INS error propagation — error state equations: Worked examples

3.5.17 · D3 · Physics › Guidance, Navigation & Control (GNC) › INS error propagation — error state equations

Definition Vocabulary jo hum parent note se lete hain

Parent note 3.5.17 un objects ko define karta hai jo hum yahan reuse karte hain. Simple words mein:

$δ x$ — error state: har cheez ka stack jo INS galat karta hai. Iske teen parts hain: $δ r$ (position error, metres), $δ v$ (velocity error, m/s), aur $ψ$ (attitude/tilt error, ek chhota angle radians mein jo batata hai ki computed level true level se kitna tilted hai).
$F$ — system matrix: ye batata hai ki har error doosron ko kaise feed karta hai (jaise velocity error, position error ko feed karta hai). Hum $F$ ko re-derive nahi karte; hum iske blocks parent note se padhte hain.
$G w$ — noise/error inputs: $w$ raw sensor errors ka stack hai (accel bias, gyro drift) aur $G$ unhe state mein map karta hai.
$b_{a}$ — accelerometer bias: constant, unwanted acceleration (in $m/s^{2}$ ) jo accelerometer true specific force mein add karta hai. Yeh often micro-g ( $μg$ ) mein quote hoti hai, jahan $1 μg = 1 0^{- 6} \times 9.8 m/s^{2}$ . Yeh error input $w$ ka accelerometer part hai.
$ϵ$ — gyro drift: constant, unwanted rotation rate (rad/s mein) jo gyro true rate mein add karta hai. Often $^{\circ} / hr$ mein quote hoti hai. Yeh $w$ ka gyro part hai.
$f^{n}$ — specific-force vector jo navigation frame mein expressed hai (North-East-Down). "Specific force" simply woh acceleration hai jo accelerometer feel karta hai; superscript $n$ ka matlab hai "nav-frame coordinates mein likha hua". Level flight mein iska dominant part gravity hai, $∣ f^{n} ∣ \approx g$ .
$ω_{i e}$ — Earth-rate: kitni tezi se Earth spin karti hai ( $7.29 \times 1 0^{- 5}$ rad/s). $ω_{e n}$ — transport-rate: local nav frame kitna muda jata hai jab vehicle curved Earth ke upar move karta hai. Dono chhote flights par bahut tiny hain — Cell G exactly kab $ω_{i e}, ω_{e n}$ terms drop karna safe hai, yeh dikhata hai.

The scenario matrix

Har INS error-growth problem in cells mein se ek hai. Hum sab ko cover karenge.

Cell	Kya vary karta hai	Kaun sa question answer karta hai
A. Accel bias, sign +	constant $b_{a} > 0$	ek positive accel bias kitni tezi se position error grow karta hai?
B. Accel bias, sign −	constant $b_{a} < 0$	kya sign sirf flip hota hai, ya kuch aur subtle hai?
C. Gyro drift	constant $ϵ$	long-term gyro error accel error se kyun worse hai?
D. Zero input (degenerate)	$b_{a} = 0, ϵ = 0$ , lekin $δ v_{0} \neq = 0$	perfect sensor lekin bad initial condition mein kya hota hai?
E. Limiting / bounded case	gravity coupling ON	horizontal errors kyun blow up nahi hote — Schuler bound
F. Real-world word problem	mixed bias + drift	1 hour mein ek real navigation-grade IMU ka total drift
G. Exam twist	"kaun sa term dominate karta hai?"	order-of-magnitude reasoning, small terms drop karo

Do quick tools jo hum lean on karenge:

Figure — INS error propagation — error state equations

Figure s01. Horizontal axis = time $t$ . Teen curves usi same constant slope se: flat black line rate hai (ek constant $c$ ); dashed black line uska first integral hai, ek straight ramp jo $t$ ki tarah grow karta hai; red curve second integral hai, ek parabola jo $t^{2}$ ki tarah grow karti hai. Ise padho jaise "har integration line ko ek power of $t$ se upar moda deta hai" — exactly wahi jo ek constant accelerometer bias ko quadratic position error mein badalta hai.

