GPS — pseudorange, trilateration, dilution of precision: Exercises

3.5.18 · D4 · Physics › Guidance, Navigation & Control (GNC) › GPS — pseudorange, trilateration, dilution of precision

Source material ke liye parent dekho: the GPS topic note. Prerequisite ideas yahan hain: Trilateration and Multilateration, Least Squares Estimation, Time of Flight and Ranging, Clock Bias and Atomic Clocks, Covariance Propagation, aur Reference Frames — ECEF and WGS84.

Poore page mein, $c = 3 \times 1 0^{8} m/s$ speed of light hai, $ρ$ ek pseudorange hai (measured, clock-contaminated distance), $r$ ek true geometric range hai, aur $b$ receiver ka clock bias seconds mein hai.

Definition Do acronyms jo Level 2 se pehle zaroor yaad hone chahiye

UERE (User-Equivalent Range Error): single pseudorange measurement ke noise ki standard deviation $σ_{UERE}$ , metres mein express ki gayi. Yeh sab kuch jo ek range ko corrupt karta hai (atmospheric delay, satellite clock error, receiver noise) — sab ko ek number mein samait leta hai. Yeh measurements ki property hai, geometry ki nahi.
DOP (Dilution of Precision): ek pure-geometry multiplier jo satellite directions se banta hai. Yeh batata hai ki position error, range error ke relative kitna amplify hota hai. Family:
- PDOP (Position DOP) — 3D position ( $x, y, z$ ).
- HDOP (Horizontal DOP) — sirf do horizontal axes.
- VDOP (Vertical DOP) — sirf up/down axis.
- TDOP (Time DOP) — sirf clock bias axis.
- GDOP (Geometric DOP) — charon ek saath (position + clock). Is poore page mein use hone wala master relation: $σ_{pos} = DOP \times σ_{UERE} (geometry \times measurement noise) .$

Level 1 — Recognition

Exercise 1.1 — Contamination ko naam do

Ek satellite $t_{t x}$ par signal bhejta hai. Receiver apni sasti quartz clock use karke arrival $t_{r x}$ par padhta hai. Tum $c (t_{r x} - t_{t x})$ compute karte ho. Kya yeh true range $r$ hai ya pseudorange $ρ$ ? Kaunsi ek quantity inhe alag karti hai?

Recall Solution

Yeh pseudorange $ρ$ hai. Receiver clock ek unknown bias $b$ se off hai, isliye uska $t_{r x}$ har satellite ke liye utni hi maatra mein galat hai. Separator woh term $c b$ hai: $ρ = r + c b .$ True range $r$ honest geometric distance $(x - x_{i})^{2} + (y - y_{i})^{2} + (z - z_{i})^{2}$ hai; pseudorange usme clock error ko metres mein convert karke add karta hai.

Exercise 1.2 — Unknowns gino

Har woh unknown list karo jo ek GPS receiver ko solve karna hota hai, aur minimum kitne satellites chahiye yeh batao.

Recall Solution

Char unknowns: $x, y, z$ (position) aur $b$ (clock bias). Char unknowns ko char independent equations chahiye, isliye minimum 4 satellites chahiye. Teen sirf geometry pin karte agar clock perfect hoti — jo hoti nahi.

Exercise 1.3 — Clock bias ko distance mein convert karo

Ek receiver clock $b = 2 μ s$ fast hai. Yeh har pseudorange mein kitne metres ka error inject karta hai?

Recall Solution

$c b = (3 \times 1 0^{8}) (2 \times 1 0^{- 6}) = 600 m .$ Yeh tool kyun? $b$ ek time hai; distance = speed × time, isliye $c$ se multiply karne par clock error metres mein convert ho jaata hai jitna woh har measurement ko corrupt karta hai. Yahi $600$ m har satellite ke pseudorange par baithti hai.

Level 2 — Application

Exercise 2.1 — Clock bias hatao

Do satellites pseudoranges dete hain $ρ_{1} = 22, 000, 000 m$ aur $ρ_{2} = 20, 500, 000 m$ . Receiver clock baad mein $b = - 1.5 μ s$ slow paayi gayi. Do true ranges nikalo.

