Transfer function — Laplace domain, poles and zeros: Question bank

3.5.27 · D5 · Physics › Guidance, Navigation & Control (GNC) › Transfer function — Laplace domain, poles and zeros

Shuru karne se pehle, har woh symbol jo tumhe neeche milega, ek baar yahan define kar diya gaya hai, taaki koi cheez use se pehle earn na ho.

Definition Laplace transform aur uski notation (is poore page mein use hogi)

Ek signal $f (t)$ ke liye jo $t \geq 0$ par defined hai, Laplace transform yeh operator hai: $L {f (t)} = F (s) = \int_{0}^{\infty} f (t) e^{- s t} d t$

$L {\cdot}$ operator hai ("transform lo"); $F (s)$ result hai, jo complex frequency $s$ ka function hai. Time-domain signals ko hum lowercase likhte hain ( $f, y, u$ ) aur unke transforms uppercase mein ( $F, Y, U$ ).
$e^{- s t}$ kernel hai. Yeh isliye kaam karta hai kyunki yeh differentiation ka eigenfunction hai, $\frac{d}{d t}$ ko " $s$ se multiply karo" mein convert kar deta hai.
Region of convergence (ROC): integral sirf tabhi converge hoti hai jab $Re (s) = σ$ itna bada ho ki $f (t) e^{- σ t}$ decay kare. Rational transfer functions ke liye ROC sabse rightmost pole ke daayein hoti hai — isliye poles woh values hain jahan $F (s)$ blow up ho jaata hai.

Aur geometry ki vocabulary, ek picture se anchored:

$s = σ + j ω$ ek complex frequency hai: real part $σ$ = growth/decay rate, imaginary part $ω$ = oscillation frequency.
"LHP" = complex plane ka left half ( $σ < 0$ ); "RHP" = right half ( $σ > 0$ ); vertical line $σ = 0$ imaginary axis hai.

Upar ka yeh map poora khel hai: koi pole is plane mein kahan baitha hai, yeh batata hai ki uska mode time mein kaise move karta hai. Neeche har trap asal mein is map ko padhne ke baare mein ek sawaal hai.

True or false — justify karo

Ek transfer function system ka response kisi bhi input ke liye poori tarah describe karta hai.

False — sirf zero initial conditions ke liye. Transfer function machine ka fingerprint hai; ek non-zero start state extra

f (0)

terms add karta hai jo transfer function jaanboojhkar chhod deta hai.

Do systems jinke poles identical hain, hamesha identically behave karte hain.

False — same poles ka matlab same modes

e^{pt}

hain, lekin alag zeros (aur gain) un modes ko alag tarah weight karte hain, isliye transient shapes aur overshoot differ kar sakti hai.

Ek pole ko left half-plane mein aur door le jaana hamesha poore system ko faster banata hai.

Zaroor nahi — yeh us mode ko speed up karta hai (uski time constant

τ = 1/∣ σ ∣

shrink hoti hai), lekin overall speed dominant (sabse slow, sabse rightmost) pole se set hoti hai. Kisi far-left non-dominant pole ko aur bhi left le jaane se response mein koi khaas farak nahi padta, kyunki sabse sluggish rightmost mode abhi bhi tail govern karta hai. Overall faster tabhi milega jab tum dominant pole ko left move karo.

Agar kisi system mein poles se zyada zeros hain, toh yeh theek aur common hai.

Practice mein False — ek physical (causal) system ko denominator degree

\geq

numerator degree chahiye (proper); poles se zyada zeros ka matlab pure differentiation hai, jo noise ko bina bound ke amplify karta hai aur realizable nahi hai.

Imaginary axis par exactly ek pole (

s = j ω

) stable response deta hai.

False — yeh marginally stable hai:

e^{j ω t}

na grow karta hai na decay, isliye system forever oscillate karta hai bina settle kiye. "Stable" bhi nahi aur "unstable" bhi nahi — knife's edge par hai.

Ek real pole oscillation produce kar sakta hai.

False — oscillation ke liye non-zero imaginary part chahiye (

ω \neq = 0

). Ek purely real pole

σ

sirf pure exponential

e^{σ t}

deta hai, monotone rise ya decay, koi ringing nahi.

Ek hi location par ek pole aur ek zero cancel karna us mode ko poori tarah remove kar deta hai.

Dangerous-true — algebraically cancel ho jaate hain, lekin agar pole unstable tha, toh woh hidden mode real hardware mein abhi bhi exist karta hai aur blow up ho sakta hai; paper par pole-zero cancellation ek real instability chhupa sakta hai.

Sabse slow pole long-term response dominate karta hai.

True — imaginary axis ke sabse paas wala pole (sabse chota

∣ σ ∣

) ki time constant sabse lambi hoti hai, isliye uska

e^{σ t}

term sabse zyada der tak rehta hai aur response ki tail dictate karta hai. Yahi "dominant pole" ka idea hai aur isliye pichle trap mein dhyan dena zaroor tha.

