Pole placement — Ackermann's formula: Question bank

3.5.34 · D5 · Physics › Guidance, Navigation & Control (GNC) › Pole placement — Ackermann's formula

Prerequisites jo open rakhne chahiye: Controllability and the controllability matrix, Cayley–Hamilton theorem, Characteristic polynomial, Eigenvalues and system stability, Controllable canonical form.

Teen objects jinpar neeche ke har trap ka daromadar hai

Traps se pehle, exact notation pin down karte hain, taaki koi bhi symbol use hone se pehle samjha ja sake.

Neeche ka figure yeh "sirf last coordinate actuated hai" structure dikhata hai, aur nearly-singular $C$ placement ko fragile kaise banata hai.

Figure — Pole placement — Ackermann's formula

True ya false — justify karo

True ya false: Ackermann's formula kisi bhi linear system ke closed-loop poles ko jahan chaaho place kar sakta hai.

False. Isko pair

(A, B)

ka controllable hona zaroori hai (taaki

C

invertible ho) aur single-input hona chahiye. Ek uncontrollable direction mein ek pole hota hai jise tum literally nahi hila sakte.

True ya false: jo polynomial tum

α (A)

mein daalo woh system ki apni characteristic polynomial hai.

False. Tum desired polynomial

α (s) = \prod (s - μ_{i})

daalo jo tumhare target poles se bani hai. Actual polynomial use karne se

α (A) = 0

milta hai Cayley–Hamilton theorem se, isliye

K = 0

— bilkul bhi feedback nahi.

True ya false: saare closed-loop poles ko zyada negative banana hamesha best design hota hai.

False. Bahut negative real parts matlab bahut fast decay, lekin uske liye bade gains

K

chahiye, isliye bada control effort

u = - K x

, jo actuators ko saturate karta hai aur measurement noise amplify karta hai. Speed aur effort ke beech trade-off hai — LQR — optimal pole placement alternative dekho.

True ya false: agar tum sign convention

u = + K x

kar do, toh Ackermann's

K

same rehta hai.

False. Formula

u = - K x

ke liye derive kiya gaya hai jo closed loop

A - B K

deta hai.

u = + K x

ke saath loop

A + B K

hai, isliye result negate karna padega (warna sign error double ho jaata hai).

True ya false: formula mein row vector

[0 \dots 01]

arbitrary hai — koi bhi row chalegi.

False. Controllable canonical form mein sirf last coordinate actuated hota hai, isliye feedback gains

α (A_{c})

ki bottom row mein aate hain. Woh specific selector row unhe read karta hai; koi bhi aur row galat entries pakad legi.

True ya false: do alag desired polynomials same gain

K

de sakti hain.

False. Ek fixed controllable

(A, B)

ke liye, alag desired coefficient sets

(a_{0}, \dots, a_{n - 1})

CCF mein alag closed-loop bottom rows produce karte hain, aur invertible maps

C^{- 1}

aur CCF change of basis one-to-one hain — isliye composite map coefficients

\to K

ek bijection hai. Alag pole sets ⟹ alag

K

True ya false: complex desired poles

K

ke entries ko complex banana par majboor karte hain.

False. Jab tak complex poles conjugate pairs mein aate hain,

α (s)

ke real coefficients hote hain,

α (A)

ek real matrix hai, aur

K

real hai. Isliye hum

- 1 \pm j

place karte hain, kabhi akela

- 1 + j

nahi.

True ya false: pole placement eigenvalues ke saath-saath eigenvectors bhi badal deta hai.

True.

A - B K

A

se genuinely alag matrix hai; iske dono eigenvalues aur eigenvectors generally change hote hain. Pole placement sirf eigenvalues (poles) guarantee karta hai — mode shapes saath mein aa jaate hain.

Error dhundo

C

singular tha, isliye maine use square banane ke liye ek column drop kiya aur use invert kiya." Galti kahan hai?

Singular

C

matlab system uncontrollable hai — poora ek direction unsteerable hai. Columns trim karke is problem ko fix nahi kar sakte; us direction mein pole fixed hai aur Ackermann simply apply nahi hota.

"Mere desired poles

- 2

aur

- 2

the, isliye maine

α (s) = (s - 2)^{2}

likha." Fix karo.

Sign slip.

- 2

par pole contribute karta hai factor

(s - (- 2)) = (s + 2)

, isliye

α (s) = (s + 2)^{2} = s^{2} + 4 s + 4

(s - 2)

use karne se pole

+ 2

par place hoga — unstable.

"Maine

α (A)

compute kiya

A

α

mein plug karke, lekin constant term

a_{0}

ko scalar

a_{0}

treat kiya." Kya galat hai?

Yaad raho

a_{0}

α (s)

ka constant coefficient hai. Ek matrix polynomial mein constant term

a_{0} I

hona chahiye, scalar times identity matrix, na ki akela scalar:

α (A) = A^{n} + \dots + a_{1} A + a_{0} I

"System mein do inputs the, isliye maine

B

ke do columns

C = [B A B \dots]

mein stack karke Ackermann apply kiya." Yeh unsafe kyun hai?

Ackermann ek single-input recipe hai; leading selector

[0 \dots 01]

aur CCF derivation scalar input assume karte hain. Multi-input placement mein extra freedom hoti hai aur alag methods chahiye hote hain.

