel bunching hai jo DOP ko explode karati hai.

Worked example Ab dono satellites almost same direction mein hain:

$\overset{u}{^}_{1} = (1, 0)$ aur $\overset{u}{^}_{2} = (cos 1^{\circ}, sin 1^{\circ}) \approx (0.9998, 0.0175)$ . DOP ka kya hota hai? Forecast: kya DOP $1.41$ ke paas rahega, ya blow up karega?

Step 1. $H = [1 0.9998 0 0.0175]$ banao. Yeh step kyun? Do nearly-parallel line-of-sight rows — parent note ka "carpenter's legs too close together" case.

Step 2. Uska determinant $det H = (1) (0.0175) - (0) (0.9998) = 0.0175$ hai — tiny, nearly zero. Yeh step kyun? Near-zero determinant matlab $H$ nearly singular hai: ise invert karne pe us tiny number se divide karna padta hai, toh errors enormously amplify hoti hain.

Step 3. $(H^{T} H)^{- 1}$ aur uska trace compute karo. Numerically trace $\approx 6555$ hai, toh $DOP = tr ((H^{T} H)^{- 1}) \approx 6555 \approx 81.$ Yeh step kyun? Yeh amplification factor ko ~1.4 (Ex 5) se ~81 tak jump karte dikhata hai — same measurements se 57× worse position error.

Verify: limiting behaviour — jaise satellites ke beech angle $\to 0^{\circ}$ , $det H \to 0$ aur DOP $\to \infty$ . Inhe exactly parallel set karna ( $\overset{u}{^}_{2} = (1, 0)$ ) $H^{T} H$ ko truly singular (determinant 0) banata hai — koi unique fix exist nahi karta. Yeh "ek direction mein horizon ke paas chaar satellites hona bura hai" ka mathematical face hai. ✓

Ex 7 — Relativity drift, dono signs [Cell E1]

Worked example Ek din mein, special relativity satellite clock ko slow karti hai aur general relativity usse speed up karti hai. Standard fractional rates diye hue, dikhao ki net drift

$+ 38 μ s/day$ ke kareeb hai. Forecast: net effect ka sign guess karo — satellite clock aage hoga ya peeche?

Step 1. Special relativity (satellite move kar raha hai, isliye uski clock slow chalti hai — negative contribution). GPS satellite ke liye $v \approx 3874 m/s$ : $\frac{Δ t _{S R}}{t} = - \frac{v ^{2}}{2 c ^{2}} = - \frac{( 3874 ) ^{2}}{2 ( 3 \times 1 0 ^{8} ) ^{2}} \approx - 8.3 \times 1 0^{- 11} .$ Yeh step kyun? Yeh Special Relativity — Time Dilation ka time-dilation fraction hai: ek second mein kitna second lost hota hai. Negative = slower.

Step 2. General relativity (satellite upar hai, weaker gravity mein, isliye uski clock fast chalti hai — positive contribution): $\frac{Δ t _{GR}}{t} = + \frac{GM}{c ^{2}} (\frac{1}{r _{E}} - \frac{1}{r _{s a t}}) \approx + 5.3 \times 1 0^{- 10} .$ Yeh step kyun? General Relativity — Gravitational Time Dilation se: upar = faster clock. Yeh SR effect se bada hai, isliye net positive hai.

Step 3. Fractions add karo (yeh same clock pe act karte hain, isliye simply sum karte hain): $\frac{Δ t _{net}}{t} = - 8.3 \times 1 0^{- 11} + 5.3 \times 1 0^{- 10} \approx + 4.45 \times 1 0^{- 10} .$ Yeh step kyun? Signs matter karte hain: SR subtract karta hai, GR add karta hai. GR jeet jaata hai, isliye net rate positive hai — satellite clock time gain karta hai.

Step 4. Net rate ko ek din ke seconds ki sankhya, $t = 86400 s$ , se multiply karo, poore din mein drift accumulate karne ke liye: $Δ t_{net} = (4.45 \times 1 0^{- 10}) (86400 s) \approx 3.8 \times 1 0^{- 5} s .$ Yeh step kyun? Step 3 ne ek dimensionless rate diya (seconds gained per second). 24 ghante baad receiver jo actual drift dekhta hai woh paane ke liye, hum din ke seconds se multiply karte hain — yeh tiny rate ko ek concrete accumulated time error mein badalta hai.

Step 5. Seconds ko microseconds mein convert karo ( $1 μ s = 1 0^{- 6} s$ , toh $1 0^{6}$ se multiply karo): $3.8 \times 1 0^{- 5} s = 3.8 \times 1 0^{- 5} \times 1 0^{6} μ s = 38 μ s .$ Yeh step kyun? Seconds yahan feel karne ke liye too coarse hain; microseconds jaana-pehchana headline number +38 μs/day deta hai aur Ex 2 ke 30 cm/ns ruler se compare karne deta hai.

Verify: sign positive hai (clock fast chalta hai), magnitude $\approx 38 μ s/day$ ✓. Ex 2 ke factor se damage cross-check karo: $38 μ s = 38000 ns$ , aur $38000 \times 0.30 m \approx 11, 400 m \approx 11 km/day$ position error agar uncorrected chhoda gaya ✓. Isliye satellite oscillators ko ground pe thoda slow pre-tune kiya jaata hai.

Ex 8 — Real-world word problem, urban canyon [Cell F1]

Worked example Ek delivery drone downtown ke tall buildings ke "canyon" mein sirf upar ek narrow strip mein sky dekh sakta hai. GPS akele ke saath yeh

3 satellites dekhta hai (enough nahi — Ex 4 ko 4 chahiye). Woh multi-constellation on karta hai, GLONASS + Galileo + BeiDou add karta hai, aur ab usi strip mein 9 satellites dekhta hai. Do questions: (a) kya ab fix mil sakta hai, aur (b) kya geometry achhi hai? Forecast: kya "zyada satellites" automatically "good position" ka matlab hai?

