Intuition The one-sentence pictureThe flight computer asks for a nozzle angle δ c \delta_c δ c deliver that angle after a small delay and only if you don't ask it to move faster than
its bandwidth allows . TVC dynamics is the story of the gap between "commanded" and "actual"
nozzle deflection — and that gap is what eats your control margin.
Definition Thrust Vector Control (TVC)A rocket steers by gimballing its engine nozzle by an angle δ \delta δ T T T torque about the vehicle centre of mass.
The actuator (a hydraulic/electromechanical servo ) is what physically rotates the nozzle.
Definition Gimbal servo modelThe servo is the transfer function relating actual deflection δ \delta δ commanded
deflection δ c \delta_c δ c second-order lag plus a pure time delay :
δ ( s ) δ c ( s ) = e − s τ ω n 2 s 2 + 2 ζ ω n s + ω n 2 \frac{\delta(s)}{\delta_c(s)} = e^{-s\tau}\,\frac{\omega_n^2}{s^2 + 2\zeta\omega_n s + \omega_n^2} δ c ( s ) δ ( s ) = e − s τ s 2 + 2 ζ ω n s + ω n 2 ω n 2
WHY second order? A servo is a motor (torque) driving an inertia (nozzle mass) against a
spring-like load and damping — mass–spring–damper ⇒ \Rightarrow ⇒ WHY a pure delay e − s τ e^{-s\tau} e − s τ Sensor sampling, computation, digital-to-analog conversion,
and valve transport lag all shift the response in time without changing its shape.
Model the nozzle as a rotational inertia J J J M a M_a M a c δ ˙ c\dot\delta c δ ˙ k δ k\delta k δ
J δ ¨ + c δ ˙ + k δ = M a J\ddot\delta + c\dot\delta + k\delta = M_a J δ ¨ + c δ ˙ + k δ = M a
The servo commands torque proportional to the error between commanded and actual angle,
M a = K ( δ c − δ ) M_a = K(\delta_c - \delta) M a = K ( δ c − δ )
J δ ¨ + c δ ˙ + k δ = K ( δ c − δ ) J\ddot\delta + c\dot\delta + k\delta = K(\delta_c - \delta) J δ ¨ + c δ ˙ + k δ = K ( δ c − δ ) J δ ¨ + c δ ˙ + ( k + K ) δ = K δ c J\ddot\delta + c\dot\delta + (k+K)\delta = K\delta_c J δ ¨ + c δ ˙ + ( k + K ) δ = K δ c
Divide by J J J
( s 2 + c J s + k + K J ) δ ( s ) = K J δ c ( s ) \Big(s^2 + \tfrac{c}{J}s + \tfrac{k+K}{J}\Big)\,\delta(s) = \tfrac{K}{J}\,\delta_c(s) ( s 2 + J c s + J k + K ) δ ( s ) = J K δ c ( s )
Now define the natural frequency and damping by matching to the standard form:
So the pure lag part is
G servo ( s ) = ω n 2 s 2 + 2 ζ ω n s + ω n 2 . G_{\text{servo}}(s)=\frac{\omega_n^2}{s^2+2\zeta\omega_n s + \omega_n^2}. G servo ( s ) = s 2 + 2 ζ ω n s + ω n 2 ω n 2 .
ω B \omega_B ω B ∣ G ( j ω ) ∣ |G(j\omega)| ∣ G ( j ω ) ∣ 1 / 2 1/\sqrt2 1/ 2 − 3 -3 − 3 faithfully follow .
