Worked examples — H∞ control — robust to uncertainty (intro)

3.5.37 · D3 · Physics › Guidance, Navigation & Control (GNC) › H∞ control — robust to uncertainty (intro)

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Scenario matrix

Kuch bhi work karne se pehle, chaliye un sabhi classes of cases ki list banate hain jo ek H∞ / small-gain problem de sakta hai. Baad ke har example mein uska cell(s) tag kiya gaya hai.

Cell	Kya cheez ise alag banati hai	Covered by
A. Flat gain	$G$ mein koi frequency dependence nahi — constant magnitude	Ex 1
B. Monotone roll-off	$\Vert G(j\omega )\Vert$ ka peak $\omega =0$ pe hai, koi resonance nahi	Ex 2
C. Resonant peak	lightly damped mode, peak band ke andar	Ex 3
D. Degenerate input	zero-energy ya DC-only disturbance	Ex 4
E. Limiting behaviour	damping $\zeta \to 0$, peak $\to \infty$	Ex 5
F. Multi-input (SVD)	matrix $G$, largest singular value chahiye, $\Vert \cdot \Vert$ nahi	Ex 6
G. Small-gain sizing	uncertainty size se robust-stability margin	Ex 7
H. Weight-ceiling / waterbed	$\Vert W_{S}S{\Vert }_{\infty }<1$ ek drawn ceiling ke roop mein, plus trade-off	Ex 8
I. Real-world word problem	rocket bending mode, design action chunno	Ex 9
J. Exam twist	small-gain + peak reading ek saath combine karo	Ex 10

Har cell A–J neeche aata hai. Agar tum yeh das kar sakte ho toh tumne poori surface dekh li hai.

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Example 1 — Flat gain (Cell A)

Forecast: number padhne se pehle andaaza lagao — ek constant amplifier jo sab kuch 4 se scale karta hai. Kya worst frequency low hai, high hai, ya... koi fark hi nahi padta?

1. Frequency response likho. $G(j\omega )=4$ har $\omega$ ke liye. Yeh step kyun? H∞ norm $|G(j\omega )|$ ka peak hai frequency pe, toh pehle dekhna hoga ki $|G|$ $\omega$ ke saath kaise change hota hai.
2. Magnitude lo. $|G(j\omega )|=4$, flat. s01 dekho — poori $\omega$ axis par height 4 pe ek bilkul seedha red line. Yeh step kyun? Ek real constant mein koi imaginary part nahi hota; uska magnitude saari frequencies par constant khud hi hota hai.
3. Supremum lo. ${\sup }_{\omega }4=4$, toh $\Vert G{\Vert }_{\infty }=4$. Yeh step kyun? Ek flat line ka peak uski apni value hoti hai. Har frequency equally "worst" hai.

Figure s01 — ek bilkul horizontal red line: koi bhi frequency special nahi hai, peak constant ke barabar hai.

Verify: Ek unit-energy sinusoid feed karo: output energy gain $=|G{|}^2=16$, toh $\Vert z{\Vert }_2/\Vert w{\Vert }_2=4$. Match karta hai. Units: dimensionless gain in, dimensionless gain out. ✓

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Example 2 — Monotone roll-off, koi resonance nahi (Cell B)

Forecast: kya sabse badi amplification high frequency par hoti hai, ya DC ($\omega =0$) par?

1. Magnitude function. $|G(j\omega )|=\frac{5}{\sqrt{{\omega }^2+4}}$. Yeh step kyun? $j\omega +2$ ka magnitude $\sqrt{{\omega }^2+2^2}$ hai; constant $5$ ko isse divide karne par har frequency par gain milti hai.
2. Dekho yeh kaise move karta hai. Jaise $\omega$ badhta hai, denominator badhta hai, toh $|G|$ monotonically girta hai. Koi interior peak nahi. s02 dekho — red curve apne left end se steadily neeche jati hai. Yeh step kyun? Koi second-order term nahi hai jo resonance create kare; ek single pole sirf attenuate karta hai jab frequency badhti hai.
3. Peak $\omega =0$ par hai. $|G(j0)|=5/2=2.5$. Isliye $\Vert G{\Vert }_{\infty }=2.5$. Yeh step kyun? Ek monotone-decreasing magnitude ke liye supremum left end par hota hai, yaani DC gain.

