Exercises — Transfer function — Laplace domain, poles and zeros

Zeros = numerator $2(s+4)=0\Rightarrow s=-4$ ke roots. Ek finite zero. Poles = denominator $(s+1)(s+5)=0\Rightarrow s=-1,s=-5$ ke roots. Do poles. Large-$s$ trend. Hum sirf highest powers kyun compare karte hain: jab $s$ bahut bada hota hai, toh polynomial ke andar har term apne top-degree term se dab jaati hai (jaise $s^2+6s+5$ mein, $s^2$ baaqi $6s$ aur $5$ ko swamp kar deta hai). Toh do polynomials ka ratio sirf unke leading terms ke ratio tak collapse ho jaata hai. Yahan $G\approx \frac{2s}{s^2}=\frac{2}{s}\to 0$. Toh high frequency par output shrink hoti hai — yeh machine ek low-pass shape hai.

(a) Dono real parts negative → stable. (b) Real part $-1<0$ hai dono ke liye (complex pair) → stable (yeh $\omega =3$ par ring karta hai lekin decay hota hai). (c) $+0.5$ par ek pole hai jiska $\sigma >0$ → unstable. (d) Exactly $s=0$ par ek pole hai jiska $\sigma =0$ — woh term $e^{0\cdot t}=1$ hai, ek constant jo kabhi decay nahi hoti. Yeh marginally stable hai (asymptotically stable nahi). Kyunki hamaara stability test strictly $\sigma <0$ maangta hai, yeh stable nahi hai.

Step 1 — har term ka Laplace lo, zero initial conditions. Kyun: parent note ka derivative rule $\dot{y}\to sY$, $\ddot{y}\to s^{2}Y$ kar deta hai, bas $y(0)=\dot{y}(0)=0$ hona chahiye. $s^{2}Y+5sY+6Y=2U.$ Step 2 — $Y$ factor out karo, kyunki hum $Y/U$ ratio chahte hain: $(s^2+5s+6)Y=2U.$ Step 3 — ratio banao: $G(s)=\frac{Y}{U}=\frac{2}{s^2+5s+6}=\frac{2}{(s+2)(s+3)}.$ Poles: $s=-2,-3$. Dono LHP mein → stable, non-oscillatory. Time response kaisi dikhti hai. Exponentials kyun aate hain: har pole $p$ inverse Laplace transform mein $e^{pt}$ term deta hai (parent note ka ${\mathcal{L}}^{-1}\{1/(s-p)\}=e^{pt}$). Toh kisi bhi bounded input ka response iss shape ka hoga $y(t)\sim c_1{e}^{-2t}+c_2{e}^{-3t}$, jahan $c_1$ aur $c_2$ constants (residues) hain — fixed numbers jo input aur partial-fraction split har pole ko assign karti hai. Yeh har decaying piece ki size set karte hain; poles $-2,-3$ rate set karte hain. Inki exact values particular input par depend karti hain, lekin chahe kuch bhi ho, dono terms decay karti hain → system settle ho jaata hai.

Coefficients ko standard denominator $s^2+2\zeta {\omega }_{n}s+{\omega }_n^2$ se term-by-term compare karo.

• Constant term: Yeh pehle kyun: ismein sirf ${\omega }_n$ hai, isliye yeh natural frequency isolate karta hai. ${\omega }_n^2=9\Rightarrow {\omega }_n=3\text{rad/s}$.
• $s$ ka coefficient: Yahan $2\zeta {\omega }_n$ kyun aata hai: standard form mein poora middle term $2\zeta {\omega }_{n}s$ hai, isliye uska numeric coefficient ($s$ ko multiply karne wala number) $2\zeta {\omega }_n$ ke barabar hona chahiye. Setting them equal, $2\zeta {\omega }_n=2\Rightarrow \zeta =\frac{2}{2\cdot 3}=\frac{1}{3}\approx 0.333$.

