Laplace transform — definition, region of convergence: Worked examples

4.6.25 · D3 · Maths › Ordinary Differential Equations › Laplace transform — definition, region of convergence

Shuru karne se pehle, ek reminder jo hum baar baar use karte hain. Variable $s$ ek complex number hai, likha jaata hai $s = σ + iω$ , jahan $σ = Re (s)$ iska "real part" hai (horizontal axis par ek plain number) aur $ω$ iska "imaginary part" hai. Infinite integral settle hogi ya nahi — yeh sirf $σ$ decide karta hai, kyunki kernel ka size hai $∣ e^{- s t} ∣ = ∣ e^{- (σ + iω) t} ∣ = e^{- σ t} \cdot = 1 ∣ e^{- iω t} ∣ = e^{- σ t} .$ $e^{- iω t}$ wala piece sirf radius $1$ ke circle par ghoomta rehta hai — uska size kabhi nahi badalat. Isliye convergence ek race hai $e^{- σ t}$ ke shrink hone aur $f (t)$ ke grow karne ke beech. Yeh picture apne zehan mein rakho.

Ek aur tool jo hum kai baar quote karte hain, toh chalo iska exact meaning yahan pin kar dete hain (parent ne bhi define kiya tha):

The scenario matrix

Is topic ka har Laplace problem in mein se kisi ek cell mein aata hai. Neeche har worked example ko uski cell(s) se tag kiya gaya hai. (Reminder: ROC = Region Of Convergence.)

#	Case class	Kya special hai	Example
A	Positive growth rate $f = e^{a t}, a > 0$	ROC boundary $s > a > 0$ par hai	Ex 1
B	Negative growth rate $f = e^{a t}, a < 0$	ROC $0$ ke left tak extend hoti hai: $s > a$	Ex 2
C	Zero growth (bounded / constant)	$a = 0$ , ROC hai $s > 0$	Ex 3
D	Oscillation (sine / cosine)	complex $a = \pm iω$ , real answer	Ex 4
E	Polynomial × exponential	shifting ya repeated IBP chahiye	Ex 5
F	Degenerate: koi transform exist nahi	kisi bhi $e^{a t}$ se zyada fast grow karta hai	Ex 6
G	Limiting case ( $a \to 0$ ek simpler formula recover karta hai)	continuity check	Ex 7
H	Piecewise / shifted "switch-on"	integral ek jump par split hoti hai	Ex 8
I	Word problem (radioactive decay signal)	story se $f (t)$ banao	Ex 9
J	Exam twist (same $F (s)$ , kaunsa ROC?)	ROC hi answer decide karta hai	Ex 10

Intuition Ek edge case jo apna paragraph deserve karta hai: exactly boundary

$σ = a$ par kya hota hai? Neeche har right-sided ROC ek strict inequality $Re (s) > a$ se likhi gayi hai. Kyun $\geq$ nahi? Boundary line $σ = a$ par race tie hoti hai: kernel ka decay $e^{- σ t}$ exactly function ki growth $e^{a t}$ ko cancel kar deta hai, kuch aise cheezon ko chhodke jo shrink nahi karti. $f = e^{a t}$ ke liye, $σ = a$ par integrand literally $e^{0} = 1$ hai, aur $\int_{0}^{\infty} 1 d t$ diverge karta hai — toh boundary excluded hai. $cos ω t$ jaisi bounded oscillation ke liye, exactly $σ = 0$ par integrand $cos ω t$ kabhi decay nahi karta; yeh hamesha oscillate karta rehta hai aur integral absolutely converge nahi karta (at best yeh wobble karta rehta hai). Toh honest statement yeh hai: transform (absolutely) converge karta hai open half-plane $Re (s) > a$ par, aur generally boundary line par nahi karta. Sirf special finite-energy cases wahan conditionally converge kar sakte hain — hum usp rely nahi karte. Strict $>$ rakhein.

Ex 1 — Cell A: positive growth rate

Worked example

$L {e^{2 t}}$ aur uska ROC Forecast (pehle guess karo): function $e^{2 t}$ ki tarah blow up karti hai. Kernel $e^{- σ t}$ ke liye race jeetnay ke liye, humein $σ$ ko $2$ se bada rakhna hoga. Guess: $F (s) = \frac{1}{s - 2}$ , ROC $Re (s) > 2$ .

