Gyroscope — angular velocity measurement, bias, noise: Exercises

3.5.14 · D4 · Physics › Guidance, Navigation & Control (GNC) › Gyroscope — angular velocity measurement, bias, noise

Shuru karne se pehle, do reminders plain words mein taaki koi bhi symbol bekar na lage:

Unit bridge jo aapko baar baar kaam aayega: $1 hour = 3600 s$ , isliye $t seconds = \frac{t}{3600} hours$ .

Kaam karte waqt neeche di gayi picture khuli rakho — yeh dono growth laws ko ek hi graph pe dikhati hai aur woh moment mark karti hai jahan slowly-ramp karta bias fast-but-flattening noise ko overtake karta hai. L3 aur L5 mein jo bhi crossover calculation hai, woh sirf yeh hai ki "ye dono curves kahan milte hain?"

Figure — Gyroscope — angular velocity measurement, bias, noise

Orange line hai bias ramp $b t$ — ek seedhi diagonal, jo hamesha badhti rehti hai. Teal curve hai noise spread $\propto t$ — yeh pehle upar shoot karta hai phir bend hokar flat ho jaata hai. Plum dot hai unka crossover $t_{c}$ : iske baayein noise bada hai, iske daayein bias jeetta hai. Jab bhi koi problem pooche "kaunsa error dominate karta hai?", aap khud ko is dot ke baayein ya daayein locate kar rahe ho.

L1 · Recognition

Problem 1.1 — Growth law ko pehchano

Ek gyro bilkul still baitha hai. Uska bias ek constant $0.03 °/ s$ hai. Bina calculate kiye batao ki integrated heading error time ke saath kaise badhta hai: linearly, $t$ ki tarah, ya bilkul nahi?

Recall Solution 1.1

Ek constant offset ko time pe integrate karo toh yeh ek ramp hai — classic $\int b d τ = b t$ . Isliye error time mein linearly badhta hai. (Noise, bias nahi, woh hai jo $t$ ki tarah badhti hai.) Answer: linearly, $\propto t$ .

Problem 1.2 — Spec sheet padhna

Ek datasheet do numbers list karti hai: Bias = 0.01 °/s aur ARW = 0.3 °/√h. Kaunsa zyada wait karne se average out ho sakta hai, aur kaunsa nahi?

Recall Solution 1.2

ARW zero-mean high-frequency noise hai. Zyada samples average karne se random dhakke cancel ho jaate hain → aap ise reduce kar sakte ho (yeh Allan curve ka slope $- 1/2$ region hai, dekho Allan Variance Analysis).
Bias ek low-frequency offset hai jiska non-zero mean hai; ek constant ko average karo toh constant hi milta hai. Aap ise average karke hataa nahi sakte. Answer: ARW ko average kiya ja sakta hai; bias ko nahi.

L2 · Application

Problem 2.1 — Bias ramp

Ek drone ke gyro ka bias $b = 0.04 °/ s$ hai aur woh $90$ seconds ke liye dead-reckon karta hai (sirf gyro, koi GPS nahi). Heading error kya hoga?

Recall Solution 2.1

Step 1 (WHAT): bias law use karo $Δ θ = b \cdot t$ . WHY: ek constant offset integrate karo toh ramp milta hai. Step 2: $Δ θ = 0.04 °/ s \times 90 s = 3.6°$ . Answer: $3.6°$ . Ek "tiny" $0.04°/ s$ dedh minute mein ek real error ban jaata hai — isliye ek Inertial Measurement Unit (IMU) GPS ya magnetometer fuse karta hai.

Problem 2.2 — ARW wander

Ek gyro ka $ARW = 0.6 °/ h$ hai. Pure integration ke baad (a) $1$ hour, (b) $9$ hours ke baad expected angle uncertainty kya hai?

