H∞ control — robust to uncertainty (intro): Exercises

3.5.37 · D4 · Physics › Guidance, Navigation & Control (GNC) › H∞ control — robust to uncertainty (intro)

Shuru karne se pehle, ek reminder un symbols ki jo hum use karenge, simple shabdon mein:

Definition Panch symbols jo tumhe yaad hone chahiye

$∣ G (j ω) ∣$ — system ka magnitude ek pure sine wave ke response mein jiska frequency $ω$ hai. Yeh ek seedha positive number hai: "yeh particular sine wave kitne times stretch hoti hai?"
$∥ w ∥_{2}$ — signal $w (t)$ ka 2-norm (energy): $∥ w ∥_{2} = \int_{0}^{\infty} w (t)^{⊤} w (t) d t$ . Simple words mein: signal ki squared size ko puri time par add karo aur square root lo — yeh measure karta hai total energy jo ek gust carry karta hai. Finite $∥ w ∥_{2}$ matlab signal eventually khatam ho jaata hai.
$\overset{σ}{ˉ} (\cdot)$ — ek matrix ka largest singular value (Singular Value Decomposition se). Single-input single-output system ke liye yeh bas $∣ G (j ω) ∣$ hai. Socho "sabse bada stretch factor jo ek matrix kisi bhi input direction par apply karta hai."
$∥ G ∥_{\infty}$ — us stretch factor ka peak saari frequencies par: $sup_{ω} \overset{σ}{ˉ} (G (j ω))$ . System ki ek bhi worst amplification. Equivalently worst-case energy gain $sup_{w} ∥ z ∥_{2} /∥ w ∥_{2}$ (output energy per unit input energy).
$S = \frac{1}{1 + L}$ — sensitivity, jahan $L$ open-loop gain hai. Yeh reference (ya disturbance) se tracking error tak ka transfer function hai. Chota $∣ S ∣$ = choti error.

Level 1 — Recognition

Exercise 1.1 (L1)

Ek stable system $G$ ke liye $∥ G ∥_{\infty} = 4$ hai. Ek sentence mein, yeh kya physical fact batata hai?

Recall Solution

YEH kya kehta hai: koi sinusoidal disturbance exist karta hai jiska output energy $∥ z ∥_{2}$ , input energy $∥ w ∥_{2}$ se $4$ guna hai (yaad karo $∥ \cdot ∥_{2}$ = signal ki total energy, $\int w^{⊤} w d t$ , symbols box mein define kiya gaya hai), aur koi bhi disturbance $4$ se zyada amplify nahi hoti. KYUN: $∥ G ∥_{\infty} = sup_{w} ∥ z ∥_{2} /∥ w ∥_{2}$ — literally worst-case energy gain, output-energy aur input-energy ka biggest ratio har possible disturbance par. Toh $4$ amplification par ek ceiling hai aur ek aisi ceiling jo actually reach bhi hoti hai. In words: "sabse buri sine wave chaar guna loud ho jaati hai; koi cheez isse buri nahi hoti."

Exercise 1.2 (L1)

Inmein se kaunsa scalar stable $G$ ke liye $∥ G ∥_{\infty}$ ke barabar hai? (a) $∣ G (0) ∣$ , (b) $\frac{1}{2 π} \int ∣ G (j ω) ∣ d ω$ , (c) $sup_{ω} ∣ G (j ω) ∣$ .

Recall Solution

Answer: (c). (a) DC gain hai — response sirf $ω = 0$ par (ek constant input). Yeh har doosri frequency ko ignore karta hai. (b) frequency par ek average hai. Averages peak ko smooth kar dete hain. (c) peak hai, aur peak wahi matter karta hai kyunki sabse buri disturbance apni energy exactly wahin concentrate karti hai jahan $∣ G ∣$ sabse bada hota hai (parent note se Parseval argument).