Ek zero add karna kabhi bhi overshoot cause nahi kar sakta.

False — ek zero response mein ek derivative-like term add karta hai; imaginary axis ke paas wala zero ek fast kick inject kar sakta hai jo final value ko overshoot kar deta hai, chahe system otherwise well-damped ho.

Error dhundho

L {f^{'} (t)} = s F (s)

, hamesha."

Poora rule hai

s F (s) - f (0)

(defining integral par integration by parts se).

f (0)

drop karna sirf tabhi legal hai jab transfer functions zero initial conditions assume karein — yeh kehna zaroori hai, chupke se assume nahi karna.

"Is system mein RHP mein ek pole hai lekin RHP mein ek zero bhi hai, isliye yeh balance ho jaate hain aur system stable hai."

Sirf poles stability decide karte hain. RHP pole ka matlab ek

e^{+ σ t}

term hai jo bina bound ke grow karta hai; koi bhi zero us growth ko exponent mein cancel nahi kar sakta — zero sirf coefficients ko reweight karta hai.

G (s) = \frac{5}{s ^{2} - s - 6}

ke sabhi coefficients negative hain... ruko, iska

- 6

hai, lekin leading coefficient positive hai isliye yeh stable hai."

Leading coefficient ka sign kuch nahi batata. Isko factor karo:

(s - 3) (s + 2)

, jisse

s = + 3

par ek pole milta hai (RHP) → unstable.

D (s)

mein koi missing ya negative coefficient red flag hai (dekho Stability - Routh-Hurwitz criterion).

\frac{s + 3}{( s + 1 ) ( s + 2 )}

mein

s = - 3

par zero dono poles se left mein hai, isliye yeh system ko slow karta hai."

Zeros decay rates set nahi karte — poles karte hain.

s = - 3

par zero shape karta hai ki response kaise shuru hoti hai (uski early curvature), na ki kitni tezi se settle hoti hai.

s

sirf ek dummy variable hai, isliye uski units matter nahi karti."

s

ki units

1/ s

(inverse seconds) hain kyunki yeh

\frac{d}{d t}

ki jagah khadha hai. Iska real part ek rate hai aur imaginary part ek frequency — isliye real/imaginary axes wale plane par poles plot karna physically meaningful hai.

"Transfer function mujhe bata deta hai ki ek moving vehicle se input

u (t) = sin (2 t)

ke liye output kya hoga."

Seedha nahi — agar vehicle ki pehle se non-zero velocity/position hai, toh woh initial conditions response terms add karti hain jo transfer function chhod deta hai. Yeh sirf rest se forced response deta hai.

Why questions

Ek pole time domain mein

e^{pt}

kyon ban jaata hai?

Kyunki ek teen-step chain hai (figure s02 dekho): (1) partial fractions

G (s)

\sum_{k} \frac{A _{k}}{s - p _{k}}

ke roop mein rewrite karte hain; (2) har simple term

\frac{1}{s - p}

, defining integral se, exactly

e^{pt}

ka transform hai (check karo:

\int_{0}^{\infty} e^{pt} e^{- s t} d t = \frac{1}{s - p}

); (3) isliye term-by-term invert karne par

\sum_{k} A_{k} e^{p_{k} t}

milta hai. Pole location

p

exponent pick karta hai, coefficient

A_{k}

weight pick karta hai.

Hum transfer function define karte waqt zero initial conditions par kyun insist karte hain?

Taaki answer sirf system par depend kare, na ki kisi ek particular starting state par. Transfer function machine ka reusable fingerprint hona chahiye.

e^{- s t}

Laplace integral ke liye sahi kernel kyun hai?

Yeh differentiation ka eigenfunction hai (

\frac{d}{d t} e^{- s t} = - s e^{- s t}

), jo exactly derivatives ko

s

se multiplication mein convert karta hai — calculus ko algebra mein badal deta hai.

Zeros transient shape karte hain lekin stability nahi — kyun?

Zeros kabhi exponents

e^{p_{k} t}

mein appear nahi hote — woh poles se aate hain. Zeros sirf coefficients

A_{k}

set karte hain jo har mode ko multiply karte hain, isliye woh shape/weighting change karte hain, yeh nahi ki koi cheez grow karegi ya nahi.

Ek right-half-plane zero "wrong-way" initial motion kyun cause kar sakta hai?

Ek RHP zero pehle ek opposite sign wala response term add karta hai, isliye output pehle apni final value se door jaata hai phir correct hota hai — non-minimum-phase aircraft jo pull up karne par pehle sink karta hai.

Transfer function ko prefer kyun karte hain ODE har baar solve karne ke bajaaye?

