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

3.5.45 · D3 · Physics › Guidance, Navigation & Control (GNC) › TVC dynamics — gimbal servo bandwidth, time delay

Ye page ek drill sheet hai. Parent note ne machinery banayi thi: second-order servo lag, bandwidth formula, aur phase khaane wala delay. Yahan hum har tarah ka case us machinery pe daalnge — normal numbers, edge cases, zeros, limits, ek word problem, aur ek exam trap — aur har ek ko poori tarah grind karenge.

Shuru karne se pehle, teen tools ka ek reminder jo hum baar baar use karte hain, taaki koi symbol bina wajah na aaye:

Recall Teen formulas jo hum reuse karte hain (parent se)

• Natural frequency & damping: ${\omega }_n=\sqrt{\frac{k+K}{J}}$, $\zeta =\frac{c}{2\sqrt{(k+K)J}}$. ${\omega }_n$ hai nozzle kitni tezi se jhoom sakta hai; $\zeta$ hai kitna ring karta hai (0 = pure ringing, 1 = koi overshoot nahi).
• Bandwidth: ${\omega }_B={\omega }_n\sqrt{(1-2{\zeta }^2)+\sqrt{(1-2{\zeta }^2{)}^2+1}}$. Sabse tezi wali wiggle jise servo abhi bhi full size ke $1/\sqrt{2}$ ke andar follow karta hai.
• Phase jo ek pure delay $\tau$ khata hai frequency $\omega$ par: $\Delta \varphi =-\omega \tau$ radians $=-57.3\omega \tau$ degrees. ($57.3=180/\pi$ bas radians→degrees ka converter hai.)

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Scenario matrix

Is topic mein jo bhi problem poochi ja sakti hai woh in cells mein se kisi ek mein aati hai. Neeche ke examples tagged hain us cell se jo wo cover karte hain.

Cell	Kya khaas hai	Covered by
A · Nominal under-damped	$0<\zeta <1$, ordinary numbers, ${\omega }_B>{\omega }_n$	Ex 1
B · Sweet-spot damping	$\zeta =1/\sqrt{2}\approx 0.707$ to ${\omega }_B={\omega }_n$ exactly	Ex 2
C · Over-damped / critical	$\zeta \ge 1$, sluggish, ${\omega }_B<{\omega }_n$	Ex 3
D · Degenerate: zero delay	$\tau =0$ — delay kuch nahi khata	Ex 4 (part)
E · Limiting: zero damping	$\zeta \to 0$ — resonance, bandwidth blow up ho jaata hai	Ex 4
F · Delay phase-budget	diya hua ${\omega }_c,\tau$: kitna margin gaya	Ex 5
G · Word problem (real vehicle)	hardware ko numbers mein translate karo, decide karo safe/unsafe	Ex 6
H · Exam twist: "fix delay ke liye gain badhao"	trap — gain badhane se delay aur bura ho jaata hai	Ex 7
I · Padé wrong-way sign	first-order Padé RHP-zero, output pehle galat direction mein jaata hai	Ex 8

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Example 1 — Cell A: nominal under-damped servo

1. ${\omega }_n=\sqrt{(k+K)/J}=\sqrt{1800/0.5}=\sqrt{3600}=60\text{rad/s}$. Ye step kyun? ${\omega }_n$ sirf stiffness-to-inertia ratio se fix hota hai — "spring vs. mass" ki contest swing speed set karti hai.
2. $\zeta =\frac{c}{2\sqrt{(k+K)J}}=\frac{30}{2\sqrt{1800\cdot 0.5}}=\frac{30}{2\sqrt{900}}=\frac{30}{60}=0.5$. Kyun? $\zeta$ actual damping $c$ ko critical amount $2\sqrt{(k+K)J}$ se compare karta hai. Yahan ye exactly aadha hai, to ring karta hai.
3. $1-2{\zeta }^2=1-2(0.25)=0.5$. Phir ${\omega }_B=60\sqrt{0.5+\sqrt{0.5^2+1}}=60\sqrt{0.5+\sqrt{1.25}}=60\sqrt{0.5+1.118}=60\sqrt{1.618}=76.3\text{rad/s}.$ Kyun? Under-damped systems roll off hone se pehle thoda resonate karte hain, to $-3$ dB point ${\omega }_n$ se upar hota hai.