Recall Solution 2.2

Step 1 (WHAT): $σ_{θ} = ARW \cdot t$ use karo jahan $t$ hours mein ho. WHY: integrated white noise ek random walk hai jiska spread $t$ ki tarah badhta hai. (a) $σ_{θ} = 0.6 \times 1 = 0.6°$ . (b) $σ_{θ} = 0.6 \times 9 = 0.6 \times 3 = 1.8°$ . Answer: $0.6°$ aur $1.8°$ . Note karo ki error teen guni karne ke liye $9 \times$ zyada time laga — yeh $t$ ki pehchaan hai.

Problem 2.3 — ARW ko per-second feel mein convert karo

$ARW = 0.6 °/ h$ lo. Sirf $t = 100 s$ ke baad expected wander kya hai?

Recall Solution 2.3

Step 1 (WHAT + WHY): hum $σ_{θ} = ARW \cdot t$ use karte hain. WHY the $t$ : noise independent random chote kicks ki ek stream hai; inhe integrate karna ek random walk hai, aur random walk ka spread steps ki sankhya (ya elapsed time) ke square root ki tarah badhta hai kyunki kicks partly cancel ho jaate hain ek hi direction mein push karne ki bajaye. Isliye 2.2 waala wahi $t$ law yahan bhi apply hota hai — sirf naya kaam ek unit conversion hai. Step 2: time ko hours mein convert karo: $100 s = 100/3600 h = 0.02 \overline{7} h$ . Step 3: $σ_{θ} = 0.6 \times 0.02 \overline{7} = 0.6 \times 0.1667 \approx 0.10°$ . Answer: $\approx 0.10°$ . Chhota, kyunki $t$ short times pe gentle hota hai.

L3 · Analysis

Problem 3.1 — Crossover time: noise kab dominate karna band karta hai?

Ek gyro ka bias $b = 0.005 °/ s$ aur $ARW = 0.5 °/ h$ hai. Kis time $t_{c}$ par bias error aur ARW spread barabar ho jaate hain? $t < t_{c}$ ke liye kaunsa dominate karta hai?

Recall Solution 3.1

Step 1 (unhe equal karo, WHAT): hum chahte hain $b t = ARW t /3600$ , jahan baayein taraf $t$ seconds mein hai aur daayein taraf ka $t$ hours mein convert hai. ARW ko per-root-second likhein: $0.5 °/ h = 0.5/ 3600 °/ s = 0.5/60 = 0.008 \overline{3} °/ s$ . Step 2 (solve karo): $b t = k t$ jahan $k = 0.0083 \overline{3}$ . $t$ se divide karo: $b t = k \Rightarrow t = \frac{k}{b} = \frac{0.008333}{0.005} = 1.667 \Rightarrow t_{c} = 1.66 7^{2} \approx 2.78 s .$ Step 3 (interpret karo, WHY): $\sim 2.8 s$ se neeche $t$ noise term bada hota hai (yeh instantly dominate karta hai, phir flatten ho jaata hai); uske upar linear bias ramp overtake kar leta hai aur bina bound ke badhta rehta hai. Answer: $t_{c} \approx 2.8 s$ ; $t < t_{c}$ ke liye noise dominate karta hai, $t > t_{c}$ ke liye bias dominate karta hai. Upar crossover figure dekho.

Problem 3.2 — Allan deviation slope padhna

Ek log-log Allan deviation plot par, aap short averaging times $τ$ par slope $- 1/2$ ka ek seedha segment measure karte ho, phir ek flat bottom, phir slope $+ 1/2$ ka ek segment. Har region ke peeche ka physical process identify karo.

Recall Solution 3.2

Slope $- 1/2$ : Angle Random Walk (white noise) — averaging longer helps, error giri.
Flat bottom (slope $0$ ): bias instability — woh floor jisko aap average karke paar nahi kar sakte.
Slope $+ 1/2$ : rate random walk / slowly drifting bias — averaging ulta asar karta hai, drift jeet jaata hai. Yeh bathtub shape (dekho Allan Variance Analysis) ek hi plot se parent note ke dono dushmanon ko alag karta hai. Underlying stochastic model ka link: Random Walk & Wiener Process.

