Worked examples — Linear Quadratic Regulator (LQR) — Riccati equation, optimal gains

3.5.35 · D3 · Physics › Guidance, Navigation & Control (GNC) › Linear Quadratic Regulator (LQR) — Riccati equation, optimal

Shuru karne se pehle, hum — general Riccati equation se, seedha is page pe — woh scalar formula dobara banate hain jis par hum baar baar bharosa karte hain, taaki neeche ka kuch bhi parent note ko yaad rakhne par depend na kare.

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Scenario matrix

Har LQR problem inhi cells mein se kisi ek mein aata hai. Har column ek knob hai; har row us knob ki ek value hai. Neeche ke worked examples batate hain ki woh kaunse cell(s) mein aate hain.

Cell	Plant ka sign / structure	Weight regime	Kya stress-test hota hai
A	Scalar, stable ($a<0$)	balanced $q=r$	baseline, positive root
B	Scalar, unstable ($a>0$)	balanced	control ko stability banana padega
C	Scalar, marginal ($a=0$)	balanced	integrator edge case
D	Scalar	cheap control $r\to 0$	limiting: gain blast ho jaata hai
E	Scalar	expensive control $r\to \infty$	limiting: gain fade hota hai, plant pe depend karna padega
F	2-state double integrator	vary $\rho$	LQR = PD, pole placement
G	2-state	$Q$ degenerate (rank-deficient)	$\sqrt{Q}$ ki observability zaroori
H	Any	$Q\to$ huge vs $R$	aggressiveness / sirf ratio ka rule
I	Uncontrollable plant	any	recipe fails — pehchanna zaroori
J	Word problem (satellite)	design task	spec ke liye $\rho$ choose karo

Ab hum A–J cover karte hain.

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Worked examples

Example 1 — Cell A: stable scalar plant, balanced weights

Forecast: plant pehle se hi stable hai ($a=-2<0$). Kya tumhe lagta hai LQR phir bhi pole ko aur left mein kheenchega, ya chhodega? Padhne se pehle andaza lagao.

1. Scalar Riccati likho. $2ap-\frac{b^2}{r}{p}^2+q=0\Rightarrow -4p-p^2+1=0.$ Ye step kyun? Humne upar $2ap-\frac{b^2}{r}{p}^2+q=0$ derive kiya tha; bas $a=-2,b=1,q=1,r=1$ substitute kiya.

2. Quadratic solve karo, positive root rakho. $p^2+4p-1=0$ likhte hain, toh $p=\frac{-4+\sqrt{16+4}}{2}=-2+\sqrt{5}\approx 0.236$. Positive root kyun? $V=p{x}^2$ ek cost hai aur $\ge 0$ honi chahiye, isliye $p>0$ zaroori hai; doosra root $-2-\sqrt{5}\approx -4.24$ negative hai aur reject ho jaata hai.

3. Gain compute karo. $K=\frac{b}{r}p=p\approx 0.236$.

4. Closed-loop pole. $a-bK=-2-0.236=-2.236$.

Verify: open-loop pole tha $-2$; closed loop hai $-2.236$, left half-plane mein aur andar. Toh LQR ek pehle se stable plant ko bhi thoda aur push karta hai — thoda control kharach karke thoda kam error karta hai. Sanity check: $a-bK<0$, stable. ✔

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Example 2 — Cell B: unstable scalar plant

Forecast: ek unstable plant apne aap bhaag jaata hai. Kya ek fixed gain $K$ use control kar sakta hai, aur pole kitna left ja sakta hai?

1. Numbers ke saath Riccati. $2ap-\frac{b^2}{r}{p}^2+q=6p-4p^2+4=0$. Standard quadratic form mein laane ke liye (positive leading coefficient), dono taraf $-1$ se multiply karo: $4p^2-6p-4=0$, phir $2$ se divide karo: $2p^2-3p-2=0$. Ye manipulation kyun? $-1$ se multiply karne par $p^2$ coefficient positive ho jaata hai toh quadratic formula saaf padhta hai; common factor $2$ se divide karne par numbers chhote ho jaate hain. Dono operations roots nahi badlte (ek equation "$=0$" ko nonzero constant se multiply karne par solutions waahi rehte hain).