Recall Solution

Clock offset metres mein: $c b = (3 \times 1 0^{8}) (- 1.5 \times 1 0^{- 6}) = - 450 m$ . $r_{i} = ρ_{i} - c b = ρ_{i} - (- 450) = ρ_{i} + 450.$

$r_{1} = 22, 000, 000 + 450 = 22, 000, 450 m$ .
$r_{2} = 20, 500, 000 + 450 = 20, 500, 450 m$ . Yahan add kyun? Slow clock $t_{r x}$ ko chhota padhata hai ⇒ distance underestimate hoti hai ⇒ humein missing $450$ m wapas add karna hoga. Yahi correction dono par lagti hai — shared-bias insight.

Exercise 2.2 — 1D pseudorange fix

Ek number line par unknown position $x$ wala receiver, unknown bias $b$ ke saath, do satellites dekhta hai. Satellite A at $x_{A} = 0$ deta hai $ρ_{A} = c b + ∣ x - 0∣$ ; satellite B at $x_{B} = 100 m$ deta hai $ρ_{B} = c b + ∣ x - 100∣$ . Measured: $ρ_{A} = 340 m$ , $ρ_{B} = 280 m$ , aur receiver do satellites ke beech lie karta hai ( $0 < x < 100$ ). $x$ aur $c b$ nikalo.

Recall Solution

Satellites ke beech, $∣ x - 0∣ = x$ aur $∣ x - 100∣ = 100 - x$ . To: $ρ_{A} = c b + x = 340, ρ_{B} = c b + (100 - x) = 280.$ Subtract karo: $ρ_{A} - ρ_{B} = x - (100 - x) = 2 x - 100 = 60$ , se $x = 80 m$ . Back-substitute: $c b = 340 - 80 = 260 m$ , yaani $b = 260/ (3 \times 1 0^{8}) \approx 8.67 \times 1 0^{- 7} s$ . Do equations kyun? Do unknowns ( $x$ aur $b$ ) ko do satellites chahiye — "4 unknowns ke liye 4 satellites" ka 1D echo.

Exercise 2.3 — PDOP se position error

Ek receiver ka $PDOP = 2.5$ hai (yaad karo PDOP = Position Dilution of Precision, 3D position ke liye geometry multiplier) aur har pseudorange ka $σ_{UERE} = 4 m$ hai (UERE = User-Equivalent Range Error, per-measurement noise). Expected position error $σ_{pos}$ kya hai?

Recall Solution

$σ_{pos} = PDOP \times σ_{UERE} = 2.5 \times 4 = 10 m .$ Multiply kyun? DOP geometric amplification factor hai (satellite directions se); UERE raw per-measurement noise hai (signals se). Final error = geometry × noise, aur yeh independent factors hain.

Level 3 — Analysis

Exercise 3.1 — 2D geometry matrix banao

Ek flat 2D world mein ek receiver origin par baitha hai. Do satellites line-of-sight unit directions $e_{1} = (1, 0)$ aur $e_{2} = (cos 6 0^{\circ}, sin 6 0^{\circ}) = (0.5, 0.866)$ par hain. Is toy ke liye clock column ignore karo (2 unknowns: $x, y$ ). $G$ likho, $G^{⊤} G$ compute karo, aur DOP $= tr (G^{⊤} G)^{- 1}$ nikalo.

Do directions aur unke beech ka angle dekhne ke liye Figure s01 dekho.

Figure — GPS — pseudorange, trilateration, dilution of precision

Recall Solution

$G = [1 0.5 0 0.866], G^{⊤} G = [1^{2} + 0. 5^{2} \cdot 1 \cdot 0 + 0.5 \cdot 0.866 0^{2} + 0.86 6^{2}] = [1.25 0.433 0.433 0.75] .$ Determinant: $det = 1.25 \cdot 0.75 - 0.43 3^{2} = 0.9375 - 0.1875 = 0.75$ . Inverse: $Q = \frac{1}{0.75} [0.75 - 0.433 - 0.433 1.25] = [1.0 - 0.577 - 0.577 1.667]$ . Trace of $Q$ : $1.0 + 1.667 = 2.667$ . To $DOP = 2.667 \approx 1.63$ . Trace kyun? $Q$ ke diagonal entries per-axis variance multipliers hain; unhe sum karke square root lene se combined position DOP milta hai.

