LQG — LQR + Kalman filter, separation principle: Question bank

3.5.36 · D5 · Physics › Guidance, Navigation & Control (GNC) › LQG — LQR + Kalman filter, separation principle

Quick symbol reminder, taaki koi cheez bina matlab ke use na ho:

$x$ = true state (jo hum control karna chahte hain); $\overset{x}{^}$ = uska humara estimate; $e = x - \overset{x}{^}$ = estimation error.
$u = - K \overset{x}{^}$ = control input, estimate se banaya gaya, regulator gain $K$ ke saath.
$y = C x + v$ = noisy measurement; $L$ = Kalman gain (kitni strongly hum $y$ ki taraf correct karte hain).
$A - B K$ = controller (regulator) closed loop; $A - L C$ = estimator error dynamics.
$W$ = process-noise covariance, $V$ = sensor-noise covariance.
CARE = Continuous-time Algebraic Riccati Equation (regulator ke liye solve ki gayi steady-state Riccati equation).

Figure — LQG — LQR + Kalman filter, separation principle

True or false — justify

Separation ka matlab hai ki controller aur estimator poles dono designs ka union hote hain.

True. Joint matrix

[A - B K 0 B K A - L C]

block upper-triangular hai (figure mein red

0

block dekho), isliye uski eigenvalues exactly

eig (A - B K) \cup eig (A - L C)

hain.

Separation ka matlab hai ki process aur sensor noise ab tumhe kuch cost nahi karte.

False. Separation gain design aur pole placement ko decouple karta hai, performance ko nahi. Off-diagonal

B K e

term estimation error ko state mein inject karta hai, isliye LQG cost ideal full-state LQR cost se strictly zyada hoti hai.

LQG, LQR ke guaranteed

\geq 6 0^{\circ}

phase margin ko inherit karta hai.

False. Doyle (1978): "LQG has no guaranteed margins." Kalman filter insert karne se LQR ki robustness khatam ho sakti hai; usse recover karne ki koshish ke liye LQG-LTR Loop Transfer Recovery use karte hain.

Q

R

ke relative zyada bada banana ek faster, zyada aggressive regulator produce karta hai.

True. Bada

Q

state error ko heavily penalize karta hai, isliye optimizer error ko jaldi khatam karne ke liye bada control effort accept karta hai,

A - B K

poles ko aur left push karta hai.

Sensor noise

V

ko bada karne se Kalman gain

L

bada hota hai.

False — isse $L$ chota hota hai. Algebraically

L = Σ C^{⊤} V^{- 1}

hai, aur

V^{- 1}

shrink hota hai jab

V

badhta hai, isliye

L \to 0

: ek noisy sensor innovation

y - C \overset{x}{^}

ko unreliable bana deta hai, aur hum model prediction par zyada depend karte hain.

Kalman filter aur LQR do alag tarah ki equations se solve hote hain.

False. Dono ek algebraic Riccati Equation solve karte hain; filter wali equation control wali ki exact dual hai

(A, B, Q, R) \to (A^{⊤}, C^{⊤}, W, V)

ke under.

Agar true system nonlinear hai, toh LQG ab bhi exactly optimal hai.

False. LQG mein "L" Linear ke liye hai; optimality sirf linear dynamics, Gaussian noise, aur quadratic cost ke liye proven hai. Nonlinear plants par yeh zyada se zyada ek achha local approximation hai.

Same numbers ke liye, scalar LQR gain aur scalar Kalman gain identical aate hain.

True (jab data dual ho). CARE

A^{⊤} P + P A - P B R^{- 1} B^{⊤} P + Q = 0

with

A = 1, B = 1, Q = 1, R = 1

hai

2 P - P^{2} + 1 = 0

; filter Riccati

A Σ + Σ A^{⊤} - Σ C^{⊤} V^{- 1} C Σ + W = 0

with

C = 1, W = 1, V = 1

term-for-term dual

2Σ - Σ^{2} + 1 = 0

hai — same equation, same positive root

1 + 2

, isliye

K = L = 1 + 2

Spot the error

"Optimal control hai

u = - K y

: bas measurement ko gain ke through directly feed karo."

Do jagah galat hai:

y

mein sensor noise

v

hoti hai, aur

y = C x

sirf state ka ek projection dekhta hai, isliye unmeasured components invisible hote hain. Sahi hai

u = - K \overset{x}{^}

filtered estimate use karke.

"Kyunki closed loop poles separate hote hain, estimation error kabhi state tak nahi pahunchti."

Zero block error ki row mein baitha hai, matlab

K

e

ko corrupt nahi kar sakta — yeh nahi ki

e

x

tak nahi pahunch sakta. State row mein

B K e

term dikhata hai ki error does drive the state.

"Hum

u = - K x

ko running estimator mein plug karte hain, true

x

use karke."

Estimation ka poora point yahi hai ki hamare paas

x

nahi hota; implementable law hai

u = - K \overset{x}{^}

x

use karna full-state LQR hoga, jo LQG problem nahi hai.

"CARE hai

A^{⊤} P + P A + P B R^{- 1} B^{⊤} P + Q = 0

Quadratic term subtracted hota hai:

A^{⊤} P + P A - P B R^{- 1} B^{⊤} P + Q = 0

. Woh minus sign hi closed loop ko stabilize karta hai; plus sign galat (anti-stabilizing)

P

deta.

"Bada

R

controller ko aggressive banata hai kyunki hum control ki zyada care karte hain."