2. Positive root. $p=\frac{3+\sqrt{9+16}}{4}=\frac{3+5}{4}=2$. Positive root kyun? Cost non-negativity wahi hai; $p=2>0$, doosra root $-\frac{1}{2}$ reject ho jaata hai.

3. Gain. $K=\frac{b}{r}p=2\cdot 2=4$.

4. Closed loop. $a-bK=3-2\cdot 4=3-8=-5<0$.

Verify: $+3$ pe shuru hua (bhaag raha tha), $-5$ pe khatam hua (decay ho raha hai). Controller ne pole ka sign flip kar diya — bilkul wahi promise jo LQR karta hai ki koi bhi controllable plant stable hoga. ✔

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Example 3 — Cell C: marginal plant ($a=0$ integrator)

Forecast: $a=0$ ke saath plant na badhta hai na ghatata hai — bas hold karta hai. Kya formula phir bhi clean answer deta hai?

1. Riccati. $2(0)p-\frac{1}{4}{p}^2+1=0\Rightarrow -\frac{1}{4}{p}^2+1=0\Rightarrow p^2=4$. Ye step kyun? $2ap$ term zero ho jaata hai kyunki $a=0$ hai; quadratic sirf ek pure square mein collapse ho jaati hai.

2. Positive root. $p^2=4$ ke roots hain $p=+2$ aur $p=-2$; hum $p=2$ rakhte hain. $p=-2$ reject kyun? Bilkul wahi reason jaise har scalar case mein: $V=p{x}^2$ cost-to-go hai aur kabhi negative nahi ho sakti, isliye $p\ge 0$ chahiye. Root $-2$ kisi bhi $x\ne 0$ ke liye $V<0$ banata, jo meaningless hai, isliye discard ho jaata hai.

3. Gain. $K=\frac{b}{r}p=\frac{1}{4}\cdot 2=0.5$.

4. Closed loop. $a-bK=0-0.5=-0.5<0$.

Verify: $a=0$ ke saath Riccati padhti hai $\frac{b^2}{r}{p}^2=q$, toh $p=\sqrt{qr}/b$ (yahan $b>0$ lete hain). Numerically $p=\sqrt{1\cdot 4}/1=2$, step 2 se match. ✔ Clean edge case, stable.

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Example 4 — Cell D & E: do limiting regimes (cheap vs expensive control)

Forecast: agar control almost free hai ($r\to 0$), toh gain bada hona chahiye ya chhota? Agar control almost forbidden hai ($r\to \infty$), toh hum kis pole par phanse rahenge?

1. Apne box se general root lo. Dobara derive kiya hua scalar formula $p(r)=\frac{ar+\sqrt{a^2{r}^2+qr{b}^2}}{b^2}$ upar $2ap-\frac{b^2}{r}{p}^2+q=0$ par quadratic formula apply karke mila tha. $a=b=q=1$ substitute karo: $p(r)=r+\sqrt{r^2+r}.$ Ye step kyun? Ye $p$ (aur isliye $K$) ko single knob $r$ ke explicit function ke roop mein express karta hai, taaki hum seedha limits le sakein instead of har $r$ ke liye quadratic dobara solve karne ke.

2. $r$ ke function mein gain. $K(r)=\frac{b}{r}p=\frac{r+\sqrt{r^2+r}}{r}=1+\sqrt{1+\frac{1}{r}}$. Aise kyun likhein? $r$ se divide karne par (aur $r$ root ke andar kheenchne par: $\sqrt{r^2+r}/r=\sqrt{1+1/r}$) woh $\frac{1}{r}$ term isolate ho jaata hai jiska limit directly padha ja sakta hai.

3. Cheap control, $r\to 0^{+}$. $\frac{1}{r}\to \infty$, toh $K\to \infty$. Kyun? Control almost free hai, toh optimizer unlimited effort kharach karta hai → infinite gain → pole $a-bK\to -\infty$ (instant correction). Ye D-cell limit hai.

