Controllability matrix — rank test: Exercises — Hinglish deep dive

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

Intuition Is page ko kaise use karein

Ye problems ek ladder ki tarah chadhti hain. L1 sirf pieces ko pehchanne ke liye hai; L5 mein ideas invent aur combine karne padte hain. Har problem ka full worked solution ek collapsible callout mein chhupa hua hai — pehle khud try karo, phir reveal karo. Ye page seedha parent note par build karta hai.

Pehle, taaki koi symbol par na atke, yahan har letter diya gaya hai — ek picture ke saath:

$x$ hai state vector: numbers ki poori list jo system ko abhi describe karti hai (positions, velocities, angles, …). Geometrically, $x$ ek $n$ -dimensional space mein ek point hai; jaise jaise time badhta hai yeh ek path trace karta hai.
$u$ hai input (control): woh numbers jo aap choose karte ho system ko push karne ke liye — thruster levels, voltages. Geometrically, $u$ choose karta hai ki kaun si push directions on ki jaati hain.
$\overset{x}{˙} = A x + B u$ woh rule hai jisse point $x$ move karta hai: $A$ batata hai ki $x$ apne aap kaise drift karta hai, $B$ batata hai ki aapka push $u$ ise kaise nudge karta hai.
$A$ hai $n \times n$ system matrix (state apne aap kaise evolve hoti hai).
$B$ hai $n \times m$ input matrix (aapke controls state ko kaise push karte hain).
$C = [B A B A^{2} B \dots A^{n - 1} B]$ hai controllability matrix.
$rank (C)$ = uske columns ki independent directions ki sankhya. Controllable $⟺ rank = n$ .

Poora game geometric hai: aapke inputs sirf $C$ ke columns ke span ke andar hi motion produce kar sakte hain. Controllable ka matlab hai ki woh span poore state space ko fill karta hai — kuch bhi locked out nahi hai. Neeche di gayi picture woh mental image hai jo har problem mein saath rakhni hai.

Figure — Controllability matrix — rank test

Level 1 — Recognition

Recall Solution L1.1

(a) $x$ mein 2 rows hain, isliye $n = 2$ . (b) $u$ ek single scalar hai ( $B$ mein ek column), isliye $m = 1$ . (c) $C$ hai $n \times (n \cdot m) = 2 \times (2 \cdot 1) = 2 \times 2$ . (d) Aap $B, A B, \dots, A^{n - 1} B$ compute karte ho, yani $A^{1} B$ tak. Yeh 2 blocks hain: $B$ aur $A B$ . (Cayley–Hamilton kehta hai ki $A^{2} B$ kuch naya nahi deta — dekho Cayley–Hamilton theorem.)

Recall Solution L1.2

Doosri row poori zero hai, isliye columns sirf $[a 0]$ form ke vectors tak pahunch sakte hain. Yeh ek direction span karta hai, isliye $rank (C) = 1 < 2 = n$ . Controllable nahi. Dead direction hai $x_{2}$ -axis $[01]$ : koi bhi input combination kabhi wahan motion produce nahi kar sakta.

Level 2 — Application

Recall Solution L2.1

Step 1 — $A B$ (sirf yahi ek block chahiye kyunki $n = 2$ ): $A B = [0 - 1 2 - 3] [01] = [2 - 3] .$ Step 2 — stack karo: $C = [B A B] = [01 2 - 3] .$ Step 3 — determinant (square $C$ ke liye sabse fast full-rank test): $det C = 0 \cdot (- 3) - 2 \cdot 1 = - 2 \neq = 0.$ Nonzero determinant $\Rightarrow rank = 2 = n$ . ✅ Controllable.

Recall Solution L2.2

Step 1: $A B = [40 0 - 1] [30] = [120] .$ Step 2: $C = [30120] .$ Step 3: $det = 3 \cdot 0 - 12 \cdot 0 = 0$ . Doosri row poori zero $\Rightarrow rank = 1 < 2$ . ❌ Controllable nahi. Physically: $A$ diagonal hai, isliye dono modes ek dusre se baat nahi karte, aur $B$ ki doosri entry $0$ hai, isliye input kabhi mode 2 ko touch nahi karta. Mode 2 bina input ke $\overset{x}{˙}_{2} = - x_{2}$ ki tarah evolve karta hai — untouchable. (Aisi systems ko split karne ke liye Kalman decomposition dekho.)

