Worked examples — LQG — LQR + Kalman filter, separation principle

3.5.36 · D3 · Physics › Guidance, Navigation & Control (GNC) › LQG — LQR + Kalman filter, separation principle

Koi bhi number se pehle: teen chhote reminders taaki koi symbol bina wajah na ho.

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The scenario matrix

Neeche har cell ko kam se kam ek worked example cover karta hai. Ise checklist ki tarah padho.

#	Case class	Kya special hai	Hit by
A	Stable scalar plant ($A<0$)	drift pehle se help karta hai	Ex 1
B	Unstable scalar plant ($A>0$)	control ko zaroor stabilize karna hai	Ex 2
C	Cheap control limit ($R\to 0$)	gain blast ho jaata hai, fast poles	Ex 3
D	Expensive control limit ($R\to \infty$)	gain $\to$ minimal, mirror trick	Ex 3
E	Zero state penalty ($Q=0$) stable plant par	degenerate: kuch mat karo	Ex 4
F	Kalman: clean vs noisy sensor ($V$ small/large)	sensor pe trust vs model pe trust	Ex 5
G	2×2 plant, full separation, poles = union	matrix case, koi coupling nahi	Ex 6
H	Real-world word problem (spacecraft rate damping)	physics $\to$ matrices mein translate karo	Ex 7
I	Exam twist: uncontrollable / degenerate $B$	Riccati ka koi stabilizing $P$ nahi	Ex 8

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Ex 1 — Cell A: stable scalar plant

Forecast: plant pehle se decay kar raha hai ($A=-1$). Kya optimal controller zyada push karega ya halka sa nudge dega? Pole ka sign pehle guess karo.

1. Scalars ke liye CARE likho. Scalars mein sab commute karta hai, to $A^{\top }P+PA=2AP=-2P$ aur $PB{R}^{-1}{B}^{\top }P=P^2$. $-2P-P^2+2=0\Rightarrow P^2+2P-2=0.$ Yeh step kyun? CARE hi HJB equation se optimality condition hai; ise solve karna poora LQR problem hai.
2. Positive root lo (value function $x^{\top }Px$ ek bowl honi chahiye, to $P>0$): $P=\frac{-2+\sqrt{4+8}}{2}=-1+\sqrt{3}\approx \mathrm{0.732.}$ Yeh step kyun? Do algebraic roots mein se sirf $P⪰0$ real cost-to-go aur stabilizing gain deta hai.
3. Gain aur closed loop: $K=R^{-1}{B}^{\top }P=P\approx 0.732$, to $A-BK=-1-0.732=-\mathrm{1.732.}$ Yeh step kyun? $u=-Kx$ state-feedback law hai; $A-BK$ controlled system ka pole hai.

Verify: plant pehle se stable tha ($-1$) aur optimum ne use sirf aur stable banaya ($-1.732<0$) — modest push kyunki plant khud apni help karta hai. ✅ Units: $P,K$ yahan dimensionless hain (sab matrices scalar).

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Ex 2 — Cell B: unstable scalar plant

Forecast: ek unstable plant jiska $A=2$ hai use zero se neeche kheenchna zaroori hai. Kya $K$ exceed karega $2$ ko? Guess karo.

1. CARE: $2AP-P^2+Q=4P-P^2+1=0\Rightarrow P^2-4P-1=0$. Yeh step kyun? Same scalar CARE template; sirf $A=2$ badla.
2. Positive root: $P=2+\sqrt{5}\approx 4.236$. Yeh step kyun? Stabilizing solution ke liye $P>0$ chuno.
3. Gain: $K=P=4.236$. Closed loop: $A-BK=2-4.236=-2.236$. Yeh step kyun? Controller ko $+2$ drift overcome karna hai aur usse bhi zyada taaki negative pole milega.

Verify: sach mein $K=4.236>2$, bilkul waisa jaisa ek unstable plant maangta hai — controller $+2$ ko "cancel" karta hai aur $-2.236$ decay add karta hai. ✅

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Ex 3 — Cells C & D: cheap vs expensive control limits

Forecast: free control ke saath hum infinitely aggressive pole expect karte hain; costly control ke saath mildest stabilizing pole. Unstable $A=1$ ke liye woh mildest pole kya hoga?

1. $R$ rakh ke general scalar CARE: $2P-P^2/R+Q=0\Rightarrow P^2-2RP-RQ=0$, to $P=R+\sqrt{R^2+RQ},K=\frac{P}{R}=1+\sqrt{1+\frac{Q}{R}}.$ Yeh step kyun? $R$ symbolic rakhne se clean limits lena aasaan ho jaata hai badhaate se zyada.
2. Cheap control $R\to 0$: $K=1+\sqrt{1+Q/R}\to \infty$; pole $A-BK\to -\infty$. Yeh step kyun? Free control matlab rokne ki koi wajah nahi — infinite gain, instant correction. (Physically impossible, isliye hum kabhi $R=0$ nahi set karte.)
3. Expensive control $R\to \infty$: $K\to 1+\sqrt{1}=2$; pole $A-BK\to 1-2=-1$. Yeh step kyun? Yeh famous mirror-image result hai: sabse kam-effort stabilizer unstable pole $+1$ ko imaginary axis ke across sirf $-1$ par reflect karta hai.