4. Expensive control, $r\to \infty$. $\frac{1}{r}\to 0$, toh $K\to 1+\sqrt{1}=2$; pole $a-bK\to 1-2=-1$. $K\to 0$ kyun nahi? Kyunki $a=1$ unstable hai — tum kharcha karna kabhi poora nahi rok sakte, warna state bhaag jaati hai. Controller us sabse saste gain par settle karta hai jo phir bhi stabilize kare, plant ki apni rate ko mirror karte hue. Ye mirror-image pole ($-|a|$) ek famous LQR fact hai (cheap-vs-expensive limit).

5. Figure.

   

   Horizontal axis control weight $r$ hai (0 se 8 tak); vertical axis optimal gain $K$ hai. Lavender curve $K(r)=1+\sqrt{1+1/r}$ left par $+\infty$ se ghoom kar right par mint dashed asymptote $K=2$ tak aa jaata hai; coral dot value $K(1)\approx 2.414$ mark karta hai.

Verify: $r=1$ par, $K=1+\sqrt{2}\approx 2.414$ — parent note ke worked number se exactly match. ✔ Do asymptotes ($K\to \infty$ aur $K\to 2$) ise sahi bracket karte hain.

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Example 5 — Cell H: sirf ratio $Q/R$ matter karta hai (aggressiveness)

Forecast: dono penalties ko same factor se scale karna — kya controller zyada aggressive, kam aggressive, ya unchanged hoga?

1. Ratio ke terms mein general gain. $a=b=1$ ke saath positive root (box se) hai $p=r+\sqrt{r^2+qr}$, toh $K=\frac{p}{r}=1+\sqrt{1+\frac{q}{r}}.$ Ye step kyun? $r$ ko square root se bahar nikalo: $\sqrt{r^2+qr}=r\sqrt{1+q/r}$. $r$ se divide karne ke baad, sirf ratio $q/r$ bachta hai — koi standalone $q$ ya $r$ nahi rehta.

2. Case $(1,1)$: $q/r=1$, toh $K=1+\sqrt{2}\approx 2.414$.

3. Case $(100,100)$: $q/r=1$ phir se, toh $K=1+\sqrt{2}\approx 2.414$. Identical.

Verify: dono gains numerically equal hain kyunki $K$ sirf $q/r$ par depend karta hai. Ye parent-note ki mistake #1 confirm karta hai: sirf ratio aggressiveness set karta hai, dono ko scale karna no-op hai. ✔

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Example 6 — Cell F: double integrator, LQR = PD controller

Forecast: LQR yahan PD controller ban jaata hai. Kaunsa damping ratio expect karna chahiye?

1. CARE ko entry by entry expand karo (yahan kiya gaya hai, assume nahi). $A^{\top }P+PA-PB{R}^{-1}{B}^{\top }P+Q=0$ aur $B^{\top }P=[p_2,p_3]$ ke saath, toh $PB{R}^{-1}{B}^{\top }P=\frac{1}{\rho }\left[\begin{array}{cc}{p}_2^2& p_2{p}_3\\ p_2{p}_3& p_3^2\end{array}\right]$, teen scalar equations hain:

   • $(1,1)$: $1-p_2^2/\rho =0$,
   • $(1,2)$: $p_1-p_2{p}_3/\rho =0$,
   • $(2,2)$: $2p_2-p_3^2/\rho =0$. Ye step kyun? Ek symmetric $2\times 2$ matrix equation exactly teen independent scalar equations deti hai — ek per distinct entry — jinhe hum order mein solve karte hain.

2. Solve karo. $(1,1)$ se: $p_2=\sqrt{\rho }$. $(2,2)$ se: $p_3=\sqrt{2\rho p_2}=\sqrt{2}{\rho }^{3/4}$. $(1,2)$ se: $p_1=p_2{p}_3/\rho$. $\rho =1$ ke saath: $p_2=1,p_3=\sqrt{2},p_1=\sqrt{2}$.

3. Gain. $K=\frac{1}{\rho }[p_2,p_3]=[1,\sqrt{2}]$. Toh $k_p=1$ (position/spring), $k_d=\sqrt{2}$ (velocity/damper). Ye PD kyun hai? $u=-k_p{x}_1-k_d{x}_2=-(\text{spring})x_1-(\text{damper})x_2$ — textbook PD form (dekho PID Controllers, Pole Placement).