Yeh ek differential equation ko ek polynomial ratio mein convert kar deta hai jo ek baar read off ho jaata hai; phir tum algebraically aur pole locations inspect karke kisi bhi input ka response predict kar sakte ho, bina dobara integrate kiye.

Edge cases

Neeche ka har edge case figure s01 ke complex-plane map par ek special jagah hai — pole-response gallery (figure s03) dikhati hai ki yeh time mein kaisa dikhta hai.

Origin par ek pole,

s = 0

, physically kya matlab rakhta hai?

Yeh

e^{0 \cdot t} = 1

contribute karta hai, ek constant — yeh integrator hai. Output hold/accumulate karta hai decay karne ke bajaaye; marginally stable, aur feedback loops mein steady-state error khatam kar deta hai (dekho Block diagrams and feedback loops).

Repeated pole ke saath kya hota hai, jaise

\frac{1}{( s + 1 ) ^{2}}

Partial fractions ek plain exponential ke saath saath ek

t e^{- t}

term bhi produce karte hain.

t

ka extra factor response ko pehle rise karwata hai phir decay (yeh instantly jump nahi karta), lekin exponential envelope stability decide karta hai — abhi bhi LHP mein, abhi bhi settle ho raha hai.

Agar denominator ek constant hai (koi

s

nahi), jaise

G (s) = \frac{s + 1}{4}

Koi finite poles nahi hain, isliye koi dynamic modes nahi hain. Likha jaaye toh

Y (s) = \frac{1}{4} (s + 1) U (s)

, jo time mein

y (t) = \frac{1}{4} (\overset{u}{˙} (t) + u (t))

hai — input ki ek scaled copy plus uski derivative. Toh ek step input ek scaled step ke upar ek instantaneous spike deta hai (

\overset{u}{˙}

term): memoryless, aur akela

s

ise improper/non-physical banata hai.

Complex poles kis arrangement mein aane chahiye, aur kyun?

Conjugate pairs mein

σ \pm j ω

, kyunki system ke coefficients real hote hain. Do

e^{(σ \pm j ω) t}

terms combine hokar ek real

e^{σ t} cos (ω t + ϕ)

banaate hain — ek decaying (ya growing) oscillation bina kisi leftover imaginary part ke.

ζ = 0

second-order poles

s = - ζ ω_{n} \pm j ω_{n} 1 - ζ^{2}

ke saath kya karta hai?

Real part vanish ho jaata hai, poles imaginary axis par

s = \pm j ω_{n}

par reh jaate hain — undamped,

ω_{n}

par forever ringing. Marginally stable, decay aur growth ke beech ki boundary (dekho Second-order systems - damping ratio and natural frequency).

ζ > 1

(overdamping) un poles ke saath kya karta hai?

Square root

1 - ζ^{2}

imaginary ban jaata hai, isliye

j ω_{n} 1 - ζ^{2}

ek real

\pm ω_{n} ζ^{2} - 1

ban jaata hai — do real negative poles deta hai. Bilkul koi oscillation nahi, sirf do decaying exponentials; system bina overshoot ke crawl karte hue apne target tak pahunchta hai.

Kya hota hai jab ek stable pole imaginary axis ki taraf drag kiya jaata hai?

Uski time constant

τ = 1/∣ σ ∣

bina bound ke badhti hai; mode kabhi zyada slowly decay karta hai, aur

σ = 0

par decay bilkul ruk jaati hai — marginal stability ki onset.

Poora page do rules mein

Figure s03 tumhara visual cheat-sheet hai: left column ek pole ko plane par plot karta hai, right column woh time-response dikhata hai jo woh produce karta hai. Left-to-right padho aur neeche ke do rules obvious ho jaayenge — tum literally stability ko poles mein dekh sakte ho.

Recall Self-test: kya tumne pattern pakda?

Upar ke har trap do rules mein reduce hota hai. Unhe name karo. Rule 1 ::: Poles stability decide karte hain (sirf wahi exponents $e^{pt}$ mein baithte hain; LHP = decaying = stable, RHP = growing = unstable, imaginary axis = marginal). Rule 2 ::: Zeros (aur gain) sirf transient shape karte hain (woh coefficients reweight karte hain, kabhi growth create ya remove nahi karte).

Connections

Parent topic — $G (s)$ , poles aur zeros ka poora build-up
Laplace transform — $L {\cdot}$ operator aur kernel jo upar define kiya, depth mein
Stability - Routh-Hurwitz criterion — $D (s)$ ko factor kiye bina pole side (LHP vs RHP) decide karta hai
Root locus — ek plane-plot jo dikhata hai ki poles kaise migrate karte hain jab ek gain knob ghoomta hai, taaki tum real time mein stability jeetna ya khoona dekh sako
Second-order systems - damping ratio and natural frequency — jahan $ζ$ aur $ω_{n}$ poles place karte hain
Block diagrams and feedback loops — signal paths ke saath transfer functions kaise multiply karte hain