Laplace transform — definition, region of convergence: Question bank

4.6.25 · D5 · Maths › Ordinary Differential Equations › Laplace transform — definition, region of convergence

Intuition Is page ko kaise use karein

Neeche har line ek Question ::: Answer reveal hai. Right side ko cover karo, pehle poora answer reason ke saath socho, phir check karo. Ye conceptual traps hain — koi heavy computation nahi. Ye un chaar cheezoon ko probe karte hain jo log Laplace transform ke baare mein galat samajhte hain: kernel $e^{- s t}$ kya kar raha hai, Region of Convergence (ROC) right half-plane kyun hoti hai, transform kab exist karna fail karta hai, aur $Re (s)$ ka role kya hai.

Shuru karne se pehle reminders:

$L {f (t)} = F (s) = \int_{0}^{\infty} e^{- s t} f (t) d t$ — defining integral.
$s = σ + iω$ , jahan $σ = Re (s)$ real part hai aur $ω$ imaginary part hai.
Exponential order $a$ ka matlab hai $∣ f (t) ∣ \leq M e^{a t}$ bade $t$ ke liye — $f$ kitni fast grow kar sakti hai uski ek ceiling.
ROC = $s$ ka woh set jahan ye integral finite answer deta hai; ye ek half-plane $Re (s) > a$ hoti hai.

Agar koi question sting kare toh prerequisite refreshers: Improper Integrals, Exponential Order and Growth Rates. Parent: Laplace Transform — Definition & ROC.

True or false — justify karo

f (t) = e^{5 t}

ka transform

s = 3

ke liye exist karta hai (treating

s

as real).

False. ROC hai

Re (s) > 5

, aur

3 < 5

, toh integral

\int_{0}^{\infty} e^{- (3 - 5) t} d t = \int_{0}^{\infty} e^{2 t} d t

blow up ho jaata hai. Tum growth ke andar ho, decay mein nahi.

Agar

f (t)

saare

t \geq 0

ke liye bounded hai, toh uski ROC hai

Re (s) > 0

True. Bounded ka matlab hai

∣ f (t) ∣ \leq M = M e^{0 \cdot t}

, toh

f

exponential order

a = 0

ki hai; half-plane hai

Re (s) > 0

Do alag time functions kabhi bhi same algebraic formula

F (s)

share nahi kar sakte.

False. Wo formula share kar sakte hain lekin alag ROCs ke saath — ROC transform ki identity ka part hai, isliye inversion ke liye ye matter karta hai.

F (s) = \frac{1}{s - a}

apne aap mein ek complete answer hai.

False. Bina condition

Re (s) > a

ke ye sirf ek algebraic expression hai; transform tabhi meaningful hai jahan defining integral actually converge karta hai.

t

mein har polynomial ka Laplace transform hota hai.

True. Koi bhi polynomial

e^{t}

se slower grow karta hai (actually kisi bhi

e^{a t}

se slower jab

a > 0

ho), toh ye exponential order

0

ka hai aur uski ROC hai

Re (s) > 0

f (t) = e^{t^{2}}

ka Laplace transform bahut bade

s

ke liye exist karta hai.

False. Ye har ek

e^{a t}

se faster grow karta hai, toh

e^{- s t} e^{t^{2}} \to \infty

har ek

s

ke liye; koi bhi ROC exist nahi karta.

Laplace transform ki ROC ek vertical strip ho sakti hai (left aur right dono taraf bounded).

False yahan use hone wale one-sided (

0

\infty

tak) transform ke liye — ye hamesha ek right half-plane hoti hai. (Two-sided transforms strips de sakte hain, lekin wo ek alag cheez hai.)

Integral ka convergence

s

ke imaginary part

ω

par depend karta hai.

False. Kyunki

∣ e^{- s t} ∣ = e^{- σ t}

sirf

σ = Re (s)

par depend karta hai, oscillating factor

e^{- iω t}

ki magnitude

1

hoti hai aur ye convergence ko kabhi affect nahi karta.

