Exercises — TVC dynamics — gimbal servo bandwidth, time delay

KYA karte hain: standard denominator se term-by-term match karte hain. KYUN: standard form ek template hai — numbers named slots mein fit ho jaate hain, koi derivation nahi chahiye.

• Constant term ${\omega }_n^2=2500\Rightarrow {\omega }_n=\sqrt{2500}===50\text{rad/s}==$.
• Middle term $2\zeta {\omega }_n=50\Rightarrow \zeta =\frac{50}{2\times 50}===0.5==$.

Numerator ${\omega }_n^2=2500$ ke barabar hai, jo unit DC gain confirm karta hai ($s=0$ par, $G=2500/2500=1$).

KYA/KYUN: $j\omega$ axis par (yaani $s=j\omega$ par frequency response evaluate karte hue) ek pure delay ka magnitude exactly 1 hota hai — yeh sirf phase rotate karta hai. Yeh qualifier important hai: imaginary axis se door, general complex-$s$ plane mein, $|e^{-s\tau }|$ 1 nahi hota. Frequency-response magnitude aur general-$s$ magnitude alag cheezein hain; "hamesha 1" wala claim sirf $j\omega$ axis ke baare mein hai.

• (a) $|e^{-j\omega \tau }|===1==$ kisi bhi real $\omega$ ke liye (exponent purely imaginary hai, isliye yeh unit-modulus rotation hai).
• (b) Phase $=-\omega \tau =-30\times 0.01=-0.30\text{rad}$. Degrees mein $=-0.30\times 57.2958===-17.2^{\circ }==$.

Step 1 — ${\omega }_n$ (KYUN pehle: yeh baaki sab ke liye scale set karta hai). ${\omega }_n=\sqrt{\frac{k+K}{J}}=\sqrt{\frac{4500}{0.5}}=\sqrt{9000}=94.87\text{rad/s}.$

Step 2 — $\zeta$. $\zeta =\frac{c}{2\sqrt{(k+K)J}}=\frac{30}{2\sqrt{4500\times 0.5}}=\frac{30}{2\sqrt{2250}}=\frac{30}{2\times 47.43}=\mathrm{0.316.}$

Step 3 — ${\omega }_B$. Compute $1-2{\zeta }^2=1-2(0.316{)}^2=1-0.2=0.8$. ${\omega }_B=94.87\sqrt{0.8+\sqrt{0.8^2+1}}=94.87\sqrt{0.8+\sqrt{1.64}}=94.87\sqrt{0.8+1.2806}.$ ${\omega }_B=94.87\sqrt{2.0806}=94.87\times 1.4425===136.8\text{rad/s}==.$

Kyunki $\zeta <1/\sqrt{2}$ hai, system lightly damped hai, isliye ${\omega }_B>{\omega }_n$ (roll off hone se pehle resonate karta hai).

KYA/KYUN: delay phase $\Delta \varphi =-{\omega }_c\tau$ hai jo crossover ${\omega }_c$ par evaluate hota hai (upar define kiya), kyunki crossover woh jagah hai jahan $-180^{\circ }$ stability limit judge ki jaati hai. $\Delta \varphi =-{\omega }_c\tau =-20\times 0.025=-0.50\text{rad}=-0.50\times 57.2958=-28.6^{\circ }.$ Loss ki size (jo hum "kharach karte hain") $==28.6^{\circ }==$ hai — same magnitude, positive kyunki hum quote kar rahe hain kitna margin consume hua. Remaining margin $=50^{\circ }-28.6^{\circ }===21.4^{\circ }==$ — thin hai ($30^{\circ }$ comfort line se neeche).

Step 1 — $\zeta =0.707$ identity use karo. $\zeta =1/\sqrt{2}$ par, $1-2{\zeta }^2=0$, isliye ${\omega }_B={\omega }_n\sqrt{0+\sqrt{0+1}}={\omega }_n.$ Isliye seedha ${\omega }_n=120\text{rad/s}$. Exactly isliye hum 0.707 par design karte hain — algebra collapse ho jaata hai.