1. Integral likho aur exponents merge karo. $F (s) = \int_{0}^{\infty} e^{- s t} e^{2 t} d t = \int_{0}^{\infty} e^{- (s - 2) t} d t .$ Yeh step kyun? Do exponentials multiply hoke exponents add karte hain; merge karne se product ek single decaying exponential ban jaati hai jise hum already integrate karna jaante hain.

2. Integrate karo. $= [\frac{- 1}{s - 2} e^{- (s - 2) t}]_{0}^{\infty} .$ Yeh step kyun? $\int e^{k t} d t = \frac{1}{k} e^{k t}$ ; yahan $k = - (s - 2)$ .

3. Dono limits evaluate karo.

Upper limit $t \to \infty$ : size hai $e^{- (σ - 2) t} \to 0$ sirf tab jab $σ - 2 > 0$ , yaani $Re (s) > 2$ .
Lower limit $t = 0$ : $e^{0} = 1$ . $F (s) = 0 - (\frac{- 1}{s - 2}) = \frac{1}{s - 2}, Re (s) > 2.$

Verify: growth rate $a = 2$ exactly ROC boundary ke roop mein aata hai — parent ke rule se match karta hai "ROC = right half-plane past the growth rate." Forecast confirmed. ✓

Ex 2 — Cell B: negative growth rate

Worked example

$L {e^{- 3 t}}$ aur uska ROC Forecast: ab function decay karti hai. Kernel ko zyada help nahi karni padegi. Hum $σ$ ko $- 3$ tak le ja sakte hain. Guess: $F (s) = \frac{1}{s + 3}$ , ROC $Re (s) > - 3$ .

1. Exponents merge karo: $F (s) = \int_{0}^{\infty} e^{- s t} e^{- 3 t} d t = \int_{0}^{\infty} e^{- (s + 3) t} d t .$ Yeh step kyun? Same "add exponents" move; yahan dono decays ek dusre ko reinforce karti hain.

2. Integrate karo aur evaluate karo. $= [\frac{- 1}{s + 3} e^{- (s + 3) t}]_{0}^{\infty} .$ Yeh step kyun? Hum standard exponential integral $\int e^{k t} d t = \frac{1}{k} e^{k t}$ use karte hain jahan $k = - (s + 3)$ hai, toh antiderivative $\frac{- 1}{s + 3} e^{- (s + 3) t}$ hai. Phir upper limit $\to 0$ tab iff $Re (s + 3) > 0 ⟺ Re (s) > - 3$ , aur lower limit $t = 0$ par $\frac{- 1}{s + 3}$ deta hai.

3. Collect karo: $F (s) = \frac{1}{s + 3}, Re (s) > - 3.$

Verify: abscissa of convergence $a = - 3$ hai, jo imaginary axis ke left hai. Yeh Ex 1 se key contrast hai: ROC boundary growth rate ke sign ke saath chalta hai. ✓

Figure — Laplace transform — definition, region of convergence

Upar ki figure $s$ -plane par dono ROCs ko shaded half-planes ki tarah dikhati hai. Notice karo ki $e^{2 t}$ ka shaded region (magenta boundary $σ = 2$ par) poora right mein hai, jabki $e^{- 3 t}$ (violet boundary $σ = - 3$ par) bahut left tak pahunchta hai. Dono rightward opening half-planes hain — woh shape kabhi nahi badlati.

Ex 3 — Cell C: zero growth (constant)

Worked example

$L {5}$ Forecast: ek constant bilkul grow nahi karta ( $a = 0$ ), toh ROC honi chahiye $Re (s) > 0$ . Kyunki parent ne dikhaya tha $L {1} = \frac{1}{s}$ , linearity se $\frac{5}{s}$ milna chahiye.

1. Constant bahar nikalo: $F (s) = \int_{0}^{\infty} e^{- s t} \cdot 5 d t = 5 \int_{0}^{\infty} e^{- s t} d t .$ Yeh step kyun? Integral linear hai — ek constant multiplier seedha pass ho jaata hai, toh hum known $L {1}$ par reduce ho jaate hain.

