Visual walkthrough — Observability — when KF can estimate state

3.5.23 · D2 · Physics › Guidance, Navigation & Control (GNC) › Observability — when KF can estimate state

Hum sirf ek kahani apne paas rakhte hain: tum ek signal $y(t)$ dekh sakte ho, aur tum uss chhupi hui starting condition $x(0)$ ka andaaza lagaana chahte ho jisne use produce kiya. Bus itna hi. Neeche ka sab kuch usi ek wish se nikalta hai.

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Step 1 — Chhupi hui state aur chhhoti si window

KYA. Ek box ka picture banao. Andar numbers ki ek list hai jo time ke saath badlti rehti hai — yeh state hai, jise $x(t)$ likhte hain. Agar box mein ek ghoomta hua satellite ho, toh $x$ ho sakta hai "current angle" aur "current spin rate." Tum box nahi khol sakte. Tumhare paas sirf ek chhhoti si window hai jo tumhe ek combined reading $y(t)$ dikhati hai.

KYO. Estimation exactly yahi situation hai: sensors tumhe kabhi bhi poori internal sach nahi dete, sirf ek filtered jhaaank milti hai. Kisi bhi maths se pehle, humein yeh tay karna hoga ki hum kya dekh sakte hain aur kya nahi.

PICTURE. Figure mein, violet cloud woh sab kuch hai jo box ke andar rehta hai (state). Orange window sirf ek patli beam bahar aane deti hai — woh beam $y(t)$ hai. Magenta gear ko dekho jo wall ke peeche poori tarah chhupi hai: hum poora page yeh poochte hue bitaayenge ki "kya hum phir bhi us chhupi hui gear ko figure out kar sakte hain?"

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Step 2 — Chhupi hui numbers kaise chalti hain, aur window unhe kaise padhti hai

KYA. Do rules hamare box ko govern karte hain.

1. State apne aap evolve hoti hai: $\dot{x}=Ax$. Dot ka matlab hai "rate of change per second." Matrix $A$ kehti hai ki har chhupi hui number doosron ko kaise nudge karti hai.
2. Window ek fixed lens hai: $y=Cx$. Matrix $C$ kehti hai ki chhupi hui numbers ka kaun sa combination bahar leakta hai.

$\underset{\text{state kitni tezi se badlti hai}}{\underbrace{\dot{x}}}=\underset{\text{mixing rule}}{\underbrace{A}}\underset{\text{chhupi hui numbers}}{\underbrace{x}},\underset{\text{jo tum dekhte ho}}{\underbrace{y}}=\underset{\text{window}}{\underbrace{C}}\underset{\text{chhupi hui numbers}}{\underbrace{x}}$

Sirf yeh do matrices kyon aur kuch nahi? Humne full state-space form se input term $Bu$ hata diya. Reason: inputs $u$ woh cheezein hain jo tum command karte ho, isliye tumhe unka effect pehle se pata hai aur tum use subtract kar sakte ho. Sirf unforced motion ($\dot{x}=Ax$) test karta hai ki kya window khud kaafi achhi hai. Observability pair $(A,C)$ ki property hai — mixing rule plus window — $B$ ki kabhi nahi.

PICTURE. Figure mein $A$ ko violet arrows ke roop mein dikhaya gaya hai chhupi hui gears ke beech (kaun kise push karta hai) aur $C$ ko single orange arrow ke roop mein gears se window tak. Agar kisi gear ka koi path nahi hai — na arrows, na window — jo orange beam tak pahunche, toh problem aane waali hai.

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Step 3 — Film ko aage chalana: time $t$ par state

KYA. Starting list $x(0)$ di hui ho, toh rule $\dot{x}=Ax$ ka exactly ek solution hai: $x(t)=e^{At}x(0).$ Object $e^{At}$ propagator hai: ise start do, yeh film ko time $t$ tak fast-forward kar deta hai.

Exponential kyon aur koi simple cheez kyon nahi? Ek single number ke liye, $\dot{x}=ax$ solve hoke $x(t)=e^{at}x(0)$ deta hai — exponential woh function hai jiska growth rate apne aap ka $a$ times hai. Matrices bhi same tarah behave karti hain jab $e^{At}=I+At+\frac{1}{2}(At{)}^2+\cdots$ define kar lo. Hum yeh tool isliye choose karte hain kyunki yeh "kaunsa function hai jiska slope $A$ times khud hai?" ka unique answer hai — jo exactly equation $\dot{x}=Ax$ hai.

