TVC dynamics — gimbal servo bandwidth, time delay: Question bank

Before the questions, let's pin down every symbol these traps rely on, because the whole point of a concept trap is that it hides in an undefined letter.

Definition The nomenclature — read this first

$ω$ (Greek "omega") ::: a frequency — how many radians per second an oscillating command wiggles through. Big $ω$ = fast wiggling. Every " $ω$ -something" below is a specific frequency we care about.
$s$ ::: the Laplace variable, a bookkeeping stand-in for "the operation of differentiating in time." A transfer function written in $s$ lets us treat a differential equation like ordinary algebra. When we test a steady oscillation of frequency $ω$ , we set $s = j ω$ .
$j$ ::: the imaginary unit, $j = - 1$ (engineers write $j$ , not $i$ ). It is the machinery that turns " $s$ " into a rotation, so $e^{- j ω τ}$ is a pure phase-shift of size $ω τ$ with no change in size.
$ω_{n}$ ::: the natural frequency — how fast the servo can swing the nozzle; $ω_{n} = (k + K) / J$ .
$ω_{B}$ ::: the bandwidth — the highest frequency the servo can still faithfully follow, defined as where its gain has fallen to $1/ 2$ ( $- 3$ dB) of the DC value.
$ω_{c}$ ::: the loop's gain-crossover frequency — where the open-loop gain passes through 1; it's the frequency at which stability is decided (see Bode plots & phase margin).
$ζ$ ::: the damping ratio — rings ( $ζ < 1$ ), settles cleanly ( $ζ \approx 0.7$ ), crawls ( $ζ > 1$ ).
$τ$ ::: the pure time delay — seconds between the computer deciding an angle and the nozzle starting to move.

The figure below is the mental picture for the whole page — glance back at it whenever a trap mentions "delay," "phase," or "magnitude."

Figure — TVC dynamics — gimbal servo bandwidth, time delay

Recall What to read off the figure (alt text / caption)

Left panel ::: A yellow commanded step in nozzle angle; the blue actual response starts only after the red band (the delay $τ$ ), then rings before settling — showing the "commanded vs actual" gap. Right panel ::: For a pure delay, the green magnitude curve stays pinned at 1 for all frequencies while the red phase curve slides steadily downward ( $- ω τ$ ) — the visual proof that delay touches phase only, never amplitude.

True or false — justify

True or false: A pure delay

e^{- s τ}

makes the servo's output smaller at high frequency.

False. Its magnitude is

∣ e^{- j ω τ} ∣ = 1

at every frequency — it touches only phase, never amplitude (right panel of the figure).

True or false: For a delay, the phase lag

ω τ

grows without bound as frequency rises.

True.

∠ e^{- j ω τ} = - ω τ

is linear in

ω

, so it keeps sliding more negative forever — unlike a second-order lag which levels off at

- 180°

True or false: Bandwidth

ω_{B}

always equals natural frequency

ω_{n}

False. They coincide only at

ζ = 1/ 2 \approx 0.707

. For

ζ = 0.5

ω_{B} \approx 1.27 ω_{n}

; for

ζ = 1

ω_{B} \approx 0.64 ω_{n}

True or false: Raising loop gain is a valid cure for a delay-induced stability problem.

False. Gain lifts the crossover frequency

ω_{c}

, and phase lost is

ω_{c} τ

— a higher

ω_{c}

spends more phase, making it worse.

True or false: A higher servo bandwidth is unconditionally good for the vehicle.

False. If

ω_{n}

climbs into the structural bending or slosh frequencies, the servo starts exciting flexible modes and TVC turns into an oscillator — see Structural bending modes & notch filters.

True or false: The first-order Padé approximation of a delay has a right-half-plane zero.

True. It places a zero at

s = + 2/ τ

; that RHP zero is the mathematical fingerprint of "output first moves the wrong way," the destabilising essence of delay — see Padé approximation of transport delay.

True or false: An under-damped servo (

ζ < 0.707

) has bandwidth below its natural frequency.

False. Under-damped systems resonate slightly before rolling off, so they still track faithfully past

ω_{n}

; hence

ω_{B} > ω_{n}

for

ζ < 0.707

True or false: At

ζ = 0.707

the servo's frequency response has a flat resonant peak (no overshoot in magnitude).

True.

ζ = 1/ 2

is the maximally-flat (Butterworth) condition where the magnitude curve rolls off with no peak, which is exactly why it's the design sweet spot.

True or false: If we double the actuator authority

K

(with

K ≫ k

), the DC gain

K / (k + K)

moves further from unity.

False. Larger

K

drives

K / (k + K)

toward 1, which is why a strong inner loop is assumed to have unit steady gain.

Spot the error

"The delay eats

ω_{c} τ

radians of gain margin." — what's wrong?

It eats phase margin, not gain margin. Delay's magnitude is 1, so it can't touch gain;

ω_{c} τ

is a phase angle.

"Since

ζ = 0.7

gives

ω_{B} = ω_{n}

, I'll just always assume they're equal." — what's wrong?

The equality holds only at that one value of

ζ

. Always check the actual damping ratio first, because at

ζ = 0.5

the bandwidth is 27% higher than

ω_{n}

"A delay of 20 ms is fine because 20 ms is tiny." — what's the missing consideration?