Figure s02 — red magnitude curve 2.5 se shuru hoti hai aur sirf neeche jaati hai; peak sabse left wala point hai.

Verify: DC gain $=G(0)=5/2=2.5$; aur $|G(j\omega ){|}^2=25/({\omega }^2+4)$ tab sabse bada hota hai jab ${\omega }^2=0$. ✓

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Example 3 — Band ke andar resonant peak (Cell C)

Forecast: ab peak DC par nahi hai. Tumhare hisaab se yeh kahan hai — ${\omega }_n=2$ se neeche, uspe, ya upar?

1. Magnitude. $|G(j\omega )|=\frac{10}{\sqrt{(4-{\omega }^2{)}^2+(0.4\omega {)}^2}}$. Yeh step kyun? $s=j\omega$ substitute karo: $s^2=-{\omega }^2$ toh real part $4-{\omega }^2$ hai, imaginary part $0.4\omega$ hai.
2. Resonant frequency. Second-order response ka peak ${\omega }_r={\omega }_n\sqrt{1-2{\zeta }^2}=2\sqrt{1-0.02}\approx 1.98$ rad/s par hota hai — ${\omega }_n$ se thoda neeche. Yeh step kyun? Denominator ko minimize karna (simply $\omega ={\omega }_n$ set karna nahi) true peak ko thoda left shift karta hai jab damping hoti hai.
3. Peak magnitude. $\Vert G{\Vert }_{\infty }\approx \frac{10/4}{2\zeta \sqrt{1-{\zeta }^2}}=\frac{2.5}{2(0.1)\sqrt{0.99}}\approx 12.56$. Yeh step kyun? Standard resonant-peak formula; numerator $10/4=2.5$ DC gain hai, aur chhota $\zeta$ peak ko tall banata hai. s03 dekho — $\omega \approx 2$ ke paas tall red spike.

Figure s03 — $\omega =2$ ke paas ek tall red resonant spike jo dashed DC-gain line 2.5 se kaafi upar hai.

Verify: DC gain $=10/4=2.5$; peak $\approx 12.56$ jo DC gain ka $\approx 5\times$ hai, ek $\zeta =0.1$ mode ke consistent. $\omega \approx 2$ par ek gust $\sim 12.6\times$ amplify hoti hai. ✓

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Example 4 — Degenerate input: zero-energy aur DC-only disturbances (Cell D)

Forecast: (b) ke liye, ek constant $\omega =0$ par baithta hai. Kya tumhe peak gain 12.56 milti hai, ya DC gain 2.5?

1. Case (a): $w=0$. Toh $\Vert w{\Vert }_2=0$, aur ratio $\Vert z{\Vert }_2/\Vert w{\Vert }_2$ hai $0/0$ — undefined, aur yahi theek wajah hai kyun parent note ka supremum ${\sup }_{w\ne 0}$ likha jaata hai. Yeh step kyun? H∞ norm deliberately zero signal ko exclude karta hai; ek robustness margin tabhi kuch matlab rakhti hai jab disturbance actually energy carry kare.
2. Case (b): $w=1$ hamesha ke liye. Ek constant finite-energy nahi hai: ${\int }_0^{\infty }{1}^{2}dt=\infty$. Toh yeh 2-norm world ke bahar hai. Yeh step kyun? $\Vert w{\Vert }_2^2={\int }_0^{\infty }{w}^{\top }wdt$ diverge karta hai; H∞ theory finite-energy (fading) disturbances ke liye banai gayi hai, un signals ke liye nahi jo kabhi khatam nahi hote.
3. Kaunsa number actually constant par laagoo hota hai? Steady-state response, jo DC gain $|G(0)|=2.5$ se govern hoti hai: output $2.5$ pe settle karta hai. Yeh ek DC gain statement hai, H∞ statement nahi. Yeh step kyun? Step input ka final-value behaviour sirf $\omega =0$ pick off karta hai, toh woh $|G(0)|$ dekhta hai, resonant peak kabhi nahi.

