Visual walkthrough — Transfer function — Laplace domain, poles and zeros

3.5.27 · D2 · Physics › Guidance, Navigation & Control (GNC) › Transfer function — Laplace domain, poles and zeros

Hum poore walkthrough mein ek concrete machine follow karenge: ek mass jo spring pe hang kar rahi hai thodi friction ke saath, jise describe kiya gaya hai $\ddot{y}+3\dot{y}+2y=u(t).$ Symbols se ghabrao mat — pehla step hi unme se har ek ko ek picture pe define karega.

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Step 1 — Machine draw karo aur har symbol ko naam do

KYA. Ek block of mass ek spring pe baitha hai. Ek haath use ek force se push karta hai. Hum dekhte hain ki woh kitna door move karta hai.

KYUN. Kisi bhi algebra se pehle, humein pata hona chahiye ki har letter physically kya hai. Ek symbol jise tum picture pe point nahi kar sakte, woh ek aisa symbol hai jise tum baad mein galat use karoge.

PICTURE.

• $t$ — time, seconds mein. Woh horizontal cheez jo tick karti rehti hai.
• $y(t)$ — the output: block time $t$ pe rest se kitna door move hua hai. (Woh green measuring arrow.)
• $u(t)$ — the input: haath jo force apply karta hai (woh yellow push arrow). "$u$" control engineering mein standard naam hai us command ke liye jo tum bhejte ho.
• $\dot{y}$ — $\frac{dy}{dt}$ ka shorthand, yaani velocity: $y$ kitni tezi se change ho rahi hai. Single dot matlab "ek time-derivative."
• $\ddot{y}$ — do dots: $\frac{d^{2}y}{d{t}^2}$, yaani acceleration: velocity kitni tezi se change ho rahi hai.

Problem yeh hai: kisi naye push $u(t)$ ke liye $y(t)$ jaanna ho toh har baar yeh differential equation phir se solve karni padti hai. Yahi dard hai jise hum abhi khatam karne wale hain.

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Step 2 — Bachne ka raasta: calculus ki jagah algebra

KYA. Hum Laplace transform apply karte hain — ek operation $\mathcal{L}$ jo ek time-signal $f(t)$ leta hai aur ek naye variable $s$ ka naya function $F(s)$ return karta hai.

KYUN. Kyunki ek magic property ki wajah se (jo next step mein built hai): is naye world mein, derivative lena $s$ se multiply karne ke barabar ho jaata hai. Derivatives ek differential equation ka mushkil hissa hoti hain; agar woh simple multiplication mein badal jaayein, toh poori equation ordinary algebra mein collapse ho jaati hai.

PICTURE.

Left room (time domain) ko socho — springs, dots, aur calculus se bhara hua. Darwaza $\mathcal{L}$ right room ($s$-domain) ki taraf le jaata hai, jahan wohi springs sirf polynomials hain. Hum easy room mein solve karte hain, phir wapas nikal aate hain.

Aaj humein integral hand se compute bhi nahi karna — hume sirf iska ek consequence chahiye, jo Step 3 hai.

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Step 3 — Woh ek rule jo saara kaam karta hai: $\frac{d}{dt}\to s$

KYA. Hum dikhate hain ki derivative $f^{\prime }(t)$ ko transform karne se $sF(s)$ milta hai (zero start ke saath).

KYUN. Yeh woh hinge hai jis par poora method ghoomta hai. Agar hum trust kar sakein ki "derivative → multiply by $s$", toh Step 1 ki har derivative ek power of $s$ ban jaayegi, aur ODE ek polynomial ban jaayega.

PICTURE.

Definition se shuru karke aur integration by parts use karke (product rule ka reverse — standard trick jab integral ke andar $f^{\prime }(t)e^{-st}$ jaisa product ho):

$\mathcal{L}\{f^{\prime }(t)\}=\underset{\text{boundary term}=-f(0)}{\underbrace{[f(t)e^{-st}{]}_0^{\infty }}}+s\underset{=F(s)}{\underbrace{{\int }_0^{\infty }f(t)e^{-st}dt}}=sF(s)-f(0)$

• $[f(t)e^{-st}{]}_0^{\infty }$ — do ends pe value. $t=0$ pe yeh $-f(0)$ chhodta hai; $t\to \infty$ pe yeh vanish ho jaata hai bas probe jeete tab (neeche dekho).
• $f(0)$ — initial condition: signal kahaan se shuru hua tha.
• $sF(s)$ — payoff: derivative $s$ times transform ban gayi.

