Observability — when KF can estimate state: Exercises

3.5.23 · D4 · Physics › Guidance, Navigation & Control (GNC) › Observability — when KF can estimate state

Intuition Is page kaise use karein

Har problem neeche clearly stated hai, phir ek collapsible solution follow karta hai jise aap sirf try karne ke baad khol sakte ho. Upar se neeche karo: levels ek doosre pe build karte hain. Jo kuch bhi chahiye tha woh parent note mein bana tha; agar koi symbol unfamiliar lagey, woh note usay define karta hai.

Jin objects ke saath hum kaam karte hain. State ek column of numbers hai $x = x_{1} x_{2} ⋮ x_{n}$ — sab kuch jo system ek instant mein "hai" (positions, velocities, biases...). Hum $x (0)$ likhte hain state ke liye time zero pe, woh starting snapshot jo hum recover karne ki koshish kar rahe hain. System evolve karta hai $\overset{x}{˙} = A x$ (koi inputs nahi) aur hum measure karte hain $y = C x$ .

Woh ek tool jis pe hum rely karte hain: observability matrix $O = C C A C A^{2} ⋮ C A^{n - 1}, system observable ⟺ rank (O) = n .$ Yahan $n$ states ki number hai ( $x$ ki size), $A$ $n \times n$ dynamics matrix hai, aur $C$ measurement matrix hai (ek row har sensor ke liye). "Rank" ka matlab bas hai: kitni rows genuinely independent hain — kitni information carry karti hain jo baaki se already nahi mil sakti.

Neeche wala figure poore page ka ek map hai, koi formula nahi. Har box exercise ladder ka ek level hai; colour batata hai kitna mushkil hai (chalk-blue = warm-up, pale-yellow = analysis, chalk-pink = mastery). Har box ke andar, italic line us level ka flavour of question hai. Arrows intended climb order dikhate hain. Upar ki single line moral hai: chahe aap kisi bhi box mein ho, actual computation hamesha wohi teen moves hai — $O$ build karo, independent rows gino, $n$ se compare karo.

Figure — Observability — when KF can estimate state

Ise kaise padhein: bottom-left box se start karo (L1 Recognition — " $O$ build karo, rank gino") aur arrows follow karo upar-right ki taraf L5 Mastery tak ("kya KF converge karega?"). Notice karo ki har box ka question wordier aur more physical hota jaata hai jaise aap climb karte ho, lekin neeche ka tool kabhi nahi badlta. Agar aap kabhi kisi exercise ke andar lost feel karo, is ladder pe apna box dhundho aur poochho: "main kis rung pe hun, aur kya mera current step waqai sirf rank test ka ek disguised roop hai?"

Level 1 — Recognition

Goal: $n$ read off karo, $O$ mechanically build karo, rank gino.

Recall Solution 1.1

KYA karte hain: hum sirf $C$ aur $C A$ chahiye kyunki $n = 2$ hai (hum $C A^{n - 1} = C A^{1}$ tak stack karte hain). KYUN: har extra row starting snapshot $x (0)$ ke baare mein ek aur equation hai (woh state at time zero jo upar introduce ki gayi); Cayley–Hamilton (dekho Cayley–Hamilton Theorem) kehta hai $A^{n - 1}$ pe rokne se kuch miss nahi hota.

Row 1: $C = [10]$ . Row 2: $C A = [10] [0 - 2 1 - 3] = [01]$ . $O = [1001] .$ Dono rows independent directions mein point karti hain (yeh identity hai), toh $rank (O) = 2 = n$ . Answer: observable. ✅

Recall Solution 1.2

KYUN sirf do rows: $n = 2$ , toh Cayley–Hamilton ke hisab se hum $C$ aur $C A$ stack karte hain aur ruk jaate hain — $C A^{2}$ in ka ek combination hoga aur koi information add nahi karega. $C = [01]$ , aur $C A = [01] [3005] = [05]$ . $O = [0015] .$ Row 2 bas $5 \times$ row 1 hai — koi nayi direction nahi. Pehla column sab zeros hai, toh rows ka koi combination kabhi bhi state 1 ko "reach" nahi kar sakta. $rank (O) = 1 \neq = 2$ . Answer: NOT observable. State $x_{1}$ decoupled hai ( $A$ diagonal hai) aur unmeasured hai, toh yeh kabhi $y$ mein leak nahi hota. ❌

Level 2 — Application

Goal: real GNC dynamics plug in karo; rank ko physically interpret karo.

Recall Solution 2.1

KYUN sirf do rows: $n = 2$ , toh test ko exactly $C$ aur $C A$ chahiye (Cayley–Hamilton stack ko $C A^{n - 1}$ pe cap karta hai). $C = [01]$ , $C A = [01] [0010] = [00]$ . $O = [0010], rank = 1 \neq = 2.$ Answer: NOT observable. Interpretation: velocity kabhi position pe depend nahi karti (position dynamics mein wapas feed nahi hoti — $A$ ka pehla column zero hai), toh velocity dekhne se aapko pata nahi chalta ki cart kahan hai. Parent note ke Example 1 se contrast karo, jahan aap position measure karte ho aur velocity $A_{12} = 1$ ke through leak hoti hai. Yahan information flow galat direction mein chal raha hai. Ek KF ki $x_{1}$ pe variance kabhi shrink nahi hogi.

