H∞ control — robust to uncertainty (intro): Question bank

3.5.37 · D5 · Physics › Guidance, Navigation & Control (GNC) › H∞ control — robust to uncertainty (intro)

Shuru karne se pehle, is bank mein use hone wale key symbols plain words mein:

$w$ — exogenous inputs: har woh cheez jo system ko bahar se push kar rahi hai (disturbances, reference commands, sensor noise). Socho "woh hawa aur hiccups jinhein tum control nahi karte."
$z$ — performance outputs: woh cheez jo tum chhoti rakhna chahte ho (tracking error, control effort). Socho "woh hiccups jinhe par tumhara grade lagta hai."
$T_{z w}$ — closed-loop map $w$ se $z$ tak; $z = T_{z w} w$ . Yeh woh ek object hai jise H∞ synthesis "chhota" banane ki koshish karta hai.
$∥ G ∥_{\infty}$ — system ka gain ka peak saari frequencies ke upar; woh sabse bada factor jis se koi bhi single sinusoid amplify ho sakta hai.
$S$ — sensitivity, reference (ya output disturbance) se tracking error tak ka map; chhota $∣ S ∣$ matlab achha tracking. Dekho Sensitivity and Complementary Sensitivity (S+T=1).
$T$ — complementary sensitivity, $S + T = 1$ har frequency par.
$Δ$ — ek unknown lekin size-bounded "uncertainty block" jo sab kuch represent karta hai jo model ne galat kiya; dekho Model Uncertainty in GNC.
$M$ — woh transfer function jise uncertainty $Δ$ "dekhta" hai jab woh closed loop mein wapas jhankta hai.

True or false — justify karo

Real vehicles kabhi apne model se match nahi karte, isliye nominal plant par great LQR margins wala controller automatically robust hai

False — LQR margins sirf modelled plant ke liye guaranteed hain; ek alag plant (added phase lag, ek unmodelled mode) woh margins zero tak erode kar sakta hai. Robustness ko plants ke ek set ke against design karna padta hai, jo H∞ ka kaam hai.

∥ G ∥_{\infty}

system ka DC gain

∣ G (0) ∣

hai

False — yeh supremum over all frequencies hai,

ω = 0

par ki value nahi. Ek lightly damped mode ka resonant peak uske DC gain se kaafi zyaada ho sakta hai (Bode figure dekho), aur wahi peak H∞ norm report karta hai.

Agar

∥ T_{z w} ∥_{\infty} < γ

hai, to koi bhi disturbance outputs mein

γ

se zyaada amplify nahi hoti

True — yaad karo

T_{z w}

exogenous inputs

w

se performance outputs

z

tak ka map hai; inequality uska worst-case energy gain

∥ z ∥_{2} /∥ w ∥_{2}

bound karta hai, aur peak worst case hone ki wajah se, har disturbance (sabse nasty wali bhi) zyaada se zyaada

γ

se amplify hoti hai.

H∞ norm ko chhota karne ka matlab hai loop gain ko har jagah bada karna

False — gain ko har jagah bada karna impossible hai kyunki

S + T = 1

; tum ek hi frequency par

∣ S ∣

aur

∣ T ∣

dono ko tiny force nahi kar sakte. H∞ gain shape karta hai (jahan tum track karte ho wahan high, jahan roll off ho wahan low), use uniformly badhata nahi.

Small-gain condition

∥ M ∥_{\infty} < 1

real plant ki stability ke liye necessary aur sufficient hai

Half-true — yeh saari unstructured

Δ

with

∥Δ ∥_{\infty} \leq 1

ke against stability ke liye necessary aur sufficient hai (guarantee poori ball par quantify hoti hai). Us ek specific true

Δ

ke liye yeh sirf sufficient hai, kabhi necessary nahi: kuch individual plants jinhein yeh reject karta hai woh actually stable hain.

Parseval's theorem wahi hai jo "time mein worst energy gain" ko "frequency mein peak gain" mein translate karne deta hai

True — Parseval's Theorem kehta hai signal energy time ya frequency dono mein same compute hoti hai, isliye energy-gain ratio frequency-by-frequency analyse ho sakta hai, jahan

G

simple multiplication ki tarah kaam karta hai aur uska peak bound govern karta hai.

