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

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The scenario matrix

Before working anything, let us list every class of case an H∞ / small-gain problem can hand you. Each later example is tagged with the cell(s) it covers.

Cell	What makes it distinct	Covered by
A. Flat gain	$G$ has no frequency dependence — constant magnitude	Ex 1
B. Monotone roll-off	$\Vert G(j\omega )\Vert$ peaks at $\omega =0$, no resonance	Ex 2
C. Resonant peak	lightly damped mode, peak inside the band	Ex 3
D. Degenerate input	zero-energy or DC-only disturbance	Ex 4
E. Limiting behaviour	damping $\zeta \to 0$, peak $\to \infty$	Ex 5
F. Multi-input (SVD)	matrix $G$, need largest singular value not $\Vert \cdot \Vert$	Ex 6
G. Small-gain sizing	robust-stability margin from an uncertainty size	Ex 7
H. Weight-ceiling / waterbed	$\Vert W_{S}S{\Vert }_{\infty }<1$ as a drawn ceiling, plus the trade-off	Ex 8
I. Real-world word problem	rocket bending mode, choose the design action	Ex 9
J. Exam twist	combine small-gain + peak reading in one go	Ex 10

Every cell A–J appears below. If you can do all ten you have seen the whole surface.

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Example 1 — Flat gain (Cell A)

Forecast: guess the number before reading — a constant amplifier scaling everything by 4. Is the worst frequency low, high, or... does it even matter?

1. Write the frequency response. $G(j\omega )=4$ for every $\omega$. Why this step? The H∞ norm is the peak of $|G(j\omega )|$ over frequency, so we must first see how $|G|$ varies with $\omega$.
2. Take the magnitude. $|G(j\omega )|=4$, flat. Look at s01 — a dead-level red line at height 4 across the whole $\omega$ axis. Why this step? A real constant has no imaginary part; its magnitude is the constant itself, at all frequencies.
3. Take the supremum. ${\sup }_{\omega }4=4$, so $\Vert G{\Vert }_{\infty }=4$. Why this step? The peak of a flat line is its own value. Every frequency is equally "worst."

Figure s01 — a perfectly horizontal red line: no frequency is special, the peak equals the constant.

Verify: Feed a unit-energy sinusoid: output energy gain $=|G{|}^2=16$, so $\Vert z{\Vert }_2/\Vert w{\Vert }_2=4$. Matches. Units: dimensionless gain in, dimensionless gain out. ✓

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Example 2 — Monotone roll-off, no resonance (Cell B)

Forecast: does the biggest amplification happen at high frequency, or at DC ($\omega =0$)?

1. Magnitude function. $|G(j\omega )|=\frac{5}{\sqrt{{\omega }^2+4}}$. Why this step? $j\omega +2$ has magnitude $\sqrt{{\omega }^2+2^2}$; dividing the constant $5$ by it gives the gain at each frequency.
2. See how it moves. As $\omega$ grows, the denominator grows, so $|G|$ falls monotonically. No interior peak. See s02 — the red curve slides steadily downhill from its left end. Why this step? There is no second-order term to create resonance; a single pole only attenuates as frequency rises.
3. Peak is at $\omega =0$. $|G(j0)|=5/2=2.5$. Hence $\Vert G{\Vert }_{\infty }=2.5$. Why this step? For a monotone-decreasing magnitude the supremum sits at the left end, the DC gain.

Figure s02 — red magnitude curve starts at 2.5 and only ever descends; the peak is the leftmost point.

Verify: DC gain $=G(0)=5/2=2.5$; and $|G(j\omega ){|}^2=25/({\omega }^2+4)$ is largest when ${\omega }^2=0$. ✓

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Example 3 — Resonant peak inside the band (Cell C)

Forecast: the peak is not at DC now. Where do you think it sits — below, at, or above ${\omega }_n=2$?

