Visual walkthrough — Controllability matrix — rank test

3.5.32 · D2 · Physics › Guidance, Navigation & Control (GNC) › Controllability matrix — rank test

Neeche sab kuch zero se build kiya gaya hai. Agar koi symbol aata hai, pehle draw kiya gaya tha.

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Step 1 — "State" aur "input" actually dikhte kaise hain

KYA. Kisi bhi matrix se pehle, state ko ek single dot ki tarah picture karo jo ek room mein rehti hai. Agar system mein $n=2$ numbers hain usse describe karne ke liye (maan lo position $x_1$ aur velocity $x_2$), toh dot ek flat 2D floor par rehti hai. Uske coordinates HI state hain.

KYUN. Controllability ek geographic sawaal hai — "kya yeh dot floor ke har point tak pahunch sakti hai?" — isliye pehle agree karna zaroori hai ki state ek point hai, aur state ka change ek arrow hai (ek direction jisme dot move kar sakti hai).

PICTURE. Floor poora plane ${\mathbb{R}}^2$ hai. Dot origin par baithti hai (system "at rest"). Teen example target points mark hain — sawaal yeh hai ki kya hum dot ko har ek tak le ja sakte hain.

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Step 2 — Ek push sirf $B$ ke columns ke along move karti hai

KYA. Drift ko ek second ke liye freeze karo (pretend karo $A=0$). Tab $\dot{x}=Bu$. Tum $u$ mein jo bhi numbers dial karo, resulting motion arrow $Bu$ hai — $B$ ke columns ka ek blend.

KYUN. Hume jaanna hai ki bilkul pehle wale available directions kya hain. Single input ($m=1$) ke saath, $B$ ek column hai, isliye teri sirf instantaneous move hai "us ek arrow ke along aage ya peechhe." Yeh sab kuch reachable ka seed hai.

PICTURE. Origin se, arrow $B$ ek taraf point karta hai; $u$ positive ya negative dial karna dot ko us ek line pe upar-neeche slide karta hai. Us line se bahar sab kuch, is instant ke liye, unreachable hai.

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Step 3 — Drift push ko ek NAYI direction mein bend karti hai: enter $AB$

KYA. Ab drift wapas on karo. Jab dot $B$ ke along thodi si move karti hai, tab room ka field $A$ us displacement ko pakad ke kahin aur push karta hai. Jis rate se push-direction bend hoti hai woh $AB$ hai.

KYUN. Yahi controllability ka poora secret hai. Ek single steering arrow $B$ sirf ek line deta lagta hai — lekin agar $A$ us arrow ko rotate karke fresh direction $AB$ mein le jaaye, tab push karke, wait karke, aur phir push karke, tum ek doosri dimension tak reach kar sakte ho. Hume $AB$ chahiye kyunki woh pehli nayi direction hai jo drift free mein manufacture karti hai.

PICTURE. $B$ ke along shuru karo (blue). Field apply karo: blue arrow ke har point ko $A$ thoda nudge karta hai, arrow ko $AB$ ki taraf tip kar deta hai (pink). Blue + pink ab plane ka poora patch open karte hain, sirf ek line nahi.

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Step 4 — Rippling jari rakho: $A^{2}B,A^{3}B,\dots$

KYA. Repeat karo. $AB$ ko field mein ek baar aur feed karo $A^{2}B$ paane ke liye; $A^{3}B$ ke liye phir. Har ek tumhari original single push ka agla ripple hai.

KYUN. $n>2$ states ke liye, do directions room fill karne ke liye enough nahi hain. Hum field se ek aur fresh direction maangte rehte hain. Saari directions ki list jo ek push eventually reach kar sakti hai woh growing family hai: $B,AB,A^{2}B,A^{3}B,\dots$

PICTURE. Arrows ki ek chain bahar ki taraf fanning out: blue $B$, pink $AB$, yellow $A^{2}B$, har ek field se aur tip hota. 3D room mein yeh teen teen independent axes ke along point kar sakte hain — poora space fill kar sakte hain.

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Step 5 — List $A^{n-1}B$ par RUKTI hai (Cayley–Hamilton)

KYA. Ripples hamesha ke liye nayi directions nahi dete rehte. Jab tak tumne $A^{n-1}B$ tak compute kar liya, bilkul agla wala $A^{n}B$ sirf pehle ke arrows ka ek recycled blend hai — koi nayi direction nahi.

KYUN. Cayley–Hamilton theorem kehta hai ki har $n\times n$ matrix $A$ apni characteristic equation follow karta hai, jo tumhe $A^n$ ko $I,A,A^2,\dots ,A^{n-1}$ ke mix ke roop mein rewrite karne deta hai. Toh $A^{n}B$ un arrows mein wapas fold ho jaata hai jo tumhare paas pehle se hain. Isliye parent ka formula $A^{n-1}B$ par rukta hai — aage jaana waste chalk hai.

