Worked examples — TVC dynamics — gimbal servo bandwidth, time delay

This page is a drill sheet. The parent note built the machinery: the second-order servo lag, the bandwidth formula, and the phase-eating delay. Here we throw every kind of case at that machinery — normal numbers, edge cases, zeros, limits, a word problem, and an exam trap — and grind each one out fully.

Before we start, one reminder of the three tools we keep reaching for, so no symbol appears unearned:

Recall The three formulas we reuse (from the parent)

• Natural frequency & damping: ${\omega }_n=\sqrt{\frac{k+K}{J}}$, $\zeta =\frac{c}{2\sqrt{(k+K)J}}$. ${\omega }_n$ is how fast the nozzle can swing; $\zeta$ is how much it rings (0 = pure ringing, 1 = no overshoot).
• Bandwidth: ${\omega }_B={\omega }_n\sqrt{(1-2{\zeta }^2)+\sqrt{(1-2{\zeta }^2{)}^2+1}}$. The fastest wiggle the servo still follows within $1/\sqrt{2}$ of full size.
• Phase eaten by a pure delay $\tau$ at frequency $\omega$: $\Delta \varphi =-\omega \tau$ radians $=-57.3\omega \tau$ degrees. ($57.3=180/\pi$ is just the radians→degrees converter.)

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The scenario matrix

Every problem this topic can ask lives in one of these cells. The examples below are tagged with the cell they cover.

Cell	What makes it special	Covered by
A · Nominal under-damped	$0<\zeta <1$, ordinary numbers, ${\omega }_B>{\omega }_n$	Ex 1
B · Sweet-spot damping	$\zeta =1/\sqrt{2}\approx 0.707$ so ${\omega }_B={\omega }_n$ exactly	Ex 2
C · Over-damped / critical	$\zeta \ge 1$, sluggish, ${\omega }_B<{\omega }_n$	Ex 3
D · Degenerate: zero delay	$\tau =0$ — delay eats nothing	Ex 4 (part)
E · Limiting: zero damping	$\zeta \to 0$ — resonance, bandwidth blows up	Ex 4
F · Delay phase-budget	given ${\omega }_c,\tau$: how much margin lost	Ex 5
G · Word problem (real vehicle)	translate hardware into numbers, decide safe/unsafe	Ex 6
H · Exam twist: "add gain to fix delay"	trap — raising gain makes delay worse	Ex 7
I · Padé wrong-way sign	first-order Padé RHP-zero, output goes wrong way first	Ex 8

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Example 1 — Cell A: nominal under-damped servo

1. ${\omega }_n=\sqrt{(k+K)/J}=\sqrt{1800/0.5}=\sqrt{3600}=60\text{rad/s}$. Why this step? ${\omega }_n$ is fixed purely by the stiffness-to-inertia ratio — the "spring vs. mass" contest sets the swing speed.
2. $\zeta =\frac{c}{2\sqrt{(k+K)J}}=\frac{30}{2\sqrt{1800\cdot 0.5}}=\frac{30}{2\sqrt{900}}=\frac{30}{60}=0.5$. Why? $\zeta$ compares actual damping $c$ to the critical amount $2\sqrt{(k+K)J}$. Here it's exactly half, so it rings.
3. $1-2{\zeta }^2=1-2(0.25)=0.5$. Then ${\omega }_B=60\sqrt{0.5+\sqrt{0.5^2+1}}=60\sqrt{0.5+\sqrt{1.25}}=60\sqrt{0.5+1.118}=60\sqrt{1.618}=76.3\text{rad/s}.$ Why? Under-damped systems resonate a little before rolling off, so the $-3$ dB point sits above ${\omega }_n$.

Verify: Units: $\sqrt{\text{(N⋅m/rad)}/({\text{kg⋅m}}^2)}=\sqrt{1/{\text{s}}^2}=\text{rad/s}$ ✔. $\zeta$ is dimensionless ✔. Ratio ${\omega }_B/{\omega }_n=1.27$ — matches the parent's "$\zeta =0.5\Rightarrow {\omega }_B\approx 1.27{\omega }_n$".

