Observability matrix — rank test: Exercises — Hinglish deep dive

3.5.33 · D4 · Physics › Guidance, Navigation & Control (GNC) › Observability matrix — rank test

Everything here rests on the parent note Observability matrix — rank test. Keep this recipe in view:

Level 1 — Recognition

Kya tum $O$ bana sakte ho aur sabse aasaan cases mein uska size aur rank padh sakte ho?

L1.1 — Size of $O$

Ek system mein $n = 4$ states aur $p = 2$ outputs hain. Kuch compute kiye bina, $O$ ki dimensions batao aur woh number batao jo "full rank" ko equal hona chahiye.

Recall Solution

$O$ mein $n = 4$ blocks stack hote hain ( $C, C A, C A^{2}, C A^{3}$ ), har ek ka size $p \times n = 2 \times 4$ hota hai. Toh $O$ hai $(p n) \times n = (2 \cdot 4) \times 4 === 8 \times 4 ==$ . Full column rank ka matlab hai $rank (O) = n === 4 ==$ . Kaisa dikhta hai: ek tall skinny slab, 8 rows lambi, 4 columns chaudi — hume sirf 4 independent rows chahiye saare 4 states ko "dekhne" ke liye.

L1.2 — Ek diagonal system padho

$A = [3005], C = [10] .$ $O$ banao aur observability decide karo.

Recall Solution

$C A = [10] [3005] = [30]$ . $O = [1300] .$ Doosra column poora zeros hai ⇒ $rank = 1 < 2 = n$ ⇒ not observable. Kyun: $C$ sirf state 1 ko touch karta hai; state 2 apne eigenvector par akela rehta hai, kabhi $y$ ko influence nahi karta. $x_{0} = [01]$ unobservable hai.

L1.3 — Doosra diagonal slot padho

Same $A$ jaisa L1.2 mein hai lekin $C = [01]$ . Observable hai?

Recall Solution

$C A = [05]$ , toh $O = [0015]$ . Pehla column zero hai ⇒ rank $1$ ⇒ not observable. L1.2 ka symmetric case: ek single scalar output ek diagonal channel par sirf us ek mode ko hi dekh sakta hai.

Level 2 — Application

Coupled systems ke liye $O$ compute karo aur kernel interpret karo.

L2.1 — Chain of integrators

$A = [0010], C = [01] .$ Yahan $C$ doosre state ko measure karta hai. Observable hai?

Recall Solution

$C A = [01] [0010] = [00]$ . $O = [0010], rank = 1 < 2 \Rightarrow not observable .$ Interpret karo: state 1, state 2 ko feed karta hai ( $A$ mein "1" ke zariye), lekin output state 2 ko dekhta hai jabki state 2 ki dynamics ( $\overset{x}{˙}_{2} = 0$ ) state 1 ki info kabhi $y$ tak nahi pahunchati. Parent ke Example 1 se compare karo, jahan $C = [10]$ observable tha — "upstream" state measure karne se derivatives "downstream" wale ko reveal karte hain, lekin ulta nahi. Chain ki direction matter karti hai.

L2.2 — Genuinely coupled $A$

$A = [0 - 2 1 - 3], C = [10] .$

Recall Solution

$C A = [10] [0 - 2 1 - 3] = [01]$ . $O = [1001] = I \Rightarrow rank = 2 = n \Rightarrow observable .$ Kyun kaam karta hai: yeh ek position/velocity pair hai. $y = x_{1}$ position hai; $\overset{y}{˙} = C A x = x_{2}$ velocity hai. Do measurements, do states, khatam.

L2.3 — Unobservable direction explicitly dhundho

$A = [1011], C = [01] .$ Observability decide karo; agar not observable hai, toh $ker O$ mein ek vector batao aur verify karo ki woh $y \equiv 0$ deta hai.

Recall Solution

$C A = [01] [1011] = [01]$ . $O = [0011], rank = 1 \Rightarrow not observable .$ Kernel: $O x = 0$ solve karo $\Rightarrow x_{2} = 0$ , $x_{1}$ free. Toh $x_{0} = [10] \in ker O$ . Invisibility verify karo: $x_{2} (0) = 0$ ke saath, $A$ mein coupling "1" ke zariye $x_{2}$ mein push karta hai, lekin output seedha check karte hain. $y = x_{2}$ . $x_{0} = (1, 0)$ se start karke: $\overset{x}{˙}_{2} = x_{2}$ , aur $x_{2} (0) = 0 \Rightarrow x_{2} (t) = 0$ har $t$ ke liye. Kyunki $\overset{x}{˙}_{2}$ sirf $x_{2}$ par depend karta hai ( $A$ ki doosri row $[01]$ hai, koi $x_{1}$ term nahi), $x_{1}$ ki value kabhi output tak nahi pohonchti. Isliye $y (t) \equiv 0$ . ✅

Level 3 — Analysis

Brute force ki jagah structure, eigenvalues, aur duality use karo.

L3.1 — Repeated eigenvalue, scalar output

$A = [4004], C = [a b], (a, b) \neq = (0, 0) .$ Dikhao ki scalar $C$ ki koi bhi choice ise observable nahi banati.

