Controllability matrix — rank test: Question bank

3.5.32 · D5 · Physics › Guidance, Navigation & Control (GNC) › Controllability matrix — rank test

Intuition Is page ko kaise use karein

Yeh ek trap-hunting deck hai Controllability matrix — rank test ke liye. Neeche har item wahan target karta hai jahan intuition aapko quietly galat raasta dikhata hai. Question padho, answer cover karo, apni reasoning ek sentence mein zor se bolo, phir reveal karo. Agar tumhari reasoning match karti hai — chahe verdict bhi match kare — tum jeet gaye. Agar tumne sirf "true/false" guess kiya bina kisi why ke, toh yeh miss count hoga.

Hamare notation ki reminder (parent note mein poori tarah build ki gayi hai):

$A$ = $n \times n$ system matrix jo batata hai ki state apne aap kaise drift karti hai.
$B$ = $n \times m$ input matrix jo batata hai ki tumhare actuators kahaan push karte hain.
$C = [B A B A^{2} B \dots A^{n - 1} B]$ = controllability matrix; "controllable" ka matlab hai $rank (C) = n$ .

Definition Hum kis system ki baat kar rahe hain — continuous ya discrete?

Jab tak koi question explicitly "discrete" na kahe, is page par sab kuch continuous-time system $\overset{x}{˙} = A x + B u$ ke baare mein hai (state smoothly time mein change hoti hai, $\overset{x}{˙}$ = rate of change). Wahi matrix $C$ aur wahi rank- $n$ test dono worlds mein controllability decide karte hain — lekin dono domains mein kitni jaldi target tak pahuncha ja sakta hai, yeh alag hai:

Continuous-time $\overset{x}{˙} = A x + B u$ : agar rank $= n$ hai, toh tum kisi bhi finite time $T > 0$ mein target tak pahunch sakte ho, chahe woh kitna bhi chhota ho (badi control effort ki cost par).
Discrete-time $x_{k + 1} = A x_{k} + B u_{k}$ (state step by step jump karta hai): agar rank $= n$ hai, toh tum zyada se zyada $n$ steps mein kisi bhi target tak pahunch sakte ho — aur generally tum isse kam mein nahi kar sakte, kyunki $k$ steps ke baad tum sirf $B, A B, \dots, A^{k - 1} B$ span kar sakte ho. Edge cases section is discrete-time difference par explicitly wapas aata hai.

Pehle ideas ko picture karo

True ya false — justify karo

Har reachable direction

B, A B, \dots, A^{n - 1} B

ke columns ka koi combination hoti hai.

True — matrix exponential ki power series Cayley–Hamilton theorem se un blocks par collapse ho jaati hai, isliye unke span ke bahar kuch bhi kabhi reachable nahi hota.

Ek controllable system stable zaroor hogi.

False — controllability kehta hai tum har mode ko move kar sakte ho; ek unstable mode phir bhi controllable ho sakta hai, aur yeh exactly woh case hai jahan feedback use rescue kar sakta hai.

Continuous-time

\overset{x}{˙} = A x + B u

ke liye, agar

rank (C) = n

hai toh tum kisi bhi time

T > 0

mein target tak pahunch sakte ho, chahe woh kitna bhi chhota ho.

True — continuous-time controllability har positive

T

ke liye reachability deta hai (chhote

T

ki cost huge control effort hai, lekin phir bhi possible hai).

Zyada inputs add karna (

B

ko widen karna) controllability kabhi reduce nahi kar sakta.

True — extra columns

C

ke column span ko sirf enlarge ya preserve kar sakte hain, isliye rank drop nahi ho sakta; zyada actuators system ko kabhi lock nahi karte.

Ek single-input system (

m = 1

) kabhi controllable nahi ho sakta jab

n > 1

ho.

False — Example 1 ka double integrator

m = 1

n = 2

hai aur controllable hai kyunki

A

ek push ko dono directions mein mix kar deta hai.

Agar

B = 0

ho toh system har

A

ke liye uncontrollable hai.

True — koi actuator nahi hone se har block

B, A B, \dots

zero column hai, isliye

rank (C) = 0 < n

; tumhare paas push karne ke liye kuch nahi hai.

(A, B)

ki controllability unchanged rehti hai agar hum

A

A^{T}

se replace kar dein.

Generally False —

A^{T}

ek alag dynamics describe karta hai; transpose par jaana observability (the dual) ka recipe hai, controllability ka nahi.