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Step 4 — Machine ko doorway se push karo

KYA. Step 3 use karke $\ddot{y}+3\dot{y}+2y=u$ ka har term transform karo.

KYUN. Ab calculus gayab ho jaata hai. Har dot ek power of $s$ ban jaati hai.

PICTURE.

Term by term, zero initial conditions ke saath ($Y(s)$, $y$ ka transform hai, $U(s)$, $u$ ka):

$\underset{\ddot{y}\to s^{2}Y}{\underbrace{s^{2}Y}}+\underset{3\dot{y}\to 3sY}{\underbrace{3sY}}+\underset{2y\to 2Y}{\underbrace{2Y}}=\underset{u\to U}{\underbrace{U}}$

• $\ddot{y}$ (do dots) → $s^{2}Y$ (do factors of $s$),
• $3\dot{y}$ (ek dot) → $3sY$ (ek factor of $s$),
• $2y$ (koi dot nahi) → $2Y$ (koi $s$ nahi),
• $u$ → $U$.

Note karo: dots ki sankhya $s$ ki power ban gayi. Ek second-order machine ek $s^2$ deta hai. Yeh coincidence nahi — yeh Step 3 do baar apply karna hai.

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Step 5 — Machine ko isolate karo: ratio $G(s)=Y/U$ banao

KYA. Left side se $Y$ factor out karo aur divide karke output-over-input nikalo.

KYUN. Hum ek aisa single object chahte hain jiski meaning ho "kisi bhi input $U$ ke liye, output $Y=G\cdot U$ hai." Woh object hai transfer function — machine ka fingerprint.

PICTURE.

$Y$ factor karo: $(s^2+3s+2)Y=U$

Dono sides ko $U$ aur $(s^2+3s+2)$ se divide karo:

Ab $Y(s)=G(s)U(s)$: fingerprint ko kisi bhi input se multiply karo aur output mil jaata hai. ODE dobara solve nahi karni. Yeh parent note ka central result hai — aur humne ise ek hanging block se build kiya.

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Step 6 — Denominator ko roots mein crack karo: poles

KYA. Bottom polynomial ko factor karo aur dhundho ki woh kahaan zero hota hai.

KYUN. Kyunki zero se divide karne par $G$ infinity ho jaata hai, aur — jaise Step 8 dikhata hai — woh blow-up points exactly machine ke natural behaviours hain. Unhe ek naam deserve hai: poles.

PICTURE.

Factor karo: $s^2+3s+2=(s+1)(s+2)$ Har factor ko zero set karo: $s+1=0\Rightarrow s=-1,s+2=0\Rightarrow s=-2$

Kyunki $s=\sigma +j\omega$ ek complex number hai, hum har pole ko ek 2-D map pe dot ki tarah plot karte hain: horizontal axis $\sigma$ (real part), vertical axis $\omega$ (imaginary part). Hamare dono poles negative-real axis par $-1$ aur $-2$ par baithe hain. Un spots ko yaad rakho — agli steps unhe meaning denge.

• $\sigma$ (real part) — ek mode kitni tezi se grows ya decays karta hai.
• $\omega$ (imaginary part) — ek mode kitni tezi se oscillate karta hai.

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Step 7 — Fraction ko single-pole pieces mein split karo (partial fractions)

KYA. $G(s)=\frac{1}{(s+1)(s+2)}$ ko one-pole chunks $\frac{A}{s+1}+\frac{B}{s+2}$ ki sum mein todho, aur numbers $A,B$ solve karo.

KYUN. Hum abhi har pole ko time room mein wapas le jaane wale hain. Lekin hum sirf ek single pole $\frac{1}{s-p}$ ko invert karna jaante hain. Toh pehle us product ko aise singletons ki sum mein kaatna hoga. Woh kaatna partial fractions kehlata hai.

PICTURE.

