Exercises — INS error propagation — error state equations

Kya: $\mathbf{\psi }$ attitude/tilt error hai, radians mein measure hota hai. Picture: INS ek computed North-East-Down frame carry karta hai. Agar uski bookkeeping thodi galat hai, toh woh computed frame true frame se thoda door rotate ho jaata hai. $\mathbf{\psi }$ woh chhota arrow (rotation vector) hai jo us turn ko describe karta hai — uski length tilt angle hai, uski direction woh axis hai jiske baare mein tum twist karoge. $\delta \mathbf{r}$ metres mein hai, $\delta \mathbf{v}$ metres/second mein hai, $\mathbf{\psi }$ radians mein hai.

Position ← velocity: identity block $I$ top row mein, second column. Yeh exactly $\dot{\delta \mathbf{r}}=\delta \mathbf{v}$ encode karta hai. Attitude → velocity: $-[{\mathbf{f}}^n\times ]$ block (middle row, third column). Ek tilt $\mathbf{\psi }$ bade specific-force vector ${\mathbf{f}}^n$ ko galat resolve karta hai, aur woh mis-projection velocity error ke roop mein appear hoti hai. $F$ ko is tarah kyun padhen: row $=$ "main kaun hoon", column $=$ "mujhe kaun push karta hai". Entry $F_{ij}$ batata hai state $j$ kitni strongly state $i$ ki rate ko drive karta hai. (Velocity diagonal par $-[(2{\mathbf{\omega }}_{ie}+{\mathbf{\omega }}_{en})\times ]$ block Coriolis/transport reshuffling of velocity error hai jaise frame ghoomti hai.)

Symbols ka matlab: ${\mathbf{\omega }}_{in}={\mathbf{\omega }}_{ie}+{\mathbf{\omega }}_{en}$ inertial ke relative nav frame ki total turn rate hai (Earth spin plus transport rate); $C_b^n$ body-frame gyro error ko nav coordinates mein rotate karta hai.

1. Transport-rate coupling $-{\mathbf{\omega }}_{in}\times \mathbf{\psi }$: jaise vehicle rotating Earth ke upar move karta hai, nav frame khud ${\mathbf{\omega }}_{in}$ rate se turn karta hai, aur tilt vector us turn se carry hota hai.
2. Gyro error $-C_b^n\delta {\mathbf{\omega }}_{ib}^b$: gyros ka drift aur noise, body frame se nav frame mein rotate kiya gaya. Yeh har instant mein error ka fresh injection hai — long-term drift ki root cause.

Step 1 — bias convert karo. $b_a=300\times 9.8\times 10^{-6}=2.94\times 10^{-3}{\text{m/s}}^2$. Step 2 — velocity (pehla integral). $\dot{\delta v}=b_a$ constant $\Rightarrow \delta v(t)=b_{a}t$. $\delta v(120)=2.94\times 10^{-3}\times 120=0.3528\text{m/s}$. Linear kyun? Ek constant ka ek integration $t$ mein straight ramp deta hai. Step 3 — position (doosra integral). $\delta r(t)=\frac{1}{2}{b}_a{t}^2$. $\delta r(120)=\frac{1}{2}\times 2.94\times 10^{-3}\times 120^2=21.168\text{m}$. Quadratic kyun? Position source ka second integral hai, toh $t^2$.

Step 1 — drift ko rad/s mein convert karo. $ϵ=0.01\times \frac{\pi }{180}=1.7453\times 10^{-4}\text{rad/s}$. Step 2 — tilt (drift ka pehla integral). $\psi (t)=ϵt$, toh $\psi (60)=1.7453\times 10^{-4}\times 60=1.0472\times 10^{-2}$ rad ($\approx 0.6^{\circ }$). Step 3 — velocity (overall doosra integral). $\dot{\delta v}=-gϵt\Rightarrow \delta v(t)=-\frac{1}{2}gϵt^2$. $\delta v(60)=-\frac{1}{2}\times 9.8\times 1.7453\times 10^{-4}\times 60^2=-3.0788\text{m/s}$. Step 4 — position (overall teesra integral). $\delta r(t)=-\frac{1}{6}gϵt^3$. $\delta r(60)=-\frac{1}{6}\times 9.8\times 1.7453\times 10^{-4}\times 60^3=-61.576\text{m}$. Cubic kyun? drift → tilt (1) → velocity (2) → position (3): teen integrations, isliye $t^3$.