Do systems jinke

C

invertible column operations tak same hain, unka controllability verdict same hoga.

True — controllability ek rank statement hai, aur rank invertible matrices se multiply karne par invariant rehta hai, isliye yes/no answer preserve hota hai.

Ek uncontrollable system phir bhi stabilizable ho sakti hai.

True — stabilizability sirf demand karta hai ki unstable modes controllable hon; harmless (stable) uncontrollable modes allowed hain, aur yahi Stabilizability ka poora point hai.

Rank-

n

test discrete-time system

x_{k + 1} = A x_{k} + B u_{k}

ke liye bhi controllability decide karta hai.

True — wahi

C

aur wahi rank condition apply hoti hai; sirf timing alag hai, kyunki discrete-time

T > 0

ki jagah zyada se zyada

n

steps mein target tak pahunchta hai.

Spot the error

A

3 \times 3

hai, isliye main safety ke liye

C = [B A B A^{2} B A^{3} B]

banaunga."

A^{3} B

block waste hai — Cayley–Hamilton

A^{3}

I, A, A^{2}

ke combination ke roop mein likhta hai, isliye yeh koi nai direction add nahi karta;

A^{n - 1} B = A^{2} B

par ruk jao.

det C \neq = 0

hai, isliye controllable hai — maine ek square block check kiya."

m > 1

ke liye,

C

n \times (nm)

hai aur square nahi hai, isliye "the determinant" undefined hai; tumhe

rank (C) = n

check karna hoga, jaise row reduction se, na ki ek single determinant se.

B

mein koi zero entry nahi hai, isliye system controllable hai."

Full

B

kuch guarantee nahi karta — controllability is baare mein hai ki

A

powers mein

B

ko kaise rotate karta hai; tumhe phir bhi

C

ka rank compute karna hoga (ek full

B

phir bhi rank

< n

de sakta hai agar mixing degenerate ho).

C

ki Row 2 poori zeros hai, isliye state

x_{2}

uncontrollable hai."

Raw coordinates mein

C

ki zero row ek lost direction signal karti hai, lekin tum blindly labelled state

x_{2}

ko blame nahi kar sakte — coordinates change karne ke baad untouched direction states ka mixture ho sakta hai; rank batata hai ki ek mode lost hai, na ki kaun sa physical variable.

"System unstable hai, isliye woh controllable nahi ho sakti."

Stability aur controllability independent hain; instability free evolution describe karta hai, controllability describe karta hai ki inputs kya kar sakte hain — ek unstable system often wahi hoti hai jis par tum sabse zyada controllable rehna chahte ho taaki tum poles place kar sako.

"Maine rank

n

paaya, isliye sab poles apne aap settle ho jaayenge."

Rank

n

ka matlab hai tum feedback design kar sakte ho poles ko kahin bhi move karne ke liye; yeh open-loop poles ke baare mein kuch nahi kehta, jo loop actually close karne tak unstable region mein baithe reh sakte hain.

"Discrete-time: rank

n

hai, isliye main ek hi step mein target tak pahunch sakta hoon."

Ek step sirf

B

ke columns span karta hai; generally tumhe

B, A B, \dots, A^{n - 1} B

build up karne ke liye

n

steps tak chahiye, isliye ek step claim karna error hai — ek step sirf tab kaafi hai jab

B

akele already rank

n

ho.

Why questions

Reachable set sirf powers

B

A^{n - 1} B

tak kyun collapse hoti hai, na ki

A

ke saari infinite powers tak?

Kyunki Cayley–Hamilton

A^{n}

I, A, \dots, A^{n - 1}

ke combination ke roop mein likhta hai; har higher power unhi directions ko recycle karta hai (doosri figure dekho), isliye span

A^{n - 1} B

par grow karna band kar deta hai.

Target rank exactly

n

kyun hota hai aur zyada nahi?

State

R^{n}

mein rehti hai, isliye uska reachable span at most

n

-dimensional ho sakta hai; rank

= n

ka matlab hai columns already poora state space fill kar lete hain aur pahunchne ke liye kuch bacha nahi hai.

Ek diagonal

A

jismein

B

ki matching row mein zero ho, woh ek uncontrollable mode kyun deta hai?

Ek diagonal

A

kabhi modes mix nahi karta, isliye ek mode jiska

B

-entry zero hai woh kisi input se push nahi hota aur kisi neighbour se feed nahi hota — woh hamesha ke liye apne aap drift karta rehta hai, untouchable.

