Question bank — Linear Quadratic Regulator (LQR) — Riccati equation, optimal gains

False. Sirf ratio $Q/R$ hi aggressiveness set karta hai; dono ko same factor se scale karne par $J$ ka minimizer unchanged rehta hai, isliye $K$ identical hota hai.

True. $P$ quadratic value function $V=x^{\top }Px$ ki matrix hai; ek quadratic form mein sirf uska symmetric part contribute karta hai, isliye hum construction se $P=P^{\top }$ lete hain.

False. Stability optimality nahi hai — ek pehle se stable plant phir bhi sluggish ya fuel-wasteful ho sakta hai, aur LQR woh control dhundhta hai jo combined error-plus-effort cost ko minimize kare.

False. Zyada $R$ control effort ko zyada penalize karta hai, isliye optimizer kam actuation use karta hai, jo slower aur gentler response deta hai. $Q$ badhane se speed up hoti hai.

False. $R$ strictly positive definite hona chahiye taaki $R^{-1}$ exist kare $K=R^{-1}{B}^{\top }P$ mein aur cost strictly convex ho $u$ mein; $R$ mein ek zero direction infinite free control allow kar dega.

True. $(A,B)$ controllable aur $(A,\sqrt{Q})$ observable hone par, CARE ka ek unique symmetric $P⪰0$ hota hai jo $A-BK$ ko Hurwitz banata hai; baki algebraic roots non-stabilizing hote hain.

True (standard conditions ke under). Stabilizing Riccati solution precisely wahi define ki jaati hai jo $A-BK$ ko Hurwitz render kare, isliye har closed-loop eigenvalue ka negative real part hoga.

False. Infinite-horizon steady state mein $\dot{P}=0$ hota hai, isliye $P$ aur hence $K$ constant hote hain — ek single fixed gain jo har waqt apply hota hai.

True. Law $u=-Kx$ ko har state chahiye; jab sirf outputs measure hoti hain to $x$ estimate karna padta hai (jaise ek Kalman Filter), jo LQG Control produce karta hai.

False. Origin se cost-to-go zero hai aur problem $x\to -x$ ke under symmetric hai, isliye $V$ ek pure quadratic $x^{\top }Px$ hai jisme koi linear ya constant term nahi hota.

$R^{-1}$ missing hai: correct gain $K=R^{-1}{B}^{\top }P$ hai. Yahi factor exactly isliye hai kyunki $R$ invertible hona chahiye.

Quadratic term negative hona chahiye: $A^{\top }P+PA-PB{R}^{-1}{B}^{\top }P+Q=0$. Woh minus sign wahi hai jo optimizer control karke "bachata" hai; plus sign ka koi stabilizing solution nahi hota.

Tumhe stabilizing root pick karni hogi — woh unique $P⪰0$ jo $A-BK$ ko Hurwitz banaye. $P/⪰0$ wala root ek aisa gain deta hai jo plant ko destabilize kare.

Ye poore trajectory par integral ${\int }_0^{\infty }(x^{\top }Qx+u^{\top }Ru)dt$ minimize karta hai, terminal value nahi; running cost har instant par error ko effort ke against trade karta hai.

Quadratic cost hi woh cheez hai jo value function ko quadratic aur optimal control ko linear ($u=-Kx$) banati hai. Ek linear (L1) cost us clean closed form ko tod deta hai aur bang-bang-type control deta hai.

Sirf $Q⪰0$ (positive semidefinite) required hai; tum sirf kuch states ko penalize kar sakte ho. Strict condition $R\succ 0$ par hai, plus uniqueness ke liye $(A,\sqrt{Q})$ observability chahiye.

Ye ek PD controller deta hai: $K=[k_p,k_d]$ position aur velocity par act karta hai. Koi integral term nahi hai kyunki standard LQR cost mein koi integral-of-error state nahi hoti.

Quadratic cost ek quadratic value function $V=x^{\top }Px$ force karti hai; HJB bracket ko $u$ ke upar minimize karna ek quadratic ko $u$ mein minimize karna hai, jiska stationary point $x$ mein linear hota hai.

Steady state mein cost-to-go change karna band ho jaata hai, isliye $\dot{P}=0$ hota hai aur differential Riccati ODE ek purely algebraic matrix equation mein collapse ho jaata hai jisme koi time derivative nahi hoti.

Term $-PB{R}^{-1}{B}^{\top }P$ unknown $P$ mein quadratic hai, isliye linear dynamics ke bawajood $P$ ke liye equation nonlinear hai.

Closed-loop pole $a-bK=a-\frac{b^2}{r}p$ positive Riccati root ke liye hamesha negative hota hai, isliye feedback $a$ ke sign ke barabar unstable dynamics ko override karta hai.

Ek quadratic form $x^{\top }Qx$ mein sirf $Q$ ka symmetric part contribute karta hai, isliye koi bhi antisymmetric part waste hota hai; hum inhe generality ke bina loss ke symmetric lete hain.

Controllability guarantee karta hai ki tum har state ko move kar sakte ho; $\sqrt{Q}$ ki observability ensure karti hai ki har unstable mode cost se actually penalized ho, isliye optimizer ke paas usse damp karne ka reason ho — milke ye ek unique stabilizing $P$ dete hain.

HJB bracket $u$ mein convex hai kyunki $R\succ 0$ hai, isliye uska single stationary point global minimum hai.

Infinite-horizon steady-state gain constant hai, isliye poori optimization ek CARE solve karne par reduce ho jaati hai pehle se; runtime sirf sasta product $u=-Kx$ hai.

$u=-K\cdot 0=0$. Koi error nahi matlab koi push ki zaroorat nahi; controller exactly setpoint par idle karta hai.

Gain $K\to 0$, isliye controller almost kuch nahi karta aur closed loop open-loop dynamics approach karta hai — theek hai agar plant pehle se stable ho, catastrophic agar nahi ho.

Controller bahut zyada aggressive ho jaata hai, bahut large gains aur fast poles ke saath; limit mein $R^{-1}$ blow up karta hai, isliye $R$ strictly positive hona zaroori hai taaki $K$ well defined rahe.

Kuch bhi error penalize nahi karne par, $\int u^{\top }Rudt$ minimize karne ka sabse sasta tarika $u=0$ hai; lekin agar $(A,\sqrt{Q})$ unobservable ho to solution unstable modes ko uncontrolled chod sakta hai, isliye $Q=0$ degenerate hai aur generally use nahi hota.

Nahi. Gain $K$ har starting point ke liye same hai; achieve ki gayi value $J$ equals $x(0{)}^{\top }Px(0)$ hai, jo $x(0)$ ke saath scale karti hai, lekin control rule nahi karta.

Nahi. Agar ek unstable mode ko $u$ influence nahi kar sakta, to koi bhi feedback use left half-plane mein move nahi kar sakta, isliye koi stabilizing $P⪰0$ exist nahi karta — unstable part ki controllability essential hai.

Closed loop test karo: optimal wahi $P⪰0$ hai jiske liye $A-BK$ ke saare eigenvalues left half-plane mein hain (Hurwitz); doosra ek anti-stable closed loop deta hai.