Verify: $G(0)=10/(0+0+4)=2.5$, toh unit step $2.5$ pe settle karta hai; peak $12.56$ ek DC input ke liye irrelevant hai. Distinction (energy-gain peak vs. DC gain) yahi toh pura lesson hai. ✓

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Example 5 — Limiting behaviour: $\zeta \to 0$ (Cell E)

Forecast: kya peak level off hoti hai, ya infinity ki taraf bhaag jaati hai?

1. Peak formula $\zeta$ mein. $\Vert G{\Vert }_{\infty }\approx \frac{2.5}{2\zeta \sqrt{1-{\zeta }^2}}$. Yeh step kyun? Same resonant-peak expression jaise Ex 3, ab $\zeta$ ke function ke roop mein rakha gaya hai taaki hum $\zeta$ ko $0$ ki taraf push kar sakein.
2. Shrinking values plug karo. $\zeta =0.1\Rightarrow \approx 12.56$; $\zeta =0.05\Rightarrow \approx 25.03$; $\zeta =0.02\Rightarrow \approx 62.5$. Yeh step kyun? $\zeta$ ko aadha karne se peak roughly double hoti hai — denominator mein $2\zeta$ dominate karta hai.
3. Limit lo. Jaise $\zeta \to 0$, $\Vert G{\Vert }_{\infty }\to \infty$. Mode ek undamped oscillator ban jaata hai infinite peak gain ke saath. s04 dekho: red curves taller aur sharper hoti jaati hain jaise damping khatam hoti hai. Yeh step kyun? Ek undamped pole imaginary axis par baithta hai; exactly apni natural frequency par response ka koi bound nahi hota. Robust control ko kabhi loop ko aisi mode excite nahi karne deni chahiye.

Figure s04 — teen red resonance curves; jaise $\zeta$ 0.1 se 0.02 tak girta hai, peak roughly har baar double hoti hai aur ek spike mein narrow ho jaati hai.

Verify: $2.5/(2\cdot 0.02\cdot \sqrt{1-0.02^2})\approx 62.5$, aur sequence $12.56\to 25.03\to 62.5$ diverge karta hai jaise $\zeta \downarrow 0$. ✓

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Example 6 — Multi-input system: magnitude nahi, singular value use karo (Cell F)

Forecast: scalar ke liye hum $|G|$ lete. Matrix ke liye "gain" depend karta hai kis direction mein input push karta hai. Kisi bhi unit-length input per unit output length ki sabse badi value kya hai?

1. Largest singular value kyun. $\bar{\sigma }(G)={\max }_{\Vert x\Vert =1}\Vert Gx\Vert$ — sabse bada stretch jo $G$ kisi bhi unit vector pe laagoo karta hai. Yeh "peak gain" ka multi-input generalization hai. Yeh step kyun? Page ke upar ki definition Singular Value Decomposition use karti hai; ek matrix ke liye gain direction-dependent hoti hai, toh hume worst direction chahiye.
2. $G^{\top }G$ banao. $G^{\top }G=\left[\begin{array}{cc}3& 4\\ 0& 0\end{array}\right]\left[\begin{array}{cc}3& 0\\ 4& 0\end{array}\right]=\left[\begin{array}{cc}25& 0\\ 0& 0\end{array}\right]$. Yeh step kyun? $G$ ke singular values $G^{\top }G$ ke eigenvalues ke square roots hain — standard SVD route.
3. Eigenvalues aur root. Eigenvalues $\{25,0\}$, toh singular values $\{5,0\}$, giving $\bar{\sigma }=5$. Yeh step kyun? Largest eigenvalue $25$ ka root $5$ hai; input direction $x=(1,0{)}^{\top }$ output length $\sqrt{3^2+4^2}=5$ produce karta hai, ek 3-4-5 triangle.

Verify: $\Vert G(1,0{)}^{\top }\Vert =\Vert (3,4{)}^{\top }\Vert =\sqrt{9+16}=5$, aur koi bhi unit input isse beat nahi karta ($x=(0,1{)}^{\top }$ $0$ deta hai). Toh $\bar{\sigma }=5$. ✓

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Example 7 — Small-gain sizing: uncertainty kitna bada ho sakta hai? (Cell G)

Forecast: parent note ne $\pm 30\%\Rightarrow \Vert M{\Vert }_{\infty }<3.33$ kiya tha. Compute karne se pehle yahan ka answer guess karo.

