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

Before you start, the key symbols this bank uses, in plain words:

$w$ — the exogenous inputs: everything pushing on the system from outside (disturbances, reference commands, sensor noise). Think "the wind and the wobbles you don't control."
$z$ — the performance outputs: the things you want kept small (tracking error, control effort). Think "the wobbles you get graded on."
$T_{z w}$ — the closed-loop map from $w$ to $z$ ; $z = T_{z w} w$ . It is the single object H∞ synthesis tries to make "small."
$∥ G ∥_{\infty}$ — the peak of the system's gain over all frequencies; the biggest factor by which any single sinusoid can be amplified.
$S$ — the sensitivity, the map from reference (or output disturbance) to tracking error; small $∣ S ∣$ means good tracking. See Sensitivity and Complementary Sensitivity (S+T=1).
$T$ — the complementary sensitivity, $S + T = 1$ at every frequency.
$Δ$ — an unknown but size-bounded "uncertainty block" standing in for everything the model got wrong; see Model Uncertainty in GNC.
$M$ — the transfer function the uncertainty $Δ$ "sees" when it looks back into the closed loop.

True or false — justify

Real vehicles never match their model, so a controller with great LQR margins on the nominal plant is automatically robust

False — LQR margins are guaranteed for the modelled plant only; a plant that differs (added phase lag, an unmodelled mode) can erode those margins to zero. Robustness must be designed against a set of plants, which is H∞'s job.

∥ G ∥_{\infty}

is the DC gain

∣ G (0) ∣

of the system

False — it is the supremum over all frequencies, not the value at

ω = 0

. A lightly damped mode can have a resonant peak far above its DC gain (see the Bode figure), and that peak is what the H∞ norm reports.

∥ T_{z w} ∥_{\infty} < γ

, then no disturbance is amplified into the outputs by more than

γ

True — recall

T_{z w}

is the map from exogenous inputs

w

to performance outputs

z

; the inequality bounds its worst-case energy gain

∥ z ∥_{2} /∥ w ∥_{2}

, and since the peak is the worst case, every disturbance (including the nastiest) is amplified by at most

γ

Making the H∞ norm small means making the loop gain large everywhere

False — large gain everywhere is impossible because

S + T = 1

; you cannot force both

∣ S ∣

and

∣ T ∣

tiny at the same frequency. H∞ shapes gain (high where you track, low where you roll off), it does not crank it up uniformly.

The small-gain condition

∥ M ∥_{\infty} < 1

is necessary and sufficient for stability of the real plant

Half-true — it is necessary and sufficient for stability against all unstructured

Δ

with

∥Δ ∥_{\infty} \leq 1

(the guarantee quantifies over the whole ball). For the one specific true

Δ

it is only sufficient, never necessary: some individual plants it rejects are actually stable.

Parseval's theorem is what lets us translate "worst energy gain in time" into "peak gain in frequency"

True — Parseval's Theorem says signal energy is the same computed in time or in frequency, so the energy-gain ratio can be analysed frequency-by-frequency, where

G

acts as simple multiplication and its peak governs the bound.

A finite-energy disturbance (

∥ w ∥_{2} < \infty

) is a signal that eventually decays pointwise to zero

False — finite energy (

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

) only bounds the total energy; it does not force

w (t) \to 0

at every instant (an

L_{2}

signal can have ever-narrower spikes that never vanish pointwise). The correct picture is "a gust whose total energy is finite," not "a signal guaranteed to settle."

Because H∞ optimises the worst case, the resulting controller is always the best choice for the typical disturbance too

False — worst-case optimisation can be pessimistic; it may sacrifice average-case performance to defend against a disturbance nature rarely produces. LQG optimises the average (expected) case; the two answer different questions.

\overset{σ}{ˉ} (G (j ω))

and

∣ G (j ω) ∣

are the same thing for every system

False — they coincide only for scalar (single-input single-output) systems. For multivariable systems

\overset{σ}{ˉ}

is the largest singular value (top curve in the singular-value figure), capturing the worst input direction as well as the worst frequency.

