Pole placement — Ackermann's formula: Visual walkthrough

3.5.34 · D2 · Physics › Guidance, Navigation & Control (GNC) › Pole placement — Ackermann's formula

Hum ek target chase kar rahe hain. Hamare paas ek machine hai $\overset{x}{˙} = A x + B u$ jiska natural behaviour humein pasand nahi, aur hum ek feedback rule $u = - K x$ choose karna chahte hain taaki combined machine bilkul waise behave kare jaise hum specify karte hain. Neeche sab kuch $K = [0 \dots 01] C^{- 1} α (A)$ ke pieces earn karta hai.

Step 1 — "Pole" kya hota hai, motion ke roop mein drawn

KYA. Ek system $\overset{x}{˙} = A x$ (abhi koi input nahi) time mein evolve hota hai. Chhota dot matlab hai "rate of change." State $x$ sirf numbers ki ek list hai jo machine ko abhi describe karta hai (position, velocity, ...). Matrix $A$ kehta hai ki current state change in state ko kaise push karta hai.

KYUN. Kuch bhi move karne se pehle humein jaanna chahiye ki ek pole kya control karta hai. Poori motion $e^{λ t}$ terms se bani hai, jahan har $λ$ (ek eigenvalue) ek pole hai. Pole ek akela number hai jo motion ka ek "flavour" set karta hai.

PICTURE. Teen poles, teen motions.

$λ > 0$ (real, positive) $\Rightarrow$ $e^{λ t}$ badhta hai — red curve bhaag jaata hai: unstable.
$λ < 0$ $\Rightarrow$ $e^{λ t}$ zero tak decay karta hai: stable. Zyada negative = faster decay.
$λ$ complex with negative real part $\Rightarrow$ decaying wiggle: oscillation jo khatam ho jaati hai.

Step 2 — Feedback matrix ko rebuild karta hai

KYA. Hum ek control input add karte hain. Rule $u = - K x$ kehta hai: poori state dekho, gains ki ek row $K = [k_{1} \dots k_{n}]$ se multiply karo, aur push back karo. Substitute karne par: $\overset{x}{˙} = A x + B (- K x) = nayi machine (A - B K) x .$

KYUN. Har symbol ka ek kaam hai:

$B$ — woh column jo kehta hai ki single push $u$ state mein kahaan enter karta hai.
$K$ — woh gains jo hum choose kar sakte hain, har state coordinate ke liye ek.
$A - B K$ — closed-loop matrix. Iske poles woh hain jo machine actually dikhayegi.

$K$ badalne se literally matrix reshape hoti hai, aur matrix reshape karne se poles move hote hain. Yahi hamaara ek lever hai.

PICTURE. State $x$ $A$ se flow karta hai; ek copy tap hoti hai, $- K$ se scale hoti hai, aur input mein feed back hoti hai.

Step 3 — Target ko ek polynomial mein package karo

KYA. $n$ alag poles track karne ki jagah, unhe ek single desired characteristic polynomial mein multiply karo: $α (s) = (s - μ_{1}) (s - μ_{2}) \dots (s - μ_{n}) = s^{n} + a_{n - 1} s^{n - 1} + \dots + a_{1} s + a_{0} .$

KYUN. Har factor $(s - μ_{i})$ exactly ek target pole par vanish karta hai. Unhe multiply karne se roots (poles kahaan hain) ko coefficients $a_{0}, \dots, a_{n - 1}$ (numbers jo ek matrix mein add ho sakte hain) mein trade kiya jaata hai. Coefficients woh hain jo ek matrix formula consume kar sakta hai; roots nahi.

PICTURE. Number line par roots coefficient list mein collapse ho jaate hain.

Left: chosen poles $μ_{i}$ red mein marked.
Right: wahi information coefficient row $[a_{0} a_{1} \dots a_{n - 1} 1]$ ke roop mein.

Goal restated: $K$ choose karo taaki $det (s I - (A - B K)) = α (s)$ .

Step 4 — Easy world: controllable canonical form

KYA. Ek special coordinate system hai, Controllable canonical form (CCF), jismein pole placement trivial hai. CCF mein matrices aisi dikhti hain: $A_{c} = 00 - a_{0}^{'} 10 - a_{1}^{'} 01 - a_{2}^{'}, B_{c} = 001 .$

KYUN yeh form. Do features ise magical banate hain:

Actual characteristic coefficients $a_{0}^{'}, a_{1}^{'}, \dots$ exactly $A_{c}$ ki bottom row par hote hain.
$B_{c}$ bottom par $1$ ke alawa sab zeros hai, isliye input sirf last row ko touch karta hai.

Isliye feedback $K_{c} = [k_{1} k_{2} k_{3}]$ sirf bottom row ko change karta hai: $A_{c} - B_{c} K_{c} ki bottom row hai [- (a_{0}^{'} + k_{1}) - (a_{1}^{'} + k_{2}) - (a_{2}^{'} + k_{3})] .$

PICTURE. Feedback vector surgically last row par land karta hai; upar sab kuch untouched rehta hai.

Desired coefficients $a_{i}$ hit karne ke liye, simply $a_{i}^{'} + k_{i + 1} = a_{i}$ solve karo, yani $k_{i + 1} = a_{i} - a_{i}^{'} .$ CCF mein, pole placement subtraction hai. Catch yeh hai: tumhari real machine almost kabhi CCF mein nahi hoti. Steps 5–6 ise fix karte hain.