Recall Solution

$C A = [4 a 4 b] = 4 C$ . $O = [a 4 a b 4 b] .$ Row 2 $= 4 \times$ row 1 hai, toh rows parallel hain ⇒ $rank = 1 < 2$ har $(a, b)$ ke liye. Kyun fundamental hai: $A = 4 I$ ka eigenvalue $4$ hai geometric multiplicity $2$ ke saath — poora 2-D eigenspace. Ek single output do identical, decoupled modes ko distinguish nahi kar sakta (parent ka Example 3 dekho). PBH test ke zariye, eigenvalue $λ = 4$ par hume $rank [A - 4 I C] = 2$ chahiye, lekin $A - 4 I = 0$ hai, sirf ek row $C$ bachti hai ⇒ rank $1$ . Tumhe $p \geq 2$ outputs chahiye.

L3.2 — Duality check

$A = [0010]$ , $C = [10]$ diya hua hai (parent ka Example 1, observable). Duality use karke ek controllable pair batao, phir verify karo.

Recall Solution

Duality: $(A, C)$ observable $⟺ (A^{⊤}, C^{⊤})$ controllable. Yahan $A^{⊤} = [0100], B := C^{⊤} = [10] .$ Controllability matrix $C = [B A^{⊤} B]$ . $A^{⊤} B = [0100] [10] = [01]$ . $C = [1001], rank = 2 \Rightarrow controllable .✅$ Consistent hai: observable original $\Rightarrow$ controllable dual, jaisa promise kiya tha.

L3.3 — PBH cross-check

$A = [2003]$ , $C = [11]$ ke liye, do tarikhon se observability decide karo: rank test aur PBH test.

Recall Solution

Rank test: $C A = [23]$ , $O = [1213]$ . Determinant $= 1 \cdot 3 - 1 \cdot 2 = 1 \neq = 0$ ⇒ rank $2$ ⇒ observable. PBH: har eigenvalue $λ \in {2, 3}$ check karo. $rank [A - λ I C] = 2$ chahiye.

$λ = 2$ : $001011$ — rows $(0, 1)$ aur $(1, 1)$ independent hain ⇒ rank $2$ . ✔
$λ = 3$ : $- 1 01 001$ — rows $(- 1, 0)$ aur $(1, 1)$ independent hain ⇒ rank $2$ . ✔ Dono eigenvalues pass karte hain ⇒ observable. Dono methods agree karte hain. Distinct eigenvalues + ek $C$ jo har eigenvector ko "touch" kare ⇒ observable.

Figure — Observability matrix — rank test

Level 4 — Synthesis

Systems ko spec ke hisaab se design karo, aur observable/unobservable split banao.

L4.1 — Diagonal $A$ ke liye $C$ design karo

$A = diag (1, 2, 3)$ hai. Sabse chhoti output rows ki sankhya aur ek explicit $C$ dhundho jo $(A, C)$ ko observable banaye.

Recall Solution

Eigenvalues $1, 2, 3$ distinct hain, har ek ka 1-D eigenspace hai (coordinate axes). Koi modal degeneracy nahi, toh ek single output ( $p = 1$ ) kaam kar sakta hai bas agar woh teeno eigenvectors ko touch kare. $C = [111]$ choose karo. $O = 111124139 .$ Yeh nodes $1, 2, 3$ mein ek Vandermonde matrix hai; uska determinant $= (2 - 1) (3 - 1) (3 - 2) = 2 \neq = 0$ ⇒ rank $3$ ⇒ observable. Smallest hai $p = 1$ . ✅ Kyun Vandermonde aata hai: row $k$ hai $C A^{k - 1} = (1 \cdot λ_{1}^{k - 1}, 1 \cdot λ_{2}^{k - 1}, 1 \cdot λ_{3}^{k - 1})$ — eigenvalues ki powers. Distinct nodes ⇒ nonzero determinant.

L4.2 — Prescribed unobservable state wala system banao

Ek 2-state $(A, C)$ construct karo jiska unobservable subspace exactly $span {(1, 1)^{⊤}}$ ho.

Recall Solution

Hum chahte hain $ker O = span {(1, 1)^{⊤}}$ , yaani rank $1$ , aur $(1, 1)^{⊤}$ se $y \equiv 0$ mile. $C = [1 - 1]$ choose karo taki $C (1, 1)^{⊤} = 0$ ho. Ab ensure karna hai ki $C A (1, 1)^{⊤} = 0$ bhi ho (taki poori line kernel mein rahe — ek $A$ -invariant subspace). $A = [1001] = I$ choose karo. Tab $C A = C = [1 - 1]$ hai. $O = [11 - 1 - 1], rank = 1, ker = span {(1, 1)^{⊤}} .✅$ Check karo: $O (1, 1)^{⊤} = (0, 0)^{⊤}$ aur $(1, - 1)^{⊤}$ row space span karta hai, toh sirf $(1, 1)$ direction invisible hai. Yeh ek mini Kalman decomposition hai: observable part $(1, - 1)$ direction hai, unobservable part $(1, 1)$ hai.