Level 2 — Application

Exercise 2.1 (L2)

Lightly damped mode $G (s) = \frac{9}{s ^{2} + 0.6 s + 9}$ ke liye ( $ω_{n} = 3$ , $ζ = 0.1$ ), resonant-peak formula se $∥ G ∥_{\infty}$ estimate karo: $M_{r} = \frac{∣ G ( 0 ) ∣}{2 ζ 1 - ζ ^{2}} .$

Recall Solution

Step 0 — check karo ki formula apply bhi hoti hai. Kyun? Yeh resonant-peak formula maan leti hai ki ek genuine peak exist karta hai, jo sirf $0 < ζ < \frac{1}{2} \approx 0.707$ ke liye hota hai. Isse zyada damping mein magnitude sirf monotonically roll off karti hai (koi peak nahi), aur tab $∥ G ∥_{\infty} = ∣ G (0) ∣$ hota hai. Yahan $ζ = 0.1 < 0.707$ , toh ek true resonant peak exist karta hai — achha hai. Step 1 — $ω_{n}, ζ$ identify karo. Kyun? Standard form hai $\frac{ω _{n}^{2} K}{s ^{2} + 2 ζ ω _{n} s + ω _{n}^{2}}$ . Match karte hain: $ω_{n}^{2} = 9 \Rightarrow ω_{n} = 3$ ; $2 ζ ω_{n} = 0.6 \Rightarrow ζ = 0.1$ ; DC gain $∣ G (0) ∣ = 9/9 = 1$ . Step 2 — plug in karo. Peak formula kyun? Kyunki $∥ G ∥_{\infty}$ , $∣ G (j ω) ∣$ ka peak hai, aur lightly damped second-order mode ke liye peak $ω_{n}$ ke paas hota hai jiska height $M_{r}$ hota hai. $M_{r} = \frac{1}{2 ( 0.1 ) 1 - 0.01} = \frac{1}{0.2 \times 0.99499} \approx 5.025.$ Toh $∥ G ∥_{\infty} \approx 5.03$ : $ω \approx 3$ rad/s par ek resonant disturbance lagbhag paanch guna amplify hoti hai. Figure mein kya dekhna hai: lavender curve $∣ G (j ω) ∣$ hai. Notice karo yeh low frequency par $1$ ke paas rehti hai (mint dotted DC line), phir $ω \approx 3$ par ek sharp coral dot tak rise karti hai — us dot ki height ( $\approx 5.03$ ) poora H∞ norm hai. Sab kuch left aur right mein chota hai; sirf us single spike ki height matter karti hai. Isliye "peak, not average" sahi reading hai.

Figure — H∞ control — robust to uncertainty (intro)

Exercise 2.2 (L2)

Actual actuator gain $1 \pm 25%$ hai. Uncertainty ko $0.25 \tilde{Δ}$ likho jahan $∥ \tilde{Δ} ∥_{\infty} \leq 1$ , phir Small-Gain Theorem se largest nominal loop peak $∥ M ∥_{\infty}$ dhundho jo robustly stable guarantee kare.

Recall Solution

Step 1 — normalise karo. Kyun? Small-Gain Theorem uncertainty ko unit ball $∥ \tilde{Δ} ∥_{\infty} \leq 1$ ke andar chahta hai, toh hum size $0.25$ bahar nikaalte hain: real uncertain block $0.25 \tilde{Δ}$ hai. Step 2 — small-gain apply karo. Aise sab blocks ke liye stability tabhi hogi jab loop-gain product ek se kam ho: $∥ 0.25 M ∥_{\infty} < 1 ⟺ ∥ M ∥_{\infty} < \frac{1}{0.25} = 4.$ Toh $\pm 25%$ ke andar koi bhi actuator error tolerate ki jaati hai agar nominal loop ka peak gain us channel mein $4$ se kam ho. YAHAN kya mila: ek scalar inequality infinite family of plants ko certify karti hai.

Level 3 — Analysis

Exercise 3.1 (L3)

Tumne sensitivity weight $W_{S} (s)$ place kiya aur $∥ W_{S} S ∥_{\infty} < 1$ achieve kiya. $ω = 1$ rad/s par weight ka magnitude $∣ W_{S} (j 1) ∣ = 20$ hai. Yeh $∣ S (j 1) ∣$ par kya ceiling force karta hai, aur us frequency par tracking error ke liye iska kya matlab hai?