Ek finite-energy disturbance (

∥ w ∥_{2} < \infty

) woh signal hai jo pointwise eventually zero par decay ho jaata hai

False — finite energy (

\int_{0}^{\infty} w^{⊤} w d t < \infty

) sirf total energy bound karta hai; yeh

w (t) \to 0

ko har instant par force nahi karta (ek

L_{2}

signal mein ever-narrower spikes ho sakti hain jo pointwise kabhi vanish nahi hoti). Sahi picture hai "ek gust jiska total energy finite hai," na ki "ek signal jo settle hone ki guarantee rakhta hai."

Kyunki H∞ worst case optimise karta hai, resulting controller typical disturbance ke liye bhi hamesha best choice hota hai

False — worst-case optimisation pessimistic ho sakta hai; woh ek aise disturbance ke against defend karne ke liye average-case performance sacrifice kar sakta hai jo nature rarely produce karta hai. LQG average (expected) case optimise karta hai; dono alag sawaalon ke jawab dete hain.

\overset{σ}{ˉ} (G (j ω))

aur

∣ G (j ω) ∣

har system ke liye same cheez hain

False — yeh sirf scalar (single-input single-output) systems ke liye coincide karte hain. Multivariable systems ke liye

\overset{σ}{ˉ}

sabse bada singular value hai (singular-value figure mein top curve), jo worst input direction ko bhi capture karta hai worst frequency ke saath.

H∞ norm aur small-gain theorem sirf continuous time (

s

-domain) mein sense make karte hain

False — dono discrete time mein bhi carry over karte hain:

G (j ω)

ki jagah

G (e^{j θ})

ko unit circle

θ \in [- π, π]

par evaluate karo, aur

∥ G ∥_{\infty}

woh circle ke around peak gain ban jaata hai. Small-gain bilkul same padhta hai (

∥ M ∥_{\infty} < 1

), yahi wajah hai ki sampled-data flight controllers same robustness machinery use kar sakte hain.

Spot the error

ω_{0}

frequency par disturbance reject karne ke liye, bas

∣ S (j ω_{0}) ∣

ko bahut chhota karo — aur hum yeh har

ω

par kar sakte hain."

Error hai "har

ω

par." Waterbed law

\int_{0}^{\infty} ln ∣ S ∣ d ω = 0

(typical plants) ka matlab hai kahin dabaya gaya area kahin aur uthega; tum sensitivity sirf redistribute kar sakte ho, eliminate nahi (waterbed figure dekho).

"Maine ±30% actuator error ko

0.3 \tilde{Δ}

ke roop mein normalise kiya jisme

∥ \tilde{Δ} ∥_{\infty} \leq 1

hai, isliye robust stability ko

∥0.3 M ∥_{\infty} < 1

chahiye, yaani

∥ M ∥_{\infty} < 0.3

Algebra ulta hai:

∥0.3 M ∥_{\infty} < 1 \Rightarrow ∥ M ∥_{\infty} < 1/0.3 \approx 3.33

. Chhoti uncertainty

M

par bound ko relax karti hai, tighten nahi.

"Small-gain kehta hai stability iff

∥ M ∥_{\infty} ∥Δ ∥_{\infty} < 1

, isliye main actual

∥Δ ∥_{\infty} = 0.3

use karunga aur

∥ M ∥_{\infty} < 1/0.3

require karunga."

Reasoning sahi number tak pahunchti hai lekin ek subtly galat wajah se: theorem poori ball

∥Δ ∥_{\infty} \leq 1

ke liye stated hai, isliye hum

Δ

ko pehle block mein normalise karte hain aur

∥ M^{'} ∥_{\infty} < 1

demand karte hain jahan

M^{'}

0.3

absorb kar liya hai. Same inequality, lekin guarantee ball mein har

Δ

ko cover karti hai, na sirf us worst one ko jo tumne plug in kiya.