Ab window upar se lagao. Sensor actually jo record karta hai woh hai: $y(t)=C{e}^{At}x(0).$

PICTURE. Figure mein same starting dot $x(0)$ (magenta) ko ek curved violet trajectory par $e^{At}$ ke saath le jaate hue dikhaya gaya hai, aur har instant par orange window ek point sample karti hai $y(t)$ curve banane ke liye jo right side par hai.

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Step 4 — Derivatives se aur equations nikalna

KYA. Ek snapshot $y(0)=Cx(0)$ aksar bahut kam hota hai. Toh hum dekhte hain ki $y$ kaise badlta hai — uska slope, uski curvature, aur aage bhi — exactly start $t=0$ par:

$y(0)=Cx(0),\dot{y}(0)=CAx(0),\ddot{y}(0)=C{A}^{2}x(0),\dots$

Har line same unknown $x(0)$ mein ek fresh linear equation hai.

Derivatives kyon? Derivative poochti hai "signal abhi kis direction mein aur kitni tezi se ja raha hai?" $y=C{e}^{At}x(0)$ ko differentiate karne par har baar ek factor of $A$ neeche aata hai (kyunki $e^{At}$ ka slope $A{e}^{At}$ hai), aur $t=0$ par propagator $e^{A\cdot 0}=I$ cleanly vanish ho jaata hai. Toh $k$-th derivative crisp row $C{A}^{k}x(0)$ deti hai. Hum derivatives isliye use karte hain kyunki yeh time ke saath signal ki shape ko static equations ki stack mein convert karne ka natural tarika hai.

PICTURE. Figure $y$ curve par $t=0$ par zoom karta hai: orange dot $y(0)$ (height) hai, magenta tangent line $\dot{y}(0)$ (slope) hai, violet curving arc $\ddot{y}(0)$ (bending) hai. Har geometric feature = ek equation = $x(0)$ ke baare mein ek aur clue.

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Step 5 — Hum kyon rukते hain: Cayley–Hamilton pile ko cap karta hai

KYA. Hum forever differentiate karte reh sakte the: $C{A}^3,C{A}^4,\dots$. Lekin hum $C{A}^{n-1}$ par ruk jaate hain. Cayley–Hamilton theorem kehta hai ki har $n\times n$ matrix $A$ apni khud ki characteristic equation satisfy karti hai, jo rearrange hoke deta hai: $A^n=c_{n-1}{A}^{n-1}+\cdots +c_{1}A+c_{0}I.$ Toh $A^n$ sirf earlier powers $I,A,\dots ,A^{n-1}$ ka recipe hai. $C$ se multiply karo: row $C{A}^n$ woh rows ka blend hai jo tum pehle se likh chuke ho.

Yeh kyon matter karta hai. Yeh promise karta hai ki test finite hai. Iske bina, "forever differentiate karo" kabhi terminate nahi hoti. Cayley–Hamilton guarantee karta hai: $n$ rows ke baad, koi naya direction of information kabhi appear nahi ho sakta. Row $n$, row $n+1$, infinity tak — sab redundant.

PICTURE. Figure rows $C,CA,\dots$ ko horizontal bars ke roop mein stack karta hai. Pehli $n$ bars fresh colours hain; jis moment hum $C{A}^n$ pahunche, bar ko earlier bars ka shadow blend dikhaya gaya hai — literally same colour mix — yeh dikhate hue ki yeh kuch naya nahi add karta.

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Step 6 — Ek matrix $\mathcal{O}$ mein stack karna

KYA. Saari $n$ useful equations ko ek single tall matrix equation mein line up karo:

$$\underset{\mathcal{O}\text{(observability matrix)}}{\underbrace{\left[\begin{array}{c}C\\ CA\\ C{A}^2\\ ⋮\\ C{A}^{n-1}\end{array}\right]}}x(0)=\left[\begin{array}{c}y(0)\\ \dot{y}(0)\\ \ddot{y}(0)\\ ⋮\\ y^{(n-1)}(0)\end{array}\right].$$

Term by term. Har block $C{A}^k$ ek "clue lens" hai: $C$ raw window hai, aur har extra $A$ ka matlab hai "aur dynamics ko dekhne se pehle $k$ baar pot hilane do." Right-hand column pure recorded data hai ($y$ ki values aur derivatives $t=0$ par). Sirf unknown $x(0)$ hai.

Stack kyon karo? Ek single lens state ke directions miss kar sakta hai. Bahut saari lenses stack karna (har ek $A$ se alag tarah se stirred) humara best possible combined view hai. Agar yeh stack phir bhi $x(0)$ pin down nahi kar sakti, toh koi cheez kabhi nahi kar sakti.

PICTURE. Figure literally orange-tinted lens rows ko tall violet block mein stack karta hai jise $\mathcal{O}$ label kiya gaya hai, magenta unknown $x(0)$ right par hai aur data column neeche.