Recall Solution L2.3

Step 1: $A B = [4 - 1] .$ Step 2: $C = [11 4 - 1] .$ Step 3: $det = 1 \cdot (- 1) - 4 \cdot 1 = - 5 \neq = 0 \Rightarrow rank = 2$ . ✅ Controllable. Input ab dono modes ko push karta hai, aur kyunki dono modes ke different eigenvalues hain ( $4$ aur $- 1$ ), $B$ aur $A B$ alag alag directions mein point karte hain aur plane span karte hain.

Level 3 — Analysis

Recall Solution L3.1

Step 1: $A B = [1 λ] .$ Step 2: $C = [11 1 λ] .$ Step 3: $det C = 1 \cdot λ - 1 \cdot 1 = λ - 1.$ Uncontrollable exactly jab $det = 0$ , yani $λ = 1$ . Kyun: $λ = 1$ par dono diagonal modes ka eigenvalue same ho jaata hai. Repeated eigenvalues aur single input ke saath, input dono modes ko distinguish nahi kar sakta — $B$ aur $A B = λ B$ parallel ho jaate hain. Yahi PBH test (Popov–Belevitch–Hautus) ka intuition hai: ek mode uncontrollable hota hai jab $B$ ek repeated/shared eigen-direction ke "flat" ho jaata hai.

Neeche di gayi figure dekho: jaise jaise $λ$ $3$ se $1$ ki taraf slide karta hai, vector $A B$ $B$ ke kareeb aata jaata hai. $λ = 1$ par dono coincide ho jaate hain — $C$ ke dono columns ek line par lie karte hain, isliye unka span ek line hai, plane nahi, aur rank $1$ tak drop ho jaata hai.

Recall Solution L3.2

Step 1: $A B = [24] = 2 B .$ ( $A = 2 I$ , isliye $A B = 2 B$ — $B$ ke parallel!) Step 2: $C = [1224] .$ Step 3: $det = 1 \cdot 4 - 2 \cdot 2 = 0 \Rightarrow rank = 1$ . ❌ Controllable nahi. Reachable subspace = span of $B = [12]$ , yani line $x_{2} = 2 x_{1}$ . Unreachable target: us line ke bahar koi bhi point, jaise $x_{f} = [10]$ . Check karo: kya $[10]$ $[12]$ ka multiple hai? Nahi. Isliye yeh origin se kabhi reach nahi ho sakta. Deep reason: jab $A = 2 I$ (scalar times identity), toh har power $A^{k} B$ sirf $B$ ka ek scalar multiple hai — saare blocks ek line par rehte hain. Ek single input kabhi us line se bahar nahi ja sakta.

Neeche di gayi figure mein, magenta line reachable subspace $x_{2} = 2 x_{1}$ hai; $B$ (navy) aur $A B = 2 B$ (orange) dono us par lie karte hain, jabki violet target $(1, 0)$ line se door hai — provably out of reach.

Level 4 — Synthesis

Recall Solution L4.1

Yahan $n = 3$ hai, isliye humein $B, A B, A^{2} B$ chahiye. $A B$ : $A B = 000100010001 = 010 .$ $A^{2} B = A (A B)$ : $A^{2} B = 000100010010 = 100 .$ Stack: $C = 001010100 .$ Determinant: yeh ek permutation (anti-diagonal) matrix hai; $det = - 1 \neq = 0 \Rightarrow rank = 3 = n$ . ✅ Controllable. Kyun kaam karta hai: single input bottom mode ko hit karta hai, aur $A$ ki har power ise chain mein upar shift karti hai — "ripple" $B \to A B \to A^{2} B$ teeno directions fill kar deta hai. Isliye chains of integrators controllability ki poster child hain, aur isliye Pole placement & state feedback yahan freely har pole assign kar sakta hai.