G (s) = 10/ (s^{2} + 0.4 s + 4)

ka DC gain

10/4 = 2.5

hai, isliye

∥ G ∥_{\infty} = 2.5

DC gain peak nahi hai. Kyunki

ζ = 0.1

chhota hai,

ω_{n} = 2

ke paas ek resonance hai jo magnitude ko

\approx 2.5/ (2 ζ) \approx 12.6

tak lift kar deta hai; H∞ norm woh resonant peak hai, DC value nahi.

"Small-gain ne meri design reject kar di, isliye meri design unstable honi chahiye."

Small-gain fail hona instability ka proof nahi hai — yeh stability ke liye sufficient hai, necessary nahi. μ-synthesis jaisi structured analysis same design ko

Δ

ki known shape exploit karke certify kar sakti hai.

"Maine ek bada weight

W_{S}

har jagah lagaya taaki

∣ S ∣

saari frequencies par tiny force ho, tracking guarantee karo."

Tum

∥ W_{S} S ∥_{\infty} < 1

satisfy nahi kar sakte agar

W_{S}

sab

ω

par bada hai: woh

∣ S ∣ < 1/∣ W_{S} ∣

har jagah demand karta hai, jo waterbed constraint violate karta hai.

W_{S}

sirf us low-frequency band mein bada hona chahiye jis par tumhein actually concern hai aur elsewhere roll off karna chahiye.

Why questions

Hum

∥ G ∥_{\infty}

define karte waqt frequency par average ki jagah supremum kyun lete hain?

Kyunki worst disturbance apni energy us single most-amplified frequency par concentrate karne ke liye free hai, isliye robustness peak se govern hoti hai, mean se nahi — averaging us failure mode ko chhhupa deta jisse hum darte hain.

Ek norm (ek number) bound karna infinite family of plants kyun certify karta hai?

Uncertainty ball

{Δ : ∥Δ ∥_{\infty} \leq 1}

infinitely many plants contain karta hai, lekin ek inequality

∥ M ∥_{\infty} < 1

un sab ke liye ek saath loop-gain-

\geq 1

rule out kar deta hai — norm family ke worst member ko summarise karta hai.

Hum directly

∥ S ∥_{\infty}

minimise karne ki jagah weighting functions

W_{i}

kyun insert karte hain?

Weights encode karte hain kahaan performance matter karta hai (low

ω

par tracking, high

ω

par roll-off, control effort limits). Weighted norm minimise karna ek single H∞ problem ko inhe automatically trade off karne deta hai; ek raw

∥ S ∥_{\infty}

mein frequency-dependent priorities ka koi notion nahi hota.

Small-Gain Theorem ka heart geometric-series argument kyun hai?

Loop mein circulate ho raha ek signal har lap mein

M Δ

se multiply hota hai; total output

\sum_{n} (M Δ)^{n}

hai. Agar

∥ M ∥_{\infty} ∥Δ ∥_{\infty} < 1

hai to yeh series ek bounded value par converge hoti hai (stable); agar

\geq 1

hai to koi

Δ

ise diverge kara deta hai (unstable).

Worst-case energy gain frequency-response peak ke barabar kyun hota hai, sirf us se neeche nahi?

Upper bound "integral se peak bahar nikalo" us disturbance choose karne se achieve hota hai jiska energy peak frequency par pile up hota hai — isliye sup attain ho jaata hai aur inequality equality ban jaati hai.

μ

-synthesis plain small-gain se kam conservative kyun hai?

Small-gain

Δ

ko ek single full block treat karta hai jo har channel ko har doosre se couple kar sakta hai; μ-synthesis

Δ

ki known block-diagonal structure respect karta hai, un couplings ko forbid karta hai jo physics rule out karti hai, isliye woh kam safe plants reject karta hai.

Multivariable systems ke liye H∞ worst input ki direction ki kyun parwah karta hai?

Kyunki

\overset{σ}{ˉ}

Singular Value Decomposition se aata hai: gain frequency aur input direction dono par depend karta hai, aur worst case disturbance ko highest-gain singular direction ke saath align karta hai — ek scalar magnitude yeh miss kar deta.

Discrete-time H∞ norm imaginary axis ki jagah unit circle par kyun padha jaata hai?

Sampling stable region ko left half

s

-plane se

z

-plane mein unit circle ke andar map karta hai; frequency axis (imaginary axis) unit circle

z = e^{j θ}

par map hoti hai, isliye "saari frequencies sweep karo" jo peak define karta tha, woh us circle ke around

θ

ka sweep ban jaata hai.