The H∞ norm and small-gain theorem are ideas that only make sense in continuous time (the

s

-domain)

False — both carry over to discrete time: replace

G (j ω)

G (e^{j θ})

evaluated on the unit circle

θ \in [- π, π]

, and

∥ G ∥_{\infty}

becomes the peak gain around that circle. Small-gain reads identically (

∥ M ∥_{\infty} < 1

), which is exactly why sampled-data flight controllers can use the same robustness machinery.

Spot the error

"To reject a disturbance at frequency

ω_{0}

, just make

∣ S (j ω_{0}) ∣

very small — and we can do that at every

ω

The error is "at every

ω

." The waterbed law

\int_{0}^{\infty} ln ∣ S ∣ d ω = 0

(typical plants) means area pushed down somewhere must pop up elsewhere; you can only redistribute sensitivity, not eliminate it (see the waterbed figure).

"I normalised the ±30% actuator error as

0.3 \tilde{Δ}

with

∥ \tilde{Δ} ∥_{\infty} \leq 1

, so robust stability needs

∥0.3 M ∥_{\infty} < 1

, i.e.

∥ M ∥_{\infty} < 0.3

The algebra is backwards:

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

. Smaller uncertainty relaxes the bound on

M

, it does not tighten it.

"Small-gain says stability iff

∥ M ∥_{\infty} ∥Δ ∥_{\infty} < 1

, so I'll use the actual

∥Δ ∥_{\infty} = 0.3

and require

∥ M ∥_{\infty} < 1/0.3

The reasoning reaches the right number but for a subtly wrong reason: the theorem is stated for the whole ball

∥Δ ∥_{\infty} \leq 1

, so we normalise

Δ

into the block first and demand

∥ M^{'} ∥_{\infty} < 1

where

M^{'}

absorbs the

0.3

. Same inequality, but the guarantee covers every

Δ

in the ball, not just the worst one you plugged in.

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

has DC gain

10/4 = 2.5

, so

∥ G ∥_{\infty} = 2.5

The DC gain is not the peak. Because

ζ = 0.1

is small there is a resonance near

ω_{n} = 2

that lifts the magnitude to

\approx 2.5/ (2 ζ) \approx 12.6

; the H∞ norm is that resonant peak, not the DC value.

"Small-gain rejected my design, so my design must be unstable."

Small-gain failing is not proof of instability — it is sufficient for stability, not necessary. A structured analysis like μ-synthesis may certify the same design by exploiting the known shape of

Δ

"I put a big weight

W_{S}

everywhere to force

∣ S ∣

tiny at all frequencies, guaranteeing tracking."

You cannot satisfy

∥ W_{S} S ∥_{\infty} < 1

W_{S}

is large at all

ω

: that demands

∣ S ∣ < 1/∣ W_{S} ∣

everywhere, which violates the waterbed constraint.

W_{S}

must be large only in the low-frequency band you actually care about and roll off elsewhere.

Why questions

Why do we take the supremum over frequency rather than an average, when defining

∥ G ∥_{\infty}

Because the worst disturbance is free to concentrate its energy at the single most-amplified frequency, so robustness is governed by the peak, not the mean — averaging would hide the very failure mode we fear.

Why does bounding a norm (one number) certify an infinite family of plants?

The uncertainty ball

{Δ : ∥Δ ∥_{\infty} \leq 1}

contains infinitely many plants, but one inequality

∥ M ∥_{\infty} < 1

rules out loop-gain-

\geq 1

for all of them at once — the norm summarises the worst member of the family.

Why do we insert weighting functions

W_{i}

instead of directly minimising

∥ S ∥_{\infty}

Weights encode where performance matters (tracking at low

ω

, roll-off at high

ω

, control effort limits). Minimising the weighted norm lets a single H∞ problem trade these off automatically; a raw

∥ S ∥_{\infty}

has no notion of frequency-dependent priorities.

Why is the geometric-series argument the heart of the Small-Gain Theorem?

A signal circulating the loop is multiplied by

M Δ

each lap; the total output is

\sum_{n} (M Δ)^{n}

. If

∥ M ∥_{\infty} ∥Δ ∥_{\infty} < 1

this series converges to a bounded value (stable); if

\geq 1

some

Δ

makes it diverge (unstable).

Why does the worst-case energy gain equal the frequency-response peak, not merely stay below it?

