Exercises — LQG — LQR + Kalman filter, separation principle

• $w$ = process noise: physics pe random pushes — wind gusts, unmodelled dynamics, thrust jitter.
• $v$ = measurement noise: sensor reading ke andar random error.
• $W$ = $w$ ki strength (covariance); $V$ = $v$ ki strength.
• Kyunki $L=\Sigma C^{\top }{V}^{-1}$ aur $\Sigma$, $W$ ke saath barhta hai: bada $W$ → bada $L$ (sensor pe zyada trust karo, kyunki model kick khaata reh raha hai). Bada $V$ → chota $L$ (model pe trust karo, sensor kharab hai).

$R$ badhao (effort ko zyada penalise karo). Bada $R$ matlab $R^{-1}$ chota hai, toh $K=R^{-1}{B}^{\top }P$ shrink hoga → chote control moves → gentler, slower response. $Q$ badhana ulta karega (aggressive tracking).

Bottom-left $0$ block. Yeh batata hai ki estimation error $e$ apni khud ki physics ($A-LC$) se evolve karta hai aur controller gain $K$ se affect nahi hota. Kyunki matrix block-upper-triangular hai, iske eigenvalues sirf union $\text{eig}(A-BK)\cup \text{eig}(A-LC)$ hain — toh do designs decouple ho jaate hain.

Scalar CARE se shuru karo $2AP-\frac{B^2{P}^2}{R}+Q=0$ (upar derive kiya: $A^{\top }P+PA=2AP$ kyunki scalars commute karte hain, aur $PB{R}^{-1}{B}^{\top }P=B^2{P}^2/R$). $A=2,B=1,R=1,Q=3$ substitute karo: $2(2)P-\frac{1^2{P}^2}{1}+3=0\Rightarrow 4P-P^2+3=0\Rightarrow P^2-4P-3=0.$ Quadratic formula: $P=\frac{4+\sqrt{16+12}}{2}=\frac{4+\sqrt{28}}{2}=2+\sqrt{7}\approx 4.6458$ (positive root rakho — $P$ $>0$ hona chahiye). $K=R^{-1}{B}^{\top }P=1\cdot 1\cdot P=2+\sqrt{7}\approx \mathrm{4.6458.}$ Closed loop: $A-BK=2-(2+\sqrt{7})=-\sqrt{7}\approx -2.6458<0$ ✅ stable. Negative kyun? LQR ka poora point hi yeh hai ki closed loop decay kare; ek positive $A$ (unstable open loop) ko feedback negative pull karta hai.

Yeh L2.1 ka dual hai identical numbers ke saath ($A\to A^{\top }=2$, $C^{\top }=1$, $W=3$, $V=1$). Scalar FARE hai $2A\Sigma -\frac{C^2{\Sigma }^2}{V}+W=0$ (same collapse: $A\Sigma +\Sigma A^{\top }=2A\Sigma$, aur $\Sigma C^{\top }{V}^{-1}C\Sigma =C^2{\Sigma }^2/V$). $A=2,C=1,V=1,W=3$ substitute karo: $2(2)\Sigma -\frac{1^2{\Sigma }^2}{1}+3=0\Rightarrow 4\Sigma -{\Sigma }^2+3=0\Rightarrow \Sigma =2+\sqrt{7}\approx \mathrm{4.6458.}$ $L=\Sigma C^{\top }{V}^{-1}=2+\sqrt{7}\approx \mathrm{4.6458.}$ Estimator pole $A-LC=2-(2+\sqrt{7})=-\sqrt{7}\approx -2.6458<0$ ✅.