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

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

1. ${\omega }_n=60\text{rad/s}$ (unchanged — sirf $k+K$ aur $J$ par depend karta hai). Kyun? Damping ${\omega }_n$ ko affect nahi karta; ye sirf ringing ko affect karta hai.
2. $\zeta =1/\sqrt{2}\approx 0.7071$ by construction. Kyun? Humne choose kiya $c$ ko is value tak paunchane ke liye; ye classic control target hai.
3. $1-2{\zeta }^2=1-2\cdot \frac{1}{2}=0$. To ${\omega }_B=60\sqrt{0+\sqrt{0+1}}=60\sqrt{1}=60\text{rad/s}={\omega }_n.$ Kyun? Jab $1-2{\zeta }^2=0$ hota hai to formula ${\omega }_B={\omega }_n$ tak collapse ho jaata hai — bandwidth natural frequency ke barabar hoti hai, isliye engineers $\zeta \approx 0.7$ design karte hain.

Verify: ${\zeta }^2=1/2$ ko $x^2-2x(1-2{\zeta }^2)-1=0\Rightarrow x^2-1=0\Rightarrow x=1\Rightarrow \omega ={\omega }_n$ mein plug karo ✔.

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Example 3 — Cell C: over-damped, sluggish

1. $1-2{\zeta }^2=1-2(1)=-1$. Kyun? Ab ye term negative hai — yahi cheez ${\omega }_B$ ko ${\omega }_n$ se neeche push karti hai.
2. ${\omega }_B=40\sqrt{-1+\sqrt{(-1{)}^2+1}}=40\sqrt{-1+\sqrt{2}}=40\sqrt{-1+1.414}=40\sqrt{0.414}=25.7\text{rad/s}$. Kyun? Inner $\sqrt{(1-2{\zeta }^2{)}^2+1}$ $x$ ko positive rakhta hai, lekin 1 ghatane se ye chhota ho jaata hai → bandwidth $\approx 0.64{\omega }_n$ tak girti hai.
3. Interpretation: critically damped servo commands ko saaf follow karta hai lekin sirf $25.7\text{rad/s}$ tak — apne ${\omega }_n$ se bhi slower. Kyun matter karta hai: over-damping ringing ke badle sluggishness leta hai; dono extremes cost karte hain.

Verify: Ratio $25.7/40=0.643$ — parent ke "$\zeta =1\Rightarrow {\omega }_B\approx 0.64{\omega }_n$" se match karta hai ✔.

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

1. (a) $\zeta =0$ rakho: $1-2{\zeta }^2=1$, to ${\omega }_B=50\sqrt{1+\sqrt{1^2+1}}=50\sqrt{1+\sqrt{2}}=50\sqrt{2.414}=77.7\text{rad/s}.$ Kyun? Zero damping matlab magnitude resonance par spike karti hai, to $-3$ dB point ${\omega }_n$ se kaafi upar push hota hai (kisi bhi $\zeta$ se wider).
2. (b) $\Delta \varphi =-{\omega }_c\tau =-20\times 0=0$ rad $=0°$. Ye step kyun? Phase lost $\tau$ ke proportional hoti hai. Ek perfect (zero) delay kuch nahi leta — degenerate baseline.
3. Sanity limit: jaise $\zeta \to 0$ hum ${\omega }_B={\omega }_n\sqrt{1+\sqrt{2}}\approx 1.554{\omega }_n$ tak paunchte hain, jo sabse bada possible ratio hai. Koi bhi real damping ise neeche kheench laata hai.

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

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Example 5 — Cell F: delay phase margin khata hai

1. Phase eaten $={\omega }_c\tau =25\times 0.024=0.60\text{rad}$. Ye step kyun? Ek pure delay phase $-\omega \tau$ contribute karta hai (magnitude 1 rehti hai). Frequency ko delay se multiply karo.
2. Degrees mein convert karo: $0.60\times 57.3=34.4°$. Kyun? Phase margin degrees mein quote ki jaati hai; $57.3=180/\pi$ convert karta hai.
3. Remaining margin $=50°-34.4°=15.6°$. Kyun matter karta hai: $\sim 30°$ safety floor se neeche — vehicle gusts mein ring karega. Fix karo $\tau$ ghatake (faster computer) ya ${\omega }_c$ kam karke, kabhi bhi gain add karke nahi (Ex 7 dekho).