Recall Solution

Step 1 — norm inequality ko unpack karo. $∥ W_{S} S ∥_{\infty} < 1$ matlab product $∣ W_{S} (j ω) S (j ω) ∣ < 1$ har frequency par. Kyun har frequency? Kyunki $\infty$ -norm peak hai; agar peak $1$ se kam hai, toh sab frequencies $1$ se kam hain. Step 2 — $∣ S ∣$ ke liye solve karo. $ω = 1$ par: $∣ W_{S} (j 1) ∣ ∣ S (j 1) ∣ < 1 ⟹ ∣ S (j 1) ∣ < \frac{1}{20} = 0.05.$ Interpretation: tracking error $1$ rad/s par reference/disturbance amplitude ka zyada se zyada $5%$ hai. Weight $W_{S}$ literally ceiling $1/∣ W_{S} ∣$ draw karta hai jiske neeche $∣ S ∣$ rehna chahiye. Figure mein kya dekhna hai: coral curve ceiling $1/∣ W_{S} ∣$ hai — ek wall jise $∣ S ∣$ kabhi cross nahi kar sakta. Lavender curve ek feasible $∣ S ∣$ hai; dekho kaise yeh har jagah coral ceiling ke neeche rehti hai. Mint dot $ω = 1$ mark karta hai: wahan ceiling $0.05$ par hai, toh $∣ S ∣$ $5%$ ke neeche squeeze hoti hai. Picture padhna: jahan bhi weight tall banao, ceiling low press ho jaati hai aur tight tracking force hoti hai.

Exercise 3.2 (L3)

Ek colleague kehta hai: "Chalte hain $∣ S ∣$ ko $0$ se $\infty$ tak sab frequencies par $0.01$ se neeche push karte hain." Bode sensitivity integral (waterbed) $\int_{0}^{\infty} ln ∣ S (j ω) ∣ d ω = 0$ use karke explain karo yeh kyun impossible hai.

Recall Solution

Step 0 — assumptions state karo. Pehle kyun? Identity $\int_{0}^{\infty} ln ∣ S ∣ d ω = 0$ universal nahi hai — yeh tab hold karti hai jab open-loop transfer function $L$ stable ho (koi right-half-plane poles nahi), jiska relative degree kam se kam $2$ ho (taaki $L \to 0$ itni jaldi ho ki integral converge kare), aur jisme koi right-half-plane zeros nahi hon. RHP poles right-hand side ko positive value $π \sum Re (p_{i})$ tak raise kar dete hain, waterbed aur bura kar dete hain. Yahan hum benign stable, minimum-phase case assume karte hain toh right side exactly $0$ hai. Step 1 — integral padho. Bode Sensitivity Integral (waterbed) se, $ln ∣ S ∣$ ke neeche signed area $0$ par fixed hai. Jahan $∣ S ∣ < 1$ wahan $ln ∣ S ∣ < 0$ (negative area, "achha"); jahan $∣ S ∣ > 1$ wahan $ln ∣ S ∣ > 0$ (positive area, "bura"). Step 2 — contradiction argue karo. Agar $∣ S ∣ < 0.01$ har jagah ho, toh $ln ∣ S ∣ < ln (0.01) \approx - 4.6$ har jagah hoga, toh integral ek bada negative number hoga, $0$ nahi. $\int_{0}^{\infty} ln ∣ S ∣ d ω \leq \int_{0}^{\infty} (- 4.6) d ω = - \infty \neq = 0.$ Toh yeh conservation law violate karta hai: yahan $∣ S ∣$ neeche push karo, toh kahin aur $1$ se upar pop up karna hi padega — waterbed. Tum sirf $S$ ko shape kar sakte ho, globally shrink nahi kar sakte.

Level 4 — Synthesis

Exercise 4.1 (L4)

Ek first-order sensitivity weight $W_{S} (s) = \frac{s / M + ω _{b}}{s + ω _{b} A}$ design karo taaki:

steady-state error $\leq 1%$ (yaani $∣ S (0) ∣ \leq 0.01$ ),
peak sensitivity $\leq 2$ (yaani $∣ S ∣ < 2$ high frequency par),
bandwidth $ω_{b} = 5$ rad/s.

$A$ aur $M$ dhundho, aur low- aur high-frequency ceilings $1/∣ W_{S} ∣$ state karo.