Shape guess karo (har pole ke liye ek term) aur unknown weights nikalo: $\frac{1}{(s+1)(s+2)}=\frac{A}{s+1}+\frac{B}{s+2}$

• $A,B$ — weights (kitna har pole present hai).
• Cover-up trick: $A$ nikalne ke liye, dono sides ko $(s+1)$ se multiply karo aur $s=-1$ set karo (jo $B$ wale term ko khatam kar deta hai): $A={\frac{1}{s+2}|}_{s=-1}=\frac{1}{1}=1$.
• $B$ ke liye: $(s+2)$ se multiply karo, $s=-2$ set karo: $B={\frac{1}{s+1}|}_{s=-2}=\frac{1}{-1}=-1$.

Toh $G(s)=\frac{1}{s+1}-\frac{1}{s+2}.$ Single fraction do clean singletons ban gayi — dono invert karne ke liye ready hain.

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Step 8 — Pole ka matlab kya hai: impulse response $\sum A_k{e}^{p_{k}t}$ hai

KYA. Machine ko ek sharp kick (impulse) do aur uski natural motion dekho — yeh woh response hai jo batata hai ki har pole kya karta hai.

KYUN. Poles ki personality kisi aur cheez ke bina dikhane ke liye, hum sabse saaf possible input choose karte hain: unit impulse $u(t)=\delta (t)$, jiska transform simply $U(s)=1$ hai. Tab $Y(s)=G(s)\cdot 1=G(s)$, toh $G(s)$ ko directly invert karne se impulse response $y(t)$ milta hai. Isliye hum $G$ ko directly time domain mein wapas le ja sakte hain — aur kyunki impulse sirf $t=0$ par "fuse jalata" hai aur phir chhod deta hai, baad mein jo dikhta hai woh exactly system ke natural (homogeneous) modes hain, ek per pole.

PICTURE.

Step 7 ke do singletons lo aur single-pole rule ${\mathcal{L}}^{-1}\{\frac{1}{s-p}\}=e^{pt}$ use karo: $y(t)=\underset{\text{pole }s=-1}{\underbrace{(+1)e^{-1\cdot t}}}+\underset{\text{pole }s=-2}{\underbrace{(-1)e^{-2\cdot t}}}=e^{-t}-e^{-2t}$

• $A=+1$ pole $s=-1$ par → $e^{-t}$: time constant $\tau =1$ s ke saath marta hai (slow, dominant mode),
• $B=-1$ pole $s=-2$ par → $-e^{-2t}$: $0.5$ s mein marta hai (faster, jaldi fade hota hai).

Weights $A,B$ zeros/numerator se aaye (Step 7); exponents $p$ poles se aaye. Yeh puri "poles decide whether, zeros shape how" story ek hi line mein hai.

Dono exponents negative hain, toh dono modes shrink ho rahe hain → block ek baar hilta hai aur settle ho jaata hai. Stable.

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Step 9 — Doosre cases cover karo: oscillation, instability, repeated poles, aur zeros

Humne ek stable, non-oscillating machine derive ki. Lekin reader ko kabhi aisa case nahi milna chahiye jo humne skip kiya. Yeh woh cases hain jo map handle karta hai.

PICTURE.

Case A — Complex poles (oscillation). Standard second-order form $G(s)=\frac{{\omega }_n^2}{s^2+2\zeta {\omega }_{n}s+{\omega }_n^2},\text{poles }s=-\zeta {\omega }_n\pm j{\omega }_n\sqrt{1-{\zeta }^2}.$

• ${\omega }_n$ — natural frequency, $\zeta$ — damping ratio.
• Real part $-\zeta {\omega }_n<0$ → phir bhi decays.
• Imaginary part $\pm {\omega }_n\sqrt{1-{\zeta }^2}\ne 0$ → decay karte waqt rings karta hai. Poles ko further left push karo (zyada damping) → faster settling, kam ringing.

Case B — Ek pole right pe (instability). Lo $G(s)=\frac{5}{s^2-s-6}=\frac{5}{(s-3)(s+2)}$. Poles $s=+3$ aur $s=-2$ par. $+3$ par dot right pe hai → term $e^{+3t}$ hamesha badhta rahe ga → unstable, chahe doosra pole kuch bhi kare. Ek bura dot machine barbad kar deta hai — yeh directly Stability - Routh-Hurwitz criterion aur Root locus ideas mein jaata hai.

