Question bank — Observability — when KF can estimate state

False — observability ek rank property hai. Sensors jo already-visible linear combinations measure karte hain woh $\mathcal{O}$ mein rows toh add karte hain lekin rank nahi, isliye tall $C$ kuch bhi guarantee nahi karta.

False — observable hona sirf itna guarantee karta hai ki $x(0)$ ek interval ke upar eventually recoverable hai; ek poorly conditioned (weakly observable) $\mathcal{O}$ sensor noise ko kaafi der tak estimate ko duba dene deta hai.

True — hum $u=0$ set karte hain kyunki known inputs subtract ho sakte hain; sirf unforced response $y=C{e}^{At}x(0)$ hi $x(0)$ ke baare mein information carry karta hai.

False — agar unobservable modes pehle se hi stable hain (woh khud se decay karte hain) toh system detectable hai, aur yahi bounded steady-state covariance ke liye kaafi hai.

Form mein True, meaning mein False — ye duals hain: $(A,C)$ observable $⟺(A^{\top },C^{\top })$ controllable. Ek state ko bahar padhne ke baare mein hai, doosra use andar drive karne ke baare mein.

True — unka difference $x_1(0)-x_2(0)$ $\ker (\mathcal{O})$ mein pada hai aur $y(t)\equiv 0$ produce karta hai, isliye koi bhi estimator unhe distinguish nahi kar sakta.

False — observability sirf $(A,C)$ ki property hai. Noise $Q$ us blind direction mein sirf covariance ko badhaata hai; yeh kabhi koi measurement pathway nahi kholti.

False — stability $A$ ke eigenvalues ke baare mein hai, observability is baare mein hai ki $C$ unse kaise couple karta hai. Ek decaying lekin unmeasured decoupled mode stable phir bhi unobservable hota hai (woh detectable hai, observable nahi).

Extra block $C{A}^n$ Cayley–Hamilton ki wajah se redundant hai ($A^n$ lower powers ka linear combo hai), isliye yeh rank kabhi nahi badha sakta. $C{A}^{n-1}$ pe rukna shortcut nahi hai, yeh complete hai.

Observability ko full column rank $=n$ chahiye, full row rank nahi. Matrix tall hai; hum sirf yahi require karte hain ki iske columns independent hon taaki $\mathcal{O}x(0)=\text{data}$ ka unique solution ho.

Agar $A_{12}=0$ hai aur $C$ state 2 ko touch nahi karta, toh uske baare mein kuch bhi $y$ mein nahi aata; measurement update us variance ko untouched chhod deta hai, isliye wahan ek chhoti reported variance matlab filter tumse jhooth bol raha hai.

Invertibility (full rank) sirf haan/nahi wala sawaal decide karta hai. Ek near-singular $\mathcal{O}$ (bahut chhota smallest singular value) technically observable hai lekin weakly — conditioning ke liye Gramian eigenvalues check karo.

$A=\left[\begin{array}{cc}0& -1\\ 0& 0\end{array}\right]$, $C=[10]$ ke saath, hume $CA=[0-1]$ milta hai, toh $\mathcal{O}=\left[\begin{array}{cc}1& 0\\ 0& -1\end{array}\right]$ ka rank 2 hai — bias angle ko drift karta hai, isliye angle dekhne se use back out kiya ja sakta hai. Dekho Gyro bias estimation / star-tracker calibration.

Rank test $T$ se independent hai; agar $x(0)\in \ker (\mathcal{O})$ hai toh yeh hamesha $y\equiv 0$ produce karta hai. Lambi window woh information conjure nahi kar sakti jo kabhi output se coupled hi nahi thi.

Har derivative $y^{(k)}(0)=C{A}^{k}x(0)$ unknown $x(0)$ ko constrain karne wali ek aur independent linear equation hai; unhe stack karna hi woh tarika hai jisse hum itni equations ikaTha karte hain ki use pin down kar sakein.

Kernel mein koi $x(0)$ har block $C{A}^{k}x(0)=0$ bana deta hai, toh $y$ ke sabhi derivatives zero ho jaate hain aur $y(t)\equiv 0$ hota hai — yeh exactly "silent" initial states ka set hai jo kisi bhi output ko invisible hain.

Measurement update $P\leftarrow (I-KC)P$ wahan kuch nahi karta (us direction mein koi output nahi hai), jabki time update process noise $Q$ daalta rehta hai aur unstable dynamics use inflate karte hain — toh yeh kabhi converge nahi karta.

Detectability sirf yehi demand karti hai ki unobservable modes stable hon, toh woh khud se shrink karte hain; filter kisi aisi cheez ke baare mein blind rahne ka afford kar sakta hai jo apne aap fade ho jati hai, isliye $P$ phir bhi bounded steady state pe aa jaata hai.

Kyunki information $y$ tak interaction $C{A}^k$ ke through pahunchti hai — dynamics $A$ hidden states ko measured ones mein kaise shuffle karta hai — na ki $C$ ke paas kitne raw channels hain usse.

Observability (aur noise input ki stabilizability) exactly woh condition hai jo guarantee karti hai ki Riccati equation ka ek unique stabilizing steady-state solution hai, taaki $P$ na diverge kare na kisi bekar value par stuck rahe.

Haan — agar uski dynamics mein coupling nonzero hai (jaise velocity position ko drive kare, $A_{12}\ne 0$), toh yeh $CA$ ke through output mein leak hota hai aur reconstruct kiya ja sakta hai, chahe koi sensor use directly na dekhe.

Unobservable — $\mathcal{O}=[c]=[0]$ ka rank 0 hai. Sirf ek state ki koi measurement nahi, sensor disconnected hai aur kuch bhi recover nahi ho sakta.

Nahi — doosra block ek sealed box mein hai jismein $A_{12}=0$ hai aur koi $C$ coupling nahi; uske states unobservable subspace banate hain aur kabhi $y$ mein appear nahi karte.

Haan — tab $\mathcal{O}$ ke pehle block mein already identity hai, jo $A$ chahe kuch bhi ho rank $n$ deta hai; sabhi states ka direct measurement trivially observable case hai.

Tab $k\ge 1$ ke liye $C{A}^k=0$ hai, toh $\mathcal{O}$ sirf $C$ tak collapse ho jaata hai. Sirf woh states observable hain jinhe $C$ touch karta hai; koi dynamics nahi toh kuch leak nahi hota, aur unmeasured states permanently blind hain.

Detectable nahi — detectability require karti hai ki unobservable modes asymptotically stable hon (strictly decaying); ek marginal (undamped, non-decaying) unobservable mode yeh fail karta hai, toh covariance bounded guarantee nahi ho sakti.

LTI system ke liye nahi — $(A,C)$ fixed constants hain, toh observability global hai. Point-dependence sirf tab aati hai jab nonlinear system ko linearize karo, jahan $A,C$ trim state ke saath change karte hain.