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

WHAT it says: there exists some sinusoidal disturbance whose output energy $\Vert z{\Vert }_2$ is $4$ times its input energy $\Vert w{\Vert }_2$ (recall $\Vert \cdot {\Vert }_2$ = the signal's total energy, $\sqrt{\int w^{\top }wdt}$, defined in the symbols box), and no disturbance is amplified by more than $4$. WHY: $\Vert G{\Vert }_{\infty }={\sup }_w\Vert z{\Vert }_2/\Vert w{\Vert }_2$ — literally the worst-case energy gain, the biggest ratio of output-energy to input-energy over every possible disturbance. So $4$ is a ceiling on amplification and a ceiling that is actually reached. In words: "the nastiest sine wave gets four times louder; nothing gets worse than that."

Answer: (c). (a) is the DC gain — the response only at $\omega =0$ (a constant input). That ignores every other frequency. (b) is an average over frequency. Averages smear the peak away. (c) is the peak, and the peak is what matters because the worst disturbance concentrates its energy exactly where $|G|$ is largest (this is the Parseval argument from the parent note).

Step 0 — check the formula even applies. Why? This resonant-peak formula assumes a genuine peak exists, which happens only for $0<\zeta <\frac{1}{\sqrt{2}}\approx 0.707$. Above that damping the magnitude just rolls off monotonically (no peak), and then $\Vert G{\Vert }_{\infty }=|G(0)|$ instead. Here $\zeta =0.1<0.707$, so a true resonant peak exists — good. Step 1 — identify ${\omega }_n,\zeta$. Why? The standard form is $\frac{{\omega }_n^{2}K}{s^2+2\zeta {\omega }_{n}s+{\omega }_n^2}$. Matching: ${\omega }_n^2=9\Rightarrow {\omega }_n=3$; $2\zeta {\omega }_n=0.6\Rightarrow \zeta =0.1$; DC gain $|G(0)|=9/9=1$. Step 2 — plug in. Why the peak formula? Because $\Vert G{\Vert }_{\infty }$ is the peak of $|G(j\omega )|$, and for a lightly damped second-order mode the peak sits near ${\omega }_n$ with height $M_r$. $M_r=\frac{1}{2(0.1)\sqrt{1-0.01}}=\frac{1}{0.2\times 0.99499}\approx \mathrm{5.025.}$ So $\Vert G{\Vert }_{\infty }\approx 5.03$: a resonant disturbance at $\omega \approx 3$ rad/s is amplified about five times. What to look at in the figure below: the lavender curve is $|G(j\omega )|$. Notice it sits near $1$ (the mint dotted DC line) at low frequency, then rises to a sharp coral dot at $\omega \approx 3$ — that dot's height ($\approx 5.03$) is the entire H∞ norm. Everything to the left and right is smaller; only that single spike's height matters. This is why "peak, not average" is the right reading.

Step 1 — normalise. Why? The Small-Gain Theorem wants the uncertainty inside the unit ball $\Vert \tilde{\Delta }{\Vert }_{\infty }\le 1$, so we peel off the size $0.25$: the real uncertain block is $0.25\tilde{\Delta }$. Step 2 — apply small-gain. Stability for all such blocks holds iff the loop-gain product stays under one: $\Vert 0.25M{\Vert }_{\infty }<1⟺\Vert M{\Vert }_{\infty }<\frac{1}{0.25}=4.$ So any actuator error within $\pm 25\%$ is tolerated provided the nominal loop's peak gain in that channel is below $4$. WHAT this bought us: one scalar inequality certifies an infinite family of plants.