2. $L {1} = \frac{1}{s}$ use karo (valid $Re (s) > 0$ ke liye): $F (s) = 5 \cdot \frac{1}{s} = \frac{5}{s}, Re (s) > 0.$ Yeh step kyun? Scratch se dobara integrate karne ke bajaye, hum pehle se earned anchor result $L {1} = \frac{1}{s}$ (parent note mein prove hua) plug in karte hain; known transform reuse karna faster hai aur uska ROC $Re (s) > 0$ unchanged carry hota hai kyunki constant $5$ se multiply karna kabhi affect nahi karta ki integral kahan converge karti hai.

Verify: $s = 1$ plug karo: $F (1) = 5$ . Directly, $\int_{0}^{\infty} 5 e^{- t} d t = 5 \cdot [- e^{- t}]_{0}^{\infty} = 5 (0 - (- 1)) = 5$ . ✓

Ex 4 — Cell D: oscillation

Worked example

$L {cos (2 t)}$ Forecast: cosine bounded hai, toh exponential order $a = 0$ hai → ROC $Re (s) > 0$ . Bounded oscillation → answer real hona chahiye aur $s^{2} + 4$ involve karna chahiye. Guess $\frac{s}{s ^{2} + 4}$ .

1. Cosine ko exponentials se rewrite karo (Euler): $cos 2 t = \frac{e ^{i 2 t} + e ^{- i 2 t}}{2} .$ Yeh step kyun? Hum already $L {e^{a t}} = \frac{1}{s - a}$ jaante hain. Cosine ko exponentials mein badalne se hum usse reuse kar sakte hain instead of do baar integrate by parts karne ke. Yeh tool isliye choose kiya kyunki yeh oscillation ko do cheezein ban deta hai jo hum already solve kar chuke hain.

2. Exponential result apply karo $a = + i 2$ aur $a = - i 2$ ke saath: $F (s) = \frac{1}{2} (\frac{1}{s - i 2} + \frac{1}{s + i 2}) .$ Yeh step kyun? Phir se linearity; har piece ek known transform hai.

3. Common denominator $(s - i 2) (s + i 2) = s^{2} - (i 2)^{2} = s^{2} + 4$ par fractions add karo: $F (s) = \frac{1}{2} \cdot \frac{( s + i 2 ) + ( s - i 2 )}{s ^{2} + 4} = \frac{1}{2} \cdot \frac{2 s}{s ^{2} + 4} = \frac{s}{s ^{2} + 4}, Re (s) > 0.$ ROC kyun? Har exponential $e^{\pm i 2 t}$ ka size $1$ hai (pure oscillation, $a = 0$ ), toh dono ko $Re (s) > 0$ chahiye; joint region hai $Re (s) > 0$ . (Exactly boundary $Re (s) = 0$ par integrand kabhi decay nahi karta — boundary edge case upar flag kiya gaya hai.)

Verify: $s = 2$ par, $F (2) = \frac{2}{4 + 4} = \frac{1}{4}$ . Aur $\frac{s}{s ^{2} + 4}$ real $s$ ke liye real hai — "real function → real transform" se match karta hai. ✓

Ex 5 — Cell E: polynomial × exponential

Worked example

$L {t e^{- t}}$ Forecast: hum jaante hain $L {t} = \frac{1}{s ^{2}}$ (ROC $Re (s) > 0$ ). $f (t)$ ko $e^{- t}$ se multiply karne se $s \to s + 1$ shift hona chahiye (decay convergence easier banata hai, ROC ko left push karta hai). Guess $\frac{1}{( s + 1 ) ^{2}}$ , ROC $Re (s) > - 1$ .

1. Setup karo aur exponential merge karo: $F (s) = \int_{0}^{\infty} t e^{- t} e^{- s t} d t = \int_{0}^{\infty} t e^{- (s + 1) t} d t .$ Yeh step kyun? Dono exponentials group karne se pata chalta hai ki poora problem bas " $L {t}$ lekin $s$ ki jagah $s + 1$ " hai — first shifting property action mein.