Wahi

A

B

ke saath uncontrollable aur doosre

B

ke saath controllable kyun ho sakta hai?

Controllability pair

(A, B)

ki property hai; actuator kahaan push karta hai yeh change karna (Example 3 vs Example 2) push ko ek aisi mode mein spread kar sakta hai jo

A

ki mixing phir poore space mein propagate kar deti hai.

Pole placement karne se pehle hum controllability kyun check karte hain?

Pole placement sirf un poles ko relocate kar sakta hai jo controllable subspace mein rehte hain; uncontrollable modes apne original poles rakhte hain chahe feedback gain kuch bhi ho.

Observability ko controllability ka "dual" kyun kaha jaata hai na ki alag idea?

Kyunki observability test transposed system par controllability test hai (

A \to A^{T}

B \to C^{T}

); ek ke baare mein har theorem transposition se doosre mein mirror ho jaata hai.

Discrete-time controllability ke saath "number of steps" kyun aata hai jabki continuous-time ke saath "any

T > 0

" aata hai?

Ek discrete system finite jumps mein move karta hai, isliye woh sirf ek aur power of

A

per step span kar sakta hai aur usse

n

steps tak chahiye; ek continuous system infinitely many intermediate instants se flow karta hai, isliye woh kisi bhi positive time mein reachable subspace fill kar leta hai.

Edge cases

n = 1

A = [a]

B = [0]

: controllable?

Nahi —

C = [0]

ka rank

0 < 1

hai; koi actuator nahi hone se single state move nahi ho sakta

a

chahe kuch bhi ho.

n = 1

A = [a]

B = [b]

jahan

b \neq = 0

ho: controllable?

Haan —

C = [b]

ka rank

1 = n

hai; ek scalar system exactly tab controllable hai jab uski akeli input entry nonzero ho.

A = 0

(zero matrix),

B = I_{n}

: controllable?

Haan —

C = [I 0 \dots 0]

already

B = I

block se rank

n

rakhta hai; har state ka apna dedicated actuator hai, isliye koi mixing bhi zaroor nahi.

A = 0

B

ek single column:

n > 1

hone par controllable?

Nahi —

A B, A^{2} B, \dots

sab zero hain, isliye

C = [B 0 \dots]

ka rank

\leq 1 < n

hai; dynamics na hone se ek push spread nahi ho sakta, ek single input sirf ek direction tak pahunch sakta hai.

Do identical decoupled modes (same eigenvalue) jo same scalar input se equal weights se fed hain: controllable?

Nahi — input dono copies ko identically push karta hai, isliye unka difference direction kabhi touch nahi hota; single input ke saath repeated eigenvalues generally ek mode ko unreachable chhod dete hain.

Discrete-time controllable system,

n = 3

: rest se kisi bhi target tak pahunchne ki guaranteed fewest steps kitni hain?

Zyada se zyada

3

steps —

k

steps ke baad tum

B, A B, \dots, A^{k - 1} B

span karte ho, isliye

n = 3

steps full span guarantee karte hain; kam steps tab kaafi hain jab pehle ke blocks already rank

3

tak pahunch jaayein.

Ek stable lekin uncontrollable system — kya yeh useless hai?

Zaroor nahi — agar untouchable modes stable hain toh woh apne aap settle ho jaate hain, isliye system stabilizable hai aur perfectly usable hai; tum bas un modes ko arbitrarily reshape nahi kar sakte.

Ek purely marginal mode (eigenvalue exactly

0

) jo uncontrollable hai — problem?

Stabilizability ke liye Haan — zero par ek mode na decay karta hai na fix ho sakta hai, isliye ek uncontrollable marginal ya unstable mode exactly wahi hai jo stabilizability tod deta hai.

Recall Page band karne se pehle ek-line self-test

Zor se bolo: "Controllable ka matlab hai == $rank (C) = n$ ==, yeh states ko move karne ke baare mein hai na ki settle karne ke, yeh pair $(A, B)$ par depend karta hai, $A^{n - 1} B$ ke baad higher powers Cayley–Hamilton se nothing add karte hain, aur wahi test continuous- aur discrete-time dono ke liye kaam karta hai (discrete $\leq n$ steps mein pahunchta hai)." Agar koi clause shaky laga, upar woh group dobara dekho.

Connections

Controllability matrix — rank test
Observability matrix — rank test
State-space representation
Cayley–Hamilton theorem
Pole placement & state feedback
Kalman decomposition
Matrix exponential $e^{At}$
Stabilizability