Step 1 — unpack the norm inequality. $\Vert W_{S}S{\Vert }_{\infty }<1$ means the product $|W_S(j\omega )S(j\omega )|<1$ at every frequency. Why every frequency? Because the $\infty$-norm is the peak; if the peak is under $1$, all frequencies are under $1$. Step 2 — solve for $|S|$. At $\omega =1$: $|W_S(j1)||S(j1)|<1⟹|S(j1)|<\frac{1}{20}=\mathrm{0.05.}$ Interpretation: the tracking error is at most $5\%$ of the reference/disturbance amplitude at $1$ rad/s. The weight $W_S$ literally draws the ceiling $1/|W_S|$ that $|S|$ must live under. What to look at in the figure below: the coral curve is the ceiling $1/|W_S|$ — a wall $|S|$ may never poke through. The lavender curve is a feasible $|S|$; watch how it hugs below the coral ceiling everywhere. The mint dot marks $\omega =1$: the ceiling there sits at $0.05$, so $|S|$ is squeezed under $5\%$. Reading the picture: wherever you make the weight tall, you press the ceiling low and force tight tracking.

Step 0 — state the assumptions. Why first? The identity ${\int }_0^{\infty }\ln |S|d\omega =0$ is not universal — it holds for a loop whose open-loop transfer function $L$ is stable (no right-half-plane poles), whose relative degree is at least $2$ (so $L\to 0$ fast enough for the integral to converge), and which has no right-half-plane zeros. RHP poles raise the right-hand side to a positive value $\pi \sum \text{Re}(p_i)$, making the waterbed even worse. We assume the benign stable, minimum-phase case here so the right side is exactly $0$. Step 1 — read the integral. From Bode Sensitivity Integral (waterbed), the signed area under $\ln |S|$ is fixed at $0$. Where $|S|<1$ we have $\ln |S|<0$ (negative area, "good"); where $|S|>1$ we have $\ln |S|>0$ (positive area, "bad"). Step 2 — argue the contradiction. If $|S|<0.01$ everywhere, then $\ln |S|<\ln (0.01)\approx -4.6$ everywhere, so the integral would be a large negative number, not $0$. ${\int }_0^{\infty }\ln |S|d\omega \le {\int }_0^{\infty }(-4.6)d\omega =-\infty \ne 0.$ So it violates the conservation law: push $|S|$ down here, it must pop up above $1$ somewhere else — the waterbed. You can only shape $S$, not shrink it globally.

Step 1 — low frequency ($\omega \to 0$). Why look here? Steady-state tracking is set by DC. $W_S(0)=\frac{{\omega }_b}{{\omega }_{b}A}=\frac{1}{A}.$ The ceiling on $|S|$ is $1/|W_S(0)|=A$. We want $|S(0)|\le 0.01$, so choose $A=0.01$. Step 2 — high frequency ($\omega \to \infty$). Why look here? The peak-sensitivity spec lives at high $\omega$. $W_S(\infty )=\frac{s/M}{s}=\frac{1}{M}.$ The ceiling there is $1/|W_S(\infty )|=M$. We want $|S|<2$, so choose $M=2$. Step 3 — find the exact crossover, then justify the $\approx$. Why? Bandwidth is where the weight's magnitude passes through $1$; we must derive that frequency, not assert it. Set $|W_S(j\omega )|=1$, i.e. $|j\omega /M+{\omega }_b{|}^2=|j\omega +{\omega }_{b}A{|}^2$: $\frac{{\omega }^2}{M^2}+{\omega }_b^2={\omega }^2+{\omega }_b^2{A}^2⟹{\omega }^2\left(1-\frac{1}{M^2}\right)={\omega }_b^2\left(1-A^2\right).$ Solve exactly: ${\omega }_{\text{cross}}={\omega }_b\sqrt{\frac{1-A^2}{1-1/M^2}}.$ Plug in $M=2,A=0.01,{\omega }_b=5$: $1-A^2=0.9999$ and $1-1/M^2=1-0.25=0.75$, so ${\omega }_{\text{cross}}=5\sqrt{\frac{0.9999}{0.75}}=5\times 1.1547\approx 5.77\text{rad/s}.$ Why the earlier "$\approx {\omega }_b$" is only rough: the correction factor $\sqrt{(1-A^2)/(1-1/M^2)}$ equals $1$ only in the ideal limit $M\to \infty ,A\to 0$. With $M=2$ the $1/M^2=0.25$ term is not negligible — it pulls the crossover about $15\%$ above ${\omega }_b$. So the honest statement is: the crossover is $\approx 5.77$ rad/s, near but not equal to ${\omega }_b=5$; the tiny $A$ barely moves it, the modest $M$ moves it noticeably. Result: $W_S(s)=\frac{s/2+5}{s+5(0.01)}=\frac{0.5s+5}{s+0.05}.$ Ceilings: low-freq $|S|\le A=0.01$ (1% error ✓), high-freq $|S|\le M=2$ (peak $\le 2$ ✓). If a controller achieves $\Vert W_{S}S{\Vert }_{\infty }<1$, all three specs are met simultaneously. What to look at in the figure below: the lavender curve is the ceiling $1/|W_S|$. It starts pinned near the mint line at $A=0.01$ (1% floor for tracking), climbs through $1$ near ${\omega }_b$ (the butter dotted line), and flattens out at the coral line $M=2$ (the peak allowance). Reading it: the shape of this ceiling is exactly the performance spec drawn as a wall.