1. Unit ball par normalize karo. $\Delta =0.4\tilde{\Delta }$ likho jahan $\Vert \tilde{\Delta }{\Vert }_{\infty }\le 1$. Yeh step kyun? Small-Gain Theorem ek unit-bounded block ke liye stated hai; $0.4$ ko loop mein pull karo.
2. Small-gain apply karo. Saare $\tilde{\Delta }$ ke liye stability iff $\Vert 0.4M{\Vert }_{\infty }<1$, yaani $\Vert M{\Vert }_{\infty }<1/0.4=2.5$. Yeh step kyun? $\Vert 0.4M{\Vert }_{\infty }=0.4\Vert M{\Vert }_{\infty }$, aur loop tab safe hai jab uske around gains ka product $1$ se neeche rahe.
3. Margin padho. Jab tak nominal loop ka peak uncertain channel par $2.5$ se neeche rahe, koi bhi actuator error $\pm 40\%$ ke andar tolerate hogi. Yeh step kyun? Ek inequality plants ki poori infinite family certify karti hai — small-gain ki taakat.

Verify: $1/0.4=2.5$; aur badi uncertainty ($40\%>30\%$) tighter margin deti hai ($2.5<3.33$), jo correct direction hai. ✓

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Example 8 — Weight ceiling aur waterbed (Cell H)

Forecast: agar weight 3 hai, toh allowed sensitivity $3$ hai, $1/3$ hai, ya $1$?

1. Norm ko pointwise ceiling mein badlo. $\Vert W_{S}S{\Vert }_{\infty }<1\Rightarrow |W_S(j\omega )||S(j\omega )|<1\Rightarrow |S(j\omega )|<1/|W_S(j\omega )|$ har $\omega$ par. Yeh step kyun? H∞ norm product ke peak ko bound karta hai; kyunki yeh peak par $1$ se neeche hai, toh har jagah $1$ se neeche hai, isliye $|S|$ ceiling $1/|W_S|$ ke neeche baithta hai. s05 dekho — sensitivity curve ko red ceiling ke neeche rehna chahiye.
2. Plug in karo. $\omega =0.1$ par: $|S(j0.1)|<1/3\approx 0.333$. Achha tracking — error wahan reference ka sirf ek tehai hai. Yeh step kyun? Bada weight $\Rightarrow$ neecha ceiling $\Rightarrow$ chhoti sensitivity $\Rightarrow$ achha low-frequency tracking.
3. Kyun har jagah nahi. Standard assumptions ke under — ek open-loop-stable, minimum-phase plant relative degree $\ge 2$ ke saath (loop gain kaafi fast roll off kare) — Bode sensitivity integral normalized form mein hold karta hai ${\int }_0^{\infty }\ln |S(j\omega )|d\omega =0$. (Ek ya zyada unstable poles $p_k$ ke saath right side $\pi \sum \mathrm{Re}(p_k)>0$ ban jaata hai, trade badtam ho jaata hai.) Low $\omega$ par $\ln |S|$ ko bahut negative push karne se yeh kahin aur positive (yaani $|S|>1$) hone pe majboor ho jaata hai, kyunki total signed area fixed hai. Yeh step kyun? Integral $\ln |S|$ ka signed area pin karta hai: curve ko $\ln |S|=0$ line ke baare mein see-saw socho. Left par suppression ki deep trough khodne se right par matching amplification ka hump uthna zaroori hai taaki areas cancel ho ke fixed total tak aayein. s06 dekho — shaded trough below axis ka same area hai jitna shaded hump above.

Figure s05 — black $|S|$ curve red ceiling $1/|W_S|$ ke neeche tucked; jahan ceiling neecha hai, $|S|$ chhota hone pe majboor hai (achha tracking).

Figure s06 — $\ln |S|$ vs $\omega$: red shaded trough (suppression, area negative) exactly balance karta hai black hump above axis ko (amplification, area positive); unka signed areas sum zero hai.

Verify: $1/3=0.3\overline{3}$, toh $|S(j0.1)|<0.333$; aur $\int \ln |S|=0$ matlab negative area (ceiling ke neeche) exactly offset hota hai positive area se (1 ke upar ek bump). ✓

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Example 9 — Real-world word problem: rocket bending mode (Cell I)

Forecast: roughly kitni attenuation — 2 ka factor, 6 ka, ya 60 ka?