L4.3 — Ek unobservable system repair karo

$A = [2002]$ , $C = [10]$ unobservable hai (parent ka Example 2). Ek doosri output row add karo taki pair observable ho jaye, kam se kam naye nonzero entries use karke.

Recall Solution

Unobservable state $(0, 1)^{⊤}$ hai. Hume ek naya row chahiye jisme uska nonzero projection ho. Sabse sasta: $C_{2} = [01]$ add karo. $C = [1001] = I .$ Tab $O$ ke pehle do rows mein pehle se $I$ hai ⇒ rank $2$ ⇒ observable. ✅ Kyun yahan do outputs forced hain: $A = 2 I$ ka ek repeated eigenvalue hai 2-D eigenspace ke saath (L3.1). Koi scalar output ise kabhi fix nahi kar sakta — tumhe $p \geq 2$ rakhna hi hoga. Yeh designer ki laziness nahi hai; yeh ek hard structural requirement hai.

Level 5 — Mastery

Structural facts prove karo aur multiple tools ek saath use karo.

L5.1 — Similarity invariance

Prove karo: agar $\tilde{A} = T A T^{- 1}$ aur $\tilde{C} = C T^{- 1}$ ho (state coordinates ka change $z = T x$ ), toh $rank (\tilde{O}) = rank (O)$ hoga. Toh observability coordinates par depend nahi karti.

Recall Solution

Transformed matrix ka ek block row compute karo: $\tilde{C} \tilde{A}^{k} = (C T^{- 1}) (T A T^{- 1})^{k} = C T^{- 1} T A^{k} T^{- 1} = C A^{k} T^{- 1},$ $(T A T^{- 1})^{k} = T A^{k} T^{- 1}$ use karke (beech mein $T^{- 1} T$ pairs cancel ho jaate hain). Saare $k = 0, \dots, n - 1$ stack karke: $\tilde{O} = O T^{- 1} .$ Kyunki $T^{- 1}$ invertible hai, isse right-multiply karne se rank nahi badlti. Isliye $rank (\tilde{O}) = rank (O)$ . ∎ Matlab: observability system ki property hai, un basis ki nahi jisme tum ise likhte ho — exactly wahi cheez jis par Kalman decomposition rely karta hai.

L5.2 — Numeric mastery problem

$A = 00 - 6 10 - 11 01 - 6, C = [100]$ (ek companion / controllable-canonical-looking form). $det O$ compute karke observability determine karo.

Recall Solution

$C = [100]$ . $C A = [010]$ ( $A$ ki pehli row). $C A^{2} = (C A) A = [010] A = [001]$ ( $A$ ki doosri row). $O = 100010001 = I \Rightarrow det = 1 \neq = 0 \Rightarrow rank = 3 = n .$ Observable. ✅ Yeh "observer companion" pattern hai: integrators ki chain ka pehla state measure karne se successive derivatives $x_{2}, x_{3}$ reveal hote hain, toh poori chain unfold ho jaati hai.

L5.3 — Kalman-filter tie-in (conceptual + ek computation)

Ek state estimator ke liye $x (0)$ reconstruct karne ke liye, pair observable hona chahiye. $A = [0 - 1 10]$ (ek pure oscillator) aur $C = [10]$ diya hua hai, observability verify karo, aur explain karo ki "ek oscillator ki position dekhna" kya reveal karta hai.

Recall Solution

$C A = [10] [0 - 1 10] = [01]$ . $O = [1001] = I \Rightarrow rank = 2 \Rightarrow observable .✅$ Interpretation: harmonic oscillator mein, $x_{1}$ position hai aur $\overset{x}{˙}_{1} = x_{2}$ velocity hai. Position aur uska slope dekhne se kisi bhi instant par dono states milte hain, toh ek estimator true trajectory par lock kar sakta hai. Yahan eigenvalues $\pm i$ hain (undamped), phir bhi observability hold karti hai — parent ki mistake note reinforce hoti hai: observability stability se independent hai.

Recall Self-test summary (sabhi levels khatam karne ke baad reveal karo)

L1: size $= p n \times n$ ; ek zero column rank kill kar deta hai.
L2: coupling ki ek direction hoti hai — hamesha $C A^{k}$ compute karo.
L3: repeated eigenvalues ko $p \geq 2$ chahiye; distinct eigenvalues ko phir bhi $C$ ko har eigenvector touch karna hoga; PBH rank test ko cross-check karta hai.
L4: $C$ ko Vandermonde se design karo; unobservable subspace = $O$ ka $A$ -invariant kernel; repeated modes zyada sensors force karte hain.
L5: $\tilde{O} = O T^{- 1}$ ⇒ coordinate-free; observability ⟂ stability.

Connections

Observability matrix — rank test (parent)
Controllability matrix — rank test (duality, L3.2 mein use kiya)
Cayley–Hamilton theorem (isliye $A^{n - 1}$ par rukते hain)
PBH test (eigenvalue cross-check, L3.1 aur L3.3)
Kalman decomposition (observable/unobservable split, L4.2)
Kalman Filter (observability chahiye, L5.3)
Rank and null space of a matrix (kernel = unobservable states)
State-space representation