Laplace transform — definition, region of convergence

4.6.25 · Maths › Ordinary Differential Equations

1. Definition (scratch se banate hain)

Kernel $e^{- s t}$ kyun? Hum chahte hain ek "weighting factor" jo large $t$ par function ki growth ko khatam kar de taaki infinite integral ek finite number par settle ho sake. Exponential $e^{- s t}$ decay karta hai (jab $Re (s)$ kaafi bada ho), $f (t)$ ko squash kar deta hai. Iska ek magical property bhi hai — $\frac{d}{d t} e^{- s t} = - s e^{- s t}$ , aur yahi cheez derivatives ko $s$ se multiplication mein badal deti hai.

HUM KYA PAATE HAIN: ek function $F (s)$ jo " $s$ -space" mein rehti hai. $f$ ke baare mein information repackage hoti hai, khoyi nahi jaati.

KAISE COMPUTE KAREN: $f (t)$ ko integral mein daalo aur evaluate karo. Chalo ek example earn karte hain.

2. Region of Convergence (ROC)

Bound derive karna (YEH kyun kaam karta hai): $∣ F (s) ∣ = \int_{0}^{\infty} e^{- s t} f (t) d t \leq \int_{0}^{\infty} e^{- σ t} ∣ f (t) ∣ d t \leq \int_{0}^{\infty} e^{- σ t} M e^{a t} d t = M \int_{0}^{\infty} e^{- (σ - a) t} d t = \frac{M}{σ - a},$ finite hai iff $σ - a > 0$ , yaani $σ > a$ . Ho gaya — half-plane seedha inequality se nikal aati hai.

Figure — Laplace transform — definition, region of convergence

3. Common Mistakes (Steel-manned)

4. Active Recall

Recall Quick self-test (answers dhako)

Q: Kernel kya hai aur wahi kyun? → $e^{- s t}$ ; yeh decay karta hai growth ko tame karne ke liye aur $d / d t$ ko $\times s$ mein convert karta hai.
Q: $L {e^{3 t}}$ ka ROC? → $Re (s) > 3$ .
Q: Existence guarantee karne ke liye kaun si do conditions hain? → piecewise continuous + exponential order.
Q: $e^{t^{2}}$ ka transform kyun nahi hota? → kisi bhi $e^{a t}$ se tezi grow karta hai; saare $s$ ke liye integral diverge karta hai.
Q: ROC ki shape kaisi hoti hai? → ek right half-plane $Re (s) > a$ .

Recall Feynman: ek 12-saal ke bachche ko explain karo

Ek shor machate classroom ki imagination karo (time mein ek wiggly signal). Laplace transform ek special headphones ki tarah hai jo, normal sunne ki jagah, har awaaz ko ek "fade-out" filter se multiply karta hai jo time ke saath dheemedaar hoti jaati hai. Agar shor besharam loud nahi hai (itni tezi se nahi badhta), toh headphones saari faded awaazein ek hi number mein add kar deta hai har "fade speed" $s$ ke liye. Un fade speeds ki list jo sensible (finite) number deti hai wahi Region of Convergence hai — basically "kaafi tezi se fade karo aur hamesha kaam karta hai; original jitna shorgul, utni tezi se fade karna padega."

5. The 80/20 (jo sach mein own karna hai)

$F (s) = \int_{0}^{\infty} e^{- s t} f (t) d t$ — aur kyun woh kernel.
ROC half-plane $Re (s) > a$ hai jahan $a$ = growth rate.
Existence = piecewise continuous + exponential order.
Teen anchor results: $L {1} = \frac{1}{s}$ , $L {e^{a t}} = \frac{1}{s - a}$ , $L {t} = \frac{1}{s ^{2}}$ .

Connections

Inverse Laplace Transform — $F (s)$ se $f (t)$ par wapas jaana; ROC uniqueness fix karta hai.
Laplace Transform of Derivatives — $\times s$ property jo ODEs solve karne mein use hoti hai.
Solving ODEs with Laplace Transforms — asli payoff application.
Improper Integrals — ROC ke peeche convergence machinery.
Fourier Transform — special case along $Re (s) = 0$ .
Exponential Order and Growth Rates

Define the Laplace transform of

f (t)

F (s) = \int_{0}^{\infty} e^{- s t} f (t) d t

, un

s

ke liye jahan integral converge kare.

Why is the kernel chosen as

e^{- s t}

Yeh decay karta hai function ki growth tame karne ke liye (taaki infinite integral converge kare) aur satisfy karta hai

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

, differentiation ko

s

se multiplication mein badal deta hai.

What is the Region of Convergence (ROC)?

s

ka set jiske liye defining integral converge karta hai; one-sided transforms ke liye yeh ek right half-plane

Re (s) > a

hoti hai.

What is the abscissa of convergence?

Boundary value

a

aisi ki

F (s)

converge kare

Re (s) > a

ke liye; function ki exponential growth rate ke barabar hoti hai.

State the two sufficient conditions for existence of

L {f}

f

piecewise continuous hai

[0, \infty)

par aur exponential order

a

ki hai (

∣ f (t) ∣ \leq M e^{a t}

Compute

L {1}

and its ROC.

\frac{1}{s}

, with

Re (s) > 0

Compute

L {e^{a t}}

and its ROC.

\frac{1}{s - a}

, with

Re (s) > a

Compute

L {t}

and its ROC.

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

, with

Re (s) > 0

Compute

L {sin ω t}

and its ROC.

\frac{ω}{s ^{2} + ω ^{2}}

, with

Re (s) > 0

Why does

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

have no Laplace transform?

Yeh kisi bhi

e^{a t}

se tezi grow karta hai, isliye

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

har

s

ke liye aur integral diverge karta hai.

The convergence condition is on which part of

s

Re (s) = σ

par, kyunki

∣ e^{- s t} ∣ = e^{- σ t}

Bound showing convergence for exponential-order

f

∣ F (s) ∣ \leq \frac{M}{σ - a}

for

σ > a

, hence finite (converges).

1. Definition (scratch se banate hain)

2. Region of Convergence (ROC)

3. Common Mistakes (Steel-manned)

4. Active Recall

5. The 80/20 (jo sach mein own karna hai)

Connections

Concept Map