Laplace transform — definition, region of convergence: Visual walkthrough

4.6.25 · D2 · Maths › Ordinary Differential Equations › Laplace transform — definition, region of convergence

Hum sirf yeh assume karte hain: tum ek curve ke neeche ka area add kar sakte ho (integral itna hi hota hai), aur tum $2^{3} = 8$ style powers jaante ho. Har symbol use se pehle earn kiya jayega.

Step 1 — "Time ka function" ka matlab kya hota hai

KYA hum dekh rahe hain: ek simple height-versus-time graph.

KYUN hum yahan se shuru karte hain: poora Laplace idea is curve ko reshape karne wala hai, toh pehle jaanna zaroori hai ki hum kaunsi curve reshape kar rahe hain. Hamara running example hoga $f (t) = e^{0.5 t}$ — ek curve jo time ke saath grow karti hai. Woh word "grows" pakad lo; yeh story ka villain hai, aur exponent mein baitha number $0.5$ exactly woh growth-rate hai jise hum baad mein $a$ ka naam denge.

PICTURE — blue curve hamesha chadh rahi hai. Yahi chadhai problem hai: agar hum iska area infinity tak sum karne ki koshish karein, toh infinite answer milega.

Figure — Laplace transform — definition, region of convergence

Step 2 — Fade-out filter $e^{- σ t}$

KYA humne kiya: ek doosri curve introduce ki, orange decaying exponential $e^{- σ t}$ .

KYUN yahi tool aur koi nahi? Humhe kuch chahiye jo (a) $1$ se shuru ho taaki early signal distort na ho, (b) zero ki taraf shrink kare taaki growth ko control kar sake, aur (c) yeh property ho $\frac{d}{d t} e^{- σ t} = - σ e^{- σ t}$ — differentiate karne se sirf $- σ$ se multiply hota hai. Koi polynomial ya sine teeno nahi karta. Exponential woh unique tool hai jo calculus ko algebra mein badal deta hai, aur yahi poora payoff hai (dekho Laplace Transform of Derivatives).

PICTURE — orange height $1$ se shuru hota hai aur neeche slide karta hai. Bada $σ$ matlab steeper orange slide.

Step 3 — Inhe multiply karo: tug-of-war

KYA humne kiya: dono exponentials ko unke powers add karke combine kiya, $- σ t + 0.5 t = (0.5 - σ) t$ .

KYUN combine karte hain: ek single exponent ek nazar mein net behaviour bata deta hai. $(0.5 - σ)$ ka sign sab kuch decide karta hai:

Agar $σ > 0.5$ → exponent negative → product decay karta hai → area finite hai. ✓
Agar $σ < 0.5$ → exponent positive → product blow up karta hai → area infinite hai. ✗
Agar $σ = 0.5$ → exponent $0$ hai → product hamesha ke liye constant $1$ hai → area phir bhi infinite. ✗ (razor's edge)

PICTURE — teen products saath mein drawn: green ( $σ$ bada, decays), gray (borderline, flat), red ( $σ$ chhota, explodes). Yeh single picture hi ROC condition hai.

Step 4 — Integral winning curve ke neeche ka area hai

KYA humne kiya: Step 3 ke product ke neeche running area shade kiya aur dekha ki uska total kahan jaata hai jab right edge $T$ infinity ki taraf badhta hai.

KYUN integral hamare sawaal ka jawaab deta hai: hum ek number chahte hain jo diye gaye fade-speed $σ$ ke liye faded signal ko summarise kare. Saare faded heights (area) ko add karna aur $T \to \infty$ limit lena exactly woh number hai.

PICTURE — green (converging) case: $T$ tak running area badhta hai lekin $T$ badh'ne par ek finite ceiling ki taraf flatten hota hai. Red case: running area bina bound ke badhta rehta hai — koi limit nahi, koi answer nahi.

Step 5 — Ek honestly compute karo: $f (t) = e^{a t}$

Yahan growth-rate parameter apna naam earn karta hai: $a$ woh number hai jo signal ke exponent mein hai, toh $f (t) = e^{a t}$ rate $a$ se grow karta hai (hamare running example mein $a = 0.5$ ).