2. Pattern pehchano. Kyunki $\int_{0}^{\infty} t e^{- k t} d t = \frac{1}{k ^{2}}$ (from $L {t} = \frac{1}{s ^{2}}$ jahan $s$ ki jagah $k$ hai), $k = s + 1$ set karo: $F (s) = \frac{1}{( s + 1 ) ^{2}} .$ Yeh kyun work karta hai? Integral sirf combined constant $k = s + 1$ "dekhti" hai; $\frac{1}{s ^{2}}$ ki derivation ne kabhi assume nahi kiya ki $s$ special tha, toh yeh kisi bhi $k$ ke liye hold karta hai jab $Re (k) > 0$ .

3. ROC: chahiye $Re (k) = Re (s) + 1 > 0$ , yaani $Re (s) > - 1$ . $F (s) = \frac{1}{( s + 1 ) ^{2}}, Re (s) > - 1.$

Verify: $s = 1$ par: $F (1) = \frac{1}{4} = 0.25$ . $t e^{- 2 t}$ ka direct numeric integral $0$ se $\infty$ tak $\frac{1}{2 ^{2}} = 0.25$ hai. ✓

Ex 6 — Cell F: degenerate, koi transform NAHI

Worked example Kya

$L {e^{t^{2}}}$ exist karta hai? Forecast: parent ne warn kiya tha ki yeh kisi bhi $e^{a t}$ se zyada fast grow karta hai. Guess: kisi bhi $s$ ke liye koi transform nahi.

1. Kisi bhi real $σ$ ke liye integrand ka size dekho: $∣ e^{- s t} e^{t^{2}} ∣ = e^{- σ t} e^{t^{2}} = e^{t^{2} - σ t} .$ Yeh step kyun? Convergence is baat se decide hoti hai ki magnitude eventually shrink hoti hai ya nahi. Toh hum exponent $t^{2} - σ t$ examine karte hain.

2. Poochho: kya $t^{2} - σ t \to - \infty$ (decay ke liye zaroori) ya $+ \infty$ (blow-up)? Factor karo: $t^{2} - σ t = t (t - σ)$ . Kisi bhi fixed $σ$ ke liye, jaise hi $t > σ$ ho jaata hai dono factors positive hain, toh $t (t - σ) \to + \infty$ . Yeh step kyun? Yeh dikhata hai ki exponent har haad mein grow karta hai chahe $σ$ kitna bhi bada kyun na ho — $t^{2}$ term hamesha eventually linear $σ t$ ko dominate kar leti hai.

3. Isliye $e^{t^{2} - σ t} \to \infty$ , integrand grow karta hai, aur integral har $s$ ke liye diverge karta hai.

Verify (upar di gayi exponential order ki definition ke khilaf): yaad karo humein chahiye size bound $M > 0$ aur start time $T$ aisa ki $e^{t^{2}} \leq M e^{a t}$ for all $t \geq T$ . Rearrange karne par, yeh demand karta hai $\frac{e ^{t^{2}}}{e ^{a t}} = e^{t^{2} - a t} \leq M$ — yaani ratio fixed ceiling $M$ ke neeche rehe. Lekin $e^{t^{2} - a t} \to \infty$ , toh yeh eventually kisi bhi ceiling $M$ ko punch through kar deta hai jo tum name karo, kisi bhi growth rate $a$ ke liye. Koi valid $(M, a)$ pair exist nahi karta, toh $e^{t^{2}}$ exponential order ka nahi hai aur existence theorem ka hypothesis fail ho jaata hai — sahi taur par predict karta hai ki koi transform nahi. ✓ (Numerically check karne ke liye kuch nahi — yahi answer hai.)

Ex 7 — Cell G: limiting case (continuity)

Worked example

$L {e^{a t}}$ se $a \to 0$ karte hue $L {1}$ recover karo Forecast: $a = 0$ set karne se $e^{a t}$ constant $1$ ban jaata hai. Toh $\frac{1}{s - a}$ smoothly $\frac{1}{s}$ ban jaana chahiye. Guess: haan, koi surprises nahi.

1. General result se shuru karo $L {e^{a t}} = \frac{1}{s - a}$ , valid $Re (s) > a$ ke liye. Yeh step kyun? Hum machine ki consistency test karte hain: ek special case ( $a = 0$ ) ko directly-computed answer $L {1} = \frac{1}{s}$ se agree karna chahiye.