Recall Solution

Step 1 — low frequency ( $ω \to 0$ ). Yahan kyun dekhein? Steady-state tracking DC se set hoti hai. $W_{S} (0) = \frac{ω _{b}}{ω _{b} A} = \frac{1}{A} .$ $∣ S ∣$ par ceiling $1/∣ W_{S} (0) ∣ = A$ hai. Hum $∣ S (0) ∣ \leq 0.01$ chahte hain, toh $A = 0.01$ choose karo. Step 2 — high frequency ( $ω \to \infty$ ). Yahan kyun dekhein? Peak-sensitivity spec high $ω$ par rehti hai. $W_{S} (\infty) = \frac{s / M}{s} = \frac{1}{M} .$ Wahan ceiling $1/∣ W_{S} (\infty) ∣ = M$ hai. Hum $∣ S ∣ < 2$ chahte hain, toh $M = 2$ choose karo. Step 3 — exact crossover dhundho, phir $\approx$ justify karo. Kyun? Bandwidth woh jagah hai jahan weight ka magnitude $1$ se pass karta hai; humein woh frequency derive karni hai, assert nahi. $∣ W_{S} (j ω) ∣ = 1$ set karo, yaani $∣ j ω / M + ω_{b} ∣^{2} = ∣ j ω + ω_{b} A ∣^{2}$ : $\frac{ω ^{2}}{M ^{2}} + ω_{b}^{2} = ω^{2} + ω_{b}^{2} A^{2} ⟹ ω^{2} (1 - \frac{1}{M ^{2}}) = ω_{b}^{2} (1 - A^{2}) .$ Exactly solve karo: $ω_{cross} = ω_{b} \frac{1 - A ^{2}}{1 - 1/ M ^{2}} .$ $M = 2, A = 0.01, ω_{b} = 5$ plug in karo: $1 - A^{2} = 0.9999$ aur $1 - 1/ M^{2} = 1 - 0.25 = 0.75$ , toh $ω_{cross} = 5 \frac{0.9999}{0.75} = 5 \times 1.1547 \approx 5.77 rad/s .$ Kyun pehle wala " $\approx ω_{b}$ " sirf rough tha: correction factor $(1 - A^{2}) / (1 - 1/ M^{2})$ , $1$ ke barabar hota hai sirf ideal limit $M \to \infty, A \to 0$ mein. $M = 2$ ke saath $1/ M^{2} = 0.25$ term negligible nahi hai — yeh crossover ko $ω_{b}$ se lagbhag $15%$ upar kheencha hai. Toh honest statement hai: crossover $\approx 5.77$ rad/s hai, $ω_{b} = 5$ ke paas lekin equal nahi; tiny $A$ ise barely move karta hai, modest $M$ ise noticeably move karta hai. Result: $W_{S} (s) = \frac{s /2 + 5}{s + 5 ( 0.01 )} = \frac{0.5 s + 5}{s + 0.05} .$ Ceilings: low-freq $∣ S ∣ \leq A = 0.01$ (1% error ✓), high-freq $∣ S ∣ \leq M = 2$ (peak $\leq 2$ ✓). Agar ek controller $∥ W_{S} S ∥_{\infty} < 1$ achieve kare, toh teeno specs simultaneously meet hote hain. Figure mein kya dekhna hai: lavender curve ceiling $1/∣ W_{S} ∣$ hai. Yeh mint line ke paas $A = 0.01$ (tracking ke liye 1% floor) se start hoti hai, $ω_{b}$ ke paas $1$ se guzarti hai (butter dotted line), aur coral line $M = 2$ (peak allowance) par flatten ho jaati hai. Isko padhna: is ceiling ki shape exactly woh performance spec hai jo ek wall ki tarah draw ki gayi hai.

Level 5 — Mastery

Exercise 5.1 (L5)

Ek rocket loop mein nominal complementary sensitivity $T$ hai jiska peak $∥ T ∥_{\infty} = 0.8$ hai. Plant mein multiplicative output uncertainty $P = (1 + W_{Δ} Δ) P_{0}$ hai jahan $∥Δ ∥_{\infty} \leq 1$ aur interest ke band mein scalar weight $∣ W_{Δ} (j ω) ∣ = 0.5$ hai. Is uncertainty structure ke liye robust-stability condition hai $∥ W_{Δ} T ∥_{\infty} < 1$ . (a) Kya loop robustly stable hai? (b) Sabse badi uncertainty size $∣ W_{Δ} ∣$ kya hai jo yeh loop tolerate kar sakta hai? (c) Agar $∥ T ∥_{\infty}$ $2.5$ hota, toh tum kya conclude karte, aur kaunsa vault tool conservatism reduce karta hai?