Step 2 — stiffness back out karo. ${\omega }_n^2=(k+K)/J$ se: $k+K={\omega }_n^{2}J=120^2\times 0.6=14400\times 0.6===8640\text{N⋅m/rad}==.$

KYA karte hain: har $\zeta$ ko bandwidth radical mein daalo. KYUN: hum dekhna chahte hain ki damping change hone par ${\omega }_B$, ${\omega }_n$ ke relative kaise move karta hai — ek variable ($\zeta$) sweep, baaki sab fixed, isliye result sirf damping ka effect isolate karta hai.

${\omega }_B={\omega }_n\sqrt{(1-2{\zeta }^2)+\sqrt{(1-2{\zeta }^2{)}^2+1}}$ use karo.

• $\zeta =0.5$: KYA — $1-2{\zeta }^2=0.5$ (positive, isliye light damping ${\omega }_B$ ko ${\omega }_n$ se upar push karta hai). ${\omega }_B=100\sqrt{0.5+\sqrt{1.25}}=100\sqrt{0.5+1.118}=100\sqrt{1.618}===127.2\text{rad/s}==$.
• $\zeta =0.707$: KYA — $1-2{\zeta }^2=0$, term vanish ho jaata hai aur radical collapse hokar 1 ho jaata hai. ${\omega }_B=100\sqrt{0+1}===100\text{rad/s}==$.
• $\zeta =1.0$: KYA — $1-2{\zeta }^2=-1$ (negative, isliye heavy damping ${\omega }_B$ ko ${\omega }_n$ se neeche kheeench leta hai). ${\omega }_B=100\sqrt{-1+\sqrt{2}}=100\sqrt{-1+1.4142}=100\sqrt{0.4142}===64.4\text{rad/s}==$.

Trend: jaise jaise damping badhti hai, ${\omega }_B$ girta hai — heavy damping following-speed calm settling ke liye trade kar deta hai.

Figure (s01) padhna: horizontal axis damping ratio $\zeta$ hai aur vertical axis bandwidth ratio ${\omega }_B/{\omega }_n$ hai. Red curve woh ratio hai; yeh 1 se upar start hoti hai (light damping, fast following), dashed horizontal line ${\omega }_B/{\omega }_n=1$ ko exactly dotted vertical $\zeta =1/\sqrt{2}$ par cross karti hai, aur neeche continue karti hai. Teen black dots hamare teen answers hain ($1.27$, $1.00$, $0.64$) — visual proof hai ki ${\omega }_B={\omega }_n$ special isliye hai kyunki humne choose kiya us crossing point par land karna.

Delay ka phase contribution $\Delta \varphi =-{\omega }_c\tau$ hai (ek lag, isliye negative); hum iske magnitude ko loss ke roop mein quote karte hain.

• Pehle: $|\Delta {\varphi }_1|={\omega }_c\tau =15\times 0.02=0.30\text{rad}=0.30\times 57.2958===17.2^{\circ }==$.
• Baad mein: $|\Delta {\varphi }_2|={\omega }_c^{\prime }\tau =25\times 0.02=0.50\text{rad}=0.50\times 57.2958===28.6^{\circ }==$.

Synthesis point: gain badhane se ${\omega }_c$ badha, aur kyunki delay phase magnitude ${\omega }_c$ ke proportional hai, loss $17.2^{\circ }$ se $28.6^{\circ }$ tak badh gaya — gain ne ise aur bura kar diya. $j\omega$ axis par $e^{-j\omega \tau }$ ka magnitude har frequency par 1 hai, isliye gain delay ko "outrun" nahi kar sakta; sirf $\tau$ kam karna (faster computer / higher sample rate, dekho Digital control — sampling & computational delay) ya bandwidth lower karna help karta hai.

$\frac{\tau }{2}=0.02\text{s}$.

• Zero: numerator $1-0.02s=0\Rightarrow s=+\frac{2}{\tau }=+\frac{2}{0.04}===+50{\text{s}}^{-1}==$. Yeh right half-plane mein baithta hai (positive real part).
• Pole: denominator $1+0.02s=0\Rightarrow s=-\frac{2}{\tau }===-50{\text{s}}^{-1}==$ (left half-plane, stable).