Recall Solution L4.2

Single input kyun fail karta hai: $A = I$ , isliye $A B = I B = B$ . Tab $C = [B B]$ — dono columns identical hain, rank $\leq 1 < 2$ . Koi single-column $B$ jeet nahi sakta. Sabse chhota working $B$ : humein $m = 2$ columns chahiye jinka combined span $R^{2}$ ho. Lo $B = [1001] .$ Tab $A B = B$ , aur $C = [B A B] = [10011001]$ . Rectangular $C$ ka rank — upar wale definition box se minors rule use karo. Is $2 \times 4$ matrix mein $2 \times 2$ sub-block (columns 1 aur 2) $[1001]$ hai jiska $det = 1 \neq = 0$ hai, isliye $rank = 2 = n$ . ✅ Controllable. (Row reduction se bhi same milta hai: do non-zero rows rank 2 confirm karti hain.) Lesson: jab $A = I$ hota hai toh har mode ka eigenvalue same hota hai aur $A$ bilkul koi mixing nahi karta; aapko directions seedha $B$ ke through supply karni padti hain. Aapko kam se kam utne independent input columns chahiye jitna eigenvalue multiplicity hai.

Level 5 — Mastery

Recall Solution L5.1

$A B$ : $A 01 a = 1 a 0 .$ $A^{2} B = A (A B)$ : $A 1 a 0 = a 00 .$ Stack: $C = 01 a 1 a 0 a 00 .$ Determinant (bottom row ke along expand karo — sirf ek nonzero entry hai, position $(3, 1)$ mein $a$ ): $det C = a \cdot det [1 a a 0] = a (1 \cdot 0 - a \cdot a) = a (- a^{2}) = - a^{3} .$ Controllable $⟺ det \neq = 0 ⟺ - a^{3} \neq = 0 ⟺ a \neq = 0$ . Interpretation: yahan $B = 01 a$ input ko middle state mein inject karta hai (aur, agar $a \neq = 0$ ho, toh bottom state mein bhi). $a = 0$ par input sirf middle rung mein enter karta hai, isliye yeh third state kabhi reach nahi kar sakta — uncontrollable. Yeh L4.1 se alag injection point hai (jiska $B = 001$ bottom mein enter karta hai aur controllable rehta hai). Yahan koi bhi $a \neq = 0$ bottom rung ko reconnect kar deta hai aur full rank restore kar deta hai. Bilkul yahi type ka edge case hai jiske baare mein Stabilizability care karta hai.

Recall Solution L5.2

Key idea (duality): $(A, C)$ observable hai $⟺ (A^{T}, C^{T})$ controllable hai. Isliye transposed pair ki controllability matrix banao. (Yeh bridge hai Observability matrix — rank test ki taraf.) Lo $A^{T} = [- 1 2 0 - 4]$ aur $C^{T} = [10]$ . $A^{T} C^{T}$ : $[- 1 2 0 - 4] [10] = [- 1 2] .$ Stack: $C_{dual} = [10 - 1 2] .$ Determinant: $1 \cdot 2 - (- 1) \cdot 0 = 2 \neq = 0 \Rightarrow rank = 2 = n$ . Controllable dual $\Rightarrow$ observable original. ✅ Aapne same rank test reuse kiya $A \to A^{T}, B \to C^{T}$ ke saath.

Recall Solution L5.3

$A B$ : $[0 - a_{0} 1 - a_{1}] [01] = [1 - a_{1}] .$ Stack: $C = [01 1 - a_{1}] .$ Determinant: $0 \cdot (- a_{1}) - 1 \cdot 1 = - 1 \neq = 0$ saare $a_{0}, a_{1}$ ke liye. Isliye $rank = 2 = n$ hamesha. ✅ Hamesha controllable. Yeh kyun matter karta hai: companion form by construction controllable hai, isliye controllable systems ko isme transform kiya ja sakta hai — aur isliye Pole placement & state feedback companion form use karta hai feedback gains directly read off karne ke liye. Note karo ki $a_{0}$ kabhi $det C$ mein appear hi nahi kiya: yeh poles change karta hai, controllability nahi.

Recall Ek line summary

Har problem teen same moves par reduce ho gayi: $A^{k} B$ blocks compute karo $A^{n - 1} B$ tak, unhe $C$ mein stack karo, aur rank test karo (determinant jab $C$ square ho; minors ya row-reduction jab rectangular ho). Full rank $= n$ ⇒ controllable; kuch bhi kam ⇒ ek mode forever locked hai.