Hume yeh kyun chahiye: har example hai "source ko position tak pahunchne se pehle kitni baar integrate kiya jaata hai?" Integrations gino, powers of $t$ gino.

Cell A — Positive accelerometer bias, level flight

Worked example A ·

$b_{a} = + 100 μg$ , level flight, 60 s Ek navigation-grade accelerometer ka residual bias $100 μg$ hai. Ek $g \approx 9.8 m/s^{2}$ , isliye $b_{a} = 100 \times 1 0^{- 6} \times 9.8 \approx 9.8 \times 1 0^{- 4} m/s^{2} .$ Earth rotation aur gravity coupling ignore karo (short, level flight). Initial errors zero. $t = 60$ s par position error find karo.

Forecast: abhi guess karo — position error kuch centimetres jaisi hai, ya couple of metres?

Velocity error. $δ v (0) = 0$ ke saath: $\dot{δ v} = b_{a}$ , ek constant, isliye $δ v (t) = b_{a} t$ . Yeh step kyun? Velocity equation reduce ho jaati hai "rate = constant" mein; zero se ek constant integrate karne par straight line milti hai.
Position error. $δ r (0) = 0$ ke saath: $\dot{δ r} = δ v = b_{a} t$ , ek ramp, isliye $δ r (t) = \frac{1}{2} b_{a} t^{2}$ . Yeh step kyun? Position second integral hai. Ek ramp integrate karne par figure s01 ki parabola milti hai.
$t = 60$ plug in karo. $δ r = \frac{1}{2} (9.8 \times 1 0^{- 4}) (60)^{2} = \frac{1}{2} (9.8 \times 1 0^{- 4}) (3600) \approx 1.76 m .$

Verify: units $= m/s^{2} \cdot s^{2} = m$ ✓. Sanity: 60 s par velocity error $b_{a} t \approx 0.059 m/s$ hai — kuch cm/s, ek acche IMU ke liye plausible; ek minute mein yeh ~1.8 m accumulate karta hai. Position quadratically badi.

Cell B — Negative bias (kya sign matter karta hai?)

Worked example B ·

$b_{a} = - 100 μg$ , same conditions Same magnitude, opposite sign. Initial errors zero. $δ r (60)$ kya hai, aur kya structurally kuch badla?

Forecast: same size, opposite direction? Ya square sign ko khatam kar deta hai?

A ko sign ke saath repeat karo. $δ v = b_{a} t = - (9.8 \times 1 0^{- 4}) (60) = - 0.0588 m/s$ . Yeh step kyun? Linear model mein kuch bhi sign ki parwah nahi karta; woh faithfully propagate hota hai.
Position. $δ r = \frac{1}{2} b_{a} t^{2} = - 1.76 m .$ Yeh step kyun? $t^{2}$ hamesha positive hai, isliye $δ r$ mein $b_{a}$ ka sign preserve hota hai — error doosri taraf point karta hai, same magnitude.

Verify: $∣ δ r ∣$ Cell A se identical, sign flipped ✓. Lesson: ek single linear term ke liye, negative input sirf trajectory mirror karta hai. Square $b_{a}$ par nahi balki $t$ par hai, isliye woh sign rectify nahi karta — ek common trap.

Cell C — Constant gyro drift (gyros kyun worse hain)

Worked example C · gyro drift

$ϵ = 0.01^{\circ} / hr$ , 60 s Ek gyro $0.01^{\circ} / hr$ drift karta hai (navigation grade). Convert karo: $ϵ = 0.01 \times \frac{π}{180} rad / 3600 s = 4.848 \times 1 0^{- 8} rad/s .$ Gravity coupling ON $\dot{δ v} = - g ψ$ ke zariye, $g = 9.8$ ke saath (upar definition box dekho). Initial errors zero. $δ r (60)$ find karo.