Compute ∣ G ( j ω ) ∣ 2 |G(j\omega)|^2 ∣ G ( j ω ) ∣ 2 s = j ω s=j\omega s = j ω s 2 = − ω 2 s^2=-\omega^2 s 2 = − ω 2 ∣ G ∣ 2 = ω n 4 ( ω n 2 − ω 2 ) 2 + ( 2 ζ ω n ω ) 2 |G|^2=\frac{\omega_n^4}{(\omega_n^2-\omega^2)^2+(2\zeta\omega_n\omega)^2} ∣ G ∣ 2 = ( ω n 2 − ω 2 ) 2 + ( 2 ζ ω n ω ) 2 ω n 4
Set = 1 / 2 =1/2 = 1/2 x = ( ω / ω n ) 2 x=(\omega/\omega_n)^2 x = ( ω / ω n ) 2 ( 1 − x ) 2 + 4 ζ 2 x = 2 ⇒ x 2 − 2 x ( 1 − 2 ζ 2 ) − 1 = 0 (1-x)^2+4\zeta^2 x = 2 \;\Rightarrow\; x^2 -2x(1-2\zeta^2)-1=0 ( 1 − x ) 2 + 4 ζ 2 x = 2 ⇒ x 2 − 2 x ( 1 − 2 ζ 2 ) − 1 = 0 x = ( 1 − 2 ζ 2 ) + ( 1 − 2 ζ 2 ) 2 + 1 x=(1-2\zeta^2)+\sqrt{(1-2\zeta^2)^2+1} x = ( 1 − 2 ζ 2 ) + ( 1 − 2 ζ 2 ) 2 + 1
Intuition WHY a delay is dangerousA pure delay changes phase , not amplitude. Control loops go unstable when total phase lag
hits − 180 ° -180° − 180° spends your phase margin.
A pure delay e − s τ e^{-s\tau} e − s τ ω \omega ω ∣ e − j ω τ ∣ = 1 |e^{-j\omega\tau}|=1 ∣ e − j ω τ ∣ = 1
∠ e − j ω τ = − ω τ (radians) \angle e^{-j\omega\tau} = -\omega\tau \quad\text{(radians)} ∠ e − j ω τ = − ω τ (radians)
Worked example Example 1 — Findω n , ζ , ω B \omega_n,\ \zeta,\ \omega_B ω n , ζ , ω B J = 0.8 kg⋅m 2 J=0.8\ \text{kg·m}^2 J = 0.8 kg⋅m 2 c = 40 N⋅m⋅s c=40\ \text{N·m·s} c = 40 N⋅m⋅s k + K = 2000 N⋅m/rad k+K=2000\ \text{N·m/rad} k + K = 2000 N⋅m/rad
ω n = ( k + K ) / J = 2000 / 0.8 = 2500 = 50 rad/s \omega_n=\sqrt{(k+K)/J}=\sqrt{2000/0.8}=\sqrt{2500}=50\ \text{rad/s} ω n = ( k + K ) / J = 2000/0.8 = 2500 = 50 rad/s Why this step? Directly from ω n 2 = ( k + K ) / J \omega_n^2=(k+K)/J ω n 2 = ( k + K ) / J ζ = c 2 ( k + K ) J = 40 2 2000 ⋅ 0.8 = 40 2 ⋅ 40 = 0.5 \zeta=\dfrac{c}{2\sqrt{(k+K)J}}=\dfrac{40}{2\sqrt{2000\cdot0.8}}=\dfrac{40}{2\cdot40}=0.5 ζ = 2 ( k + K ) J c = 2 2000 ⋅ 0.8 40 = 2 ⋅ 40 40 = 0.5 Why? 1600 = 40 \sqrt{1600}=40 1600 = 40 1 − 2 ζ 2 = 1 − 0.5 = 0.5 1-2\zeta^2=1-0.5=0.5 1 − 2 ζ 2 = 1 − 0.5 = 0.5 ω B = 50 0.5 + 0.25 + 1 = 50 0.5 + 1.118 = 50 1.618 = 63.6 rad/s \omega_B=50\sqrt{0.5+\sqrt{0.25+1}}=50\sqrt{0.5+1.118}=50\sqrt{1.618}=63.6\ \text{rad/s} ω B = 50 0.5 + 0.25 + 1 = 50 0.5 + 1.118 = 50 1.618 = 63.6 rad/s Why? Under-damped systems have ω B > ω n \omega_B>\omega_n ω B > ω n
Worked example Example 2 — How much phase does a delay eat?Loop crossover ω c = 15 rad/s \omega_c=15\ \text{rad/s} ω c = 15 rad/s τ = 20 ms \tau=20\ \text{ms} τ = 20 ms
Phase lost = ω c τ = 15 × 0.02 = 0.30 rad = 17.2 ° =\omega_c\tau=15\times0.02=0.30\ \text{rad}=17.2° = ω c τ = 15 × 0.02 = 0.30 rad = 17.2° Why this step? ∠ e − j ω c τ = − ω c τ \angle e^{-j\omega_c\tau}=-\omega_c\tau ∠ e − j ω c τ = − ω c τ → \to → × 57.3 \times57.3 × 57.3