Step 5 — Cayley–Hamilton gains ko bottom row mein load karta hai

KYA. Cayley–Hamilton theorem kehta hai ki har matrix apna khud ka characteristic polynomial satisfy karta hai: agar $χ$ , $A_{c}$ ki actual char. poly. hai, toh $χ (A_{c}) = 0$ (zero matrix). Ab desired polynomial $α$ ko matrix $A_{c}$ mein feed karo: $α (A_{c}) = A_{c}^{n} + a_{n - 1} A_{c}^{n - 1} + \dots + a_{1} A_{c} + a_{0} I .$

KYUN yeh zero nahi hai. $α$ aur $χ$ mein $A_{c}$ ki same powers hain lekin alag coefficients ( $a_{i}$ vs $a_{i}^{'}$ ). Subtract karne par, $χ (A_{c}) = 0$ se: $α (A_{c}) = α (A_{c}) - χ (A_{c}) = \sum_{i} (a_{i} - a_{i}^{'}) A_{c}^{i} .$ Har surviving coefficient exactly woh gain $a_{i} - a_{i}^{'} = k_{i + 1}$ hai jo Step 4 se tha. $A_{c}$ ki staircase shape ki wajah se, yeh precisely bottom row mein land karte hain. Isliye selector row unhe pluck kar leti hai: $[0 \dots 01] α (A_{c}) = K_{c} .$

PICTURE. $α (A_{c})$ ek full matrix hai, lekin sirf uski last row gains carry karti hai; $[0 \dots 01]$ arrow use read off karta hai.

$α (A_{c})$ — desired polynomial matrix par evaluate kiya gaya.
$[0 \dots 01]$ — actuated (last) coordinate select karta hai; isliye zeros-aur-ek ki row appear hoti hai. Yeh hai "bottom row padho."

Step 6 — Return ticket: $C^{- 1}$ answer ko ghar le jaata hai

KYA. Tumhari real $(A, B)$ CCF ka ek rotated/stretched version hai. Coordinates ka change $z = T x$ hai, aur dikhaya ja sakta hai ki transform hai: $T = \tilde{C} C^{- 1},$ jahan $C = [B A B A^{2} B \dots A^{n - 1} B]$ tumhari machine ki controllability matrix hai aur $\tilde{C}$ CCF wali hai.

KYUN $C$ appear hota hai. $C$ woh directions list karta hai jahan single input $A$ ke repeated applications se state ko push kar sakta hai: $B$ (abhi push karo), $A B$ (woh push ek step baad kahaan drift karta hai), $A^{2} B$ , aur aage. Agar yeh $n$ columns poore space ko span karte hain, toh har direction reachable hai — system controllable hai aur $C$ invertible hai. CCF gain $K_{c}$ ko $T$ se push back karke aur simplify karne par, canonical pieces cancel ho jaate hain aur sirf $C^{- 1}$ bachta hai: $K = [0 \dots 01] C^{- 1} α (A) .$

PICTURE. CCF easy room hai; $C^{- 1}$ tumhare coordinates mein wapas jaane ka darwaza hai.

$α (A)$ — target dynamics, tumhare coordinates mein compute kiya gaya ab (same $α$ , matrix $A$ ).
$C^{- 1}$ — easy room se wapas aane ka return ticket.
$[0 \dots 01]$ — bottom row padho.

Step 7 — Degenerate case: jab yeh toot jaata hai

KYA. Agar $det C = 0$ , toh $C^{- 1}$ exist nahi karta aur poora formula collapse ho jaata hai.

KYUN. Zero determinant matlab columns $B, A B, \dots, A^{n - 1} B$ space ko span nahi karte — state ka koi direction input se kabhi reach nahi kiya ja sakta. Tum woh pole move nahi kar sakte jo tum steer nahi kar sakte.

PICTURE. Reachable columns ek line par collapse ho jaate hain; poora direction stranded ho jaata hai.

Ek-picture summary

Formula ko right-to-left ek assembly line ki tarah padho: desired poles → coefficients → $α (A)$ → $C^{- 1}$ se un-transform karo → $[0 \dots 01]$ se bottom row padho → $K$ bahar aata hai.

Recall Feynman: poora walkthrough plain words mein

Tumhare paas ek wobbly machine hai. "Poles" woh secret numbers hain jo decide karte hain ki woh bhaagti hai, dheere settle hoti hai, ya kaampti hai. Tum better numbers choose karna chahte ho. Pehle tum woh numbers likhte ho jo tum chahte ho ek tidy polynomial mein — iske coefficients sirf ek shopping list hain. Phir tum ek moment ke liye pretend karte ho ki tumhari machine ek super-tidy form mein hai jahan push sirf ek row ko affect karta hai; wahaan poles fix karna plain subtraction hai, aur ek theorem (Cayley–Hamilton) guarantee karta hai ki sahi push-strengths us ek row mein line up hoti hain — tum unhe ek "last row ki taraf point karo" vector $[0 \dots 01]$ se grab karte ho. Tumhari real machine tidy nahi hai, isliye tum ek return ticket kharido — inverse controllability matrix $C^{- 1}$ — jo tidy answer ko tumhare real coordinates mein wapas le jaata hai. Teen pieces multiply karo aur $K$ bahar aata hai, woh exact push-per-measurement. Ek rule: tumhara single push har wobble tak pahunch sakna chahiye, warna return ticket exist nahi karta.