1. Magnitude. $|G(j\omega )|=\frac{10}{\sqrt{(4-{\omega }^2{)}^2+(0.4\omega {)}^2}}$. Why this step? Substitute $s=j\omega$: $s^2=-{\omega }^2$ so the real part is $4-{\omega }^2$, the imaginary part is $0.4\omega$.
2. Resonant frequency. The peak of a second-order response is at ${\omega }_r={\omega }_n\sqrt{1-2{\zeta }^2}=2\sqrt{1-0.02}\approx 1.98$ rad/s — just below ${\omega }_n$. Why this step? Minimising the denominator (not simply setting $\omega ={\omega }_n$) shifts the true peak slightly left when there is damping.
3. Peak magnitude. $\Vert G{\Vert }_{\infty }\approx \frac{10/4}{2\zeta \sqrt{1-{\zeta }^2}}=\frac{2.5}{2(0.1)\sqrt{0.99}}\approx 12.56$. Why this step? The standard resonant-peak formula; the numerator $10/4=2.5$ is the DC gain, and small $\zeta$ makes the peak tall. See s03 — the tall red spike near $\omega \approx 2$.

Figure s03 — a tall red resonant spike near $\omega =2$ towering over the dashed DC-gain line at 2.5.

Verify: DC gain $=10/4=2.5$; peak $\approx 12.56$ is $\approx 5\times$ the DC gain, consistent with a $\zeta =0.1$ mode. A gust at $\omega \approx 2$ is amplified $\sim 12.6\times$. ✓

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

Forecast: for (b), a constant sits at $\omega =0$. Do you get the peak gain 12.56, or the DC gain 2.5?

1. Case (a): $w=0$. Then $\Vert w{\Vert }_2=0$, and the ratio $\Vert z{\Vert }_2/\Vert w{\Vert }_2$ is $0/0$ — undefined, which is exactly why the parent note's supremum is written ${\sup }_{w\ne 0}$. Why this step? The H∞ norm deliberately excludes the zero signal; a robustness margin only means something for a disturbance that actually carries energy.
2. Case (b): $w=1$ forever. A constant is not finite-energy: ${\int }_0^{\infty }{1}^{2}dt=\infty$. So it lives outside the 2-norm world entirely. Why this step? $\Vert w{\Vert }_2^2={\int }_0^{\infty }{w}^{\top }wdt$ diverges; the H∞ theory is built for finite-energy (fading) disturbances, not signals that never die.
3. What number does apply to a constant? The steady-state response, governed by the DC gain $|G(0)|=2.5$: the output settles to $2.5$. That is a DC gain statement, not an H∞ statement. Why this step? The final-value behaviour of a step input picks off $\omega =0$ only, so it sees $|G(0)|$, never the resonant peak.

Verify: $G(0)=10/(0+0+4)=2.5$, so a unit step settles to $2.5$; the peak $12.56$ is irrelevant to a DC input. The distinction (energy-gain peak vs. DC gain) is the whole lesson. ✓

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

Forecast: does the peak level off, or run away to infinity?

1. Peak formula in $\zeta$. $\Vert G{\Vert }_{\infty }\approx \frac{2.5}{2\zeta \sqrt{1-{\zeta }^2}}$. Why this step? Same resonant-peak expression as Ex 3, now kept as a function of $\zeta$ so we can push $\zeta$ toward $0$.
2. Plug the shrinking values. $\zeta =0.1\Rightarrow \approx 12.56$; $\zeta =0.05\Rightarrow \approx 25.03$; $\zeta =0.02\Rightarrow \approx 62.5$. Why this step? Halving $\zeta$ roughly doubles the peak — the $2\zeta$ in the denominator dominates.
3. Take the limit. As $\zeta \to 0$, $\Vert G{\Vert }_{\infty }\to \infty$. The mode becomes an undamped oscillator with infinite peak gain. Look at s04: red curves grow taller and sharper as damping vanishes. Why this step? An undamped pole sits on the imaginary axis; at exactly its natural frequency the response has no bound. Robust control must never let the loop excite such a mode.

Figure s04 — three red resonance curves; as $\zeta$ drops from 0.1 to 0.02 the peak roughly doubles each time and narrows into a spike.

Verify: $2.5/(2\cdot 0.02\cdot \sqrt{1-0.02^2})\approx 62.5$, and the sequence $12.56\to 25.03\to 62.5$ diverges as $\zeta \downarrow 0$. ✓

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Example 6 — Multi-input system: use the singular value, not the magnitude (Cell F)

Forecast: for a scalar we'd take $|G|$. For a matrix, the "gain" depends on which direction the input pushes. What is the biggest output length per unit input length?