If your design phase margin was 45 ° 45° 45° ≈ 28 ° \approx28° ≈ 28° Why it matters: under 30 ° 30° 30°
Worked example Example 3 — Bandwidth budget rule of thumbDesigners want servo bandwidth ω n ≳ 5 × \omega_n \gtrsim 5\times ω n ≳ 5 × ω c \omega_c ω c ω c = 15 \omega_c=15 ω c = 15 ω n ≳ 75 \omega_n\gtrsim75 ω n ≳ 75 Why? At ω c = ω n / 5 \omega_c=\omega_n/5 ω c = ω n /5 ≈ 2 ζ ( ω c / ω n ) ⋅ 57.3 \approx 2\zeta(\omega_c/\omega_n)\cdot57.3 ≈ 2 ζ ( ω c / ω n ) ⋅ 57.3 ≈ 2 ( 0.7 ) ( 0.2 ) ( 57.3 ) ≈ 16 ° \approx 2(0.7)(0.2)(57.3)\approx16° ≈ 2 ( 0.7 ) ( 0.2 ) ( 57.3 ) ≈ 16°
Common mistake "Delay just makes the response slower, so I'll add more gain."Why it feels right: more gain speeds up other lags, so surely it fixes delay too.
The fix: A pure delay is pure phase , magnitude = 1 =1 = 1 ω c \omega_c ω c increases ω c τ \omega_c\tau ω c τ more phase. Delay problems are fixed by reducing τ \tau τ lowering bandwidth, never by brute gain.
Common mistake "Higher servo bandwidth is always better."Why it feels right: high ω n \omega_n ω n The fix: If ω n \omega_n ω n structural bending / slosh frequencies, the servo
starts driving flexible modes — TVC becomes an oscillator. Bandwidth must sit below the first
bending mode (or you need a notch filter). Fast, but not too fast.
ω B \omega_B ω B ω n \omega_n ω n Why it feels right: for ζ = 0.7 \zeta=0.7 ζ = 0.7 The fix: They coincide only at ζ = 1 / 2 \zeta=1/\sqrt2 ζ = 1/ 2 ζ = 0.5 \zeta=0.5 ζ = 0.5 ω B ≈ 1.27 ω n \omega_B\approx1.27\omega_n ω B ≈ 1.27 ω n ζ = 1 \zeta=1 ζ = 1 ω B ≈ 0.64 ω n \omega_B\approx0.64\omega_n ω B ≈ 0.64 ω n ζ \zeta ζ
Recall Feynman: explain to a 12-year-oldImagine steering a boat with an oar. Your brain (the computer) says "turn the oar 10°."
But your arm (the servo) takes a moment to react and can't twist super fast. If you keep
yelling "left! right! left!" quicker than your arm can move, the boat wobbles and might tip.
So engineers make sure the arm is fast enough (bandwidth) and that the delay between
deciding and doing is tiny. TVC dynamics is just: don't ask the nozzle to do more than its
muscle can, and account for its reaction time.
"BAND-DELAY: Big-N Answers, Delayed Eats Loop's phase, Ask-Yes-below-bending."
B andwidth ↑ with N (ω n = ( k + K ) / J \omega_n=\sqrt{(k+K)/J} ω n = ( k + K ) / J DELAY E ats phase: Δ ϕ = − ω c τ \Delta\phi=-\omega_c\tau Δ ϕ = − ω c τ Keep bandwidth below bending modes.