Kyunki $0<\zeta <1$ hai, system underdamped hai (yeh ring karta hai). Poles: $s=-\zeta {\omega }_n\pm j{\omega }_n\sqrt{1-{\zeta }^2}=-1\pm j3\sqrt{1-\frac{1}{9}}=-1\pm j3\cdot \frac{2\sqrt{2}}{3}=-1\pm j2\sqrt{2}.$ Toh $s=-1\pm j2.828$. Real part $-1$ (decay hota hai), imaginary part $2.828$ (ring karta hai) — stable, oscillatory.

Figure neeche padho: do purple × marks yahi poles hain. Coral arrow real part $\sigma =-1$ hai (envelope kitni jaldi decay hoti hai); slate arrow imaginary part $\omega =2.828$ hai (yeh kitni jaldi ring karta hai). Dono mint LHP band ke andar hain, jo stability ek nazar mein confirm karta hai.

Forecast (80/20 shortcut) — aur uski precondition. Standard form $s^2+2\zeta {\omega }_{n}s+{\omega }_n^2$ mein $s$ ka coefficient $+2\zeta {\omega }_n$ hai. Monic second-order polynomial ke liye, shortcut "agar $s$-coefficient ka sign bata de stability" tabhi valid hai jab constant term positive ho, yaani ${\omega }_n^2>0$. Precondition kyun matter karti hai: agar constant term negative hota, toh roots already real opposite signs wale hote (ek guaranteed RHP mein) chahe middle coefficient kuch bhi kare — isliye pehle constant term check karo. Yahan constant term $+5>0$ hai ✓, toh shortcut apply hota hai. $s$-coefficient $-2$ hai — negative — toh hum instability smell karte hain. Quadratic formula se confirm karo: $s=\frac{2\pm \sqrt{(-2{)}^2-4\cdot 5}}{2}=\frac{2\pm \sqrt{4-20}}{2}=\frac{2\pm \sqrt{-16}}{2}=1\pm 2j.$ Real part $=+1>0$ → poles RHP mein → term $e^{+t}(\dots )$ badhta hai → UNSTABLE (yeh $\omega =2$ par oscillate karta hai aur saath mein blow up bhi hota hai).

Har pole $p$ ek term $e^{pt}$ deta hai jiska time constant $\tau =1/|p|$ hai (apni value ka $1/e$ tak decay hone ka time).

• Pole $-1$: ${\tau }_1=1\text{s}$.
• Pole $-10$: ${\tau }_2=0.1\text{s}$. $-10$ pole ka term ($e^{-10t}$) essentially $0.3\text{s}$ tak vanish ho jaata hai, jabki $-1$ pole ka term ($e^{-t}$) seconds tak rehta hai. Toh slow pole $s=-1$ (imaginary axis ke sabse paas) long-term response mein dominate karta hai. Time constants ka ratio ${\tau }_1/{\tau }_2=10$: slow mode $10\times$ zyada time tak rehta hai.

Figure neeche padho: purple curve slow mode $e^{-t}$ hai, coral curve fast mode $e^{-10t}$. Notice karo ki coral wala near zero crash kar chuka hai (uska shaded area gone) jab purple wala barely muda bhi nahi — woh visual gap hi nearest pole ka domination hai.

Stability identical hai — dono ke poles LHP mein $-1,-2$ par hain. Zeros exponents mein kabhi enter nahi karte, isliye yeh stability nahi tod sakte.

Mechanism, quantitatively. Partial fractions kyun: yeh $G$ ko one-pole pieces $\frac{A_1}{s+1}+\frac{A_2}{s+2}$ ki sum mein split karte hain, jinka har ek $A_k{e}^{p_{k}t}$ mein invert hota hai. Residue $A_k$ us decaying mode ka weight hai.

$G_1(s)=\frac{2}{(s+1)(s+2)}$ ke liye, cover-up se milta hai $A_1=\frac{2}{(s+2)}{|}_{s=-1}=\frac{2}{1}=2,A_2=\frac{2}{(s+1)}{|}_{s=-2}=\frac{2}{-1}=-2,$ toh impulse response $2e^{-t}-2e^{-2t}$ hai. $t=0$ par yeh $2-2=0$ hai, aur smoothly ek single hump tak rise karta hai — apne khud ke peak se koi overshoot nahi, dono same-signed exponentials bas cleanly add aur subtract hote hain.