Edge cases

∥ G ∥_{\infty}

kya hoga agar

G (j ω) \to \infty

kisi frequency par (imaginary axis par ek pole)?

Woh

+ \infty

hai — H∞ norm sirf stable systems ke liye finite hota hai jisme koi imaginary-axis poles nahi hote; ek undamped resonance (

ζ = 0

) unbounded peak deta hai, yaani us frequency par ek infinitesimal disturbance unbounded output yield karta hai.

Agar

∥ M ∥_{\infty} = 1

exactly (boundary par) hai to robust stability ke baare mein kya pata chalta hai?

Yeh knife-edge hai: strict inequality

∥ M ∥_{\infty} < 1

stability guarantee karta hai, lekin exactly

1

par ek worst-case

Δ

exist karta hai (gain exactly

1

, right phase) jo loop gain ko unity tak pahuncha ke marginally destabilise kar deta hai — isliye

∥ M ∥_{\infty} = 1

safe nahi hai.

Ek disturbance ke liye H∞ framework ka kya hoga jo hamesha ke liye constant (DC) hai?

Ek constant ki infinite energy hoti hai (

\int_{0}^{\infty}

diverge karta hai), isliye yeh finite-energy signal nahi hai;

2

-norm energy-gain interpretation break ho jaata hai. Aise steady disturbances DC/integral action se handle hote hain, H∞ energy argument se directly nahi.

ζ = 0

(undamped mode) par resonant peak kya hai, aur physically uska kya matlab hai?

Peak

\to \infty

(denominator

2 ζ \to 0

), yaani

ω_{n}

par exactly ek sinusoid energy ko bina kisi loss ke pump karta hai aur amplitude unbounded badh jaata hai — ek bending mode jise zaroor damp ya notch karna padega, warna koi finite H∞ controller exist nahi karta.

Agar

∣ S (j ω) ∣ = 1

kisi frequency par hai to disturbance wahan kya kar raha hai?

Woh seedha pass ho raha hai bina kisi amplification ya attenuation ke (

S + T = 1

jisme

S = 1, T = 0

) — woh crossover-ish band jahan loop ka essentially koi authority nahi; waterbed excess attenuation kahin aur

∣ S ∣ > 1

ke hump mein push kar deta hai.

Agar true actuator error sirf ±5% hai lekin humne ±30% ke liye design kiya?

Design abhi bhi robust hai (chhota ball bade ke andar fit ho jaata hai), bas conservative hai — humne shayad performance sacrificed karke us uncertainty ke against defend kiya jo wahan hai hi nahi, exactly woh over-modelling cost jiske baare mein parent ka mistake box warn karta hai.

Kya

∥ W_{S} S ∥_{\infty} < 1

aur

∥ W_{T} T ∥_{\infty} < 1

dono ek hi frequency par hold ho sakte hain?

Aggressive weights ke saath nahi: kyunki

S + T = 1

hai, kisi bhi

ω

par tum

∣ S ∣

aur

∣ T ∣

dono ko chhota nahi bana sakte, isliye dono weights ke non-overlapping dominant bands hone chahiye (

W_{S}

low

ω

par bada,

W_{T}

high

ω

par bada) — crossover unavoidable compromise region hai.

Discrete time mein H∞ norm Nyquist frequency ke paas kaise behave karta hai?

Sweep

θ \in [- π, π]

par run hoti hai, aur

θ = \pm π

Nyquist frequency hai (sample rate ki aadhi); Nyquist ke paas fold hone wala resonance abhi bhi peak mein contribute karta hai, isliye sampled-data designs ko poori unit circle check karni padti hai, sirf low

θ

nahi — aliasing ek peak chhhupa sakta hai jo continuous model ne kabhi nahi dikhaya.

Recall Traps ka one-line summary

Peak, average nahi; saari frequencies aur directions ka supremum (discrete time mein unit circle); small-gain ek fixed plant ke liye sufficient (conservative, necessary nahi) hai; aur waterbed "har jagah chhota $∣ S ∣$ " forbid karta hai.