Exercises — IMU — integrated accelerometer + gyroscope

• Sensor A ($\text{rad/s}$) gyroscope hai, output ${\mathbf{\omega }}_b$. Yeh tumhe angular rate batata hai — abhi box har body axis ke baare mein kitni tezi se ghoom raha hai.
• Sensor B (${\text{m/s}}^2$) accelerometer hai, output ${\mathbf{f}}_b$. Yeh tumhe specific force batata hai — spring push per unit mass, yaani gravity ke alawa har force. Units hint hai: turning radians per second mein measure hoti hai; push per mass metres per second squared mein measure hoti hai.

(i) Gyro: ${\mathbf{\omega }}_b=(0,0,0)$ — kuch bhi nahi ghoom raha. (ii) Accel: true acceleration $\mathbf{a}=0$ (move nahi kar raha), toh $\mathbf{f}=\mathbf{a}-\mathbf{g}=0-(0,0,-9.81)=(0,0,+9.81).$ Yeh $+9.81{\text{m/s}}^2$ upar ki taraf read karta hai. Ek maatra spring force table ka normal push hai, aur woh push upar ki taraf point karta hai. Magnitude $9.81$.

Free fall mein ek maatra force gravity hai, toh $\mathbf{a}=\mathbf{g}$. Tab $\mathbf{f}=\mathbf{a}-\mathbf{g}=\mathbf{g}-\mathbf{g}=(0,0,0).$ Accelerometer zero read karta hai — koi spring stretch nahi hoti kyunki proof mass case ke saath saath girta hai. Gyro ${\mathbf{\omega }}_b=(0,0,0)$ read karta hai kyunki woh turn nahi kar raha. (Yahi zero-$g$ reading hai jo phones mein "free-fall detected" feature trigger karta hai.)

Pattern (parent note se) with ${\omega }_x=0.2,{\omega }_y=-0.1,{\omega }_z=0.3$: $[{\mathbf{\omega }}_b{]}_{\times }=\left(\begin{array}{ccc}0& -{\omega }_z& {\omega }_y\\ {\omega }_z& 0& -{\omega }_x\\ -{\omega }_y& {\omega }_x& 0\end{array}\right)=\left(\begin{array}{ccc}0& -0.3& -0.1\\ 0.3& 0& -0.2\\ 0.1& 0.2& 0\end{array}\right).$ $\mathbf{v}=(1,0,0{)}^{\top }$ se multiply karo — woh sirf pehla column pick karta hai: $[{\mathbf{\omega }}_b{]}_{\times }\mathbf{v}=(0,0.3,0.1).$ Cross-check definition se $\mathbf{\omega }\times \mathbf{v}=({\omega }_y{v}_z-{\omega }_z{v}_y,{\omega }_z{v}_x-{\omega }_x{v}_z,{\omega }_x{v}_y-{\omega }_y{v}_x)=(0,0.3,0.1)$. ✓ Skew matrix bas "cross product, matrix ke roop mein packaged" hai.

Reading mein koi $y$-component nahi hai, toh box $y$-axis ke baare mein tilt hai. Ek still sensor ke liye jo $y$ ke baare mein angle $\varphi$ se pitched hai, body-frame gravity reaction hai $(g\sin \varphi ,0,g\cos \varphi )$. s02 dekho: $f_x$ tilt triangle ki woh leg hai jo $\varphi$ ke saamne hai ("opposite"), $f_z$ woh leg hai jo uske saath hai ("adjacent"). $\tan \varphi =\frac{f_x}{f_z}=\frac{4.905}{8.496}=0.5774\Rightarrow \varphi =30^{\circ }.$ $\arctan$ kyun? $\tan$ ek angle ko opposite/adjacent ratio mein turn karta hai; $\arctan$ ise ulta chalaata hai — "kaunse angle ka yeh ratio hai?" Check: $g\sin 30=4.905$ ✓, $g\cos 30=8.496$ ✓. Kyun plain $\arctan$ kaafi nahi — quadrant problem. $\arctan$ hamesha $(-90^{\circ },+90^{\circ })$ mein hi answer deta hai, kyunki $\tan$ har $180^{\circ }$ repeat karta hai: $\tan \varphi =\tan (\varphi +180^{\circ })$. Toh ratio akela $30^{\circ }$ tilt aur $210^{\circ }$ tilt (box ulta) mein farq nahi kar sakta. Second reading ${\mathbf{f}}_b=(-4.905,0,-8.496)$ ka ratio abhi bhi $+0.5774$ hai, phir bhi dono $f_x<0$ aur $f_z<0$ — box third quadrant mein pitched hai. Fix hai two-argument $atan2(f_x,f_z)$, jo dono components ke signs dekhta hai: $atan2(4.905,8.496)=+30^{\circ },atan2(-4.905,-8.496)=-150^{\circ }.$ Toh hamesha $\varphi =atan2(f_x,f_z)$ use karo; plain $\arctan (f_x/f_z)$ tab hi safe hai jab tum pehle se jaante ho $f_z>0$ (box roughly upright).