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Example 2 — Cell B: the sweet spot $\zeta =1/\sqrt{2}$

1. ${\omega }_n=60\text{rad/s}$ (unchanged — depends only on $k+K$ and $J$). Why? Damping doesn't touch ${\omega }_n$; it only affects ringing.
2. $\zeta =1/\sqrt{2}\approx 0.7071$ by construction. Why? We chose $c$ to hit this value; it's the classic control target.
3. $1-2{\zeta }^2=1-2\cdot \frac{1}{2}=0$. So ${\omega }_B=60\sqrt{0+\sqrt{0+1}}=60\sqrt{1}=60\text{rad/s}={\omega }_n.$ Why? When $1-2{\zeta }^2=0$ the formula collapses to ${\omega }_B={\omega }_n$ — bandwidth equals natural frequency, the reason engineers design to $\zeta \approx 0.7$.

Verify: Plug ${\zeta }^2=1/2$ into $x^2-2x(1-2{\zeta }^2)-1=0\Rightarrow x^2-1=0\Rightarrow x=1\Rightarrow \omega ={\omega }_n$ ✔.

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Example 3 — Cell C: over-damped, sluggish

1. $1-2{\zeta }^2=1-2(1)=-1$. Why? Now the term is negative — this is what pushes ${\omega }_B$ below ${\omega }_n$.
2. ${\omega }_B=40\sqrt{-1+\sqrt{(-1{)}^2+1}}=40\sqrt{-1+\sqrt{2}}=40\sqrt{-1+1.414}=40\sqrt{0.414}=25.7\text{rad/s}$. Why? The inner $\sqrt{(1-2{\zeta }^2{)}^2+1}$ keeps $x$ positive, but subtracting 1 shrinks it → bandwidth drops to $\approx 0.64{\omega }_n$.
3. Interpretation: a critically damped servo follows commands cleanly but only up to $25.7\text{rad/s}$ — slower than its own ${\omega }_n$. Why it matters: over-damping trades ringing for sluggishness; both extremes cost you.

Verify: Ratio $25.7/40=0.643$ — matches parent's "$\zeta =1\Rightarrow {\omega }_B\approx 0.64{\omega }_n$" ✔.

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Example 4 — Cells D & E: degenerate zero-damping and zero-delay limits

1. (a) Put $\zeta =0$: $1-2{\zeta }^2=1$, so ${\omega }_B=50\sqrt{1+\sqrt{1^2+1}}=50\sqrt{1+\sqrt{2}}=50\sqrt{2.414}=77.7\text{rad/s}.$ Why? Zero damping means the magnitude spikes at resonance, so the $-3$ dB point pushes far above ${\omega }_n$ (widest of any $\zeta$).
2. (b) $\Delta \varphi =-{\omega }_c\tau =-20\times 0=0$ rad $=0°$. Why this step? Phase lost is proportional to $\tau$. A perfect (zero) delay steals nothing — the degenerate baseline.
3. Sanity limit: as $\zeta \to 0$ we approach ${\omega }_B={\omega }_n\sqrt{1+\sqrt{2}}\approx 1.554{\omega }_n$, the largest possible ratio. Any real damping pulls it back down.

Verify: $\sqrt{1+\sqrt{2}}=1.5538$, times $50=77.69$ ✔. And $20\times 0=0$ trivially ✔.

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Example 5 — Cell F: delay eats the phase margin

1. Phase eaten $={\omega }_c\tau =25\times 0.024=0.60\text{rad}$. Why this step? A pure delay contributes phase $-\omega \tau$ (magnitude stays 1). Multiply frequency by delay.
2. Convert to degrees: $0.60\times 57.3=34.4°$. Why? Phase margin is quoted in degrees; $57.3=180/\pi$ converts.
3. Remaining margin $=50°-34.4°=15.6°$. Why it matters: Below the $\sim 30°$ safety floor — the vehicle will ring in gusts. Fix by reducing $\tau$ (faster computer) or lowering ${\omega }_c$, never by adding gain (see Ex 7).