Ulta hai. Bada

R

control effort ko penalize karta hai, isliye optimizer gentle inputs use karta hai — slower, fuel-saving, less aggressive.

Σ

state hai; filter usse estimate karta hai."

Σ = E [e e^{⊤}]

error covariance hai, filter Riccati equation se ek fixed matrix. Estimated state

\overset{x}{^}

hai;

Σ

batata hai ki woh estimate kitna uncertain hai.

Why questions

Hum quadratic cost kyun choose karte hain, naaki error ki absolute value?

Squaring har term ko positive rakhta hai, bade deviations ko disproportionately punish karta hai, aur — decisively — minimizing control ko state mein linear banata hai, jo clean feedback law

u = - K x

deta hai.

Cost-to-go ke liye kuch solve karne se pehle hum

V (x) = x^{⊤} P x

kyun guess karte hain?

Quadratic guess, quadratic running cost aur linear dynamics ka natural match hai; ise HJB equation mein substitute karne par self-consistently ek quadratic reproduce hota hai, isliye guess kaam karke justify hoti hai.

Single

0

block "separation ka poora mathematical content" kyun hai?

Ek block-triangular matrix ki eigenvalues sirf uske diagonal blocks ki eigenvalues hoti hain, isliye woh zero controller aur estimator spectra ko cleanly split karne par majboor karta hai — koi cross terms nahi, koi coupled optimization nahi.

R ≻ 0

(strictly positive) kyun hona chahiye, jabki

Q ⪰ 0

sirf semidefinite ho sakta hai?

R^{- 1}

appear hota hai

K = R^{- 1} B^{⊤} P

mein; agar

R

singular hota toh woh inverse exist nahi karta.

Q

sirf states ko weight karta hai, isliye kuch states par zero penalty allow karna theek hai.

Kalman filter ko LQR ka dual kyun kaha jata hai, sirf "similar" naaki?

Error matrix

A - L C

ka wahi algebraic form hai jaisa

A - B K

ka, exact swap

(A, B) \to (A^{⊤}, C^{⊤})

ke under, isliye ek Riccati equation solve karne se dusri relabelling se solve ho jaati hai — yeh ek precise duality hai, loose analogy nahi.

Kalman filter insert karne se LQR ki robustness destroy hone ka risk kyun hota hai?

Estimator apni khud ki dynamics aur lag loop mein add karta hai; guaranteed LQR margins direct state feedback ke liye proven the, aur woh proof tab lagu nahi rehta jab

\overset{x}{^}

x

ki jagah le leta hai.

Edge cases

Agar system observable nahi hai toh kya hoga?

Tab kuch state modes

y

mein koi fingerprint nahi chhodte, isliye koi bhi

L

unki estimation error ko zero tak nahi le ja sakta — filter Riccati equation ka koi stabilizing

Σ

nahi hota aur LQG fail ho jaata hai.

Agar system controllable (ya stabilizable) nahi hai toh kya hoga?

Tab koi unstable mode

u

se move nahi ho sakta, isliye koi

K

A - B K

ko stabilize nahi karta — CARE mein koi stabilizing

P ⪰ 0

nahi hota aur LQG ka LQR wala hissa ill-posed ho jaata hai.

Agar sensor-noise covariance

V

singular (not invertible) ho toh kya hoga?

Kalman gain

L = Σ C^{⊤} V^{- 1}

V^{- 1}

chahiye, isliye singular

V

filter Riccati equation ko ill-posed bana deta hai — iska matlab hai ki koi measurement direction perfectly noise-free hai, wahan infinite gain demand karta hai. Ya toh aap us channel ko remove karo (exact algebraic constraint treat karo) ya

V

mein ek tiny regularizing floor add karo.

Agar process-noise covariance

W

singular (non-invertible) ho toh kya hoga?

W

khud invert nahi hota, isliye filter Riccati equation immediately nahi toot ti; lekin singular

W

ka matlab hai kuch state directions ko koi excitation nahi milti, aur unique stabilizing

Σ ≻ 0

tabhi guaranteed hai jab

(A, W^{1/2})

stabilizable ho — warna unexcited unstable modes

Σ

ko undefined chhod dete hain, LQR side ki controllability requirement ko mirror karte hue.

Agar process noise

W \to 0

ho (perfect model, koi gusts nahi) toh kya hoga?

Filter prediction par increasingly trust karta hai:

L \to

small, estimator open-loop propagation ki taraf relax karta hai kyunki chase karne ke liye bahut kam disturbance hai.

Agar sensor noise

V \to 0

ho (ek perfect sensor

C x

ka) toh kya hoga?

L

flawless measurement ki taraf hard correct karne ke liye blow up karta hai; estimator poles far left push hote hain, effectively observed states ko almost instantly reconstruct karte hue.

Agar

W \to 0

aur

V \to 0

simultaneously ho toh kya hoga?

Koi noise nahi matlab koi estimation error nahi;

\overset{x}{^} \to x

aur LQG plain full-state LQR mein collapse ho jaata hai, uski cost aur margins recover karte hue.

Agar hum Kalman estimate use karein ek aise system mein jo stable hai lekin sirf detectable hai, observable nahi?

Detectability guarantee karta hai ki unobservable modes already stable hain, isliye filter phir bhi ek stabilizing estimator deta hai chahe un modes ko precisely estimate na kar sake — LQG well-posed rehta hai.