4. Closed-loop matrix. $A-BK=\left[\begin{array}{cc}0& 1\\ -1& -\sqrt{2}\end{array}\right]$. Characteristic poly: $s^2+\sqrt{2}s+1=0$. Kyun? $\det (sI-(A-BK))=s(s+\sqrt{2})+1$.

5. Poles. $s=\frac{-\sqrt{2}\pm \sqrt{2-4}}{2}=-\frac{1}{\sqrt{2}}\pm j\frac{1}{\sqrt{2}}$.

6. Figure.

   

   Axes complex $s$-plane hain: horizontal = real part $\text{Re}(s)$, vertical = imaginary part $\text{Im}(s)$. Do lavender crosses closed-loop poles hain $-1/\sqrt{2}\pm j/\sqrt{2}$ par; woh coral unit circle ($|s|=1$) par left half-plane mein $45^{\circ }$ par baithe hain — damping ratio $\zeta =\cos 45^{\circ }=1/\sqrt{2}$.

Verify: ${\omega }_n=|s|=\sqrt{(1/\sqrt{2}{)}^2+(1/\sqrt{2}{)}^2}=1$, aur $\zeta =1/\sqrt{2}\approx 0.707$ — "critically nice" GNC damping. ✔

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Example 7 — Cell F (twist): double integrator par expensive control

Forecast: zyada expensive control ka matlab hona chahiye chhote gains aur dheemi (lower-frequency) response. Kitne factor se?

1. Example 6 ke general entries mein $\rho =16$ daalo. $p_2=\sqrt{\rho }=4$, $p_3=\sqrt{2}{\rho }^{3/4}=\sqrt{2}\cdot 8=8\sqrt{2}$. Kyun? ${\rho }^{3/4}=16^{3/4}=(2^4{)}^{3/4}=2^3=8$.

2. Gain. $K=\frac{1}{\rho }[p_2,p_3]=\frac{1}{16}[4,8\sqrt{2}]=[0.25,0.5\sqrt{2}]=[0.25,0.7071]$.

3. Natural frequency. Is plant par LQR ke liye, ${\omega }_n=\sqrt{k_p}=\sqrt{1/\sqrt{\rho }}={\rho }^{-1/4}=16^{-1/4}=0.5$. Kyun? Closed-loop poly hai $s^2+k_{d}s+k_p$, toh ${\omega }_n=\sqrt{k_p}=\sqrt{0.25}=0.5$.

Verify: Example 6 se compare karein (${\omega }_n=1$), $\rho$ $16\times$ badha aur ${\omega }_n$ $16^{1/4}=2\times$ gira (1 se 0.5 tak). Dheema aur gentle, exactly waise jaisa "expensive control" predict karta hai. Damping check: $\zeta =k_d/(2{\omega }_n)=0.7071/(2\cdot 0.5)=0.7071=1/\sqrt{2}$ — same damping ratio (is problem ki ek scale-invariance). ✔

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Example 8 — Cell G: degenerate $Q$ jo phir bhi theek hai

Forecast: $Q$ ka ek zero eigenvalue hai, toh hum velocity ke baare mein kuch bhi directly penalize nahi kar rahe. Kya solution phir bhi dono states ko stabilize karta hai?

1. Requirement batao. Ek stabilizing $P⪰0$ exist karta hai iff $(A,B)$ controllable aur $(A,\sqrt{Q})$ observable ho (Controllability and Observability). Ye step kyun? Ek degenerate $Q$ theek hai jab tak penalized quantity dynamics ke through har unstable mode ko "dekhti" hai.

2. $\sqrt{Q}$ chunno. Upar define kiye anusaar, $\sqrt{Q}=C=\left[\begin{array}{cc}1& 0\\ 0& 0\end{array}\right]$, matlab output $=x_1$ (position).

3. $(A,C)$ ki Observability. $\mathcal{O}=\left[\begin{array}{c}C\\ CA\end{array}\right]=\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right]$, rank 2 = full. Ye humein kyun bachaata hai? Bhale hi hum velocity directly penalize nahi karte, velocity position se observable hai $CA$ ke through (position ki rate velocity reveal karti hai). Toh cost phir bhi ek runaway velocity "feel" karti hai.