2. Limit lo: $lim_{a \to 0} \frac{1}{s - a} = \frac{1}{s - 0} = \frac{1}{s} .$ ROC boundary $Re (s) > a$ slide karke $Re (s) > 0$ ho jaati hai. Yeh step kyun? Kyunki $\frac{1}{s - a}$ parameter $a$ ki continuous function hai (jab tak $s \neq = a$ ), $a \to 0$ lena legitimate hai aur simply $a = 0$ substitute karta hai; yeh exactly woh hai jaise hum dono formulas ko shared value $a = 0$ par agree karne ki confirm karte hain instead of sirf assert karne ke.

Verify: $s = 3$ par: $\frac{1}{3 - a} \to \frac{1}{3}$ jab $a \to 0$ , aur $L {1} (3) = \frac{1}{3}$ . Perfect match — formula family $a$ mein continuous hai. ✓

Ex 8 — Cell H: piecewise / switch-on

Worked example Signal

$t = 0$ par switch on, $t = 1$ par off: $f (t) = {10 0 \leq t < 1 t \geq 1$ Forecast: finite-duration signal — integral sirf $[0, 1)$ par run karti hai, ek finite interval, toh yeh sab $s$ ke liye converge karti hai (koi infinite tail nahi ladne ke liye, toh koi growth condition nahi). Answer hona chahiye $\frac{1 - e ^{- s}}{s}$ .

1. Kyunki $f = 0$ after $t = 1$ , infinite integral split karo aur empty tail drop karo: $F (s) = \int_{0}^{1} e^{- s t} \cdot 1 d t + \int_{1}^{\infty} e^{- s t} \cdot 0 d t = \int_{0}^{1} e^{- s t} d t .$ Yeh step kyun? Ek piecewise function ko piece by piece integrate karna padta hai; doosra piece kuch contribute nahi karta.

2. Finite interval par integrate karo: $= [\frac{- 1}{s} e^{- s t}]_{0}^{1} = \frac{- 1}{s} e^{- s} - (\frac{- 1}{s}) = \frac{1 - e ^{- s}}{s} .$ Yeh step kyun? Hum phir se $\int e^{- s t} d t = \frac{- 1}{s} e^{- s t}$ use karte hain, lekin ab ek finite interval par, toh hum simply do endpoints $t = 1$ aur $t = 0$ substitute karte hain — infinity-tak limit ki koi tension nahi, isliye yahan koi convergence condition nahi aati.

ROC — poora plane. Kyunki dono endpoints finite hain, integral $\int_{0}^{1} e^{- s t} d t$ ek bounded, continuous function ka plain integral hai ek bounded interval par — yeh har complex $s$ ke liye ek finite number produce karta hai, koi exception nahi. Khaas taur par $s = 0$ par: closed form $\frac{1 - e ^{- s}}{s}$ dekhne mein $\frac{0}{0}$ lagta hai, lekin yeh sirf algebra ka artefact hai (ek removable singularity jo analytic continuation se fix hoti hai); actual integral wahan $\int_{0}^{1} 1 d t = 1$ hai, perfectly finite.

$s = 0$ par removable singularity ko sahi se handle karna. Closed form $\frac{1 - e ^{- s}}{s}$ $s = 0$ par as written undefined hai (division by zero), phir bhi humne abhi dekha ki transform wahan genuinely $1$ ke barabar hai. Dono facts sach hain aur compatible hain: $s = 0$ ek removable singularity hai — formula mein ek hole jo underlying function smoothly fill in karta hai. Hum fill-in value ek limit se confirm karte hain ( $e^{- s} \approx 1 - s$ small $s$ ke liye use karte hue): $lim_{s \to 0} \frac{1 - e ^{- s}}{s} = lim_{s \to 0} \frac{1 - ( 1 - s + \dots )}{s} = lim_{s \to 0} \frac{s - \dots}{s} = 1,$ jo direct integral $\int_{0}^{1} 1 d t = 1$ se match karta hai. Toh hum formula ko continuity se extend karte hain, $F (0) := 1$ define karte hue. Isliye honest closed form yeh two-line definition hai: $F (s) = \frac{1 - e ^{- s}}{s} (s \neq = 0), F (0) = 1; ROC = poora s -plane .$