Verify: $Q=1$ par, $R=1$ lo → $K=1+\sqrt{2}\approx 2.414$ (parent Example 1 se match karta hai); $R=100$ lo → $K=1+\sqrt{1.01}\approx 2.005\to 2$. Limits parent ke answer ko bracket karte hain. ✅

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Ex 4 — Cell E: degenerate $Q=0$ stable plant par

Forecast: $Q=0$ matlab "state error free hai." Ek pehle se stable plant par, kya control effort kharach karne ki koi wajah hai?

1. CARE with $Q=0$: $-2P-P^2=0\Rightarrow P(P+2)=0$. Yeh step kyun? Same template $Q=0$ ke saath; factor karo ise.
2. $P⪰0$ chuno: roots hain $P=0$ aur $P=-2$. Sirf $P=0$ admissible hai. Yeh step kyun? Cost-to-go negative nahi ho sakti; $P=-2$ reject hai.
3. Gain: $K=P=0$, to $u^{\star }=0$. Closed loop $=A=-1$. Yeh step kyun? Koi state penalty nahi aur stable plant ke saath, optimal action hai kuch mat karo aur plant ko free mein decay karne do.

Verify: $P=0\Rightarrow J=x^{\top }Px=0$: ek decay karte plant aur koi penalties nahi ke saath, total cost sach mein zero hai. ✅

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Ex 5 — Cell F: Kalman clean vs noisy sensor

Forecast: kaunsa sensor humein measurement par lean karta hai (bada $L$) aur kaunsa humein model trust karne deta hai (chhota $L$)?

1. Scalars ke liye FARE: $2A\Sigma -{\Sigma }^2{C}^2/V+W=-2\Sigma -{\Sigma }^2/V+1=0$, yaani ${\Sigma }^2+2V\Sigma -V=0\Rightarrow \Sigma =-V+\sqrt{V^2+V}.$ Yeh step kyun? FARE dual hai CARE ka (swap karo $A\to A^{\top }$, $B\to C^{\top }$, $Q\to W$, $R\to V$).
2. Clean sensor $V=0.01$: $\Sigma =-0.01+\sqrt{0.0001+0.01}=-0.01+\sqrt{0.0101}\approx 0.0905$. Phir $L=\Sigma C{V}^{-1}=0.0905/0.01\approx 9.05$. Yeh step kyun? Tiny $V$ matlab sensor trustworthy hai — bada $L$ estimate ko $y$ ki taraf zyada kheenchta hai.
3. Noisy sensor $V=100$: $\Sigma =-100+\sqrt{10000+100}=-100+\sqrt{10100}\approx 0.4988$. Phir $L=0.4988/100\approx 0.004988$. Yeh step kyun? Huge $V$ matlab sensor jhooth bolta hai — chhota $L$ filter ko apna model trust karne kehta hai.

Verify: $L(V=0.01)\approx 9.05$ roughly $1800\times$ bada hai $L(V=100)\approx 0.005$ se — sensor quality ke saath monotone, exactly jaisa "bada $V$ → chhota $L$" predict karta hai. ✅

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Ex 6 — Cell G: 2×2 plant, separation poles ka union deta hai

Forecast: kyunki sab kuch diagonal hai, har axis ek independent scalar problem hai. Predict karo ki 4 poles sirf ek scalar answer ki do copies hain, ek har axis ke liye.

1. Diagonal ⇒ do scalar CAREs. Axis 1: $-2P-P^2+1=0\Rightarrow P=-1+\sqrt{2}$, $K_1=P\approx 0.414$, pole $-1-0.414=-1.414$. Yeh step kyun? Ek diagonal system state-feedback problem ko independent channels mein decouple karta hai.
2. Axis 2: $-4P-P^2+1=0\Rightarrow P=-2+\sqrt{5}\approx 0.236$, $K_2\approx 0.236$, pole $-2-0.236=-2.236$. Yeh step kyun? Same, $A=-2$ ke saath.
3. Estimator poles duality se: $C=I,W=V=I$ CAREs ko mirror karta hai, to $L_1\approx 0.414,L_2\approx 0.236$ aur estimator poles $-1.414,-2.236$ milte hain. Yeh step kyun? FARE = CARE yahan (symmetric numbers), to estimator poles controller poles ke barabar hain.
4. Union of poles. Block-triangular structure se LQG poles hain $\{-1.414,-2.236\}\cup \{-1.414,-2.236\}.$ Yeh step kyun? $\left[\begin{array}{cc}A-BK& BK\\ 0& A-LC\end{array}\right]$ mein $0$ block eigenvalues ko union banata hai — separation principle, applied.