PICTURE. Step 4 ke arrows ka fan, lekin hona wala $A^{n}B$ (dashed grey) us region ke andar land karta hai jo already span ho chuka hai — woh koi nayi reach add nahi karta.

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Step 6 — Unhe stack karo: controllability matrix

KYA. Saare bachne wale direction-arrows ko ek bade matrix ke columns ki tarah side by side rakh do. Woh matrix $\mathcal{C}$ hai.

KYUN. "Kya yeh arrows room fill karte hain?" exactly woh sawaal hai "is matrix ke kitne independent columns hain?" — aur woh number rank hai. Toh Steps 1–5 ki geometry ek clean algebra check ban jaati hai.

PICTURE. Individual arrows ek grid ke labelled columns mein slide ho jaate hain; ek caption padhta hai "rank = kitne genuinely different directions mein point karte hain."

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Step 7 — Degenerate case: rank $<n$ (ek locked room)

KYA. Kabhi kabhi saare ripples ek chhoti flat mein squash rahe jaate hain — plane ke andar ek line, 3D ke andar ek plane. Tab $\mathrm{rank}(\mathcal{C})<n$ aur room ka kuch hissa sealed off hai.

KYUN. Hume failure dekhna zaroori hai, sirf success nahi. Agar $A$ kabhi $B$ ko uski apni line se nahi modhta (e.g. $B$ diagonal $A$ ki eigen-direction hai), toh har ripple usi line par rehta hai. Dot uspar forever trapped hai.

PICTURE. Ek board par do examples.

• Left (uncontrollable): $A=\left[\begin{array}{cc}1& 0\\ 0& 2\end{array}\right],B=\left[\begin{array}{c}1\\ 0\end{array}\right]$. $B$ aur $AB=\left[\begin{array}{c}1\\ 0\end{array}\right]$ dono horizontal axis par hain — vertical mode $x_2$ ek locked door hai. Rank $=1<2$.
• Right (controllable): same $A$ lekin $B=\left[\begin{array}{c}1\\ 1\end{array}\right]$. Ab $AB=\left[\begin{array}{c}1\\ 2\end{array}\right]$ off-axis tip karta hai; blue + pink poora plane span karte hain. Rank $=2$.

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Ek-picture summary

Ek board par har idea: tumhari single push $B$; field $A$ usse ripples $AB,A^{2}B,\dots$ mein tip karta hai; Cayley–Hamilton list ko $A^{n-1}B$ par snip karta hai; arrows $\mathcal{C}$ mein stack hote hain; aur verdict — full rank room fill karta hai (controllable), squashed rank ek locked door chhodta hai (uncontrollable).

Recall Feynman: poora walkthrough plain words mein

Tumhare paas ek room mein ek dot hai aur ek lever. Lever push karo aur dot ek arrow ke along slide karti hai — woh arrow $B$ hai. Lekin room mein ek current hai: jis pal dot move karti hai, current usse ek nayi direction mein sweep karta hai, aur woh nayi direction $AB$ hai. Thodi der baad current ek aur deta hai, $A^{2}B$, aur aise hi chalte rehta hai. Toh ek lever se bhi tum dhire dhire directions ka ek bouquet collect karte ho. Magic (Cayley–Hamilton) yeh hai ki exactly $n$ ke baad current kuch naya invent karna band kar deta hai — baad mein sab rerun hai. Toh tum pehle $n$ direction-arrows ko ek grid $\mathcal{C}$ mein line up karte ho aur ek sawaal poochte ho: kya woh enough independent tareekon se point karte hain ki har kone tak pahuncha ja sake? Independent wale gino — woh count rank hai. Agar woh coordinates ki sankhya $n$ ke barabar hai, poora room tumhara hai: controllable. Agar kam hai, koi direction ek locked door hai jo current kabhi nahi kholti chahe tum lever kitna bhi hilao: uncontrollable. Yeh akela count hi poora test hai.

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Connections

• Controllability matrix — rank test — woh parent result jise yeh pictures derive karti hain.
• Cayley–Hamilton theorem — woh reason ki ripple list $A^{n-1}B$ par kyun rukti hai (Step 5).
• Matrix exponential $e^{At}$ — woh propagator jiska expansion same powers of $A$ produce karta hai.
• State-space representation — jahan se $\dot{x}=Ax+Bu$ aata hai.
• Pole placement & state feedback — har mode move karne ke liye full rank chahiye.
• Kalman decomposition — Step 7 ke "locked room" ko formalize karta hai.
• Stabilizability — softer cousin: sirf unstable locked doors ka kholna zaroori hai.
• Observability matrix — rank test — mirror sawaal, steering ki jagah dekhna.

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