GPS — pseudorange, trilateration, dilution of precision: Question bank

3.5.18 · D5 · Physics › Guidance, Navigation & Control (GNC) › GPS — pseudorange, trilateration, dilution of precision

Recall Is page par use hone wale symbols (safety ke liye reveal karo)

Yahan koi bhi notation use nahi hui jo parent ne build nahi ki. Quick reminders:

$ρ_{i}$ = satellite $i$ tak pseudorange; $r_{i}$ = true geometric distance; $b$ = receiver clock bias seconds mein; $c$ = speed of light; $σ_{UERE}$ = per-satellite range-error ka size.
Line-of-sight unit vector $e_{i} = (e_{i x}, e_{i y}, e_{i z}) = \nabla_{x} r_{i}$ : range ka gradient receiver position ke respect mein. Sign convention: kyunki $r_{i}$ badhta hai jab receiver satellite $i$ se door jaata hai, $e_{i}$ satellite se receiver ki taraf point karta hai. Iska length 1 hota hai.
Geometry matrix $G$ : ek $n \times 4$ matrix (har satellite ke liye ek row, $n \geq 4$ ). Har row hai $[e_{i x} e_{i y} e_{i z} 1]$ — teen direction components plus ek clock-bias column of 1's. Woh aakhri $1$ hai $\partial ρ_{i} / \partial (c b)$ , jo har satellite ke liye identical hai.
DOP = dilution of precision, ek geometry-only error multiplier. Named flavours $Q = (G^{⊤} G)^{- 1}$ ko split karte hain: PDOP (Position) teen position diagonals use karta hai, GDOP (Geometric) clock diagonal add karta hai, HDOP/VDOP (Horizontal/Vertical) position ko ground aur height mein split karte hain, TDOP (Time) sirf clock hai.
$q_{xx}, q_{y y}, q_{z z}, q_{tt}$ = $Q = (G^{⊤} G)^{- 1}$ ke diagonal entries (position variances aur clock variance, per unit input variance). Jaise $PDOP = q_{xx} + q_{y y} + q_{z z}$ , $GDOP = q_{xx} + q_{y y} + q_{z z} + q_{tt}$ .

True or false — justify

A pseudorange hamesha true range se kam se kam utni badi hoti hai.

False. Bias term

c b

negative ho sakta hai (slow clock), jisse

ρ_{i} = r_{i} + c b

chhota ho jaata hai

r_{i}

se; pseudorange physically absurd bhi nikal sakti hai, aur ye theek hai kyunki

b

baad mein solve hota hai.

Clock-bias correction

c b

har satellite ke liye alag number hoti hai.

False. Ek hi receiver clock hoti hai, isliye wahi

c b

har pseudorange ko identically corrupt karta hai — yahi shared-ness hai jo hume

b

ke liye ek extra equation se solve karne deti hai.

Agar receiver clock ek perfect atomic clock hoti, toh 3 satellites kaafi hote.

True. Jab

b

zero jaana jaata hai, toh sirf 3 unknowns hain

(x, y, z)

, isliye 3 true-range spheres point ko pin kar dete hain — 4th satellite sirf clock error compute karne ke liye exist karta hai.

DOP is par depend karta hai ki pseudorange measurements kitne noisy hain.

False. DOP

(G^{⊤} G)^{- 1}

se banta hai, jisme sirf satellite directions hoti hain. Measurement noise poori tarah

σ_{UERE}

mein rehti hai; final error unka product hai

σ_{pos} = PDOP \times σ_{UERE}

Lower DOP hamesha better hota hai.

True. DOP tumhare range error par ek multiplier hai, isliye chhota matlab kam amplification; achhi spread-out geometry se PDOP ≈ 1–2 milta hai, clustered satellites se 10+ milta hai.

5th satellite add karna geometry ko sirf help kar sakta hai, kabhi hurt nahi.

True (geometry ke liye).

G

mein extra rows

(G^{⊤} G)^{- 1}

ke entries ko sirf shrink kar sakti hain, isliye DOP badh nahi sakta; zyada independent directions intersection ko sharpen karte hain.

Line-of-sight unit vector

e_{i}

satellite se receiver ki taraf point karta hai.