Exercise 3.2 — Angle squeeze karo, DOP explode hote dekho

3.1 repeat karo lekin satellites sirf $1 0^{\circ}$ apart hain: $e_{1} = (1, 0)$ , $e_{2} = (cos 1 0^{\circ}, sin 1 0^{\circ})$ . Dikhao ki determinant collapse karta hai aur DOP blow up karta hai. Figure s02 wide aur narrow geometries compare karta hai.

Recall Solution

$cos 1 0^{\circ} \approx 0.9848, sin 1 0^{\circ} \approx 0.1736$ . $G^{⊤} G = [1 + 0.984 8^{2} 0.1710 0.9848 \cdot 0.1736 0.173 6^{2}] = [1.9698 0.1710 0.1710 0.0301] .$ $det = 1.9698 \cdot 0.0301 - 0.171 0^{2} = 0.05929 - 0.02924 = 0.03015 = sin^{2} 1 0^{\circ}$ . Trace shortcut — kyun $tr Q = \frac{tr ( G ^{⊤} G )}{det}$ ek $2 \times 2$ ke liye: Kisi bhi $2 \times 2$ matrix $M = [a b b d]$ ke liye, inverse hai $M^{- 1} = \frac{1}{det} [d - b - b a]$ . Iske diagonal entries $d / det$ aur $a / det$ hain, to unka sum hai $\frac{a + d}{det} = \frac{tr ( M )}{det}$ . Yeh sirf ek $2 \times 2$ convenience hai (swap-diagonal inverse); higher dimensions mein actually invert karna padta hai. Yahan apply karte hain: $tr Q = \frac{tr ( G ^{⊤} G )}{det} = \frac{1.9698 + 0.0301}{0.03015} = \frac{1.9999}{0.03015} \approx 66.3.$ $DOP = 66.3 \approx 8.14$ — versus $6 0^{\circ}$ par $1.63$ . Kyun? $det (G^{⊤} G) = sin^{2} (θ)$ do unit directions ke liye jo $θ$ apart hain. Jaise $θ \to 0$ directions parallel ho jaati hain, $det \to 0$ , $Q$ entries blow up karti hain — spheres ek glancing angle par cut karti hain aur errors smear ho jaate hain. Wide spread = small DOP = good.

Level 4 — Synthesis

Exercise 4.1 — Full 2D linearized least-squares step

Ek 2D receiver ka current guess $x_{0} = (0, 0)$ hai, bias guess $c b_{0} = 0$ ke saath. Teen satellites $S_{1} = (0, 10)$ , $S_{2} = (10, 0)$ , $S_{3} = (- 8, - 6)$ par hain (units: km). Guess par predicted pseudoranges hain $ρ_{i}^{(0)} = r_{i} (x_{0}) + c b_{0}$ . Measured residuals hain $Δ ρ = (0.30, 0.30, 0.30)$ km (sab identical). Geometry matrix $G$ columns $(e_{i x}, e_{i y}, 1)$ ke saath use karte hue, heavy arithmetic ke bina argue karo ki $Δ x$ kya hona chahiye, phir clock component confirm karo.

Recall Solution

Pehle sign convention (explicitly flag karo, 3B1B-style). Hum parent ki definitions exactly follow karte hain, to koi sign guess karne ke liye nahi chhoda:

Residual: $Δ ρ_{i} = ρ_{i}^{measured} - ρ_{i}^{(0)}$ (measured minus predicted). Ek positive residual matlab measured pseudorange guess ki prediction se zyaada lambi thi.
Line-of-sight entries: $e_{i x} = (x - x_{i}) / r_{i}$ — parent ka convention, satellite se receiver ki taraf point karta hai. Guess $(0, 0)$ par evaluate kiya gaya.
Clock column entry $= + 1$ , kyunki $\partial ρ_{i} / \partial (c b) = + 1$ ; ek positive $c Δ b$ har predicted range ko lambaa karta hai.