$G_2(s)=\frac{4(s+0.5)}{(s+1)(s+2)}$ ke liye, zero at $-0.5$ residues ko reweight karta hai: $A_1=\frac{4(s+0.5)}{(s+2)}{|}_{s=-1}=\frac{4(-0.5)}{1}=-2,A_2=\frac{4(s+0.5)}{(s+1)}{|}_{s=-2}=\frac{4(-1.5)}{-1}=6.$ Ab response $-2e^{-t}+6e^{-2t}$ hai. Notice karo ki numbers badal gaye jabki poles nahi: fast mode ka weight $-2$ se $+6$ tak jump kar gaya kyunki zero origin ke paas hai aur near-pole residues amplify karta hai. Bada fast-mode weight output ko jaldi shoot up karata hai aur phir final value ki taraf relax karta hai — woh early spike hi overshoot hai. Toh zero location literally residue arithmetic set karta hai, aur residues overshoot set karte hain.

RHP zero ($s=+0.5$): cover-up numerator $\propto (s-0.5)$ ke saath dobara karo. Residue signs is tarah flip hote hain ki initial slope galat direction mein point kare: response pehle apni final direction ke opposite move karta hai. Yeh non-minimum-phase "wrong-way" motion hai — aircraft ki stick upar kheecho aur pehle yeh neeche dip karta hai badhne se pehle. Phir bhi stable (poles unchanged), lekin transient ulta shuru hota hai.

Step 1 — standard form se shuru karo (choose kiya kyunki yeh directly ${\omega }_n,\zeta$ ko parametrise karta hai): $G(s)=\frac{{\omega }_n^2}{s^2+2\zeta {\omega }_{n}s+{\omega }_n^2}.$ Numerator ${\omega }_n^2$ exactly wahi hai jo $G(0)={\omega }_n^2/{\omega }_n^2=1$ banata hai, free mein unity DC gain deta hai. Step 2 — numbers plug karo: ${\omega }_n^2=16$, $2\zeta {\omega }_n=2(0.5)(4)=4$: $G(s)=\frac{16}{s^2+4s+16}.$ Step 3 — poles: $s=-\zeta {\omega }_n\pm j{\omega }_n\sqrt{1-{\zeta }^2}=-2\pm j4\sqrt{1-0.25}=-2\pm j4\cdot \frac{\sqrt{3}}{2}=-2\pm j2\sqrt{3}.$ Toh $s=-2\pm j3.464$. LHP → stable. DC gain verify karo: $G(0)=\frac{16}{0+0+16}=1.$ ✓

Requirement (i): "DC block karo" matlab $G(0)=0$, yaani $s=0$ par zero → numerator mein $s$ factor hoga. Requirement (ii): $-3,-8$ par poles → denominator $(s+3)(s+8)$, aur dono rakhne hain, isliye numerator mein kuch $(s+3)$ ya $(s+8)$ cancel nahi kar sakta. Requirement (iii): high-frequency gain $\to 1$. Pehle sabse simple numerator try karo jo (i) satisfy kare: $G(s)=\frac{ks}{(s+3)(s+8)}?$ Large $s$ check karo: $G\approx \frac{ks}{s^2}=\frac{k}{s}\to 0$, nahi $1$. Numerator degree ($1$) denominator degree ($2$) se ek kam hai. Yeh kyun matter karta hai: high-frequency gain do leading (top-degree) terms ka ratio hai, jo nonzero sirf tab hota hai jab degrees match karein. Toh ek second zero chahiye, kahin aisa place karo jo pole par na ho (warna cancel ho jaayega). Doosra zero $s=-1$ choose karo (koi bhi value $\ne -3,-8$ chalegi): $G(s)=\frac{ks(s+1)}{(s+3)(s+8)}.$ Ab leading terms $k{s}^2/s^2=k$ hain, toh unit high-frequency gain ke liye $k=1$ set karo. Teeno requirements check karo:

• $G(0)=\frac{1\cdot 0\cdot (0+1)}{(3)(8)}=0$ ✓ (DC block karta hai),
• poles abhi bhi $s=-3$ aur $s=-8$ par ✓ (kuch cancel nahi hua — $-1\ne -3,-8$),
• $\underset{s\to \infty }{\lim }G(s)=1$ ✓. $\boxed{G(s)=\frac{s(s+1)}{(s+3)(s+8)}}.$ Yeh ek washout / high-pass shape hai (DC par zero, upar unity gain) — exactly wahi jo ek rate-damping GNC sensor use karta hai, aur yeh dono required poles rakhta hai.

(a) Transform karo, zero initial conditions. Hum initial-condition terms kyun drop kar sakte hain: derivative rule $\mathcal{L}\{f^{\prime }\}=sF(s)-f(0)$ hai, lekin ek transfer function zero ICs assume karta hai — yahan $y(0)=\dot{y}(0)=\ddot{y}(0)=0$ aur $u(0)=0$. Woh last wala right-hand side par matter karta hai: $\mathcal{L}\{\dot{u}\}=sU-u(0)=sU$ tabhi hoga jab $u(0)=0$. Toh:

• Left side: $s^{3}Y+4s^{2}Y+6sY+3Y$.
• Right side: $\mathcal{L}\{\dot{u}+2u\}=sU+2U$ ($u(0)=0$ use karke). $(s^3+4s^2+6s+3)Y=(s+2)U\Rightarrow G(s)=\frac{s+2}{s^3+4s^2+6s+3}.$ (b) Zeros: numerator $s+2=0\Rightarrow s=-2$. Ek zero, $s=-2$ par. (c) Stability. Denominator $D(s)=s^3+4s^2+6s+3$. Cubic factoring ugly hai, toh Routh–Hurwitz use karo, jo poles ke signs bina poles nikale bata deta hai. Coefficients $a_3=1,a_2=4,a_1=6,a_0=3$ sab positive hain (necessary condition pass). Array banao:

jahan $b=\frac{4\cdot 6-1\cdot 3}{4}=\frac{24-3}{4}=\frac{21}{4}=5.25$, aur last row $c=\frac{b\cdot 3-4\cdot 0}{b}=3$ hai. First column: $1,4,5.25,3$ — sab positive, toh koi sign changes nahi → RHP mein zero poles → system stable hai. Final answer. $\boxed{G(s)=\frac{s+2}{s^3+4s^2+6s+3}}$ — ek zero $s=-2$ par, teeno poles LHP mein, isliye stable. (Zero transient ko shape karta hai lekin, hamesha ki tarah, stability ko touch nahi karta.)

Time signature padho. $e^{-2t}$ term $s=-2$ par ek pole se aata hai. $t{e}^{-2t}$ term ek repeated pole at $s=-2$ (double root) ki fingerprint hai. Kyun: ${\mathcal{L}}^{-1}\{1/(s+2{)}^2\}=t{e}^{-2t}$. Toh denominator $(s+2{)}^2$ hai. Structure: $G(s)=\frac{k}{(s+2{)}^2}$ (response shape se zero ka koi evidence nahi). $k$ fix karo final value se. Final value theorem kehta hai ki step ka steady state ${\lim }_{s\to 0}s\cdot \frac{1}{s}\cdot G(s)=G(0)$ hai ($\frac{1}{s}$ unit step ka Laplace hai). $G(0)=3$ set karo: $G(0)=\frac{k}{(0+2{)}^2}=\frac{k}{4}=3\Rightarrow k=12.$ Answer: $G(s)=\frac{12}{(s+2{)}^2}$. Poles: $s=-2$ par double (LHP → stable); DC gain $G(0)=3$ ✓; response mein $e^{-2t}$ aur $t{e}^{-2t}$ ✓.

Stability decide hoti hai ::: poles ke real part ke sign se (sab $\sigma <0$ = stable); zeros kabhi isko affect nahi karte.