Recall Solution

(a) Condition check karo. $∥ W_{Δ} T ∥_{\infty}$ kyun? Multiplicative output uncertainty ke liye $Δ$ ke saath feedback-mein-loop ka forward block $W_{Δ} T$ hai; Small-Gain Theorem phir demand karta hai ki uska peak $1$ se kam ho. Kyunki $W_{Δ}$ yahan constant $0.5$ hai, $∥ W_{Δ} T ∥_{\infty} = 0.5 \times ∥ T ∥_{\infty} = 0.5 \times 0.8 = 0.4 < 1. ✓$ Robustly stable. (b) Sabse badi tolerable uncertainty. $∣ W_{Δ} ∣ ∥ T ∥_{\infty} < 1$ solve karo: $∣ W_{Δ} ∣ < \frac{1}{∥ T ∥ _{\infty}} = \frac{1}{0.8} = 1.25.$ Toh loop in frequencies par nominal plant ki 125% tak multiplicative uncertainty tolerate karta hai guarantee fail hone se pehle. (c) Agar $∥ T ∥_{\infty} = 2.5$ hota: tab $∥ W_{Δ} T ∥_{\infty} = 0.5 \times 2.5 = 1.25 > 1$ hota, toh sufficient small-gain test fail ho jaata — hum ab robust stability certify nahi kar sakte. Lekin small-gain $Δ$ ki structure ignore karta hai aur aksar conservative hota hai; woh tool jo ise tighten karta hai guarantee rakhte hue hai μ-synthesis (structured uncertainty), jo plain peak ki jagah structured singular value $μ$ use karta hai.

Exercise 5.2 (L5)

Performance aur robustness combine karo. Tumhe $∥ W_{S} S ∥_{\infty} < 1$ aur $∥ W_{Δ} T ∥_{\infty} < 1$ chahiye, jahan $S + T = 1$ (Sensitivity and Complementary Sensitivity (S+T=1) se). Ek frequency $ω_{0}$ par, maano tum $∣ S (j ω_{0}) ∣ = 0.1$ demand karte ho. Sabse chota possible $∣ T (j ω_{0}) ∣$ kya hai, aur kya yeh robustness weight $∣ W_{Δ} (j ω_{0}) ∣ = 0.5$ ke saath compatible hai?

Recall Solution

Step 1 — algebraic identity use karo. Kyun? $S$ aur $T$ free nahi hain; har frequency par $S (j ω) + T (j ω) = 1$ complex numbers ki tarah. Toh $T = 1 - S$ , aur reverse triangle inequality se magnitude neeche se bounded hai: $∣ T (j ω_{0}) ∣ = ∣1 - S (j ω_{0}) ∣ \geq 1 - ∣ S (j ω_{0}) ∣ = 1 - 0.1 = 0.9.$ Toh $∣ T (j ω_{0}) ∣$ ki sabse choti possible value $0.9$ hai. Worst-case phase remark: $∣ T ∣ = 0.9$ value tabhi attain hoti hai jab $S (j ω_{0})$ real aur positive ho (phase $0$ ). Agar $S$ mein nonzero phase ho, toh $∣ T ∣ = ∣1 - S ∣$ larger ho sakta hai — jaise agar $S = 0.1 j$ toh $∣ T ∣ = ∣1 - 0.1 j ∣ = 1 + 0.01 \approx 1.005$ . Toh $0.9$ ek floor hai, aur honest robustness check mein worst-case $∣ T ∣$ use karna padega, jo $0.9$ se zyada ho sakta hai. Step 2 — floor par robustness test karo. Sabse chote $∣ T ∣ \approx 0.9$ use karke: $∣ W_{Δ} T ∣ = 0.5 \times 0.9 = 0.45 < 1. ✓$ Compatible — tightest $∣ T ∣ = 0.9$ par bhi, robustness product $0.45$ safely $1$ se kam hai. (Agar $S$ ka phase $∣ T ∣$ ko $1.005$ tak push karta, product $0.5 \times 1.005 = 0.5025$ hota, phir bhi $1$ se kam, toh yeh design saari phases mein robust hai.) Sabak: tiny error demand karna ( $∣ S ∣$ chota) $∣ T ∣$ ko $1$ ke paas force karta hai — tum dono ko ek hi frequency par chota nahi kar sakte. Yahan toh sabse chota $∣ T ∣ = 0.9$ bhi robustness margin chhodata hai kyunki weight sirf $0.5$ hai. $∣ S ∣$ aur chota karo aur $∣ T ∣ \to 1$ ; agar $∣ W_{Δ} ∣$ tab $1/∣ T ∣ \to 1$ se zyada ho jaaye, robustness toot jaayegi. Yahi performance–robustness trade-off quantitative roop mein hai.