Verify: $s = 1$ par: $F (1) = \frac{1 - e ^{- 1}}{1} = 1 - 0.367879 \dots = 0.632120 \dots$ . Direct: $\int_{0}^{1} e^{- t} d t = 1 - e^{- 1} = 0.632120 \dots$ . Aur removable-singularity fill-in check hota hai: $lim_{s \to 0} \frac{1 - e ^{- s}}{s} = 1 = \int_{0}^{1} 1 d t$ . ✓

Ex 9 — Cell I: word problem

Worked example Ek radioactive sample ek decaying signal

$f (t) = A e^{- λ t}$ emit karta hai (amplitude $A = 4$ counts/s, decay constant $λ = 0.5 s^{- 1}$ ). Uska Laplace transform aur ROC nikalo. Forecast: decaying exponential → Cell B pattern. Expect $\frac{4}{s + 0.5}$ , ROC $Re (s) > - 0.5$ .

1. Story ko function mein translate karo: $f (t) = 4 e^{- 0.5 t}$ for $t \geq 0$ . Yeh step kyun? "Constant $λ$ ke saath decay karta hai" ka matlab hai $e^{- λ t}$ se multiply karo; amplitude $A$ $t = 0$ par starting value hai.

2. Linearity + $e^{a t}$ result use karo jahan $a = - 0.5$ : $F (s) = 4 \cdot \frac{1}{s - ( - 0.5 )} = \frac{4}{s + 0.5}, Re (s) > - 0.5.$ Yeh tool kyun? Humne already $L {e^{a t}}$ solve kar liya hai; word problem usi result ko physical units mein dressed kiya hua hai. Ise reuse karna (with $a = - λ = - 0.5$ ) re-integrating se zyada fast hai, aur constant $A = 4$ linearity se pass ho jaata hai bina ROC ko touch kiye.

3. Units sanity: $f$ ke units hain counts/s; transform variable $s$ ke units hain $s^{- 1}$ , toh $F (s) = \int e^{- s t} f d t$ ke units hain $(counts/s) \cdot s = counts$ . Aur $\frac{4}{s + 0.5}$ : numerator counts/s divided by denominator $s^{- 1}$ = counts. ✓

Verify: $s = 1.5 s^{- 1}$ par: $F = \frac{4}{1.5 + 0.5} = \frac{4}{2} = 2$ . ✓

Ex 10 — Cell J: exam twist (ROC sab decide karta hai)

Worked example Do problems same algebra

$F (s) = \frac{1}{s - 1}$ produce karte hain. Problem $P$ mein bataya gaya hai ki signal causal hai ( $f = 0$ for $t < 0$ ); uska ROC hai $Re (s) > 1$ . Ek tempting "trap answer" claim karta hai ki ROC $Re (s) < 1$ bhi ho sakta tha. Hamare right-sided definition ke liye kaunsa sahi hai, aur kyun? Forecast: haara poora course $\int_{0}^{\infty}$ (right-sided) use karta hai. Toh ROC ek right half-plane hona chahiye. Guess: $Re (s) > 1$ hi yahan ek valid hai.

1. Shape rule yaad karo. Har one-sided (right-sided) Laplace transform ek right half-plane $Re (s) > a$ par converge karta hai, kyunki convergence ki takraar sirf $\infty$ end par hoti hai, $e^{- σ t}$ se control hoti hai. Yeh step kyun? ROC ka shape is baat se force hota hai ki integral ka kaunsa end diverge kar sakta hai; ek $\int_{0}^{\infty}$ integral sirf $σ$ ko bada karke (rightward) tame ki ja sakti hai.

2. Abscissa $a$ growth rate hai. Yahan $\frac{1}{s - 1}$ exactly $e^{1 \cdot t} = e^{t}$ ka transform hai (from $L {e^{a t}} = \frac{1}{s - a}$ jahan $a = 1$ ), aur $e^{t}$ ka growth rate $1$ hai. Toh $a = 1$ aur ROC hai $Re (s) > 1$ . Yeh step kyun? ROC boundary hamesha underlying time-signal $a$ ka growth rate hoti hai (parent mein prove hua: bound $\frac{M}{σ - a}$ finite hai iff $σ > a$ ). Hum $a$ read off karte hain yeh poochh ke ki "kaunsa $e^{a t}$ is $\frac{1}{s - a}$ ko produce kiya?" — yahan denominator $s - 1$ jawab deta hai $a = 1$ .