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Step 7 — Verdict: full column rank ka matlab "solvable" hai

KYA. Ab hamare paas $\mathcal{O}x(0)=\text{data}$ hai. Iska $x(0)$ ke liye exactly ek answer kab hoga?

Rank count karta hai ki $\mathcal{O}$ ki rows kitne truly independent directions cover karti hain. Agar $\mathrm{rank}(\mathcal{O})=n$ (hidden states ki sankhya jitna wide), toh equations har direction pin down karti hain — unique solution. Agar $\mathrm{rank}(\mathcal{O})<n$, toh ek leftover direction hai jo lenses ne kabhi touch nahi kiya.

Rank kyon, "number of rows" kyon nahi? Tum sau sensors pile kar sakte ho, lekin agar woh sab same combination report karte hain, toh woh height add karte hain aur zero new independent directions — koi new rank nahi. Yahi exact trap hai jiske baare mein parent note warn karta hai.

PICTURE. Figure do cases side by side dikhata hai. Left (observable): $\mathcal{O}$ state-space ko ek single point par collapse karta hai → ek answer. Right (unobservable): starting states ka ek poora magenta arrow sab $y(t)\equiv 0$ map karte hain → window unhe distinguish nahi kar sakti.

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Step 8 — Degenerate cases jo tumhe skip nahi karne chahiye

KYA. Teen edge scenarios real GNC designs decide karte hain. Har ek figure mein apna tiny panel hai.

(a) Ek completely decoupled, unmeasured state. Agar $A$ mein ek hidden number hai jo na koi visible cheez push karta hai aur na $C$ use dekhti hai, toh har $C{A}^k$ block mein uska column zero hai. Lens beam kabhi us tak nahi pahuncha. → unobservable (parent ka Example 2).

(b) Ek hidden state jo ek visible ko push karta hai. Position–velocity with only a position sensor: velocity directly invisible hai, lekin $A_{12}\ne 0$ ka matlab hai velocity position drive karti hai, toh yeh $\dot{y}$ mein leak ho jaati hai. → observable (parent ka Example 1). Yeh gear-pushing-a-gear ki jeet hai.

(c) Full rank lekin barely — weak observability. $\mathcal{O}$ technically rank $n$ tak pahunchi, lekin ek direction almost kuch contribute nahi karta (uska beam ek whisper hai). Measurement noise phir use duba deta hai. Yeh rank se akele nahi pakda jaata; tumhe strength grade karne ke liye observability Gramian chahiye, aur yeh explain karta hai kyon detectability woh softer fallback hai jo Riccati steady state actually require karta hai.

Teeno kyon cover karo? Rank test binary hai, lekin reality "blind" se "crystal clear" tak ek spectrum par rehti hai. Ek designer jo sirf rank check karta hai woh case (c) se blindsided ho jaayega.

PICTURE. Teen stacked panels: (a) ek sealed box mein gear — koi beam nahi; (b) ek hidden gear jiska motion lit gear mein ripple karta hai; (c) ek faintly lit gear — barely observable, noise-vulnerable.

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Ek-picture summary

Is page ka sab kuch ek flow mein compress ho jaata hai: $y$ dekho → derivatives lo → $C{A}^k$ lenses ko $\mathcal{O}$ mein stack karo → uska rank check karo → ek unique $x(0)$ milta hai (ya nahi milta).

Recall Feynman: poora walkthrough simple words mein

Tumhare paas spinning gears ka ek locked box hai aur ek chhhoti si window. Tum kuch der ke liye window film karte ho. Pehla frame thoda batata hai. Picture kaise move karti hai (uski speed) aur batata hai. Woh motion kaise curve karti hai aur bhi zyada batata hai — har ek ek fresh clue hai. Lekin tum hamesha ke liye clues gain nahi karte reh sakte: $n$ ke baad (Cayley–Hamilton ka promise) har naya clue sirf ek purana clue ek disguise mein hai. Toh tum exactly $n$ clues gather karte ho, unhe $\mathcal{O}$ called ek badi clue-sheet mein stack karte ho, aur poochते ho: kya yeh clues, milke, har gear ki starting position nail down karte hain? Agar haan — full rank — box observable hai aur ek Kalman Filter sirf window se poori inside rebuild kar sakta hai. Agar kisi gear ne clues mein kahin koi fingerprint nahi chhhoda, woh $\ker (\mathcal{O})$ mein chhupta hai, aur duniya ka koi filter use dhundh nahi sakta. Aur even jab har gear koi kuch mark chhhode, noise ke through padhne ke liye bahut faint mark (weak observability) woh trap hai jo tumhe Gramian aur detectability ki taraf bhejti hai.