What two effects make up the gimbal servo transfer function? A second-order lag
ω n 2 s 2 + 2 ζ ω n s + ω n 2 \frac{\omega_n^2}{s^2+2\zeta\omega_n s+\omega_n^2} s 2 + 2 ζ ω n s + ω n 2 ω n 2 times a pure time delay
e − s τ e^{-s\tau} e − s τ .
Why is a servo modelled as second order? It's a motor torque driving a nozzle inertia against damping and stiffness → a mass–spring–damper → 2nd-order ODE.
Give ω n \omega_n ω n ω n = ( k + K ) / J \omega_n=\sqrt{(k+K)/J} ω n = ( k + K ) / J — stiffness (plus loop gain) over inertia.
Give ζ \zeta ζ ζ = c 2 ( k + K ) J \zeta=\dfrac{c}{2\sqrt{(k+K)J}} ζ = 2 ( k + K ) J c .
At what ζ \zeta ζ ζ = 1 / 2 ≈ 0.707 \zeta=1/\sqrt2\approx0.707 ζ = 1/ 2 ≈ 0.707 , since then
1 − 2 ζ 2 = 0 1-2\zeta^2=0 1 − 2 ζ 2 = 0 gives
ω B = ω n \omega_B=\omega_n ω B = ω n .
How much phase does a delay τ \tau τ ω \omega ω − ω τ -\omega\tau − ω τ radians (
= − 57.3 ω τ =-57.3\,\omega\tau = − 57.3 ω τ degrees); magnitude stays 1.
Why can't you fix delay problems by adding gain? Gain doesn't change delay's phase; it raises the crossover
ω c \omega_c ω c , increasing
ω c τ \omega_c\tau ω c τ and losing MORE phase margin.
What is the first-order Padé approximation of e − s τ e^{-s\tau} e − s τ 1 − ( τ / 2 ) s 1 + ( τ / 2 ) s \dfrac{1-(\tau/2)s}{1+(\tau/2)s} 1 + ( τ /2 ) s 1 − ( τ /2 ) s ; it introduces a RHP zero at
s = 2 / τ s=2/\tau s = 2/ τ .
Why does a RHP zero matter physically? It makes the output initially move the wrong way — the destabilising signature of transport delay.
Rule of thumb for servo bandwidth vs control-loop crossover? Make
ω n ≳ 5 ω c \omega_n\gtrsim5\,\omega_c ω n ≳ 5 ω c so servo lag looks negligible to the rigid-body loop.
Why must servo bandwidth stay below the first bending mode? Otherwise TVC drives flexible structural/slosh modes, exciting oscillations (needs a notch filter).
Second-order systems — natural frequency & damping Bode plots & phase margin Padé approximation of transport delay Structural bending modes & notch filters Rigid-body attitude control loop (autopilot) Digital control — sampling & computational delay Thrust Vector Control geometry & torque
from sampling and transport lag
Flight computer commands delta_c
Mass-spring-damper nozzle J c k
Proportional inner loop M_a=K delta_c minus delta
Natural frequency omega_n
Gap between commanded and actual
Intuition Hinglish mein samjhoDekho, rocket steer karta hai apne engine nozzle ko thoda gimbal (tilt) karke. Flight computer
bolta hai "nozzle ko δ c \delta_c δ c bandwidth — matlab
woh kitni tezi se follow kar sakta hai. Yeh ek second-order system hai (mass-spring-damper jaisa),
jiska natural frequency ω n = ( k + K ) / J \omega_n=\sqrt{(k+K)/J} ω n = ( k + K ) / J ω n \omega_n ω n
Doosri, aur zyada khatarnaak, problem hai time delay τ \tau τ e − s τ e^{-s\tau} e − s τ phase ko kha jaata hai: har frequency par − ω τ -\omega\tau − ω τ − 180 ° -180° − 180°
Ek badi galti jo log karte hain: "delay slow kar raha hai, to gain badha do." Galat! Gain
badhane se crossover frequency ω c \omega_c ω c ω c τ \omega_c\tau ω c τ τ \tau τ ω n \omega_n ω n