What matters is

ω_{c} τ

, not

τ

alone. At

ω_{c} = 15

rad/s that "tiny" delay costs

0.30

rad

\approx 17°

of phase — a big bite from a

45°

margin.

"To make the servo faster I only need to reduce its inertia

J

." — what did this overlook?

ω_{n} = (k + K) / J

, so raising stiffness

k + K

works too; and speed alone is not the goal — you must keep

ω_{n}

below the first bending mode.

e^{- s τ}

can be dropped from root-locus analysis because it's just a shift." — what's wrong?

Root-locus needs a rational transfer function; the transcendental

e^{- s τ}

must first be replaced by a Padé approximation, which introduces real poles/zeros that do alter stability.

"We set

ζ

to 0.3 for a snappy servo." — why might a designer object?

ζ = 0.3

is lightly damped: it overshoots and rings, and its resonant peak can amplify commands near

ω_{n}

, feeding energy into flexible modes. The

\approx 0.7

sweet spot avoids the peak.

Why questions

Why is a pure delay considered more dangerous to stability than an equivalent-looking slow lag?

Because its phase lag

ω τ

grows without limit, so at high frequency it can push total phase past

- 180°

where a bounded lag never could, silently crossing the instability threshold.

Where do the bandwidth factors like

1.27

(

ζ = 0.5

) and

0.64

(

ζ = 1

) come from?

From the closed form

ω_{B} = ω_{n} (1 - 2 ζ^{2}) + (1 - 2 ζ^{2})^{2} + 1

: plugging

ζ = 0.5

gives the factor

1.27

, and

ζ = 1

gives

0.64

. Intuitively, less damping lets the servo resonate past

ω_{n}

(factor

> 1

); heavy damping rolls it off early (factor

< 1

Why do designers target

ω_{n} ≳ 5 ω_{c}

(servo bandwidth well above loop crossover)?

Near crossover the second-order servo's own phase lag is

arctan (\frac{2 ζ ( ω _{c} / ω _{n} )}{1 - ( ω _{c} / ω _{n} ) ^{2}})

, which for a small ratio

ω_{c} / ω_{n}

reduces to about

2 ζ (ω_{c} / ω_{n})

rad. With

ζ \approx 0.7

and ratio

0.2

that's roughly

16°

— small enough that the servo looks "instantaneous" to the rigid-body loop.

Why is the servo modelled as second order rather than first?

It is physically a motor torque driving a nozzle inertia

J

against damping

c

and stiffness

k

— a mass–spring–damper, which is inherently second order (see Second-order systems — natural frequency & damping).

Why does adding gain to fight a delay backfire?

Gain raises the crossover frequency

ω_{c}

, and phase lost is

ω_{c} τ

; a bigger

ω_{c}

multiplies the same

τ

into a larger phase loss, eroding margin further.

Why does the Padé form use

\frac{1 - \frac{τ}{2} s}{1 + \frac{τ}{2} s}

rather than a simple pole

\frac{1}{1 + τ s}

Both match the Taylor series of

e^{- s τ}

to first order, but only the Padé form reproduces the right-half-plane zero — the "goes the wrong way first" behaviour that a lone pole would completely miss.

Why must servo bandwidth sit below the first structural bending frequency?

Because a fast servo responds to frequencies inside its passband; if bending resonances fall in that band, the actuator drives them, and TVC couples to the flexible mode instead of steering — a margin-destroying feedback loop.

Edge cases

Edge case: What happens to bandwidth as

ζ \to 0

(no damping)?

The formula gives

ω_{B} = ω_{n} 1 + 2 \approx 1.55 ω_{n}

, and the resonant peak grows toward infinity — the "bandwidth" number becomes misleading because the response wildly amplifies near

ω_{n}

Edge case: What is the phase eaten by delay at zero frequency (

ω \to 0

Zero —

ω τ \to 0

, so a steady (DC) command sees no phase penalty at all; delay only bites the fast, oscillatory commands.

Edge case: If

τ = 0

(no delay), does the Padé zero still exist?

No — the zero sits at

s = 2/ τ

, which runs off to infinity as

τ \to 0

, so the wrong-way behaviour vanishes and the delay term becomes

1

Edge case: For an over-damped servo (

ζ > 1

), what happens to bandwidth relative to

ω_{n}

It falls below

ω_{n}

(e.g.

ω_{B} \approx 0.64 ω_{n}

ζ = 1

); the servo is sluggish and rolls off earlier than its natural frequency, so it can't even track up to

ω_{n}

Edge case: What if the digital sampling rate is the dominant delay source?

Then

τ

is roughly one sample period (plus computation), so faster sampling directly shrinks

τ

and buys back phase margin — see Digital control — sampling & computational delay.

Edge case: In the limit

K ≫ k

, what does the servo DC gain become and why does it matter?

It approaches

K / (k + K) \to 1

, meaning the nozzle settles exactly at the commanded angle in steady state — the unit-gain assumption used everywhere downstream in Thrust Vector Control geometry & torque.

Recall One-line summary to lock in

Which quantity does a pure delay change, and which does it leave untouched? ::: It changes phase (by $- ω τ$ ) and leaves magnitude untouched — that's the whole reason it's a silent margin-eater.