Forecast: drift bahut tiny hai. Guess karo: ek minute mein position error micrometres, millimetres, ya centimetres?

Drift tilt build karta hai. $ψ (0) = 0$ ke saath: $\dot{ψ} = ϵ \Rightarrow ψ (t) = ϵ t$ . Yeh step kyun? Gyro error constant rate ki tarah attitude equation ko feed karta hai; tilt ramp up hota hai.
Tilt gravity ko velocity mein galat project karta hai. $\dot{δ v} = - g ψ = - g ϵ t \Rightarrow δ v = - \frac{1}{2} g ϵ t^{2}$ . Yeh step kyun? Ek tilted platform gravity ka ek slice horizontal acceleration ki tarah read karta hai — $F$ se $- [f^{n} \times]$ coupling. Woh accel case se ek aur integration hai.
Velocity se position. $δ r = - \frac{1}{6} g ϵ t^{3}$ . Yeh step kyun? Teesra integration — drift → tilt → velocity → position. Wahi extra power of $t$ hai.
Plug in karo. $δ r = - \frac{1}{6} (9.8) (4.848 \times 1 0^{- 8}) (60)^{3} = - \frac{1}{6} (9.8) (4.848 \times 1 0^{- 8}) (216000) \approx - 1.71 \times 1 0^{- 2} m .$

Verify: $\frac{1}{6} g ϵ t^{3}$ ka dimensional check, term by term: $[g] [ϵ] [t^{3}] = (\frac{m}{s ^{2}}) (\frac{rad}{s}) (s^{3}) = \frac{m \cdot rad \cdot s ^{3}}{s ^{3}} = m \cdot rad = m$ kyunki radian dimensionless hai ✓. Magnitude ek minute mein ~1.7 cm. Cubic growth: ek ghante mein yeh term accel bias se kahin tezi se explode karta hai, isliye gyro spec lambi flights dominate karta hai.

Cell D — Perfect sensors, bad initial velocity (degenerate input)

Worked example D ·

$b_{a} = 0, ϵ = 0$ , lekin $δ v_{0} = 0.1 m/s$ Yeh woh ek cell hai jo deliberately hamari zero-initial-condition convention todhta hai. IMU ideal hai — zero bias, zero drift — lekin INS ko $δ v (0) = δ v_{0} = 0.1 m/s$ velocity error ke saath initialize kiya gaya (bad GPS handoff); $δ r (0) = 0$ . Coupling ignore karo. $δ r (60)$ find karo.

Forecast: perfect sensor ke saath, kya error wahi rehta hai, shrink hota hai, ya phir bhi grow karta hai?

Velocity error frozen hai. $\dot{δ v} = 0 \Rightarrow δ v (t) = δ v_{0} = 0.1 m/s$ (constant). Yeh step kyun? Koi bias nahi matlab koi forcing nahi; velocity error bas apni initial value par rehta hai.
Position phir bhi ramp karta hai. $\dot{δ r} = δ v_{0} \Rightarrow δ r = δ v_{0} t = 0.1 \times 60 = 6 m .$ Yeh step kyun? Ek constant velocity error bhi ek linear position error mein integrate hoti hai. Perfect sensors error growth nahi rokenge agar initial state galat ho.

Verify: units $m/s \cdot s = m$ ✓. Note karo $6 m > 1.76 m$ Cell A se — yahan initial condition ne sensor bias se zyada hurt kiya. Lesson: homogeneous solution ( $F δ x$ jo non-zero $δ x_{0}$ par act karta hai) driving noise jitna hi important hai.

Cell E — Limiting case: bounded Schuler oscillation

Worked example E · horizontal error kyun blow up NAHI karta

Cells A–D mein humne gravity ka feedback drop kiya tha. Use restore karo. Ek position error $δ r$ (horizontal) local vertical ki direction change karta hai, jo tilt $ψ$ change karta hai, jo $δ v$ change karta hai, jo $δ r$ change karta hai — ek closed loop. Ise 1-D oscillator ki tarah model karo: $\ddot{δ r} = - ω_{s}^{2} δ r$ , jahan $ω_{s}^{2} = g / R$ , $R = 6.371 \times 1 0^{6}$ m Earth radius hai.