Step 1 — correct body reading. Ek still sensor world-up gravity reaction $(0,0,g)$ ko apne tilted axes mein express karke sense karta hai. "World-to-body" transpose hai $(R_b^w{)}^{\top }$: ${\mathbf{f}}_b=(R_b^w{)}^{\top }(0,0,g)=\left(\begin{array}{ccc}\cos 30& 0& -\sin 30\\ 0& 1& 0\\ \sin 30& 0& \cos 30\end{array}\right)\left(\begin{array}{c}0\\ 0\\ g\end{array}\right)=\left(\begin{array}{c}-g\sin 30\\ 0\\ g\cos 30\end{array}\right).$ Ruko — picture (s01) ke against sign check karo. Gravity reaction world $+z$ ke along point karta hai; $+30^{\circ }$ se forward pitched box mein uski body $x$-leg $+g\sin 30$ hai. Upar wale standard $R_y$ use karke, $(R_b^w{)}^{\top }$ ka top-right entry $-\sin 30$ hai, toh $f_x=-g\sin 30$ milta hai. $f_x$ ka sign bas kis taraf pitch jhauka hua hai woh encode karta hai; iska magnitude $g\sin 30=4.905$ hai aur $f_z=g\cos 30=8.496$, exactly L2.3 wale numbers. ${\mathbf{f}}_b=(g\sin 30,0,g\cos 30)=(4.905,0,8.496)$ lo, yaani opposite sense ka pitch, toh arithmetic L2.3 se match karta hai. Step 2 — world mein wapas rotate karo. $R_b^w=R_y(30^{\circ })$ apply karo. $f_x=g\sin 30,f_z=g\cos 30$ symbolically substitute karo (decimals rounding traps invite karte hain): $f_{w,x}=\cos 30f_x+\sin 30f_z=\cos 30(g\sin 30)+\sin 30(g\cos 30)=2g\sin 30\cos 30=g\sin 60,$ $f_{w,z}=-\sin 30f_x+\cos 30f_z=-\sin 30(g\sin 30)+\cos 30(g\cos 30)=g({\cos }^{2}30-{\sin }^{2}30)=g\cos 60.$ Toh is particular ${\mathbf{f}}_b$ ki naive rotation deti hai ${\mathbf{f}}_w=(g\sin 60,0,g\cos 60)=(8.496,0,4.905)$ aur ${\mathbf{a}}_w=(8.496,0,-4.905)$ — zero nahi! Box sideways accelerate karta hua lagega. Step 3 — inconsistency dhundo aur fix karo. Trap hai ek frame mismatch: ${\mathbf{f}}_b=(g\sin 30,0,g\cos 30)$ world-to-body image hai $(R_b^w{)}^{\top }$ use karke opposite sign convention ke saath jitne $R_b^w$ hum phir apply karte hain. Consistently karo: standard $R_y(30^{\circ })$ ke saath, true still reading hai ${\mathbf{f}}_b=(R_b^w{)}^{\top }(0,0,g)=(-g\sin 30,0,g\cos 30)$. Usi ko wapas rotate karo: $f_{w,x}=\cos 30(-g\sin 30)+\sin 30(g\cos 30)=0,f_{w,z}=-\sin 30(-g\sin 30)+\cos 30(g\cos 30)=g({\sin }^{2}30+{\cos }^{2}30)=g.$ Hence ${\mathbf{f}}_w=(0,0,g)=(0,0,9.81)$ aur ${\mathbf{a}}_w={\mathbf{f}}_w+\mathbf{g}=(0,0,9.81)+(0,0,-9.81)=0.✓$ Lesson (poora point): $R_b^w$ aur $(R_b^w{)}^{\top }$ ko ek matched pair ke roop mein use karna chahiye — "world-from-body" out, "body-from-world" in. Jis moment tum ek body vector ko ek transpose ke saath rotate karte ho jo match nahi karta, ek still box $g\sin (\text{mismatch})$ se accelerate karta hua lagta hai. Pairing sahi karo aur ${\mathbf{a}}_w=0$ exactly milta hai, jaise kisi cheez ke liye hona chahiye jo move nahi kar rahi.