1. Why the largest singular value. $\bar{\sigma }(G)={\max }_{\Vert x\Vert =1}\Vert Gx\Vert$ — the biggest stretch $G$ applies to any unit vector. That is the multi-input generalisation of "peak gain." Why this step? The definition at the top of the page uses the Singular Value Decomposition; for a matrix, gain is direction-dependent, so we need the worst direction.
2. Form $G^{\top }G$. $G^{\top }G=\left[\begin{array}{cc}3& 4\\ 0& 0\end{array}\right]\left[\begin{array}{cc}3& 0\\ 4& 0\end{array}\right]=\left[\begin{array}{cc}25& 0\\ 0& 0\end{array}\right]$. Why this step? Singular values of $G$ are the square roots of the eigenvalues of $G^{\top }G$ — the standard SVD route.
3. Eigenvalues and root. Eigenvalues $\{25,0\}$, so singular values $\{5,0\}$, giving $\bar{\sigma }=5$. Why this step? The largest eigenvalue $25$ has root $5$; the input direction $x=(1,0{)}^{\top }$ produces output length $\sqrt{3^2+4^2}=5$, a 3-4-5 triangle.

Verify: $\Vert G(1,0{)}^{\top }\Vert =\Vert (3,4{)}^{\top }\Vert =\sqrt{9+16}=5$, and no unit input beats it ($x=(0,1{)}^{\top }$ gives $0$). So $\bar{\sigma }=5$. ✓

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Example 7 — Small-gain sizing: how big can uncertainty be? (Cell G)

Forecast: the parent did $\pm 30\%\Rightarrow \Vert M{\Vert }_{\infty }<3.33$. Guess the answer here before computing.

1. Normalise to the unit ball. Write $\Delta =0.4\tilde{\Delta }$ with $\Vert \tilde{\Delta }{\Vert }_{\infty }\le 1$. Why this step? The Small-Gain Theorem is stated for a unit-bounded block; pull the $0.4$ into the loop.
2. Apply small-gain. Stability for all $\tilde{\Delta }$ iff $\Vert 0.4M{\Vert }_{\infty }<1$, i.e. $\Vert M{\Vert }_{\infty }<1/0.4=2.5$. Why this step? $\Vert 0.4M{\Vert }_{\infty }=0.4\Vert M{\Vert }_{\infty }$, and the loop is safe when the product of gains around it stays below $1$.
3. Read the margin. As long as the nominal loop's peak at the uncertain channel stays below $2.5$, any actuator error within $\pm 40\%$ is tolerated. Why this step? One inequality certifies the whole infinite family of plants — the power of small-gain.

Verify: $1/0.4=2.5$; and larger uncertainty ($40\%>30\%$) gives a tighter margin ($2.5<3.33$), which is the correct direction. ✓

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Example 8 — Weight ceiling and the waterbed (Cell H)

Forecast: if the weight is 3, is the allowed sensitivity $3$, $1/3$, or $1$?

1. Turn the norm into a pointwise ceiling. $\Vert W_{S}S{\Vert }_{\infty }<1\Rightarrow |W_S(j\omega )||S(j\omega )|<1\Rightarrow |S(j\omega )|<1/|W_S(j\omega )|$ at every $\omega$. Why this step? The H∞ norm bounds the product's peak; since it's below 1 at the peak, it's below 1 everywhere, so $|S|$ sits under the ceiling $1/|W_S|$. See s05 — the sensitivity curve must stay under the red ceiling.
2. Plug in. At $\omega =0.1$: $|S(j0.1)|<1/3\approx 0.333$. Good tracking — error is at most a third of the reference there. Why this step? Large weight $\Rightarrow$ low ceiling $\Rightarrow$ small sensitivity $\Rightarrow$ good low-frequency tracking.
3. Why not everywhere. Under the standard assumptions — an open-loop-stable, minimum-phase plant with relative degree $\ge 2$ (loop gain rolling off fast enough) — the Bode sensitivity integral holds in the normalised form ${\int }_0^{\infty }\ln |S(j\omega )|d\omega =0$. (With one or more unstable poles at $p_k$ the right side becomes $\pi \sum \mathrm{Re}(p_k)>0$, making the trade worse.) Pushing $\ln |S|$ far negative at low $\omega$ forces it positive (i.e. $|S|>1$) elsewhere, because the total signed area is fixed. Why this step? The integral pins the signed area of $\ln |S|$: think of the curve as a see-saw about the line $\ln |S|=0$. Digging a deep trough of suppression on the left must raise a matching hump of amplification on the right so the areas cancel to the fixed total. See s06 — the shaded trough below the axis has the same area as the shaded hump above it.