4. Consequence. Example-6 solution $K=[1,\sqrt{2}]$ valid aur stabilizing hai degenerate $Q$ ke bawajood.

Verify: observability matrix $2\times 2$ identity hai, determinant $=1\ne 0$ ⇒ full rank ⇒ LQR well-posed. ✔ (Ise Cell I se contrast karo next mein, jahan plant itself ek condition fail karta hai.)

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Example 9 — Cell I: jab recipe kaam karne se mana kar deti hai

Forecast: control $u$ sirf pehli row mein enter karta hai. Unstable mode $x_2$ hai. Kya koi gain use touch kar sakta hai?

1. Controllability matrix banao. $\mathcal{C}=[B,AB]$. $AB=\left[\begin{array}{cc}0& 0\\ 0& 1\end{array}\right]\left[\begin{array}{c}1\\ 0\end{array}\right]=\left[\begin{array}{c}0\\ 0\end{array}\right]$. Toh $\mathcal{C}=\left[\begin{array}{cc}1& 0\\ 0& 0\end{array}\right]$. Ye step kyun? $(A,B)$ controllable ⟺ $\mathcal{C}$ ka full rank $n=2$ ho (Controllability and Observability).

2. Rank. $\det \mathcal{C}=0$, rank $=1<2$. Uncontrollable.

3. Kaunsa mode diagnose karo. Uncontrollable mode $x_2$ hai, jiska eigenvalue $+1$ (unstable) hai. Ek unstable aur uncontrollable mode kisi bhi $u=-Kx$ se move nahi ho sakta. Ye LQR ko kyun khatam karta hai? Koi bhi feedback gain us pole ko push nahi kar sakta jahan control pahunch nahi sakta. Riccati equation ek stabilizing $P⪰0$ yield nahi karega.

4. Conclusion. LQR ka yahan koi stabilizing solution nahi hai; tumhe actuator ($B$) redesign karna hoga taaki woh $x_2$ mein couple ho sake.

Verify: $\det \mathcal{C}=1\cdot 0-0\cdot 0=0$ ⇒ rank-deficient ⇒ uncontrollable. Recipe sahi tarike se fail hoti hai. ✔ Ye boundary case hai jise har LQR user ko pehchanna chahiye.

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Example 10 — Cell J: satellite design word problem

Forecast: humne ${\omega }_n={\rho }^{-1/4}$ nikala tha. $0.5$ tak dheela karne ke liye, kya $\rho$ bada karte hain ya chhota?

1. Frequency relation invert karo. Example 7 se, ${\omega }_n={\rho }^{-1/4}$, toh $\rho ={\omega }_n^{-4}=(0.5{)}^{-4}=16$. Ye step kyun? Humne ${\omega }_n={\rho }^{-1/4}$ derive kiya tha; $\rho$ ke liye solve karne par required control weight milta hai. Bada $\rho$ (expensive control) ⇒ dheema response — forecast se match.

2. Gain compute karo (Example 7 reuse karo). $K=[0.25,0.7071]$.

3. Engineering reading. $k_p=0.25$ (torque per radian) = restoring "spring"; $k_d=0.707$ (torque per rad/s) = damping. Damping ratio $\zeta =1/\sqrt{2}\approx 0.707$ → lagbhag $5\%$ overshoot, ek smooth slew. Satellite ke liye ye achha kyun hai? Kam ${\omega }_n$ ⇒ gentle wheel torques ⇒ kam power aur kam structural stress, dheemai settle ke cost par. Dekho LQG Control jab state pehle estimate karni ho.

Verify: $\rho =16\Rightarrow {\omega }_n=16^{-1/4}=0.5$ ✔; $K=[0.25,0.7071]$ ✔ (bilkul Example 7 jaisa, jaisa hona chahiye — same $\rho$ hai).

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Recall Coverage checklist (kya humne har cell cover kiya?)

A ✔ (Ex 1) · B ✔ (Ex 2) · C ✔ (Ex 3) · D & E ✔ (Ex 4) · H ✔ (Ex 5) · F ✔ (Ex 6, 7) · G ✔ (Ex 8) · I ✔ (Ex 9) · J ✔ (Ex 10). Stable/unstable/marginal scalars, cheap/expensive limits, ratio-invariance, matrix PD case, degenerate-but-OK $Q$, aur uncontrollable failure — sab dikhaaya gaya.

Active Recall