Verify: $25\times 0.024=0.6$; $0.6\times 180/\pi =34.38°$; $50-34.38=15.62°$ ✔. Units: rad/s · s = rad ✔.

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Example 6 — Cell G: real-vehicle word problem

1. Required ${\omega }_n\ge 5{\omega }_c=5\times 18=90\text{rad/s}$. Hardware gives $80$. Why this step? The 5× rule keeps the servo's phase lag tiny compared to the loop's needs.
2. $80<90$: fails the rule of thumb (marginally). Why? Too-slow a servo lets its lag creep into the loop's stability budget.
3. Servo phase lag at ${\omega }_c$ (parent's small-angle estimate): $\Delta \varphi \approx 2\zeta \frac{{\omega }_c}{{\omega }_n}\times 57.3=2(0.7)\frac{18}{80}\times 57.3$. Why? For a lightly-loaded second-order lag well below ${\omega }_n$, phase $\approx 2\zeta (\omega /{\omega }_n)$ radians.
4. $=2(0.7)(0.225)(57.3)=0.315\times 57.3=18.0°$. Why it matters: 18° is above the ~16° the rule targets — confirming the hardware is slightly too slow. Either speed the servo or slow the autopilot.

Verify: $5\times 18=90$; $2\times 0.7\times 18/80=0.315$; $0.315\times 57.3=18.04°$ ✔. See also the trade-off with bending modes — you can't just crank ${\omega }_n$ up.

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Example 7 — Cell H: the exam trap "add gain to beat delay"

1. Original loss: $\Delta \varphi =15\times 0.02=0.30\text{rad}=17.2°$ (parent Ex 2). Margin left $=45-17.2=27.8°$. Why? Baseline to compare against.
2. After more gain, crossover rises to ${\omega }_c^{\prime }=24\text{rad/s}$. New loss $=24\times 0.02=0.48\text{rad}=27.5°$. Why this step? Delay's phase is $-\omega \tau$ — it grows linearly with crossover. Raising ${\omega }_c$ raises the toll.
3. Now margin $=45-27.5=17.5°$ — worse than the 27.8° we started with, and under the 30° floor. Why the fix fails: a pure delay has magnitude $1$; gain never fights it, it only pushes ${\omega }_c$ higher into more phase lag. The real cure: shrink $\tau$ (see computational delay) or lower bandwidth.

Verify: $24\times 0.02=0.48$; $0.48\times 57.3=27.5°$; $45-27.5=17.5°<27.8°$ (worse) ✔.

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Example 8 — Cell I: Padé wrong-way-first sign

1. (a) Zero where numerator $=0$: $1-\frac{\tau }{2}s=0\Rightarrow s=\frac{2}{\tau }=\frac{2}{0.04}=50$. Why this step? A positive real $s=50>0$ means a right-half-plane (RHP) zero — the mathematical mark of non-minimum-phase, "wrong-way-first" behaviour.
2. (b) Step response value at $t=0^{+}$: for $\frac{1-\frac{\tau }{2}s}{1+\frac{\tau }{2}s}$ the initial value (limit $s\to \infty$ of $s\cdot Y(s)$ for a step $1/s$) gives $\underset{s\to \infty }{\lim }\frac{1-\frac{\tau }{2}s}{1+\frac{\tau }{2}s}=\frac{-\frac{\tau }{2}}{+\frac{\tau }{2}}=-1$. Why? At the first instant the output jumps to $-1$ — the opposite direction of the commanded $+1$ before curving up to $+1$. Look at the red dip in the figure.
3. Final value: $\underset{s\to 0}{\lim }\frac{1-\frac{\tau }{2}s}{1+\frac{\tau }{2}s}=1$ — it does eventually reach the command, just after going backward first. Why it matters: that initial backward lurch is exactly the destabilising essence of transport delay, captured by a rational function root-locus tools can chew on.

Verify: RHP zero at $2/0.04=50>0$ ✔. Initial value $-1$, final value $+1$ ✔ (both computed above).

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