L4 · Synthesis

Problem 4.1 — Independent contributions ke saath total error budget

Ek gyro $t = 60 s$ ke liye dead-reckon karta hai. Uska bias $b = 0.01 °/ s$ aur $ARW = 0.4 °/ h$ hai. Bias error ko ek deterministic offset aur ARW ko ek independent random spread maanke, do (a) bias offset, (b) $1 σ$ noise spread, aur (c) total $1 σ$ error maante hue ki woh quadrature mein add hote hain.

Recall Solution 4.1

(a) Bias offset: $b t = 0.01 \times 60 = 0.6°$ . (b) ARW spread: $t = 60/3600 = 0.0167 h$ , isliye $σ_{n} = 0.4 \times 0.0167 = 0.4 \times 0.1291 \approx 0.0516°$ . (c) Quadrature sum (WHY quadrature): do independent error sources ke variances add hote hain, isliye standard deviations quadrature mein add hote hain: $σ_{tot} = (0.6)^{2} + (0.0516)^{2} = 0.36 + 0.00266 \approx 0.3627 \approx 0.602°.$ Answer: $(a) 0.6°$ , $(b) 0.052°$ , $(c) 0.602°$ . Bias budget mein completely dominate karta hai — ek Kalman Filter apni mehnat $b$ estimate karne mein lagaayega, noise mein nahi.

Problem 4.2 — Temperature-driven bias change ka effect

4.1 mein bias slowly drift karta hai kyunki $\dot{b} = n_{b}$ . Usi $60 s$ ke dauran bias linearly ramp karta hai $0.01 °/ s$ se $0.02 °/ s$ tak. Bias-induced heading error dobara calculate karo (changing bias integrate karo) aur constant-bias answer se compare karo.

Recall Solution 4.1b

Step 1 (WHAT): bias ab hai $b (t) = 0.01 + \frac{0.02 - 0.01}{60} t = 0.01 + 0.0001667 t$ (°/s). Hum ise integrate karte hain: $Δ θ = \int_{0}^{60} b (t) d t .$ Step 2 (WHY integrate): heading rate ka time-integral hoti hai, aur yahan jhoothi rate $b (t)$ hai. Step 3 (evaluate karo): linearly varying bias ke liye integral equals average bias times time: $\overset{ˉ}{b} = \frac{0.01 + 0.02}{2} = 0.015 °/ s, Δ θ = 0.015 \times 60 = 0.9°.$ Answer: $0.9°$ , jabki constant $0.01°/ s$ bias ke liye $0.6°$ tha — drift ne $0.3°$ extra add kiya. Yahi wajah hai ki "ek baar bias measure karo aur subtract karo" strategy fail hoti hai: bias run ke dauran hilta rehta hai.

L5 · Mastery

Problem 5.1 — Fusion reset interval design karo

Ek attitude estimator GPS/magnetometer resets ke beech gyro-only chalata hai. Gyro ka bias $b = 0.02 °/ s$ aur $ARW = 0.5 °/ h$ hai. Mission spec: heading error $1.0°$ se neeche rehna chahiye har waqt. Sabse lamba allowed reset interval $T$ kya hai?