Agar

f

piecewise continuous aur exponential order ka hai, toh

F (s)

exist karta hai aur

F (s) \to 0

Re (s) \to \infty

True. Ye exactly existence theorem hai; bound

∣ F (s) ∣ \leq \frac{M}{σ - a} \to 0

σ \to \infty

dono claims deliver karta hai.

Ek discontinuous

f (t)

(jaise ek step) ka Laplace transform nahi ho sakta.

False. Theorem sirf piecewise continuity (finitely many jumps) require karta hai, toh step functions bada aaram se transform ho jaate hain.

Spot the error

L {1} = [\frac{- 1}{s} e^{- s t}]_{0}^{\infty} = 0 - 0 = 0

t = 0

wala end galat hai:

e^{- s \cdot 0} = 1

0

nahi. Sahi value hai

0 - (\frac{- 1}{s}) = \frac{1}{s}

. Lower limit kabhi mat chhoddo — wo usually pura numerator carry karta hai.

"Kyunki

e^{- s t} \to 0

\infty

kisi bhi

s

ke liye, integral

\int_{0}^{\infty} e^{- s t} d t

hamesha converge karta hai."

Claim

e^{- s t} \to 0

sirf tab hold karta hai jab

Re (s) > 0

ho. Agar

σ \leq 0

toh magnitude

e^{- σ t}

decay nahi karta, toh integral diverge karta hai — convergence ek condition hai, automatic nahi.

L {e^{a t}} = \frac{1}{s - a}

toh uski ROC hai

s > a

Condition real part par hai:

Re (s) > a

, na ki

s > a

s > a

likhna silently assume karta hai ki

s

real hai aur half-plane geometry ko ignore karta hai.

f (t) = e^{a t}

grow karta hai, toh uska integral diverge karta hai aur iska koi transform nahi hai."

Half-plane

Re (s) > a

par iska transform hai; decaying kernel

e^{- s t}

growth

e^{a t}

ko precisely tab beat karta hai jab

σ > a

ho. Growth akele existence ko nahi maarta — growth har exponential se faster existence ko maarta hai.

L {t}

ke liye, boundary term

[- \frac{t}{s} e^{- s t}]_{0}^{\infty}

hai

\infty

kyunki

t \to \infty

Exponential decay linear

t

par dominate karta hai, toh

t e^{- s t} \to 0

for

Re (s) > 0

. Boundary term dono ends par vanish karta hai,

\frac{1}{s ^{2}}

chhodke.

sin (ω t)

t

badhne ke saath unbounded grow karta hai, toh ROC hai

Re (s) > ω

Sine bounded hai (

∣ sin ω t ∣ \leq 1

), exponential order

a = 0

, toh ROC hai

Re (s) > 0

. Frequency

ω

poles ko

\pm iω

par set karta hai, growth rate ko nahi.

Why questions

Kernel specifically

e^{- s t}

kyun hai, na ki, say, fixed rate ke saath

e^{- t} f (t)

Ek variable rate

s

se hum fade-speeds mein sweep kar sakte hain; aur

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

exactly woh property hai jo differentiation ko

s

se multiplication mein turn karti hai — dekho Laplace Transform of Derivatives.

ROC hamesha right side mein (bade

Re (s)

) kyun rehta hai?

Integrand

e^{(a - σ) t}

jaise behave karta hai; ye sirf tab decay karta hai jab

σ > a

ho. Bada

σ

= faster fade = safer convergence, toh "good" side

a

ke right mein hoti hai.

Exponential order ki zaroorat hi kyun hai — kya piecewise continuity kaafi nahi?

Continuity kisi bhi finite piece par behaviour control karti hai, lekin integral

\infty

tak jaata hai. Exponential order bade

t

par growth ko cap karta hai taaki tail

\int^{\infty} e^{- σ t} M e^{a t} d t

actually converge kare.

Transform invert karte waqt ROC state karna zaroori kyun hai?