True. Ye

\nabla_{x} r_{i}

hai, increasing range ki direction jab receiver move karta hai; kyunki range badhti hai jab tum satellite se door jaate ho, woh direction satellite se receiver ki taraf jaati hai, aur iska unit length hoti hai.

GDOP hamesha PDOP se bada ya barabar hota hai.

True.

GDOP = q_{xx} + q_{y y} + q_{z z} + q_{tt}

mein clock term

q_{tt} \geq 0

shaamil hai jo PDOP drop karta hai, isliye ye sirf equal ya bada ho sakta hai.

Spot the error

"Pseudorange satellite tak ki distance ke barabar hai, isliye main seedha usse position read kar sakta hoon."

Pseudo word hi warning hai:

ρ_{i} = r_{i} + c b

mein ek unknown clock offset hai jo hundreds of metres ka hota hai, isliye ye true distance nahi hai jab tak

b

estimate na ho jaaye.

"Hum GPS equations ko linearize karte hain, isliye answer sirf ek rough approximation hai."

Linearization iteratively use hoti hai: har step ek better guess par re-linearize karta hai aur

Δ x \to 0

hota jaata hai, exact nonlinear solution par converge karta hai — approximation per-step hai, final nahi.

"DOP ke naam mein 'precision' hai, isliye achha DOP matlab receiver ne accurately measure kiya."

DOP measure karta hai ki geometry kisi bhi existing error ko kitna amplify karta hai; ye measurement quality ke baare mein kuch nahi kehta. Achha DOP with noisy receiver phir bhi poor fix deta hai.

"Ek clock jo fast chalti hai woh computed range ko too short banati hai."

Ulta hai. Fast clock arrival time

t_{r x}

ko zyada bada read karaati hai, isliye

c (t_{r x} - t_{t x})

overestimate hoti hai — range too long nikalta hai, aur tum

c b

subtract karte ho ise theek karne ke liye.

"Geometry improve karne ke liye, saare satellites seedha overhead cluster kar do."

Clustering se line-of-sight vectors almost parallel ho jaate hain,

det (G^{⊤} G) \to 0

aur DOP badh jaata hai. Tumhe wide angular spread chahiye — kuch overhead, kuch horizon ke paas alag-alag directions mein.

G

mein 1's ka column sirf ek bias/padding trick hai jiska koi physical meaning nahi."

Woh column

\partial ρ_{i} / \partial (c b)

hai — clock direction. Ye encode karta hai ki bias sabhi satellites ko equally affect karta hai, aur precisely isi wajah se

b

recoverable hai.

Cov (Δ x) = (G^{⊤} G)^{- 1} G^{⊤} (σ_{UERE}^{2} I) G (G^{⊤} G)^{- 1}

se shuru karke, do geometry factors cancel ho jaate hain aur

σ_{UERE}^{2} I

bacha rehta hai."

Nahi. Middle simplify hota hai as

(G^{⊤} G)^{- 1} G^{⊤} G (G^{⊤} G)^{- 1} = (G^{⊤} G)^{- 1}

, isliye result hai

σ_{UERE}^{2} (G^{⊤} G)^{- 1}

— identity nahi. Geometry matrix survive karta hai, jo ki DOP ka poora point hai.

Why questions

GPS ko fourth satellite kyun chahiye jab 3 spheres pehle se 3D mein ek point fix kar dete hain?

Kyunki receiver clock ek fourth unknown

b

add karta hai; 4 unknowns ke saath tumhe 4 equations chahiye, aur extra wale ka kaam literally clock error compute karna hai.

Example A mein har satellite se wahi

c b

kyun subtract hota hai?

Ek physical clock ek bias produce karta hai, isliye har pseudorange identical amount se corrupted hoti hai — shared-bias insight concretely dikhaya gaya.

Clock bias

b

ko distances se compare karne se pehle

c

se kyun multiply karte hain?

b

ek time hai; speed of light

c

se multiply karne par seconds metres mein convert ho jaate hain taaki ye

ρ_{i}

aur

r_{i}

ke saath same units mein rahe.

Near-zero

det (G^{⊤} G)

bad geometry kyun signal karta hai?