Recall Solution 5.1

Step 1 (kaunsa error dominate karta hai?): deadline ke paas dono growth rates compare karo. Bias linearly badhta hai aur, jaisa L3 ne dikhaya, quickly ARW ko overtake kar leta hai. Crossover estimate karo: ARW ko $°/ s$ mein convert karo: $0.5/60 = 0.00833 °/ s$ . Crossover $t_{c} = k / b = 0.00833/0.02 = 0.4167 \Rightarrow t_{c} = 0.174 s$ . Toh second ke fraction ke baad bias completely dominate karta hai — hum $T$ size kar sakte hain sirf bias se aur ARW ko ek chhota correction maanke. Step 2 (bias-only bound, WHAT): require $b T \leq 1.0°$ : $T \leq \frac{1.0}{0.02} = 50 s .$ Step 3 (quadrature se check karo, WHY): at $T = 50 s$ , ARW spread $= 0.5 50/3600 = 0.5 \times 0.1179 = 0.0589°$ . Bias ke saath quadrature mein combine karo: $σ_{tot} = (0.02 \times 50)^{2} + 0.058 9^{2} = 1.0 + 0.00347 = 1.0017° > 1.0°.$ Yeh spec se thoda exceed karta hai. Thoda peeche hato: solve karo $(0.02 T)^{2} + (0.5 T /3600)^{2} = 1.0$ . Bias term overwhelmingly scale set karta hai, isliye $T \approx 49.9 s$ . Answer: har ~50 s mein reset karo (49.9 s noise margin include karne ke liye). Practically aap iske andar reset karoge — maano har 10 s mein — safety headroom ke liye.

Problem 5.2 — $t$ kyun? Parent ke derivation se ek-line proof

$θ = \int_{0}^{t} n d τ$ aur white-noise property $E [n (τ) n (τ^{'})] = Q δ (τ - τ^{'})$ use karke, dikhao ki integrated angle ka variance $Q t$ ke barabar hai, isliye $σ_{θ} \propto t$ .

Recall Solution 5.2

Pehle, $Q$ kya hai? $n (τ)$ gyro ki rate reading mein random noise hai (units $rad/s$ ). Symbol $E [\cdot]$ ka matlab hai "kai runs pe average." Quantity $Q$ hai noise power spectral density — physically, bandwidth ke har hertz mein kitna noise power hai, units $(rad/s)^{2} / Hz$ . Bada $Q$ matlab zyada kaampne waala gyro. White-noise property $E [n (τ) n (τ^{'})] = Q δ (τ - τ^{'})$ kehti hai ki alag instants pe do noise samples uncorrelated hain (spike $δ$ sirf tab non-zero hai jab $τ = τ^{'}$ ), aur us correlation ki height exactly $Q$ hai. Step 1 (WHAT): ek integral ka variance covariance ka double integral hai: $Var [θ] = E [\int_{0}^{t} n d τ \int_{0}^{t} n d τ^{'}] = \int_{0}^{t} \int_{0}^{t} E [n (τ) n (τ^{'})] d τ d τ^{'} .$ Step 2 (white noise use karo, WHY the delta): $E [n (τ) n (τ^{'})] = Q δ (τ - τ^{'})$ substitute karo. Delta sirf tab "fire" karta hai jab $τ = τ^{'}$ , ek integral collapse ho jaata hai: $Var [θ] = \int_{0}^{t} Q d τ = Q t .$ Step 3 (finish karo, aur ARW se connect karo): $σ_{θ} = Q t = Q \cdot t$ . Is page pe har jagah use hone wale growth law $σ_{θ} = ARW \cdot t$ se compare karo: term-by-term match karne se dikhta hai ki ARW coefficient hai hi $Q$ . Toh datasheet ka ARW number bas gyro ki noise power density ka square root hai, $°/ h$ mein re-express kiya hua. Yahi woh jagah hai jahan se poora $t$ law — aur ARW spec — aata hai. Dekho Random Walk & Wiener Process.

Recall Self-test recap (daayein side cover karo)

Bias error kaise badhta hai? ::: linearly in time, $b t$ ARW spread kaise badhta hai? ::: $t$ ki tarah, yaani $ARW \cdot t$ jahan $t$ hours mein ho ARW ko $°/ h$ se $°/ s$ mein convert karo? ::: $3600 = 60$ se divide karo Do independent errors kaise combine hote hain? ::: quadrature mein (variances add hote hain), $σ_{1}^{2} + σ_{2}^{2}$ Pehle kaunsa error calibrate karte hain? ::: bias — yeh seconds se upar dominate karta hai, linearly bina bound ke badhta hai, isliye ise hatane se accuracy mein sabse bada fayda hota hai noise ko touch karne se pehle