Accel channel: $b_a=100\times 9.8\times 10^{-6}=9.8\times 10^{-4}{\text{m/s}}^2$. $\delta r_a=\frac{1}{2}{b}_a{t}^2=\frac{1}{2}\times 9.8\times 10^{-4}\times 100^2=4.9\text{m}$. Gyro channel: $ϵ=1.7453\times 10^{-4}\text{rad/s}$. $\delta r_g=\frac{1}{6}gϵt^3=\frac{1}{6}\times 9.8\times 1.7453\times 10^{-4}\times 100^3=285.06\text{m}$. Ratio: $285.06/4.9\approx 58.2$. Gyro channel roughly $58$ ke factor se dominate karta hai. Kyun: extra integration ($t^3$ vs $t^2$) plus gravity gain $g$ gyro drift ko long-run killer banata hai.

Cross product ${\mathbf{f}}^n\times \mathbf{\psi }$ with ${\mathbf{f}}^n=(0,0,-g)$ aur $\mathbf{\psi }=(0,{\psi }_E,0)$: ${\mathbf{f}}^n\times \mathbf{\psi }=\left|\begin{array}{ccc}\mathbf{i}& \mathbf{j}& \mathbf{k}\\ 0& 0& -g\\ 0& {\psi }_E& 0\end{array}\right|=\mathbf{i}(0\cdot 0-(-g){\psi }_E)-\mathbf{j}(0)+\mathbf{k}(0)=(g{\psi }_E,0,0).$ Toh $-[{\mathbf{f}}^n\times ]\mathbf{\psi }=-(g{\psi }_E,0,0)$: yeh North velocity error drive karta hai, ${\dot{\delta v}}_N=-g{\psi }_E$. Picture / kyun: ek East-axis tilt galat-oriented North accelerometer ko gravity vector mein "lean" karta hai, toh $g$ ka ek slice North acceleration mein leak ho jaata hai. Neeche tilt figure dekho.

Flight ke dauran coupling angle compute karo: ${\omega }_{ie}t=7.292\times 10^{-5}\times 120=8.75\times 10^{-3}$ rad $\approx 0.5^{\circ }$. Kyunki yeh $\ll 1$ hai, Earth-rate rotation poori flight mein error vector ko half degree se kam reshuffle karta hai — dominant channels par $<1\%$ effect. Conclusion: haan, chhoti flight ke liye ${\mathbf{\omega }}_{ie}$ (aur comparably small ${\mathbf{\omega }}_{en}$) drop karo; sirf $-[{\mathbf{f}}^n\times ]$ coupling aur sensor-error inputs rakho. Yeh short-flight INS error ka 80/20 hai.

Loop close karo. $\dot{\delta v}=-g\psi$ ko ek baar aur differentiate karo aur $\dot{\psi }=\frac{1}{R}\delta v$ substitute karo: $\ddot{\delta v}=-g\dot{\psi }=-g\cdot \frac{1}{R}\delta v=-\frac{g}{R}\delta v.$ Yeh simple harmonic motion ki equation hai, $\ddot{\delta v}+{\omega }_s^2\delta v=0$ with ${\omega }_s=\sqrt{g/R}$. Kyunki $\delta r$ $\delta v$ ka integral hai, position error bhi same frequency par oscillate karta hai. Oscillation kyun, ramp nahi: gravity loop ko negative sign ke saath close karta hai ($\dot{\delta v}$ mein $-g$ aur $+\frac{1}{R}$ feedback milke), exactly ek spring restoring force $-kx$ ki tarah. Is feedback ke bina ($\frac{1}{R}\to 0$) loop open hai aur error $t^3$ ki tarah ramp karta hai; iske saath, error wapis swing karta hai — Schuler oscillation $\approx 84$ min par. Ek constant accel bias phir bhi ek steady offset add karta hai jise loop null nahi kar sakta, toh real error ek bounded swing plus ek slow ramp hai.

Reduced system: $\dot{\delta r}=\delta v,\dot{\delta v}=-g\psi ,\dot{\psi }=ϵ.$ Top-down integrate karo (har ek rest se):

• $\psi (t)=ϵt$ — constant drift ka pehla integral.
• $\dot{\delta v}=-gϵt\Rightarrow \delta v(t)=-\frac{1}{2}gϵt^2$ — overall doosra integral.
• $\dot{\delta r}=-\frac{1}{2}gϵt^2\Rightarrow \delta r(t)=-\frac{1}{6}gϵt^3$ — overall teesra integral. Result: $\boxed{\delta r(t)=-\frac{1}{6}gϵt^3}$ — first principles se cubic law, parent note ke Example 2 se match karta hai. Teen integrations (drift→tilt→velocity→position) $t^3$ dete hain.