Step 1. Fix ke liye count karo. Bias unknown ke saath humein $\geq 4$ satellites chahiye. GPS akele: $3 < 4$ ⇒ no fix. Multi-constellation: $9 \geq 4$ ⇒ fix possible. Yeh step kyun? Ex 4 ka 4-unknown counting rule ek hard gate hai — 4 se kam equations ke saath $x, y, z, b$ solve nahi ho sakta.

Step 2. Redundancy check karo. 4 se zyada satellites $9 - 4 = 5$ redundant measurements dete hain. Yeh step kyun? Redundancy Least Squares Estimation feed karta hai (noise ka better averaging) aur RAIM (faulty satellite pakadna) enable karta hai.

Step 3. Geometry check karo. Saare 9 ek narrow overhead strip mein hain — unki line-of-sight directions nearly parallel hain (Ex 6 jaisi). Isliye 9 satellites ke bawajood, DOP high hai aur horizontal accuracy kharab hai. Yeh step kyun? Yeh crucial nuance hai: satellite count solvability fix karta hai; satellite spread accuracy fix karta hai. Yeh alag problems hain.

Verify: answer (a) haan, fix ab possible hai ( $9 \geq 4$ ); answer (b) lekin geometry phir bhi buri hai (narrow strip ⇒ high DOP). Yeh exactly parent ki warning se match karta hai ki multi-constellation canyons mein DOP help karta hai lekin pure sky-blockage ko beat nahi kar sakta — aapko spread chahiye, sirf count nahi. Count check $9 - 4 = 5$ redundancy ✓.

Ex 9 — Exam twist: horizon pe extra travel [Cell G1, limiting value]

Worked example Exam twist. MEO altitude

$h = 20, 200 km$ par ek GPS satellite (a) bilkul upar vs (b) bilkul horizon par hai, Earth (radius $R_{E} = 6371 km$ ) par ek receiver se dekha gaya. Horizon signal kitna zyada travel karta hai, aur usse kitna extra time lagta hai? Forecast: extra distance guess karo — saikdon km, ya hazaron?

Step 1. Overhead distance sirf altitude hai: $d_{up} = h = 20, 200 km$ . Yeh step kyun? Seedha upar, receiver-to-satellite line exactly altitude hai — koi triangle nahi chahiye.

Step 2. Horizon distance Earth ke centre, receiver, aur satellite se bane right triangle pe Pythagoras use karta hai. Satellite ki Earth's centre se distance $r_{s a t} = R_{E} + h = 26, 571 km$ hai. Horizon par line of sight Earth ko tangent hai, isliye receiver par Earth ke radius se right angle banta hai: $d_{horiz} = r_{s a t}^{2} - R_{E}^{2} = 2657 1^{2} - 637 1^{2} km \approx 25, 795 km .$ Yeh step kyun? Horizon par line of sight Earth ko tangentially graze karti hai, receiver par right angle banati hai — classic tangent-line right triangle. Pythagoras teesri side deta hai.

Step 3. Extra distance: $Δ d = d_{horiz} - d_{up} \approx 25, 795 - 20, 200 = 5, 595 km .$ Yeh step kyun? Ek horizon satellite hazaron km zyada door hai — uska signal weak hai aur zyada atmosphere se guzarta hai (bada $ε_{i}$ ), isliye low-elevation satellites noisier hote hain.

Step 4. Extra travel time: $Δ t = \frac{Δ d}{c} = \frac{5.595 \times 1 0 ^{6} m}{3 \times 1 0 ^{8} m/s} \approx 0.0187 s = 18.7 ms .$ Yeh step kyun? Extra distance ko same ruler $c$ se time mein convert karo — yeh woh physical range hai jo yeh signals cover karte hain.

Verify: units: km ✓ throughout; $m / (m/s) = s$ ✓. Limiting sanity: jaise $h \to \infty$ , $d_{horiz} \to d_{up}$ aur $Δ d \to 0$ (bahut door ka satellite har direction se same distance lagta hai). Yahan $Δ d \approx 5595$ km ⇒ $Δ t \approx 18.7$ ms, Ex 1 ke 67 ms total travel ke same ballpark mein ✓.

Yeh poora drill complete hota hai: scenario matrix ka har cell (A1 se G1 tak) ab ek worked example ke saath hai. Aapne chaar knobs ko extremes pe push hote dekha hai — tiny (Ex 2) aur normal (Ex 1) time-to-distance, metres mein bias (Ex 3) aur uska zero case (Ex 4), achhi (Ex 5) aur degenerate (Ex 6) geometry, signed relativity drift (Ex 7), ek real-world multi-constellation word problem (Ex 8), aur horizon limiting-value twist (Ex 9). Exam pe ab koi bhi problem genuinely nayi shape ki nahi honi chahiye.

Recall Self-test (click to reveal)

Ek signal 80 ms mein arrive karta hai. Pseudorange km mein? ::: $c \cdot 0.080 = 2.4 \times 1 0^{7}$ m $= 24, 000$ km. 2 μs ka clock bias kitne metres inject karta hai? ::: $c \cdot 2 \times 1 0^{- 6} = 600$ m. Do satellites exactly same line-of-sight direction par — DOP kya hoga? ::: Infinite: $H^{T} H$ singular hai, koi unique fix nahi. Perfectly synchronized receiver clock ke saath 3D ke liye kitne satellites? ::: Teen — bias unknown chala gaya. Per day net relativistic satellite clock drift, aur uska sign? ::: Lagbhag $+ 38 μ s$ (fast chalta hai).