Forecast: negative feedback wala ek loop — kya error diverge karta hai, settle karta hai, ya hamesha oscillate karta hai?

Equation padho. $\ddot{δ r} = - ω_{s}^{2} δ r$ simple harmonic motion hai. Yeh step kyun? Displacement ke proportional negative feedback ek oscillator ki signature hai, runaway ki nahi.
Schuler frequency. $ω_{s} = g / R = 9.8/6.371 \times 1 0^{6} = 1.24 \times 1 0^{- 3} rad/s .$ Yeh step kyun? Yeh gravity-error loop ki natural frequency hai.
Schuler period. $T = 2 π / ω_{s} = 2 π /1.24 \times 1 0^{- 3} \approx 5064 s \approx 84.4 min .$ Yeh step kyun? $T = 2 π / ω$ frequency ko period mein convert karta hai; yeh famous 84-minute Schuler period se match karta hai.

Verify: $9.8/6.371 \times 1 0^{6} \approx 1.24 \times 1 0^{- 3}$ ✓; $2 π /1.24 \times 1 0^{- 3} \approx 5064$ s $= 84.4$ min ✓. Horizontal error isliye ek bound ke andar oscillate karta hai $t^{2}$ ki tarah runaway karne ki jagah — yahi limiting behaviour hai jo parent note mein promise kiya gaya tha. Dekho Schuler Tuning and Oscillation.

Figure s02. Horizontal axis = time jo Schuler periods mein measure hua hai (1 unit = $T \approx 84.4$ min). Vertical axis = horizontal position error, normalize kiya hua taaki oscillation amplitude $1$ ho (in these units dotted bound lines exactly $+ 1$ aur $- 1$ par baithe hain). Dashed black curve uncoupled prediction hai (gravity feedback off), $t^{2}$ ki tarah runaway karta hua. Red curve gravity-coupled truth hai: woh $t^{2}$ ki tarah diverge karne ki jagah do dotted bound lines ( $\pm 1$ ) ke beech oscillate karta hai. Figure ka point: gravity feedback on karna ek runaway ko ek bounded ~84-minute oscillation mein convert kar deta hai, apne amplitude par capped.

Cell F — Real-world word problem: nav-grade IMU, ek ghanta

Worked example F · combined bias + drift over

$t = 3600$ s Ek real navigation-grade IMU ka accel bias $b_{a} = 50 μg$ aur gyro drift $ϵ = 0.01^{\circ} / hr$ hai. Uncoupled (worst-case bound, Schuler damping ignore karo). Initial errors zero. 1 ghante baad total horizontal position error estimate karo.

Forecast: ek acche IMU ke liye ek ghante mein, order of magnitude guess karo: metres, hundreds of metres, ya kilometres?

Convert karo. $b_{a} = 50 \times 1 0^{- 6} \times 9.8 = 4.9 \times 1 0^{- 4} m/s^{2}$ ; $ϵ = 4.848 \times 1 0^{- 8} rad/s$ (jaise Cell C). Yeh step kyun? Integrate karne se pehle dono sources ko SI mein laao.
Accel term. $δ r_{a} = \frac{1}{2} b_{a} t^{2} = \frac{1}{2} (4.9 \times 1 0^{- 4}) (3600)^{2} \approx 3175 m .$ Yeh step kyun? Cell A ka quadratic law, hour-long time ke saath.
Gyro term. $δ r_{g} = \frac{1}{6} g ϵ t^{3} = \frac{1}{6} (9.8) (4.848 \times 1 0^{- 8}) (3600)^{3} \approx 3694 m .$ Yeh step kyun? Cell C ka cubic law $t = 3600$ s par apply kiya: drift se teen integrations, isliye $t^{3}$ . Note karo yeh cubic term ab quadratic accel term se aage nikal gayi — kaafi time milne par cubic hamesha jeetta hai.
Total (linear superposition). $δ r \approx 3175 + 3694 \approx 6869 m \approx 6.9 km .$ Yeh step kyun? Error model linear hai, isliye independent contributions simply add hote hain.

