Visual walkthrough — H∞ control — robust to uncertainty (intro)
Everything below assumes you have never seen a norm, a frequency response, or an integral used this way. Every symbol is earned before it is used.
Step 1 — What is a signal, and how "big" is it?
WHAT. A signal is just a number that changes over time — the strength of a wind gust on a rocket, second by second. We need a single number for "how much stuff is in this signal," because later we compare output stuff to input stuff.
WHY this tool (the integral of the square). We could measure the tallest spike, but a brief tall spike carries little "punch," while a long gentle push can carry a lot. The honest measure of total punch is energy: square the signal (so ups and downs both count as positive), then add up all those squares across time. "Add up over continuous time" is exactly what an integral means — a smooth sum. That is why we reach for it and not a plain maximum.
PICTURE. The area under the squared curve is the energy. A tall-but-brief blip and a low-but-long push can have the same shaded area.
Term-by-term: = the signal, = its instantaneous punch (always ), = total accumulated punch = the shaded area.
Step 2 — A system is a frequency-by-frequency amplifier
WHAT. A stable linear system eats an input signal and hands back an output . The magic fact: if you feed a pure sine wave at frequency (radians per second), out comes a sine wave at the same frequency, only rescaled (and shifted). The rescaling factor is a number we call — the gain at frequency .
WHY this matters. Because every signal is secretly a sum of sine waves (that is the whole idea of frequency), if we know how treats each single sine, we know how it treats everything. The symbol is just the flag "we're asking about the sine wave at frequency "; is "how many times taller does that sine come out."
PICTURE. Low-frequency sines might pass gently; near a resonance one particular sine gets blown up enormously; very high frequencies get squashed.
The tall red spike is the peak gain — remember it, it becomes the star of the show.
Step 3 — The exact question we want to answer
WHAT. Robustness asks: of all the disturbances nature could throw at us, which one gets amplified the most, and by how much? In energy terms:
WHY and not "average." Robustness is a promise against the nastiest case, not the typical one. An average could hide a catastrophic single disturbance. We want a guarantee that holds no matter what, so we take the top of the range.
PICTURE. Imagine trying thousands of input shapes; each gives a dot on a ratio-axis. is the highest dot.
The trouble: there are infinitely many possible . We cannot try them all by hand — so we need a trick to compute this supremum. That trick is the next step.
Step 4 — Parseval: energy is the same in time and in frequency
WHAT. Parseval's Theorem says the energy of a signal computed in time equals the energy computed in frequency. Write for the amount of frequency- sine hiding inside (its spectrum). Then
WHY this tool and not another. In the time domain, is a messy convolution (a smearing). But in the frequency domain, Step 2 told us acts by simple multiplication: . Parseval is the bridge that lets us stop fighting convolution and instead multiply, frequency by frequency. That is precisely the reason we invoke it here.
PICTURE. Two histograms with equal total area: one is "energy vs time," the other "energy vs frequency." Parseval says the totals match.
Step 5 — Convert the output energy into a frequency sum
WHAT. Apply Parseval to the output , then use from Step 2:
=\frac{1}{2\pi}\int_{-\infty}^{\infty}\underbrace{|G(j\omega)|^2}_{\text{gain}^2\text{ at }\omega}\ \underbrace{|W(j\omega)|^2}_{\text{input energy at }\omega}\,d\omega.$$ **WHY.** This is the key rewrite: the output energy is the input's energy spectrum, **weighted at each frequency by the gain squared**. Each slice of frequency contributes $|G|^2 \cdot |W|^2$. **PICTURE.** Take the input spectrum bars from Step 4 and multiply each bar's height by the gain-squared curve. Bars near the resonant peak grow tall; bars where $G$ is small shrink. ![[deepdives/dd-physics-3.5.37-d2-s05.png]] --- ## Step 6 — Pull out the peak (the inequality) **WHAT.** At **every** frequency, the local gain $|G(j\omega)|^2$ is no bigger than the **largest** gain anywhere, which we name $$\|G\|_\infty^2=\sup_\omega |G(j\omega)|^2 \quad(\text{the peak of the Bode curve}).$$ So replace each $|G(j\omega)|^2$ by that single biggest value — it can only make the sum larger or equal: $$\|z\|_2^2=\frac{1}{2\pi}\int |G(j\omega)|^2|W(j\omega)|^2 d\omega \ \le\ \Big(\sup_\omega|G(j\omega)|^2\Big)\underbrace{\frac{1}{2\pi}\int |W(j\omega)|^2 d\omega}_{=\ \|w\|_2^2\ \text{(Parseval again)}}.$$ Divide both sides by $\|w\|_2^2$: $$\boxed{\ \frac{\|z\|_2^2}{\|w\|_2^2}\ \le\ \|G\|_\infty^2\ \Longrightarrow\ \frac{\|z\|_2}{\|w\|_2}\le\|G\|_\infty\ }$$ **WHY.** Pulling the maximum out of an integral is the standard way to get a clean ceiling: "no frequency contributes more than its share at the worst gain." This proves the ratio **can never exceed** $\|G\|_\infty$. **PICTURE.** Flatten the wiggly gain-squared curve up to a horizontal ceiling at the peak height — the area under the flat ceiling (times the input spectrum) is never smaller than the true area. ![[deepdives/dd-physics-3.5.37-d2-s06.png]] --- ## Step 7 — The ceiling is actually reached (why it's an equality, not just $\le$) **WHAT.