Recall Solution 2.2

KYUN sirf do rows: phir se $n = 2$ , toh hum $C$ aur $C A$ stack karte hain aur ruk jaate hain — Cayley–Hamilton ke hisab se higher powers redundant hain. $C = [10]$ , $C A = [10] [00 - 1 - 0.1] = [0 - 1]$ . $O = [10 0 - 1], rank = 2 = n .$ Answer: observable. ✅ Bias $b$ ab bhi angle rate ko feed karta hai ( $A_{12} = - 1 \neq = 0$ ), toh iska footprint measured angle mein appear hota hai. Yahi mechanism Gyro bias estimation / star-tracker calibration ke peeche hai.

Level 3 — Analysis

Goal: parameters, symbolic conditions, aur degenerate values handle karo.

Recall Solution 3.1

KYUN sirf do rows: $n = 2$ ; $C$ aur $C A$ stack karo. Kyunki $O$ tab square hai ( $2 \times 2$ ), hum full rank saste mein determinant se check kar sakte hain (nonzero determinant $⟺$ full rank). $C = [1 α]$ , $C A = [1 α] [0010] = [01]$ . $O = [10 α 1], det O = (1) (1) - (α) (0) = 1.$ Determinant har $α$ ke liye $1$ hai, toh $rank = 2$ hamesha hai. Answer: saare real $α$ ke liye observable. Sensor mein extra velocity term $α$ kabhi position term se already mili information ko destroy nahi karta — doosri row $C A$ independently velocity pin down kar leti hai.

Recall Solution 3.2

KYUN sirf do rows: $n = 2$ , toh hum $C, C A$ stack karte hain aur $det O$ full-rank test ke roop mein use karte hain. $C = [c 1]$ , $C A = [c 1] [2102] = [2 c + 1 2]$ . $O = [c 2 c + 1 12], det O = 2 c - (2 c + 1) = - 1.$ Determinant $- 1$ hai, $c$ se independent. Answer: har $c$ ke liye observable — yeh kabhi observability nahi khota. Yahan trap yeh hai (neeche dekho) ki assume karo ki ek free parameter ka koi na koi bura value zaroor hoga.

Recall Solution 3.3

KYUN yahan teen rows: ab $n = 3$ , toh Cayley–Hamilton hume $C A^{n - 1} = C A^{2}$ pe rokta hai; hum $C, C A, C A^{2}$ stack karte hain. $C = [010]$ . $C A = [010] A = [0 - 2 1]$ . $C A^{2} = (C A) A = [0 - 2 1] A = [04 - 4]$ . $O = 000 1 - 2 4 01 - 4 .$ Poora pehla column zero hai — koi row state $x_{1}$ ko reach nahi karti. Toh $rank (O) \leq 2 < 3$ . Answer: NOT observable. State $x_{1}$ ( $- 1$ mode) states 2 aur 3 se fully decoupled hai aur unmeasured hai; yeh invisible hai. (Completeness ke liye: rows 2 aur 3 milke $x_{2}, x_{3}$ plane span karti hain, toh $rank = 2$ . Sirf isolated $x_{1}$ mode lost hai.)

Level 4 — Synthesis

Goal: observability ko duality, Gramians, aur detectability ke saath combine karo.

Recall Solution 4.1

Set $\overset{ˉ}{A} = A^{⊤} = [0100]$ aur $B = C^{⊤} = [10]$ . $\overset{ˉ}{A} B = [0100] [10] = [01]$ . $C = [B \overset{ˉ}{A} B] = [1001], rank = 2.$ Full rank $\Rightarrow (A^{⊤}, C^{⊤})$ controllable, $(A, C)$ observable se match karta hai. Duality confirmed. Dekho Controllability — when we can steer the state aur Detectability & Stabilizability.

Recall Solution 4.2

$A$ diagonal ke saath, $e^{A t} = [e^{- t} 0 0 e^{- 3 t}]$ , aur $C^{⊤} C = [1111]$ . Integrand ki $(i, j)$ entry hai $(C^{⊤} C)_{ij} e^{(λ_{i} + λ_{j}) t}$ jisme $λ_{1} = - 1, λ_{2} = - 3$ hain. Kyunki system stable hai ( $λ_{i} < 0$ ), yeh exponentials decay karte hain, toh $\infty$ tak integral converge karta hai: $\int_{0}^{\infty} e^{(λ_{i} + λ_{j}) t} d t = \frac{- 1}{λ _{i} + λ _{j}}$ : $(W_{o})_{11} = \frac{- 1}{- 1 - 1} = \frac{1}{2}, (W_{o})_{22} = \frac{- 1}{- 3 - 3} = \frac{1}{6} .$ Interpretation: $(W_{o})_{22} = \frac{1}{6} < (W_{o})_{11} = \frac{1}{2}$ . Faster-decaying mode ( $λ = - 3$ ) output ko influence karne mein kam time spend karta hai, toh woh less observable hai — KF use larger uncertainty ke saath recover karta hai. Yeh Riccati Equation & steady-state covariance se steady-state covariance se connect karta hai.