Kyunki same

F (s)

formula alag-alag ROCs wale alag signals se aa sakta hai; ROC select karta hai ki kaun sa signal meant hai, jo ek unique inverse guarantee karta hai — dekho Inverse Laplace Transform.

lim_{Re (s) \to \infty} F (s) = 0

har valid transform ke liye kyun hota hai?

∣ F (s) ∣ \leq \frac{M}{σ - a}

se,

σ \to \infty

bhejna bound ko

0

par drive karta hai. Physically: infinitely fast fade karo aur sirf instant

t = 0

survive karta hai, limit mein kuch contribute nahi karta.

Fourier transform ko "Laplace along a vertical line" kyun describe kiya jaata hai?

σ = 0

set karne se (toh

s = iω

)

e^{- s t}

ek pure oscillation

e^{- iω t}

ban jaata hai; ye half-plane ki boundary

Re (s) = 0

hai — dekho Fourier Transform.

Edge cases

Zero function

f (t) = 0

ki ROC kya hai?

Integral har ek

s

ke liye

0

hai, toh ROC poora complex plane hai — ye akela function hai jis par koi convergence restriction nahi.

f (t) = e^{a t}

jab

a < 0

(ek decaying exponential) ka koi special ROC hai?

Phir bhi

Re (s) > a

hai, lekin ab

a

negative hai, toh half-plane left side ka kuch part include karti hai, jaise

e^{- 3 t}

ke liye

Re (s) > - 3

. Decay boundary ko leftward push karta hai.

Exactly boundary line

Re (s) = a

par kya hota hai?

Integral typically wahan diverge karta hai (jaise

\int_{0}^{\infty} e^{0 \cdot t} d t = \infty

for

L {1}

σ = 0

), isliye ROC open half-plane

Re (s) > a

hai, boundary exclude karke.

Agar

f (t)

sirf

t \geq 0

ke liye defined hai, toh hum

t < 0

ki chinta kyun nahi karte?

One-sided transform by definition

0

\infty

tak integrate karta hai, toh negative-time values kabhi enter hi nahi karti — transform un signals ke liye bana hai jo

t = 0

par "switch on" hote hain.

Kya koi function exponential order

a = 0

ka ho sakta hai phir bhi unbounded?

Haan — jaise

f (t) = t

unbounded hai lekin

∣ t ∣ \leq M e^{0 \cdot t}

satisfy karta hai? Nahi;

t

ko strictly

a > 0

chahiye. Behtar example: koi bhi bounded function order

0

ka hai; unbounded-but-polynomial functions ko thoda positive

a

chahiye. Subtlety: order

0

ka matlab hai "constant se faster nahi," jo bade

t

ke liye boundedness force karta hai.

Agar

f

ek single point par infinite jump (ek spike) hai lekin otherwise nice hai toh?

Ek genuine infinite spike piecewise continuity todhta hai, toh existence theorem directly apply nahi hota; aise objects ko distribution theory (jaise Dirac delta) chahiye hoti hai na ki ordinary integral.

Kya

L {f}

f

ko ek isolated point par change karne se affect hota hai?

Nahi. Ek integral zero length ke set par values ke liye blind hota hai, toh

f

ko finitely many points par redefine karne se

F (s)

aur uski ROC unchanged rehti hai.

Recall Har trap ki one-line summary

Convergence sirf $Re (s)$ par depend karta hai, ROC ek open right half-plane $Re (s) > a$ hai, existence ke liye piecewise continuity plus exponential order chahiye, aur $e^{t^{2}}$ jaise functions jo saare exponentials ko outgrow karte hain unka koi transform nahi hota.

Connections

Inverse Laplace Transform — kyun ROC unique return trip ke liye zaroori hai.
Laplace Transform of Derivatives — woh $\times s$ property jiske liye kernel choose kiya gaya.
Solving ODEs with Laplace Transforms — payoff.
Improper Integrals — convergence machinery har "does it exist?" question ke neeche.
Fourier Transform — $Re (s) = 0$ boundary case.
Exponential Order and Growth Rates — existence ke peechhe growth ceiling.

#flashcard