Ek constant bias linearly integrate hota hai: $\Delta \theta =b_{g}t$. $\Delta \theta =0.05\frac{\mathrm{deg}}{\text{s}}\times 120\text{s}=6.0^{\circ }.$ Linear kyun? Ek bias false rotation ki ek constant rate hai; ek constant ka integral time mein ek straight line hai. Koi signal isse fight nahi karta, toh yeh bas accumulate hota rehta hai.

Convert: $b_g=0.05\times \frac{\pi }{180}=8.727\times 10^{-4}\text{rad/s}$. $\Delta p_{\text{tilt}}=\frac{1}{6}(9.81)(8.727\times 10^{-4})(120{)}^3.$ Compute the pieces: $(120{)}^3=1,728,000$; phir $9.81\times 8.727\times 10^{-4}=8.561\times 10^{-3}$; times $1.728\times 10^6=14,793$; 6 se divide karo $=2465\text{m}$. $\boxed{\Delta p_{\text{tilt}}\approx 2.47\times 10^3\text{m}\approx 2.5\text{km}.}$ Itna brutal kyun? Jaise s03 mein growth curve dikhata hai, gyro bias frame ko tilt karta hai ($\propto t$), gravity component ko ek ramp acceleration error ke roop mein leak karta hai ($\propto t$); ek ramp ko do baar integrate karo toh $t^3$ milta hai. Time double karo toh error $8$ guna ho jaata hai.

Orientation. $(0,0,1)$ ke liye $[{\mathbf{\omega }}_b{]}_{\times }$ hai $\left(\begin{array}{ccc}0& -1& 0\\ 1& 0& 0\\ 0& 0& 0\end{array}\right),R_1=I+0.01\left(\begin{array}{ccc}0& -1& 0\\ 1& 0& 0\\ 0& 0& 0\end{array}\right)=\left(\begin{array}{ccc}1& -0.01& 0\\ 0.01& 1& 0\\ 0& 0& 1\end{array}\right).$ Yeh $z$ ke baare mein $\approx 0.01\text{rad}\approx 0.57^{\circ }$ ka ek tiny yaw hai — exactly turning ka ek step. Acceleration. $R_0=I$ ke saath: ${\mathbf{a}}_w=I{\mathbf{f}}_b+\mathbf{g}=(0,0,9.81)+(0,0,-9.81)=(0,0,0)$. Still, level box ki zero true acceleration hai — bilkul sahi. Velocity / position. ${\mathbf{v}}_1=0+(0,0,0)(0.01)=(0,0,0)$; ${\mathbf{p}}_1=0+(0)(0.01)=(0,0,0)$. Net: box thoda rotate hua lekin move nahi kiya — ek spinning-in-place, level IMU ke liye bilkul correct. Gravity cancellation kaam kiya kyunki $g$ remove karne se pehle orientation jaana jaata tha.