Case C — Axis par (marginal). Ek pole exactly $\sigma =0$ par (e.g. $s=\pm j\omega$) na badhta hai na ghatata hai → hamesha constant amplitude par oscillate karta rehta hai. Stable aur unstable ke beech ki knife-edge.

Case D — Repeated (double) poles. Agar do poles exactly ek hi spot par land karein, e.g. $G(s)=\frac{1}{(s+1{)}^2}$, toh partial fractions ek clean $\frac{A}{s+1}$ nahi de sakti — ek doubled pole ko rule chahiye ${\mathcal{L}}^{-1}\{\frac{1}{(s-p{)}^2}\}=t{e}^{pt}$. Toh mode hai $t{e}^{-t}$: ek extra factor of $t$ jo pehle rise karta hai jab tak exponential use neeche nahi khenchta. Order $n$ ka pole $t^{n-1}{e}^{pt}$ deta hai. Qualitatively naya — lekin real part ka sign stability rule karta rehta hai (agar $p$ axis par hai, $t\cdot e^0=t$ badhta hai, toh ek repeated axis pole unstable hai, unlike Case C mein single wala).

Case E — Ek zero. $G(s)=\frac{s+3}{(s+1)(s+2)}$ ka zero $s=-3$ par hai. Yeh kisi bhi pole ko move nahi karta, toh stability change nahi kar sakta — lekin yeh coefficients $A,B$ ko reweight karta hai (Step 7 yaad karo), transient ki shape badal deta hai (overshoot, ya "wrong-way" pehli motion agar zero right par ho).

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Ek-picture summary

Poori journey ek canvas par: ODE → (doorway $\mathcal{L}$, dots powers of $s$ ban jaate hain) → ratio $G(s)=N/D$ → (bottom factor karo) → poles as dots on a map → (singletons mein split karo, impulse se kick karo, wapas walk karo) → har dot ek motion hai $e^{pt}$. Vertical line ke left = calm; right = catastrophe.

Recall Feynman retelling — plain words mein poora walkthrough

Hamare paas ek block tha ek springy, sticky spring par, aur ek annoying equation thi dots (speeds) aur double-dots (accelerations) ke saath jise har naye push ke liye fresh solve karna padta. Toh hum ek magic doorway se guzre jahan rule hai: har dot ek multiply-by-$s$ ban jaata hai — lekin sirf well-behaved signals ke liye jo doorway ke shrinking probe se faster nahi badhte (woh "region of convergence" chhota print hai). Suddenly equation mein koi calculus nahi tha — bas $s^2+3s+2$ times output equals input. Humne use single fraction mein shuffle kiya, $\frac{1}{s^2+3s+2}$: block ka ID card. Phir humne bottom ko $(s+1)(s+2)$ mein crack kiya aur do special numbers $-1$ aur $-2$ nikale — poles. Humne single fraction ko do easy pieces mein split kiya (partial fractions), block ko ek sharp kick diya taaki input bas "1" ho, aur har piece ko doorway se wapas le gaye: $-1$ par dot ek aise wiggle bana jo 1 second mein marta hai, aur $-2$ wala aadhe second mein. Dono middle line ke left → block calm ho jaata hai → good. Agar koi dot right mein ghus jaaye, uska wiggle hamesha badhta rahega → toot gaya. Ek hi spot par stacked do dots ek extra "×time" add karte hain jo ek on-the-line pole ko bhi badha deta hai. Aur zeros? Bas wiggles ke volume knobs hain — shape change karte hain lekin machine kabhi nahi tod sakte.

Connections

• Parent topic
• Laplace transform — Steps 2–3 mein use hua doorway
• Second-order systems - damping ratio and natural frequency — Case A
• Stability - Routh-Hurwitz criterion — bina factoring ke Case B
• Root locus — gain tune karne par poles kaise move karte hain
• Block diagrams and feedback loops — jahan yeh $G(s)$ aapas mein multiply hote hain
• Bode plot, State-space representation — usi machine ke doosre views