3. Trap resolve karo. Alternative $Re (s) < 1$ ek left half-plane hai. Hamare $\int_{0}^{\infty}$ definition ke liye woh shape impossible hai — ek right-sided integral kabhi sirf left mein converge nahi kar sakta, kyunki $σ$ badhana (right move karna) convergence mein hamesha help karta hai, kabhi hurt nahi karta. Left half-plane $Re (s) < 1$ actually ek alag, anticausal signal ka hai, yaani $f (t) = - e^{t}$ for $t < 0$ aur $0$ for $t \geq 0$ (ek two-sided transform, hamare course ki $\int_{0}^{\infty}$ definition ke bahar). Same algebra $\frac{1}{s - 1}$ , alag function, sirf ROC se alag kiya gaya.

Conclusion. Problem $P$ ke causal signal ke liye: $F (s) = \frac{1}{s - 1}$ with ROC $Re (s) > 1$ . Trap answer $Re (s) < 1$ reject hai kyunki yeh right-half-plane shape ke against jaata hai jo haari one-sided definition force karti hai. Exactly isliye parent insist karta hai ki hamesha half-plane state karo — algebra akela ambiguous hai; ROC tie-breaker hai.

Verify: hamare causal $e^{t}$ ke liye, $s = 2$ lo (inside $Re (s) > 1$ ): $\int_{0}^{\infty} e^{- 2 t} e^{t} d t = \int_{0}^{\infty} e^{- t} d t = 1 = \frac{1}{2 - 1}$ . ✓ Sirf right half-plane mein consistent; $s$ with $Re (s) < 1$ pick karne par original integral diverge ho jaata.

Wrap-up recall

Recall Har cell ko uske lesson se match karo

Cell A (positive growth) ::: ROC $0$ ke right se shuru hoti hai: $Re (s) > a > 0$ . Cell B (negative growth) ::: ROC $0$ ke left tak extend hoti hai: $Re (s) > a$ jahan $a < 0$ . Cell C (constant) ::: $a = 0$ , ROC $Re (s) > 0$ , transform hai $\frac{c}{s}$ . Cell D (oscillation) ::: Euler use karke $\frac{1}{s - a}$ reuse karo; real answer, ROC $Re (s) > 0$ . Cell E (poly × exp) ::: $s \to s + 1$ shift karne se $\frac{1}{( s + 1 ) ^{2}}$ milta hai. Cell F (super-fast growth) ::: exponential order nahi → kisi bhi $s$ ke liye koi transform NAHI. Cell G (limit $a \to 0$ ) ::: $\frac{1}{s - a} \to \frac{1}{s}$ ; formulas continuous hain. Cell H (finite duration) ::: jump par split karo; ==poore $s$ -plane== mein converge karta hai (including $s = 0$ removable-singularity fill-in $F (0) = 1$ ke zariye). Cell I (word problem) ::: story se $f (t) = A e^{- λ t}$ banao, phir $\frac{1}{s - a}$ reuse karo jahan $a = - λ$ . Cell J (ambiguous algebra) ::: ROC hi woh cheez hai jo same $F (s)$ wale do signals mein fark karta hai. Boundary $σ = a$ par ::: race tie hai; generally diverge karta hai, toh ROC strict-open half-plane hai.

Connections

Laplace Transform of Derivatives — Ex 5 ka shifting trick tab phir aata hai jab derivatives hit karte hain.
Solving ODEs with Laplace Transforms — yeh building-block transforms toolkit hain.
Inverse Laplace Transform — Ex 10 exactly isliye hai kyun inversion ko ROC chahiye.
Improper Integrals — Ex 1, 2, 6 mein convergence tests.
Exponential Order and Growth Rates — Cell F ki failure poori tarah explain ki gayi hai.
Fourier Transform — boundary line $Re (s) = 0$ Ex 4 mein touch ki gayi.

#flashcard