(a) Check the condition. Why $\Vert W_{\Delta }T{\Vert }_{\infty }$? For multiplicative output uncertainty the loop-in-feedback with $\Delta$ has forward block $W_{\Delta }T$; the Small-Gain Theorem then demands its peak below $1$. Since $W_{\Delta }$ is a constant $0.5$ here, $\Vert W_{\Delta }T{\Vert }_{\infty }=0.5\times \Vert T{\Vert }_{\infty }=0.5\times 0.8=0.4<1.✓$ Robustly stable. (b) Largest tolerable uncertainty. Solve $|W_{\Delta }|\Vert T{\Vert }_{\infty }<1$: $|W_{\Delta }|<\frac{1}{\Vert T{\Vert }_{\infty }}=\frac{1}{0.8}=\mathrm{1.25.}$ So the loop tolerates multiplicative uncertainty up to 125% of the nominal plant at these frequencies before the guarantee fails. (c) If $\Vert T{\Vert }_{\infty }=2.5$: then $\Vert W_{\Delta }T{\Vert }_{\infty }=0.5\times 2.5=1.25>1$, so the sufficient small-gain test fails — we can no longer certify robust stability. But small-gain ignores the structure of $\Delta$ and is often conservative; the tool that tightens this while keeping the guarantee is μ-synthesis (structured uncertainty), which uses the structured singular value $\mu$ instead of the plain peak.

Step 1 — use the algebraic identity. Why? $S$ and $T$ are not free; at every frequency $S(j\omega )+T(j\omega )=1$ as complex numbers. So $T=1-S$, and by the reverse triangle inequality the magnitude is bounded below: $|T(j{\omega }_0)|=|1-S(j{\omega }_0)|\ge |1-|S(j{\omega }_0)||=1-0.1=\mathrm{0.9.}$ So the smallest possible $|T(j{\omega }_0)|$ is $0.9$. Worst-case phase remark: the value $|T|=0.9$ is attained only when $S(j{\omega }_0)$ is real and positive (phase $0$). If $S$ has nonzero phase, $|T|=|1-S|$ can be larger — e.g. if $S=0.1j$ then $|T|=|1-0.1j|=\sqrt{1+0.01}\approx 1.005$. So $0.9$ is a floor, and the honest robustness check must use the worst-case $|T|$, which can exceed $0.9$. Step 2 — test robustness at the floor. Using the smallest $|T|\approx 0.9$: $|W_{\Delta }T|=0.5\times 0.9=0.45<1.✓$ Compatible — even at the tightest $|T|=0.9$, the robustness product $0.45$ sits safely under $1$. (If $S$'s phase pushed $|T|$ up to $1.005$, the product would be $0.5\times 1.005=0.5025$, still under $1$, so this design is robust across all phases.) The lesson: demanding tiny error ($|S|$ small) forces $|T|$ near $1$ — you cannot make both small at the same frequency. Here even the smallest $|T|=0.9$ still leaves robustness margin because the weight is only $0.5$. Push $|S|$ smaller and $|T|\to 1$; if $|W_{\Delta }|$ then exceeds $1/|T|\to 1$, robustness breaks. This is the performance–robustness trade-off made quantitative.