Level 5 — Mastery

Goal: KF consequence, detectability, aur ek fix design karne ke baare mein reason karo.

Exercises se pehle, ek definition jo parent note ne rely ki thi lekin ab hum precisely define karenge, kyunki Level 5 isi par hinge karta hai.

Recall Solution 5.1

Setup — kyun update ek scalar tak collapse hota hai. Poori covariance time update $\dot{P} = A P + P A^{⊤} + Q$ obey karti hai, jahan $Q ⪰ 0$ process-noise injection hai. Ise single hidden axis $v$ pe project karo (eigenvector jisme $A v = λ v$ ): $p = v^{⊤} P v$ likho us axis ke along variance ke liye aur $q = v^{⊤} Q v \geq 0$ usme feed hone wale noise ke liye. $A v = λ v$ use karte hue (toh $v^{⊤} A = λ v^{⊤}$ bhi), dono matrix terms mein se har ek $λ p$ contribute karta hai: $\overset{p}{˙} = v^{⊤} (A P + P A^{⊤} + Q) v = λ p + λ p + q = 2 λ p + q .$ Crucially, measurement update yahan kuch nahi karta: mode unobservable hai, toh Kalman gain $v$ ke along correct nahi kar sakta ( $C v$ koi information carry nahi karta). Scalar $\overset{p}{˙} = 2 λ p + q$ poori kahani hai.

Case $λ = - 0.5$ (stable mode). Tab $\overset{p}{˙} = - p + q$ . $\overset{p}{˙} = 0$ set karne pe steady value $p_{\infty} = q$ milti hai, ek finite number; koi bhi starting $p$ isme relax ho jaata hai. Bounded ✅. Agar hum yeh mode kabhi nahi dekhte, phir bhi yeh apne aap decay kar leta hai, toh injected noise pile up nahi ho sakta. System detectable hai: unobservable lekin stable.

Case $λ = + 0.5$ (unstable mode). Tab $\overset{p}{˙} = + p + q$ . Homogeneous part $e^{2 (0.5) t} = e^{t}$ ki tarah grow karta hai, toh $p (t) \to \infty$ — koi finite steady state exist nahi karta. Diverges ❌. Ek unobservable aur unstable mode fatal hai: filter exactly wahan blind hai jahan error explode ho rahi hai.

Woh property jo line draw karti hai: detectability. Ek Kalman Filter ka bounded steady-state covariance hoga iff har unobservable mode asymptotically stable hai (yaani $λ < 0$ ). $λ = - 0.5$ detectable hai (bounded); $λ = + 0.5$ detectable nahi hai (diverges). Yeh Kalman Filter — prediction & update Riccati recursion ka covariance behaviour hai.

Recall Solution 5.2

KYUN hum purana kaam reuse kar sakte hain: original block ${C, C A, C A^{2}}$ already $x_{2}, x_{3}$ plane span karta tha (rank 2) lekin column 1 empty chhod diya tha. Nayi row directly $x_{1}$ measure karti hai ( $g$ se scale hoke), jo exactly missing direction hai — toh hume sirf check karna hai ki yeh column 1 fill karta hai ya nahi. Do independent purani rows ke upar fresh row stack karo; $O_{new}$ ka relevant $3 \times 3$ block hai $g 00 01 - 2 001 .$ Iska determinant compute karo (pehle column ke along expand karo, kyunki sirf top entry nonzero hai wahan): $det = g \cdot det [1 - 2 01] = g \cdot (1 \cdot 1 - 0 \cdot (- 2)) = g \cdot 1 = g .$ Yeh determinant nonzero hai (aur isliye block, aur therefore $O_{new}$ , full rank $3$ hai) iff $g \neq = 0$ . Answer: har $g \neq = 0$ ke liye observable; $g = 0$ pe ab bhi unobservable (zero-gain sensor kuch measure nahi karta, column 1 pehle jaisi empty rehti hai). Yeh design lesson hai: ek decoupled mode dekhne ke liye, aapko ek sensor add karna hoga jo use nonzero gain ke saath actually touch kare — aur hamesha confirm karo ki rank $n$ tak bad gayi, parent note ki "more sensors tab tak help nahi karte jab tak woh rank add na karein" baat ko echo karte hue.

Recall Self-test checklist

Kya tum memory se kisi bhi $n = 2$ system ke liye $O$ build kar sakte ho? ::: Haan: $C$ ko $C A$ pe stack karo, $2 \times 2$ determinant check karo. Woh ek number kaunsa hai jo square $O$ ki observability decide karta hai? ::: Iska determinant — nonzero matlab full rank matlab observable. Woh kaunsi property hai jo ek KF ko unseen mode tolerate karne deti hai? ::: Detectability — unseen mode asymptotically stable hai. Ek decoupled invisible state ko rescue kaise karte hain? ::: Ek sensor add karo jo us state ko touch kare (us column mein nonzero gain) aur confirm karo ki rank $n$ tak badh jaati hai.