Symbolic: $\psi =ϵt$, toh $\dot{\delta v}=b_a-gϵt\Rightarrow \delta v=b_{a}t-\frac{1}{2}gϵt^2$, aur $\delta r(t)=\frac{1}{2}{b}_a{t}^2-\frac{1}{6}gϵt^3.$ Numbers: $b_a=200\times 9.8\times 10^{-6}=1.96\times 10^{-3}{\text{m/s}}^2$; $ϵ=0.005\times \pi /180=8.7266\times 10^{-5}\text{rad/s}$.

• Accel part: $\frac{1}{2}\times 1.96\times 10^{-3}\times 90^2=7.938\text{m}$.
• Gyro part: $\frac{1}{6}\times 9.8\times 8.7266\times 10^{-5}\times 90^3=103.90\text{m}$.
• $\delta r(90)=7.938-103.90=-95.96\text{m}$. Padho ise: ek chhota drift bhi $t=90$ s tak dominate karta hai kyunki cubic power ki wajah se — synthesis confirm karta hai ki gyro binding spec hai.

Frequency: ${\omega }_s=\sqrt{9.8/6.371\times 10^6}=1.2404\times 10^{-3}\text{rad/s}$. Period: $T_s=2\pi /{\omega }_s=5065.6\text{s}=84.4\text{min}$ — classic Schuler period. Short-time limit: $t\ll T_s$ ke liye, closed-loop solution $\sin ({\omega }_{s}t)\approx {\omega }_{s}t-\frac{1}{6}({\omega }_{s}t{)}^3$ ki tarah behave karta hai. Oscillatory closed-loop response ko $t$ mein lowest orders tak expand karna exactly polynomial (quadratic, phir cubic) growth reproduce karta hai open-loop model ki. Toh L4.1 ka open-loop cubic law simply bounded Schuler oscillation ka short-time limit hai — yeh ek swing ke pehle bend ki tarah hai jo, enough time dene par, diverge karne ki jagah wapis curve karta hai. Ek sentence mein: $t^3$ growth wahi hai jo Schuler sine spring ke error ko ghar kheenchne ka mauka milne se pehle dikhta hai.

Coupling: $-[{\mathbf{f}}^n\times ]\mathbf{\psi }=0$ jab ${\mathbf{f}}^n=0$, toh tilt velocity mein leak karna band kar deta hai true free-fall ke dauran. Lekin safe nahi: gyros drift karte rehte hain, toh $\dot{\psi }=-C_b^n\delta \mathbf{\omega }$ abhi bhi tilt $\psi$ ko grow karta hai. Jis moment specific force wapas aati hai (engine burn, re-entry drag), woh accumulated tilt suddenly ek badi force ko velocity mein re-project karta hai. Degenerate input symptom ko silence karta hai, cause ko nahi.

Inequality solve karo: $\frac{1}{6}gϵt^3\le 50\Rightarrow ϵ\le \frac{6\times 50}{9.8\times 100^3}=\frac{300}{9.8\times 10^6}=3.0612\times 10^{-5}\text{rad/s}$. ${}^{\circ }/\text{hr}$ mein convert karo: $\frac{180}{\pi }$ se multiply karo (rad→deg) aur $3600$ se (per s → per hr): $ϵ\le 3.0612\times 10^{-5}\times \frac{180}{\pi }\times 3600=6.314{}^{\circ }/\text{hr}$. Design read: tumhe $\lesssim 6{}^{\circ }/\text{hr}$ drift ka gyro chahiye — $100$ s flight ke liye comfortably low-cost MEMS-to-tactical grade. Lambi flights ($t^3$!) ise brutally tighten karti hain, isliye long-endurance INS ko navigation-grade gyros ki zaroorat hoti hai. Yeh directly Gyro and Accelerometer Error Models aur Kalman Filter for INS-GPS Integration ke estimation limits se connect karta hai.

Source se state tak integrations count karo ::: accel bias → position quadratic hai ($\frac{1}{2}{b}_a{t}^2$); gyro drift → position cubic hai ($\frac{1}{6}gϵt^3$); dono $\approx 84$ min Schuler oscillation ke short-time limbs hain.