Scalar FARE: $2A\Sigma -\frac{C^2{\Sigma }^2}{V}+W=0$. $A=0,C=1$ ke saath yeh ban jaata hai $-\frac{{\Sigma }^2}{V}+W=0\Rightarrow {\Sigma }^2=WV\Rightarrow \Sigma =\sqrt{WV}(\text{positive root}).$

• $V=4$: $\Sigma =\sqrt{1\cdot 4}=2$, toh $L=\Sigma C^{\top }{V}^{-1}=2/4=\mathrm{0.5.}$
• $V=\frac{1}{4}$: $\Sigma =\sqrt{1\cdot \frac{1}{4}}=0.5$, toh $L=0.5/0.25=2.$ Kharab sensor → chota $L=0.5$ (model pe trust karo); achha sensor → bada $L=2$ (sensor pe trust karo). Exactly yahi intuition hai.

$A-BK=-\sqrt{7},A-LC=-\sqrt{7},BK=2+\sqrt{7}.$ Matrix $M=\left[\begin{array}{cc}-\sqrt{7}& 2+\sqrt{7}\\ 0& -\sqrt{7}\end{array}\right]$. Ek block-upper-triangular (yahan sirf upper-triangular) matrix ke eigenvalues uske diagonal pe hote hain: $\{-\sqrt{7},-\sqrt{7}\}\approx \{-2.6458,-2.6458\}$. Yeh exactly $\text{eig}(A-BK)\cup \text{eig}(A-LC)$ hain. Off-diagonal $BK=2+\sqrt{7}$ kabhi eigenvalues mein nahi aaya — $0$ block yahi guarantee karta hai. ✅

State equation ek extra driving term $BKe$ carry karta hai: estimation error $e$ controller ke through true state mein leak hota hai. Bhaale hi $e\to 0$ mean mein ho, iske random fluctuations continuously $x$ mein disturbance inject karte rehte hain, toh achieved cost $J_{\text{LQG}}>J_{\text{LQR,ideal}}$ hoti hai. Separation gains jo tum compute karte ho (design) ko decouple karta hai, cost jo tum pay karte ho (performance) ko nahi. Noise phir bhi tumhe cost karaata hai.

Stability ke liye chahiye $A-BK=5-K<0\Rightarrow K>5$. LQR hamesha aisa $K$ return karta hai (yeh guarantee karta hai stability jab bhi pair controllable ho). Smallest integer $K=6$ hai, pole deta hai $5-6=-1<0$ ✅. Koi bhi LQR-optimal $K$ actually $5$ se zyada hoga; exact value $Q,R$ pe depend karti hai, lekin stability guarantee hai regardless.

Yahan $A=-1,B=1,C=1$. Controller — scalar CARE $2AP-\frac{B^2{P}^2}{R}+Q=0$. $A=-1,B=1,R=1,Q=4$ substitute karo: $2(-1)P-P^2+4=0\Rightarrow -2P-P^2+4=0\Rightarrow P^2+2P-4=0.$ $P=\frac{-2+\sqrt{4+16}}{2}=-1+\sqrt{5}\approx 1.2361$ (positive root). $K=R^{-1}{B}^{\top }P=P=-1+\sqrt{5}\approx \mathrm{1.2361.}$ Controller pole $A-BK=-1-(-1+\sqrt{5})=-\sqrt{5}\approx -\mathrm{2.2361.}$ Filter — scalar FARE $2A\Sigma -\frac{C^2{\Sigma }^2}{V}+W=0$. $A=-1,C=1,V=1,W=9$ substitute karo: $-2\Sigma -{\Sigma }^2+9=0\Rightarrow {\Sigma }^2+2\Sigma -9=0\Rightarrow \Sigma =-1+\sqrt{10}\approx \mathrm{2.1623.}$ $L=\Sigma C^{\top }{V}^{-1}=-1+\sqrt{10}\approx \mathrm{2.1623.}$ Estimator pole $A-LC=-1-(-1+\sqrt{10})=-\sqrt{10}\approx -\mathrm{3.1623.}$ Controller (clean form): numbers $\dot{\hat{x}}=A\hat{x}+Bu+L(y-C\hat{x})$ mein plug karke, $u=-K\hat{x}=-(\sqrt{5}-1)\hat{x},\dot{\hat{x}}=A\hat{x}+Bu+L(y-C\hat{x})$ jahan $A=-1,B=1,C=1,K=\sqrt{5}-1\approx 1.2361,L=\sqrt{10}-1\approx \mathrm{2.1623.}$ Closed-loop poles: $\{-\sqrt{5},-\sqrt{10}\}\approx \{-2.2361,-3.1623\}$ — union, by separation. ✅