Verify: chaaron poles strictly negative hain → closed loop stable, aur set exactly (controller poles) ∪ (estimator poles) hai. ✅

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Ex 7 — Cell H: real-world word problem (spacecraft rate damping)

Forecast: axis pehle se dheere de-spin ho raha hai ($A=-0.5$). Kya controller strong ya gentle damping add karega heavy $Q=4$ diya gaya hai?

1. Physics ko matrices mein translate karo: $A=-0.5,B=1,C=1$. Rate state hai; torque input hai. Yeh step kyun? GNC problems sirf state-space in disguise hain — pehle $A,B,C$ identify karo.
2. CARE: $2AP-P^2+Q=-P-P^2+4=0\Rightarrow P^2+P-4=0\Rightarrow P=\frac{-1+\sqrt{17}}{2}\approx 1.5616$. Gain $K=P\approx 1.5616$. Closed loop $=-0.5-1.5616=-2.0616$. Yeh step kyun? Heavy $Q=4$ tight rate control maangta hai, to hum fast pole expect karte hain — confirmed.
3. FARE: $2A\Sigma -{\Sigma }^2/V+W=-\Sigma -{\Sigma }^2+0.25=0\Rightarrow {\Sigma }^2+\Sigma -0.25=0\Rightarrow \Sigma =\frac{-1+\sqrt{2}}{2}\approx 0.2071$. Kalman gain $L=\Sigma /V\approx 0.2071$. Estimator pole $-0.5-0.2071=-0.7071$. Yeh step kyun? Gyro noise ($V=1$) process noise ke comparable hai, to ek moderate $L$ model aur sensor ko blend karta hai.
4. Controller assemble karo: $\dot{\hat{\omega }}=-0.5\hat{\omega }+u+L(y-\hat{\omega })$ chalao, phir command karo $u=-K\hat{\omega }$. Yeh step kyun? Separation hume do pieces ko directly bolt karne deta hai.

Verify: controller pole $-2.06$ (fast damping, jaisa heavy $Q$ demand karta hai) aur estimator pole $-0.707$ dono stable hain; $K$ ke units hain $[\text{torque/inertia}]/[\text{rate}]={\text{s}}^{-1}$, jo rate-damping gain se match karta hai. ✅

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Ex 8 — Cell I: exam twist, degenerate/uncontrollable $B$

Forecast: $B=0$ ke saath control kuch nahi karta. Kya LQR possibly ek unstable, uncontrollable plant ko stabilize kar sakta hai?

1. $B=0$ CARE mein daalo: $2AP-PB{R}^{-1}{B}^{\top }P+Q=2P-0+1=0\Rightarrow P=-\frac{1}{2}$. Yeh step kyun? Quadratic term vanish ho jaata hai kyunki $B=0$; hum ek linear equation ke saath bache hain.
2. Root reject karo. $P=-\frac{1}{2}<0$ ek valid cost-to-go nahi hai — koi $P⪰0$ exist nahi karta. Yeh step kyun? Koi admissible $P$ nahi matlab infinite-horizon cost $+\infty$ hai: unstable mode uncontrollable hai aur regulate nahi ho sakta.
3. Interpret karo: LQR require karta hai ki $(A,B)$ stabilizable ho. $B=0$ aur $A=1>0$ ke saath, diverging mode untouchable hai, to $J={\int }_0^{\infty }{x}^{2}dt=\infty$ kisi bhi input ke liye. Yeh step kyun? Yeh standard exam gotcha hai: Riccati solution trust karne se pehle stabilizability check karo.

Verify: sirf algebraic root $P=-1/2$ violate karta hai $P⪰0$; consistent hai $J=\infty$ ke saath ek unstable uncontrollable mode ke liye. ✅

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Recall Quick self-test

Scalar $A=2,B=1,Q=1,R=1$: $K$ kya hai? ::: $K=2+\sqrt{5}\approx 4.236$ (Ex 2) Mirror-pole limit $R\to \infty$ $A=1$ ke liye: closed-loop pole? ::: $-1$ ($+1$ reflect karo) (Ex 3) $Q=0$, $A=-1$: optimal control? ::: $u=0$, kuch mat karo (Ex 4) Huge sensor noise $V$: kya $L$ bada hoga ya chhota? ::: chhota — model trust karo (Ex 5) $B=0,A>0$: kya LQR ka solution hai? ::: nahi — uncontrollable unstable mode, $J=\infty$ (Ex 8)