** An upper bound is useless if nothing achieves it. But we can build a disturbance that **hits** the ceiling: choose $w$ whose energy is concentrated in a *narrow band of frequencies around the peak* $\omega^\star$ (where $|G|$ is largest). Then almost all of $w$'s energy sits exactly where $G$ amplifies hardest, and $$\frac{\|z\|_2}{\|w\|_2}\ \longrightarrow\ \|G\|_\infty.$$ **WHY this closes the argument.** Step 6 showed the ratio *can't exceed* $\|G\|_\infty$. This step shows it *can get arbitrarily close* to $\|G\|_\infty$. Together: the supremum **equals** $\|G\|_\infty$. That is the boxed result of the parent note, now earned. > [!formula] The result, fully earned > $$\sup_{w\neq 0}\frac{\|z\|_2}{\|w\|_2}=\|G\|_\infty=\sup_\omega|G(j\omega)|.$$ > The worst-case **energy** amplification equals the **peak** of the frequency response. **PICTURE.** A spectrum shaped like a thin spike sitting right on top of the resonant peak — the "meanest" disturbance. Its output is that spike scaled by the peak gain. ![[deepdives/dd-physics-3.5.37-d2-s07.png]] --- ## Step 8 — Edge and degenerate cases (never leave a scenario unshown) **WHAT.** We must check the corners so nothing surprises the reader. - **Zero input, $w=0$.** The ratio $\|z\|_2/\|w\|_2$ is $0/0$ — undefined, which is exactly why the supremum is taken over $w\neq 0$. No information is lost; the peak is still approached by *tiny but non-zero* inputs. - **$G$ has no resonant peak (monotone roll-off).** Then the largest gain is at $\omega=0$ (DC). The formula still holds — the "peak" is just at the left edge. *Nothing special about resonance; it's simply where the peak happens to sit.* - **Unstable $G$.** Then some sine grows without bound, $\sup_\omega|G(j\omega)|=\infty$, so $\|G\|_\infty=\infty$. The H∞ norm correctly *refuses to exist* for unstable systems — which is why H∞ synthesis first demands a **stabilising** controller. - **Vector / matrix $G$ (MIMO).** For several inputs and outputs, "gain at $\omega$" is not one number but a matrix. The right replacement for $|G(j\omega)|$ is the **largest singular value** $\bar\sigma(G(j\omega))$ from the [[Singular Value Decomposition]] — the biggest stretch factor the matrix applies to any input direction. So $\|G\|_\infty=\sup_\omega\bar\sigma(G(j\omega))$, reducing to $|G(j\omega)|$ when there is only one input and output. **PICTURE.** Three mini-panels: (a) a monotone curve with its peak at DC, (b) an unstable pole sending the curve to infinity, (c) a matrix stretching a unit circle of input directions into an ellipse whose longest axis is $\bar\sigma$. ![[deepdives/dd-physics-3.5.37-d2-s08.png]] > [!mistake] The trap this walkthrough kills > **"$\|G\|_\infty$ is the average or DC gain."** No — every step above pointed at the **peak**. The worst disturbance (Step 7) deliberately dumps its energy where $G$ is *largest*, not average. Averaging would let a single resonant gust slip through your safety promise. --- ## The one-picture summary Everything collapses into one flow: **signal energy → split into frequencies → each frequency scaled by gain → the biggest gain sets the worst case → that biggest gain *is* $\|G\|_\infty$.** ![[deepdives/dd-physics-3.5.37-d2-s09.png]] > [!recall]- Feynman retelling (plain words) > Think of your disturbance as sunlight and your system as a magnifying glass. The "energy" of the light is how much total glow there is (Step 1). A magnifying glass bends different *colours* by different amounts — that's the frequency response (Step 2). We want to know: what's the *hottest* spot this glass can make from a fixed amount of light (Step 3)? Trick: sort the light into its colours (Parseval, Step 4), and note the glass just multiplies each colour's brightness by its own factor (Step 5). No single colour is boosted more than the glass's *strongest* focusing power, so the total heat can't beat that strongest factor (Step 6). And if you shine light of *only that one best colour*, you reach exactly the maximum heat (Step 7). That maximum focusing factor — the peak — is the H∞ norm. Make it small, and no light, no gust, no disturbance can ever burn you much. Corner cases: zero light gives no ratio, a glass that focuses everything to infinity is a broken (unstable) glass, and a fancy multi-colour multi-spot glass just uses the biggest "stretch" (the largest singular value) as its focusing power. --- ## Active recall > [!recall] Quick self-test > 1. Why do we measure a signal's size as $\int w^2\,dt$ and not its tallest spike? > 2. What single fact lets us swap messy time-convolution for clean frequency-multiplication? > 3. In Step 6, why is replacing $|G(j\omega)|$ by its peak a valid *upper* bound? > 4. Step 7 built a special disturbance — what shape was it, and why does it prove equality? > 5. What happens to $\|G\|_\infty$ if $G$ is unstable, and why does H∞ care? Why is the H∞ norm the *peak* over frequency rather than an average? ::: The worst-case disturbance (Step 7) concentrates its energy at the single frequency of maximum gain, so only the peak — not the average — determines the largest possible energy amplification. Which theorem bridges time-domain energy and frequency-domain gain in the derivation? ::: [[Parseval's Theorem]] — it makes $\|w\|_2^2$ equal in both domains so $G$ can be treated as multiplication frequency-by-frequency. For a matrix (MIMO) system, what replaces $|G(j\omega)|$ as the "gain at $\omega$"? ::: The largest singular value $\bar\sigma(G(j\omega))$ from the [[Singular Value Decomposition]].