The upper bound "pull the peak out of the integral" is achieved by choosing a disturbance whose energy piles up right at the peak frequency — so the sup is attained and the inequality becomes equality.

Why is

μ

-synthesis less conservative than plain small-gain?

Small-gain treats

Δ

as a single full block that can couple every channel to every other; μ-synthesis respects the known block-diagonal structure of

Δ

, forbidding couplings that physics rules out, so it rejects fewer safe plants.

Why does H∞ care about the direction of the worst input for multivariable systems?

Because

\overset{σ}{ˉ}

comes from the Singular Value Decomposition: gain depends on both frequency and input direction, and the worst case aligns the disturbance with the highest-gain singular direction — a scalar magnitude would miss this.

Why can the discrete-time H∞ norm be read on the unit circle instead of the imaginary axis?

Sampling maps the stable region from the left half

s

-plane to the inside of the unit circle in the

z

-plane; the frequency axis (imaginary axis) maps to the unit circle

z = e^{j θ}

, so the "sweep all frequencies" that defined the peak becomes a sweep of

θ

around that circle.

Edge cases

What is

∥ G ∥_{\infty}

G (j ω) \to \infty

at some frequency (a pole on the imaginary axis)?

It is

+ \infty

— the H∞ norm is only finite for stable systems with no imaginary-axis poles; an undamped resonance (

ζ = 0

) gives an unbounded peak, meaning an infinitesimal disturbance at that frequency yields unbounded output.

What does

∥ M ∥_{\infty} = 1

exactly (the boundary) tell us about robust stability?

It is the knife-edge: the strict inequality

∥ M ∥_{\infty} < 1

guarantees stability, but at

1

there exists a worst-case

Δ

(gain exactly

1

, right phase) that makes the loop gain reach unity and marginally destabilise — so

∥ M ∥_{\infty} = 1

is not safe.

What happens to the H∞ framework for a disturbance that is a constant (DC) forever?

A constant has infinite energy (

\int_{0}^{\infty}

diverges), so it is not a finite-energy signal; the

2

-norm energy-gain interpretation breaks. Such steady disturbances are handled by DC/integral action, not by the H∞ energy argument directly.

ζ = 0

(undamped mode) what is the resonant peak, and what does that mean physically?

The peak

\to \infty

(denominator

2 ζ \to 0

), meaning a sinusoid exactly at

ω_{n}

pumps energy in with no loss and the amplitude grows without bound — a bending mode that must be damped or notched, or no finite H∞ controller exists.

∣ S (j ω) ∣ = 1

at some frequency, what is the disturbance doing there?

It is passing straight through with neither amplification nor attenuation (

S + T = 1

with

S = 1, T = 0

) — the crossover-ish band where the loop has essentially no authority; the waterbed pushes the excess attenuation elsewhere into a hump

∣ S ∣ > 1

What if the true actuator error is only

\pm 5%

but we designed for

\pm 30%

The design is still robust (a smaller ball sits inside the larger one), just conservative — we may have paid performance to defend uncertainty that isn't there, exactly the over-modelling cost the parent's mistake box warns about.

Can

∥ W_{S} S ∥_{\infty} < 1

and

∥ W_{T} T ∥_{\infty} < 1

both hold at the same frequency?

Not with aggressive weights: since

S + T = 1

, at any

ω

you cannot make both

∣ S ∣

and

∣ T ∣

small, so the two weights must have non-overlapping dominant bands (

W_{S}

big at low

ω

W_{T}

big at high

ω

) — the crossover is the unavoidable compromise region.

How does the H∞ norm behave in discrete time near the Nyquist frequency?

The sweep runs over

θ \in [- π, π]

, and

θ = \pm π

is the Nyquist frequency (half the sample rate); a resonance that folds near Nyquist still contributes to the peak, so sampled-data designs must check the whole unit circle, not just low

θ

— aliasing can hide a peak the continuous model never showed.

Recall One-line summary of the traps

Peak, not average; supremum over all frequencies and directions (unit circle in discrete time); small-gain is sufficient (conservative, not necessary) for a fixed plant; and the waterbed forbids "small $∣ S ∣$ everywhere."