"Kyunki

A - B K

char. poly

α (s)

share karta hai, uske eigenvectors companion matrix ke equal hain." Leap dhundo.

Characteristic polynomial share karna same eigenvalues guarantee karta hai, same eigenvectors nahi. Similar matrices eigenvalues share karte hain lekin coordinate-dependent eigenvectors hote hain.

"Maine exactly imaginary axis par

s = 0

par ek pole place kiya taaki response 'neutral' ho." Hidden danger kya hai?

s = 0

par pole

e^{0 \cdot t} = 1

deta hai — ek mode jo na decay karta hai na grow karta hai, marginally stable. Koi bhi disturbance permanent offset chhodti hai; Eigenvalues and system stability dekho. Genuine stability ke liye

Re (μ_{i}) < 0

chahiye.

Why questions

Poles place karne ki baat karne se pehle bhi system controllable kyun hona chahiye?

Agar state ka koi direction

u

se influence nahi ho sakta, toh koi bhi

K

usse reach nahi kar sakta, isliye uska eigenvalue frozen hai.

(A, B)

ki controllability exactly woh condition hai ki har mode reachable ho — Controllability and the controllability matrix dekho.

Cayley–Hamilton theorem derivation ko kaam karne kyun deta hai?

Woh kehta hai ki ek matrix apni khud ki characteristic polynomial satisfy karta hai, isliye actual char. poly ko

A

par evaluate karne se zero milta hai. Us zero ko

α (A)

se subtract karne par sirf coefficients ka difference bachta hai — exactly required feedback gains — baaki sab cancel ho jaata hai.

Formula mein

C^{- 1}

geometrically kyun hai?

Yeh "return ticket" hai. Pole placement Controllable canonical form mein trivial hai; CCF mein transformation

C

se bani hai, isliye

C^{- 1}

trivially-placed gains ko tumhare original coordinates mein wapas map karta hai.

Desired poles

K

ko uniquely determine karte hain (single-input) lekin LQR ko poles choose karne ki zaroorat kyun nahi?

Ackermann demand karta hai ki tum specify karo kahan har pole jaayega,

K

exactly fix karta hai. LQR — optimal pole placement alternative iske bajaye ek cost minimize karta hai aur poles optimally bahar aate hain — tum weights tune karte ho, locations nahi.

Same machinery, dualised karke, observers design kyun karta hai?

Observer error dynamics

A - L C

A - B K

ka transpose-dual hai. Ackermann ko

(A^{⊤}, C^{⊤})

par apply karne se observer gain

L

milta hai — Observer design and duality (Ackermann for observers) mirror.

Hum seedha

A

ke eigenvalues inspect karke

K

read off kyun nahi kar sakte?

A - B K

ke eigenvalues

K

par

n \geq 3

ke liye coupled, nonlinear tarike se depend karte hain; koi by-eye shortcut nahi hai. CCF detour is tangle ko "bottom row overwrite karo" mein turn karta hai, jo Ackermann automate karta hai.

Edge cases

Agar tum desired poles open-loop poles ke equal choose karo toh

K

ka kya hoga?

Tab

α = χ

hoga, actual characteristic polynomial, isliye

α (A) = 0

aur

K = 0

. Tumne kuch change nahi karne ko kaha, isliye feedback kuch bhi nahi hai — ek useful sanity check.

Scalar (

n = 1

) system

\overset{x}{˙} = a x + b u

jahan

b \neq = 0

ke liye, Ackermann kya reduce ho jaata hai?

α (s) = s - μ

ke saath, hume milta hai

α (A) = a - μ

C = b

, selector

= [1]

, isliye

K = (a - μ) / b

. Tab

a - b K = μ

— single pole exactly

μ

par land karta hai.

Tum ek repeated pole choose karo, jaise

μ = - 2

twice. Kya Ackermann tab bhi kaam karta hai?

Haan. Repeated desired poles ka matlab sirf yeh hai ki

α (s)

mein repeated factor hai; formula kabhi pole separation se divide nahi karta, isliye multiplicity theek hai (double integrator example exactly yahi karta hai).

Controllability matrix nearly singular hai (tiny determinant) lekin exactly zero nahi. Koi concern?

Haan.

C^{- 1}

tab ill-conditioned hota hai, isliye

K

bahut bada aur numerically fragile ho jaata hai — chhhoti model errors poles ko bahut hila deti hain. System "barely controllable" hai; bada control effort iska price hai.

Agar tum purpose se positive real part wale poles demand karo toh kya hoga?

Formula khushi se ek

K

compute karega jo unhe wahaan place karta hai — Ackermann stability enforce nahi karta, sirf jo poles tum maangoge.

Re (μ_{i}) > 0

choose karna deliberately unstable closed loop banata hai.

Kya

B = 0

kabhi ho sakta hai, aur formula kya kehta hai tab?

Agar

B = 0

toh input kuch bhi touch nahi karta:

C = [0 \dots 0]

singular hai, non-invertible, isliye Ackermann undefined hai. Bilkul sahi — koi actuation nahi hai toh koi bhi pole place nahi kar sakte.

Recall Ek-line summary

Ackermann single-input controllable systems ke liye exact aur unique hai, desired polynomial use karta hai, Cayley–Hamilton plus CCF round-trip ( $C^{- 1}$ ) par rely karta hai, aur sirf $A - B K$ ke eigenvalues guarantee karta hai — stability kabhi nahi unless tum stable poles maango.