Verify: har term metres mein ✓; gyro > accel cubic dominance confirm karta hai ✓. Reality check: ek uncoupled pure-inertial run kilometres per hour drift karta hai — exactly isliye real systems GPS fuse karte hain Kalman filter ke zariye. True (Schuler-damped) error chhota hoga, lekin yeh bound dikhata hai kyun aiding essential hai.

Cell G — Exam twist: kaun sa term drop kar sakte hain?

Worked example G · 2-minute missile flight — sirf jo matter kare use rakho

Vocabulary box se yaad karo ki $ω_{i e}$ Earth-rate hai ( $7.29 \times 1 0^{- 5}$ rad/s, Earth ki spin) aur $f^{n}$ nav frame mein specific-force vector hai (powered flight mein kaafi $g$ ). 120 s ki flight ke dauran, kaun sa neglect karna safe hai: Earth-rate term, ya $- [f^{n} \times]$ attitude→velocity coupling? Order of magnitude se justify karo.

Forecast: guess karo $F$ ka kaun sa block short-flight error ka 80% carry karta hai.

Flight ke dauran Earth-rate rotation ka size nikalo. $ω_{i e} t = 7.29 \times 1 0^{- 5} \times 120 = 8.75 \times 1 0^{- 3} rad \approx 0. 5^{\circ} .$ Yeh step kyun? Ek term tabhi matter karta hai jab woh run ke dauran error ko appreciably rotate kare; $0.009$ rad negligible hai, isliye $F$ mein $ω_{i e}, ω_{e n}$ terms barely act karte hain.
Specific-force coupling ka size nikalo. Powered flight mein $∣ f^{n} ∣$ kaafi $g$ ho sakta hai; yahan tak ki $1 0^{- 4}$ rad tilt se $\dot{δ v} \sim ∣ f^{n} ∣ \cdot ψ \approx 3 \times 9.8 \times 1 0^{- 4} \approx 3 \times 1 0^{- 3} m/s^{2}$ milta hai, tiny Earth-rate cross term ko dominate karta hua. Yeh step kyun? Same states par act karne wale dono $F$ blocks ki magnitudes compare karo; $[f^{n} \times]$ block orders of magnitude bada hai.
Decision. $ω_{i e}, ω_{e n}$ drop karo; $- [f^{n} \times]$ block aur sensor-error inputs rakho. Yeh step kyun? Short-flight error ka 80% usi ek coupling mein hai — $F$ ke liye 80/20 rule.

Verify: $ω_{i e} t = 8.75 \times 1 0^{- 3}$ rad $≪ 1$ ✓, isliye ise neglect karna <1% error introduce karta hai. Yeh Gyro and Accelerometer Error Models guidance se match karta hai: short flights par, sensor errors + attitude coupling dominate karte hain; Earth-rate aur transport-rate terms second order hain.

Matrix coverage check

Recall Kya humne har cell hit ki?

A (+bias) ::: Example A, quadratic, 1.76 m B (−bias) ::: Example B, sign preserved, −1.76 m C (gyro drift) ::: Example C, cubic, 60 s mein −1.71 cm D (degenerate: perfect sensor, bad init) ::: Example D, $δ v_{0}$ se linear, 6 m E (limiting/bounded) ::: Example E, Schuler oscillation, 84.4 min F (real-world) ::: Example F, ~6.9 km/hr uncoupled bound G (exam twist / drop terms) ::: Example G, $[f^{n} \times]$ rakho, Earth rate drop karo

Related: Strapdown INS Mechanization Equations · Direction Cosine Matrix and Small-Angle Rotations · Schuler Tuning and Oscillation