True body reading (L3.1 ki tarah transpose ke zariye world-to-body, matched pair) ka magnitude $g\sin 2^{\circ }$ across / $g\cos 2^{\circ }$ along split hai: ${\mathbf{f}}_b=(R_y(2^{\circ }){)}^{\top }(0,0,g)=(-g\sin 2^{\circ },0,g\cos 2^{\circ })=(-0.3424,0,9.8040).$ Agar algorithm galat se $R_0=I$ use karta hai (woh koi rotation nahi karta): ${\mathbf{a}}_w=I{\mathbf{f}}_b+\mathbf{g}=(-0.3424,0,9.8040)+(0,0,-9.81)=(-0.3424,0,-0.0060).$ Magnitude $0.342{\text{m/s}}^2$ ka ek phantom horizontal acceleration appear hota hai! Iska size exactly $g\sin 2^{\circ }$ hai: tilted axis ke along un-removed gravity component. Yahi $t^3$ drift ke peeche ka mechanism hai — ek chhoti attitude error ek real-looking horizontal push inject karti hai.

$\frac{1}{6}g{b}_g{t}^3=1$ set karo aur $t$ ke liye solve karo: $t={\left(\frac{6}{g{b}_g}\right)}^{1/3}={\left(\frac{6}{9.81\times 3\times 10^{-5}}\right)}^{1/3}={\left(2.039\times 10^4\right)}^{1/3}\approx 27.4\text{s}.$ Toh yeh gyro akela lagbhag 27 seconds tak $1\text{m}$ ke andar rehta hai. GPS update rate ke baare mein yeh kya batata hai. IMU ka error $1\text{m}$ box se ~27 s mein bahar nikal jaata hai, toh ek external fix usse kaafi pehle aana chahiye — tum kabhi last second tak nahi ruko, kyunki $t=27$ s ke paas error tezi se chad raha hai (cubic). Ek sensible rule hai comfortable margin se correct karo, jaise har kuch seconds (say $\lesssim 5$–$10$ s), taaki estimate baar baar $1\text{m}$ tak pahunchne se kaafi pehle wapas khincha jaaye. Konkretely: GPS/vision update period choose karo taaki worst-case IMU-only drift fixes ke beech budget ka ek chhota fraction rahe. Agar GPS sirf $1\text{Hz}$ ($1$ s period) par update kare toh kaafi hai; agar yeh sirf har $30$ s mein update kar sake, toh yeh gyro already budget blow kar dega fixes ke beech aur tumhe better gyro ya faster aiding sensor chahiye hogi. Design coupling hai: update period $\times$ safety margin $\ll$ drift-out time. Yahi Sensor Fusion & Kalman Filter ka kaam hai; brief outages mein (GPS-Denied Navigation) IMU tumhe carry karta hai, minutes mein nahi kar sakta.

• Accelerometer tilt estimate: correct on average (gravity ek fixed reference hai), lekin koi bhi real acceleration ya vibration ek single reading ko corrupt kar deta hai — yeh noisy but drift-free hai. Low frequency / long term par accha.
• Gyro tilt estimate: smooth aur instant-to-instant translational shake se immune, lekin uska bias isse drift karta hai $\Delta \theta =b_{g}t$ se — short-term clean, long-term galat. High frequency / short term par accha.
• Fusion fast changes ke liye gyro leta hai aur dheere dheere isse time ke saath accelerometer ke gravity reference ki taraf wapas khichta hai. Result gyro ki smoothness aur accelerometer ka zero long-term bias inherite karta hai, har sensor ki weakness cancel karta hai. Formally yahi trust-weighting hai jo Kalman filter karta hai; conceptually isliye Sensor Fusion & Kalman Filter, Dead Reckoning, aur Reference Frames — Body vs World kisi bhi real navigation stack mein sath sath appear hote hain.

Gravity subtract karne se pehle rotate karo ::: $ϵ$ angle ki tilt error $g\sin ϵ$ fake horizontal acceleration ke roop mein leak karti hai. Position drift gyro bias mein cubic kyun hai ::: Bias frame ko tilt karta hai ($\propto t$) → ramp accel error ($\propto t$) → do baar integrate karo → $t^3$. Tilt ke liye gyro + accel fuse kyun karein ::: Gyro smooth hai lekin drift karta hai; accel drift-free hai lekin noisy hai; fusion dono ki best qualities rakhta hai. atan2 woh fix karta hai jo arctan nahi kar sakta ::: arctan sirf $\pm 90^{\circ }$ span karta hai; atan2 dono component signs padh ke angle ko correct quadrant mein place karta hai.