$+50$ par right-half-plane zero fingerprint hai: yeh output ko pehle galat direction mein move karta hai (undershoot), jo transport delay ka destabilising essence hai. Higher orders ke liye dekho Padé approximation of transport delay.

Figure (s02) padhna: horizontal axis $\text{Re}(s)$ hai aur vertical axis $\text{Im}(s)$ hai — complex $s$-plane. Dayi taraf shaded band right half-plane mark karta hai. $-50$ par black "$\times$" harmless stable pole hai; red open circle $+50$ par zero hai — ise shaded (right-half) region ke andar rakhna exactly woh picture hai "output pehle galat taraf jaata hai." Imaginary axis ke baare mein mirror symmetry isliye hai kyunki first-order Padé magnitude flat rakhta hai jabki pure phase lag inject karta hai.

Step 1 — bandwidth floor (Req 1). Chahiye ${\omega }_B={\omega }_n\ge 5\times 12=60\text{rad/s}$.

Step 2 — bending ceiling (Req 2). Chahiye ${\omega }_n<180\text{rad/s}$; ek common rule ise well below rakhta hai, jaise $\le 90$. ${\omega }_n=80\text{rad/s}$ ka choice $[60,90]$ window mein comfortably baithta hai.

Step 3 — delay check (Req 3). Crossover par delay phase magnitude: $|\Delta \varphi |={\omega }_c\tau =12\times 0.015=0.18\text{rad}=0.18\times 57.2958===10.3^{\circ }==\le 12^{\circ }.✓$ (Note karo yeh sirf ${\omega }_c$ aur $\tau$ par depend karta hai, ${\omega }_n$ par nahi — isliye ${\omega }_n$ pick se independent pass karta hai.)

Step 4 — hardware from ${\omega }_n=80$, $\zeta =0.707$, $J=0.7$. $k+K={\omega }_n^{2}J=80^2\times 0.7=6400\times 0.7===4480\text{N⋅m/rad}==.$ $c=2\zeta \sqrt{(k+K)J}=2(0.707)\sqrt{4480\times 0.7}=1.414\sqrt{3136}=1.414\times 56===79.2\text{N⋅m⋅s}==.$

Verdict: ${\omega }_n=80$ rad/s se ${\omega }_B=80\ge 60$ ✓, $180$ rad/s bending mode se neeche ✓, aur delay sirf $10.3^{\circ }\le 12^{\circ }$ khaata hai ✓. Teeno requirements meet.

Bandwidth floor ($\ge 60$) satisfy hota hai, lekin ${\omega }_n=170$ servo ke apne $-3$ dB edge ko bending mode se sirf $180-170=10\text{rad/s}$ ($\approx 6\%$) neeche rakhta hai. Servo flexible structure ko resonance ke paas drive karega; chhoti gain/phase errors TVC ko ek oscillator bana deti hain. $180$ tak margin bahut thin hai — tumhe ek notch filter chahiye hoga (Structural bending modes & notch filters) ya, better, ${\omega }_n\approx 80$ par wapas aao. Fast acha hai; zyada fast structure se takraa jaata hai.

• L1.1: ${\omega }_n=50$, $\zeta =0.5$. L1.2: mag $=1$, phase $=-17.2^{\circ }$.
• L2.1: ${\omega }_n=94.87$, $\zeta =0.316$, ${\omega }_B=136.8$. L2.2: $28.6^{\circ }$ khaaye, $21.4^{\circ }$ bacha.
• L3.1: ${\omega }_n=120$, $k+K=8640$. L3.2: ${\omega }_B=127.2,100,64.4$.
• L4.1: $17.2^{\circ }\to 28.6^{\circ }$. L4.2: zero $+50$, pole $-50$.
• L5.1: ${\omega }_n=80$ valid, $|\Delta \varphi |=10.3^{\circ }$, $k+K=4480$, $c=79.2$.