1. Open vs required peak compare karo. Open-loop peak $\approx 12.6$; required peak $<2$. Yeh step kyun? $\Vert T_{zw}{\Vert }_{\infty }$ worst-case energy amplification hai (page ke upar ki definition); gust peak par concentrate karti hai, toh wahi peak neeche aani chahiye.
2. Attenuation factor. $12.6/2=6.3$, toh loop ko mode ko $\omega =2$ par kam se kam $\approx 6.3\times$ (lagbhag $16$ dB) knock down karna chahiye. Yeh step kyun? Humein closed-loop peak $12.6$ se shuru karke $2$ se neeche chahiye; ratio required suppression hai.
3. Ise command kaise karein. $\omega =2$ ke paas bade magnitude wala weight $W$ rakho aur $\Vert W{T}_{zw}{\Vert }_{\infty }<1$ require karo. Yeh ceiling $1/|W|$ ko mode par neecha draw karta hai, synthesised $K$ ko (nominal baseline ke liye LQR and LQG control dekho jo ise improve karta hai) exactly wahan notch/attenuate karne ke liye majboor karta hai. Bending frequency mein structured uncertainty tumhe μ-synthesis (structured uncertainty) ki taraf push karegi. Yeh step kyun? H∞ weights ke zariye wahan shape karta hai jahan zaroori ho; $2$ rad/s par tall weight precisely dangerous frequency par attenuation khareedta hai, waterbed ko obey karte hue un frequencies par margin kharch karke jahan koi gust energy nahi hai.

Verify: $12.6/2=6.3$; decibels mein $20{\log }_{10}(6.3)\approx 16.0$ dB. ✓

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Example 10 — Exam twist: small-gain ko peak reading ke saath combine karo (Cell J)

Forecast: $k/(s+1)$ ka peak — kya yeh DC par hai ya high frequency par? Phir sabse bada $k$ kya hai?

1. $M$ ka peak. $|M(j\omega )|=k/\sqrt{{\omega }^2+1}$, monotone-decreasing (Cell B pattern), toh peak $\omega =0$ par hai: $\Vert M{\Vert }_{\infty }=|M(0)|=k$. Yeh step kyun? Ek single stable pole monotone roll-off deta hai; supremum DC value $|M(0)|=k$ hai — hum exactly Ex 2 ka reasoning reuse kar rahe hain.
2. Small-gain apply karo. Robust stability $⟺\Vert M{\Vert }_{\infty }<1⟺k<1$. Yeh step kyun? Small-Gain Theorem: $M$–$\Delta$ interconnection saare $\Vert \Delta {\Vert }_{\infty }\le 1$ ke liye stable hai exactly jab $\Vert M{\Vert }_{\infty }<1$.
3. Boundary batao. Sabse bada safe gain $k\to 1^{-}$ hai; $k=1$ par guarantee khatam ho jaati hai (koi admissible $\Delta$ loop-gain product ko $1$ tak drive kar sakta hai, series $\sum (M\Delta {)}^n$ ki convergence todta hua). Yeh step kyun? Small-gain ek open inequality hai; equality exactly woh edge hai jahan geometric series converge karna band kar deti hai aur loop bhaag sakta hai.

Verify: $\Vert M{\Vert }_{\infty }=|M(0)|=k$; $k<1$ set karne se $\Vert M{\Vert }_{\infty }<1$ satisfy hota hai toh admissible $k$ ka supremum $1$ hai. ✓

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Recall Scenario checklist — kya tum koi nayi problem place kar sakte ho?

Flat gain? peak = constant. Monotone? peak DC par. Resonant? peak formula $\frac{\text{DCgain}}{2\zeta \sqrt{1-{\zeta }^2}}$. Zero/DC input? excluded / DC gain use karo. $\zeta \to 0$? peak $\to \infty$. Matrix? largest singular value. Uncertainty size $\delta$? $\Vert M{\Vert }_{\infty }<1/\delta$ chahiye. Weight $W$? $|S|$ par ceiling $1/|W|$, waterbed ka dhyan rakho.

Active recall