$(0, 0)$ par unit vectors compute karo. $S_{1} = (0, 10)$ ke liye: $e_{1} = (0 - 0, 0 - 10) /10 = (0, - 1)$ . $S_{2} = (10, 0)$ ke liye: $(0 - 10, 0 - 0) /10 = (- 1, 0)$ . $S_{3} = (- 8, - 6)$ ke liye, $r_{3} = 64 + 36 = 10$ : $e_{3} = (0 - (- 8), 0 - (- 6)) /10 = (0.8, 0.6)$ . $G = 0 - 1 0.8 - 1 0 0.6 111 .$ Key insight: har satellite par identical positive residual exactly ek pure clock error jaisa lagta hai — 1's ka column (sab $+ 1$ ) ise absorb kar leta hai. Model $Δ ρ = G Δ x$ mein $Δ x = (0, 0, 0.30)$ se $G Δ x = (0.30, 0.30, 0.30)$ milta hai, data match karta hai zero position change ke saath. To $Δ x = 0, Δ y = 0, c Δ b = + 0.30 km = + 300 m$ . Positive sign keh raha hai "hamare clock guess ne predictions ko 300 m chhoti banaa di ⇒ bias upar correct karo." Least-squares ka unique answer kyun hai? $G$ $3 \times 3$ aur invertible hai (satellites well spread hain), to system exact hai: $Δ x = G^{- 1} Δ ρ$ unique hai, aur humne ek solution exhibit kiya — to yahi the solution hona chahiye. Sab channels par common signal clock ke roop mein diagnose hota hai, position ke roop mein nahi — yahi 1's column ka kaam hai.

Exercise 4.2 — DOP ko error budget mein badlo

Tumhara receiver report karta hai $HDOP = 1.2$ (Horizontal DOP — do horizontal axes ke liye geometry multiplier) aur $VDOP = 2.1$ (Vertical DOP — up/down axis ke liye multiplier). Dominant error source ionospheric delay hai jo $σ_{UERE} = 5 m$ (User-Equivalent Range Error) deta hai. Horizontal aur vertical $1 σ$ errors aur total 3D position error (PDOP ke zariye) compute karo.

Recall Solution

Horizontal: $σ_{H} = HDOP \times σ_{UERE} = 1.2 \times 5 = 6 m$ . Vertical: $σ_{V} = VDOP \times σ_{UERE} = 2.1 \times 5 = 10.5 m$ . PDOP unhe combine karta hai: $PDOP = HDOP^{2} + VDOP^{2} = 1.44 + 4.41 = 5.85 \approx 2.42$ . Total: $σ_{pos} = PDOP \times σ_{UERE} = 2.42 \times 5 \approx 12.1 m$ . VDOP kyun dominate karta hai? Satellites sab tumhare upar hain — horizon ke neeche koi nahi — isliye vertical direction hamesha poorely bracketed hoti hai. Yahi structural reason hai ki GPS altitude horizontal position se worse hoti hai.

Level 5 — Mastery

Exercise 5.1 — Degenerate geometry: coplanar satellites

Maano charon satellites receiver se ek hi plane mein dikhte hain — unke line-of-sight vectors coplanar hain. $det (G^{⊤} G)$ aur DOP ke ek component ka kya hoga? Kaun sa coordinate unobservable ho jaata hai?