Scalar CARE $2AP-\frac{B^2{P}^2}{R}+Q=0$. $A=1,B=1$ ke saath: $2P-\frac{P^2}{R}+Q=0.$ (a) $Q=100,R=1$: $2P-P^2+100=0\Rightarrow P^2-2P-100=0\Rightarrow P=1+\sqrt{101}\approx \mathrm{11.0499.}$ $K=R^{-1}P=\mathrm{11.0499.}$ Pole $A-BK=1-11.0499=-10.0499$ — bahut fast. (b) $Q=1,R=100$: $2P-\frac{P^2}{100}+1=0$. $100$ se multiply karo: $200P-P^2+100=0\Rightarrow P^2-200P-100=0\Rightarrow P=100+\sqrt{10100}\approx \mathrm{200.4988.}$ $K=P/100=\mathrm{2.00499.}$ Pole $A-BK=1-2.00499=-1.00499$ — barely stable, gentle. Comment: heavy $Q$ / light $R$ ⇒ bada gain, fast (far-left) pole, bahut effort. Heavy $R$ ⇒ chota gain, sluggish pole imaginary axis ke paas, kam effort.

• Controllability of $(A,B)$ (ya kam se kam stabilisability): input $u$ state ke har unstable direction ko influence kar sakni chahiye — warna koi $K$ $A-BK$ ko stabilise nahi kar sakta.
• Observability of $(A,C)$ (ya kam se kam detectability): sensor $y$ ko har unstable direction reveal karna chahiye — warna koi $L$ $A-LC$ ko stabilise nahi kar sakta, aur $\hat{x}$ hidden unstable modes track nahi kar sakta. Dono Riccati Equation ke ek stabilising positive-(semi)definite solution hone se aate hain.

(a) Margin guarantees full-state feedback loop $A-BK$ ki property hain jo plant input pe measure ki jaati hai. Kalman filter extra dynamics ($A-LC$) measurement aur control ke beech insert karta hai, loop transfer function ko reshape karta hai; woh reshaping phase/gain margins ko arbitrarily small values tak shrink kar sakti hai. Separation gains ki optimality preserve karta hai, loop ki robustness nahi. (b) LQG/LTR: tum deliberately process-noise weight $W$ ko infinity ki taraf push karte ho (ya plant output pe design karte ho), jo Kalman gain $L$ ko itna bada kheenchta hai ki estimator almost transparent ho jaata hai — LQG loop transfer function phir full-state LQR loop shape recover kar leta hai aur uske margins wapas mil jaate hain. (Alternatively, H-infinity Control seedha ek robustness bound optimise karta hai.)

Characteristic polynomial: $\det (M-\lambda I)=\det \left[\begin{array}{cc}A-BK-\lambda & BK\\ 0& A-LC-\lambda \end{array}\right].$ Triangular matrix ke liye determinant diagonal ka product hota hai: $\det (M-\lambda I)=(A-BK-\lambda )(A-LC-\lambda ).$ Ise $0$ set karne par roots milte hain $\lambda =A-BK$ aur $\lambda =A-LC$ — off-diagonal $BK$ kabhi nahi aaya. Lower-left mein $0$ exactly yahi cheez hai jo determinant ko is tarah factor karaati hai. Yahi do lines mein separation principle hai. $■$

L1 Recognition ::: Jaano ki har symbol/piece kya hai aur kaunsa noise $L$ badhata hai. L2 Application ::: Ek scalar Riccati solve karo aur $K$, $L$, poles padho. L3 Analysis ::: Explain karo kyun poles split hote hain phir bhi performance degrade hoti hai. L4 Synthesis ::: Blank page se ek complete scalar LQG design karo. L5 Mastery ::: Existence conditions, robustness footgun batao, aur separation prove karo.