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 deta hai $80$. Ye step kyun? 5× rule servo ke phase lag ko loop ki needs ke comparison mein tiny rakhta hai.
2. $80<90$: rule of thumb fail karta hai (marginally). Kyun? Bahut slow servo uski lag ko loop ke stability budget mein creep karne deta hai.
3. Servo phase lag ${\omega }_c$ par (parent ka small-angle estimate): $\Delta \varphi \approx 2\zeta \frac{{\omega }_c}{{\omega }_n}\times 57.3=2(0.7)\frac{18}{80}\times 57.3$. Kyun? ${\omega }_n$ se kaafi neeche lightly-loaded second-order lag ke liye, phase $\approx 2\zeta (\omega /{\omega }_n)$ radians hoti hai.
4. $=2(0.7)(0.225)(57.3)=0.315\times 57.3=18.0°$. Kyun matter karta hai: 18° ~16° se upar hai jo rule target karta hai — confirming karta hai ki hardware thoda slow hai. Ya to servo speed up karo ya autopilot slow karo.

Verify: $5\times 18=90$; $2\times 0.7\times 18/80=0.315$; $0.315\times 57.3=18.04°$ ✔. Bending modes ke saath trade-off bhi dekho — tum bas ${\omega }_n$ ko crank up nahi kar sakte.

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Example 7 — Cell H: exam trap "delay fix karne ke liye gain badhao"

1. Original loss: $\Delta \varphi =15\times 0.02=0.30\text{rad}=17.2°$ (parent Ex 2). Margin bacha $=45-17.2=27.8°$. Kyun? Compare karne ke liye baseline.
2. Zyada gain ke baad, crossover ${\omega }_c^{\prime }=24\text{rad/s}$ tak rise karta hai. New loss $=24\times 0.02=0.48\text{rad}=27.5°$. Ye step kyun? Delay ki phase $-\omega \tau$ hai — ye crossover ke saath linearly badhti hai. ${\omega }_c$ badhane se toll badhta hai.
3. Ab margin $=45-27.5=17.5°$ — jo shuru ke $27.8°$ se worse hai, aur $30°$ floor ke neeche. Kyun fix fail hota hai: ek pure delay ki magnitude $1$ hoti hai; gain kabhi isse fight nahi karta, ye sirf ${\omega }_c$ ko aur phase lag mein push karta hai. Asli ilaaj: $\tau$ ghatao (dekho computational delay) ya bandwidth kam karo.

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 wahan jahan numerator $=0$: $1-\frac{\tau }{2}s=0\Rightarrow s=\frac{2}{\tau }=\frac{2}{0.04}=50$. Ye step kyun? Ek positive real $s=50>0$ matlab ek right-half-plane (RHP) zero — non-minimum-phase, "wrong-way-first" behaviour ka mathematical mark.
2. (b) Step response ki value $t=0^{+}$ par: $\frac{1-\frac{\tau }{2}s}{1+\frac{\tau }{2}s}$ ke liye initial value (step $1/s$ ke liye $s\to \infty$ of $s\cdot Y(s)$ ki limit) deti hai $\underset{s\to \infty }{\lim }\frac{1-\frac{\tau }{2}s}{1+\frac{\tau }{2}s}=\frac{-\frac{\tau }{2}}{+\frac{\tau }{2}}=-1$. Kyun? Pehle instant mein output $-1$ pe jump karta hai — commanded $+1$ ki opposite direction mein, $+1$ tak curve karne se pehle. Figure mein red dip dekho.
3. Final value: $\underset{s\to 0}{\lim }\frac{1-\frac{\tau }{2}s}{1+\frac{\tau }{2}s}=1$ — ye eventually command tak paunchta hai, bas pehle backward jaane ke baad. Kyun matter karta hai: woh initial backward lurch exactly transport delay ka destabilising essence hai, jo ek rational function mein capture hota hai jise root-locus tools process kar sakte hain.

Verify: RHP zero $2/0.04=50>0$ par ✔. Initial value $-1$, final value $+1$ ✔ (dono upar compute kiye gaye hain).

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