Recall Solution

Agar charon unit directions $e_{i}$ ek plane mein lie karte hain, to woh 3D direction space ka sirf 2D subspace span karte hain. $G$ ke position part ke columns tab linearly dependent ho jaate hain (rank $< 3$ ), isliye $G^{⊤} G$ singular hai: $det (G^{⊤} G) = 0$ . Consequently $(G^{⊤} G)^{- 1}$ exist nahi karta — formally DOP $\to \infty$ us direction mein jaata hai jo plane ke perpendicular hai. Woh perpendicular coordinate unobservable hai: in ranges ka koi combination us plane se bahar motion ko constrain nahi kar sakta. Physical picture: agar har satellite tumhare horizon plane par ho, to tumhari altitude thoda badalna sab charon ranges ko sirf same second-order amount badalta hai — first order par invisible hai. Tumhe teen-dimensional observation ke liye kam se kam ek satellite us common plane se well out (jaise directly overhead) chahiye. Yeh L3 "angle $\to 0$ " blow-up ka 3D mein continuous version hai.

Exercise 5.2 — Zero-baseline / identical satellites

Tumhare chaar satellites mein se do exactly same position aur pseudorange report karte hain (ek data-repeat glitch). Tumhare paas phir bhi "4" measurements hain. Kya tum fix pa sakte ho? $G$ ki rank ke zariye explain karo.

Recall Solution

Nahi. Do identical satellites $G$ mein do identical rows contribute karte hain. Identical rows linearly dependent hoti hain, isliye $G$ ki rank at most $3$ hai, aur $G^{⊤} G$ singular hai ( $det = 0$ ). Effectively tumhare paas sirf 4 unknowns ke liye 3 independent equations hain — underdetermined, exactly parent note ke "3 satellites" waala failure mode. "4 measurements" kyun enough nahi: jo matter karta hai woh independent rows hain (distinct geometry), raw count nahi. Ek duplicated measurement koi nayi constraint add nahi karta. Tumhare paas chaar geometrically distinct line-of-sight directions hone chahiye.

Exercise 5.3 — Common-elevation clustering TDOP bigaad deta hai

Plane-plus-clock model mein, sab satellites ko same elevation angle par rakho (sirf unka azimuth differ karta hai), to har ek clock column (all-ones column) se almost same tarah couple karta hai. Intuitively — aur near-parallel-column argument ke zariye — explain karo ki common elevation par satellites cluster karna specifically TDOP (Time DOP, DOP ka clock/time component) kyun inflate karta hai, aur cure batao.

Recall Solution

Mechanism ka statement. DOP ek coordinate mein tab blow up hota hai jab $G$ ka corresponding column nearly parallel hota hai (nearly a linear combination hota hai) doosre columns se — yeh L3 "do directions parallel ⇒ $det \to 0$ " story ka multi-column version hai. Clock column sab ones hai. TDOP tab bada hota hai jab receiver position mein shift, clock shift ko mimic kar sake, yaani jab all-ones column position columns se nearly reproducible ho.

Common elevation ise kyun trigger karta hai. Har line-of-sight unit vector local vertical se same angle banata hai (same elevation). Iska vertical position change se coupling isliye sab satellites par identical hota hai — aur ek vertical move tab har predicted range ko almost same amount se change karta hai, jo exactly wahi hai jo clock bias karta hai (sab ranges mein same $c Δ b$ add karta hai). To "tum upar/neeche gaye" aur "tumhari clock drift hui" almost same residual pattern produce karte hain; estimator inhe alag nahi kar sakta, aur clock variance $q_{tt}$ (hence TDOP $= q_{tt}$ ) inflate hota hai.

Cure. High aur low elevation satellites mix karo. Directly overhead satellite, vertical position se strongly couple karta hai lekin flat common-mode clock signature poorly reproduce karta hai; horizon ke paas wala satellite iski opposite karta hai. Woh contrast clock column (all-ones) ko position columns se no longer expressible banata hai, position–clock ambiguity todta hai, aur TDOP shrink karta hai.

Yahi sahi diagnosis kyun hai ("bas satellites add karo" nahi): same elevation par zyaada satellites add karne se columns near-parallel rehte hain aur kuch nahi hota — fix elevation mein angular diversity hai, L3 ke lesson se match karta hai ki spread, count nahi, DOP control karta hai. Dekho Covariance Propagation ki yeh ambiguity ill-conditioned $(G^{⊤} G)^{- 1}$ ke roop mein kaise appear hoti hai.