Figure s05 — black $|S|$ curve tucked beneath the red ceiling $1/|W_S|$; where the ceiling is low, $|S|$ is forced small (good tracking).

Figure s06 — $\ln |S|$ vs $\omega$: the red shaded trough (suppression, area negative) is exactly balanced by the black hump above the axis (amplification, area positive); their signed areas sum to zero.

Verify: $1/3=0.3\overline{3}$, so $|S(j0.1)|<0.333$; and $\int \ln |S|=0$ means negative area (below the ceiling) is exactly offset by positive area (a bump above 1). ✓

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

Forecast: roughly how much attenuation — a factor of 2, 6, or 60?

1. Compare open vs required peak. Open-loop peak $\approx 12.6$; required peak $<2$. Why this step? $\Vert T_{zw}{\Vert }_{\infty }$ is the worst-case energy amplification (top-of-page definition); the gust concentrates at the peak, so that peak is what must come down.
2. Attenuation factor. $12.6/2=6.3$, so the loop must knock the mode down by at least $\approx 6.3\times$ (about $16$ dB) at $\omega =2$. Why this step? We need the closed-loop peak below $2$ starting from $12.6$; the ratio is the required suppression.
3. How to command it. Place a weight $W$ with large magnitude near $\omega =2$ and require $\Vert W{T}_{zw}{\Vert }_{\infty }<1$. This draws the ceiling $1/|W|$ low at the mode, forcing the synthesised $K$ (see LQR and LQG control for the nominal baseline it improves on) to notch/attenuate exactly there. Structured uncertainty in the bending frequency itself would push you toward μ-synthesis (structured uncertainty). Why this step? H∞ shapes where it matters via weights; a tall weight at $2$ rad/s buys attenuation precisely at the dangerous frequency, obeying the waterbed by spending margin at frequencies with no gust energy.

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

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Example 10 — Exam twist: combine small-gain with a peak reading (Cell J)

Forecast: the peak of $k/(s+1)$ — is it at DC or high frequency? Then what's the biggest $k$?

1. Peak of $M$. $|M(j\omega )|=k/\sqrt{{\omega }^2+1}$, monotone-decreasing (the Cell B pattern), so the peak is at $\omega =0$: $\Vert M{\Vert }_{\infty }=|M(0)|=k$. Why this step? A single stable pole gives monotone roll-off; the supremum is the DC value $|M(0)|=k$ — we reuse exactly the Ex 2 reasoning.
2. Apply small-gain. Robust stability $⟺\Vert M{\Vert }_{\infty }<1⟺k<1$. Why this step? The Small-Gain Theorem: the $M$–$\Delta$ interconnection is stable for all $\Vert \Delta {\Vert }_{\infty }\le 1$ exactly when $\Vert M{\Vert }_{\infty }<1$.
3. State the boundary. The largest safe gain is $k\to 1^{-}$; at $k=1$ the guarantee is lost (some admissible $\Delta$ can drive the loop-gain product to $1$, breaking convergence of the series $\sum (M\Delta {)}^n$). Why this step? Small-gain is an open inequality; equality is the exact edge where the geometric series stops converging and the loop can run away.

Verify: $\Vert M{\Vert }_{\infty }=|M(0)|=k$; setting $k<1$ satisfies $\Vert M{\Vert }_{\infty }<1$ so the supremum of admissible $k$ is $1$. ✓

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Recall Scenario checklist — can you place any new problem?

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

Active recall