Tiny determinant matlab line-of-sight vectors almost linearly dependent hain (almost parallel), isliye unke intersecting spheres ek glancing angle par cut karte hain aur chhoti errors ek bade region par smear ho jaati hain — DOP blow up ho jaata hai.

DOP range error ko add kyun nahi karta, multiply kyun karta hai?

Range error se position error tak ka mapping

(G^{⊤} G)^{- 1} G^{⊤}

ke through linear hai, isliye covariance propagation input error ko ek geometry factor se scale karta hai — ek multiplicative gain, isliye

σ_{pos} = PDOP \times σ_{UERE}

Least-squares solution

(G^{⊤} G)^{- 1} G^{⊤}

kyun use karta hai instead of simply

G^{- 1}

n > 4

satellites ke saath

G

tall aur non-square hai, isliye iska koi ordinary inverse nahi hai; pseudoinverse

(G^{⊤} G)^{- 1} G^{⊤}

sabhi measurements ka best least-squares fit deta hai.

Hum

(G^{⊤} G)^{- 1}

form kar sakte hain —

G

ke baare mein kya true hona chahiye?

G^{⊤} G

invertible hoga sirf tab jab $G$ ka full column rank 4 ho, yaani kam se kam char satellites jिनके line-of-sight directions coplanar/degenerate nahi hain. Agar rank 4 se neeche gira, toh

det (G^{⊤} G) = 0

aur inverse — aur hence DOP — undefined hai.

Edge cases

Agar saare satellites sky ke ek hi taraf hों, koi overhead nahi, toh fix ka kya hoga?

Vertical geometry weak ho jaati hai: VDOP badh jaata hai isliye height estimate bahut uncertain ho jaata hai, jabki horizontal position theek reh sakti hai.

Agar sirf 3 satellites visible hain — kya position kabhi possible hai?

Sirf tab jab missing 4th constraint externally supply ki jaaye (jaise WGS84 par known altitude, ya external clock). Warna

G

ke sirf 3 rows hain, column rank 4 tak nahi pahunch sakta, aur system 4 unknowns aur 3 equations ke saath underdetermined hai.

Agar ek satellite receiver ke exactly zenith (seedha upar) par ho toh kya hoga?

Iska line-of-sight vector almost vertical hoga, VDOP mein strongly contribute karega lekin horizontal geometry mein weakly — height ke liye great, east/west ke liye poor, isliye tumhe abhi bhi spread-out companions chahiye.

Jab do satellite directions coincide ho jaayein toh DOP kya karta hai?

G

ki do rows identical ho jaati hain, isliye

G

full column rank kho deta hai,

det (G^{⊤} G) \to 0

, aur

(G^{⊤} G)^{- 1}

diverge karta hai — DOP → ∞ aur fix unresolved direction mein undefined hai.

Agar initial guess

x_{0}

bahut door ho toh iterative solver ka kya hoga?

First-order Taylor step inaccurate hogi, isliye shuruaati

Δ x

overshoot kar sakta hai, lekin har naye point par re-linearize karna phir bhi converge karta hai kyunki GPS geometry Earth's surface ke paas well-behaved hai.

Agar true clock bias exactly zero ho, toh kya 4th satellite useless ho jaati hai?

Nahi. Tumhe advance mein

b = 0

pata nahi hota; 4th equation hi solver ko discover karaati hai ki

b \approx 0

hai, aur iska extra geometry bhi DOP ko lower karta hai.

Hypothetical limit

σ_{UERE} \to 0

with DOP → ∞ mein position error kya hai?

Product

DOP \times σ_{UERE}

ek indeterminate

\infty \cdot 0

hai, isliye tum ise zero nahi keh sakte — limit carefully leni hogi. Practice mein noiseless ranges kabhi nahi hote, aur koi bhi real singular geometry (rank

< 4

) fix ko unresolved direction mein genuinely undefined chhod deti hai.

Agar true clock bias exactly zero ho, toh kya 4th satellite ka clock column redundant hai?

Nahi. Clock column

G

mein rehta hai regardless of bias value; ye encode karta hai ki solver

b

ko unknown treat karta hai. Ise remove karna ek needed column drop kar dega aur rank-4 requirement todh dega.