Term by term:

$e^{- σ t} e^{a t} = e^{- (σ - a) t}$ — exponents combine karo ( $- σ t + a t = - (σ - a) t$ ), Step 3 wali trick.
$\int_{0}^{T} e^{- (σ - a) t} d t = [\frac{- 1}{σ - a} e^{- (σ - a) t}]_{0}^{T} = \frac{1}{σ - a} (1 - e^{- (σ - a) T})$ — ek exponential ka antiderivative khud hi hai apne rate se divide hokar, dono ends par evaluate kiya.
Limit lo $T \to \infty$ : sab kuch $e^{- (σ - a) T}$ par depend karta hai.
- Agar $σ - a > 0$ toh exponent $- (σ - a) T \to - \infty$ , toh $e^{- (σ - a) T} \to 0$ , aur bracket $\to 1$ . Finite. ✓
- Agar $σ - a < 0$ toh exponent $\to + \infty$ , toh $e^{- (σ - a) T} \to \infty$ — koi limit nahi, diverges. ✗
- Agar $σ - a = 0$ toh integrand constant $1$ hai, toh $\int_{0}^{T} 1 d t = T \to \infty$ — diverges. ✗
Isliye, exactly tab valid jab $σ > a$ : $F (σ) = \frac{1}{σ - a} (1 - 0) = \frac{1}{σ - a} .$

KYA humne kiya: improper integral ko genuine limit ke roop mein evaluate kiya, dekha ki tail term $e^{- (σ - a) T}$ outcome decide karta hai.

KYUN yeh matter karta hai: growth rate $a$ literally ROC ki wall ban jaata hai. $a = 0$ set karo aur recover karo $L {1} = \frac{1}{σ}$ with ROC $σ > 0$ . (Jab Step 7 mein complex $s$ allow karte hain, yahi wall ban jaati hai line $Re (s) > a$ — lekin yahan arithmetic mein kuch bhi complex nahi chahiye.)

PICTURE — $σ$ ki number line jisme $σ = a$ par ek vertical wall hai; daayein taraf sab kuch "allowed" hai.

Step 6 — General bound: kyun yeh hamesha ek half-line hai (jaldi: half-plane)

Term by term:

$∣ F (σ) ∣ \leq \int e^{- σ t} ∣ f (t) ∣ d t$ — ek sum ka size zyada se zyada sizes ke sum ke barabar hota hai, aur yahan fade $e^{- σ t}$ ek positive real number hai.
$∣ f (t) ∣ \leq M e^{a t}$ — upar ki definition ka exponential-order cap; $a$ growth rate hai, $M$ constant cap hai.
$\int_{0}^{\infty} e^{- (σ - a) t} d t = \frac{1}{σ - a}$ — same limit computation jaise Step 5.
$\frac{M}{σ - a}$ ek finite number hai exactly tab jab $σ > a$ .

KYA humne kiya: kisi bhi well-behaved $f$ ko bound kiya, sirf exponentials ko nahi — abhi bhi sirf real knob $σ$ use karke.

KYUN: yeh prove karta hai ki allowed set har exponential order ke signal ke liye half-line $σ > a$ hai, koi case-by-case kaam nahi chahiye. Single denominator $σ - a$ poora conclusion carry karta hai. Step 7 mein hum $s$ ko complex jaane dete hain aur yeh half-line fattening hokar half-plane ban jaati hai.

PICTURE — ceiling curve $M e^{a t}$ wiggly $f (t)$ ke upar baitha hai; uske neeche, faded area guaranteed finite hai jab $σ > a$ .

Step 7 — Ab $s$ ko complex jaane do: sirf $Re (s)$ matter karta hai

Ab tak sab kuch $s$ ko real number $σ$ treat karta raha. Laplace transform ki real definition $s$ ko complex hone deti hai, toh chalo ab ise upgrade karein aur check karein ki pehle ke steps abhi bhi hold karte hain.

KYA humne kiya: $s$ ko real part $σ$ aur imaginary part $ω$ mein pela, aur kernel ko match karne ke liye split kiya.

KYUN: kyunki $∣ e^{- iω t} ∣ = 1$ , spinning part area ki size kabhi nahi badlaata — woh sirf rotate karta hai. Convergence poori tarah sirf $σ = Re (s)$ se decide hoti hai — exactly woh number jiske baare mein Steps 5–6 ki inequalities theen. Toh woh steps kabhi "galat nahi the ki $σ$ use kiya": $σ$ hai hi $Re (s)$ . Condition $σ > a$ simply padhi jaati hai $Re (s) > a$ , aur kyunki $ω$ free hai, number line par half-line complex $s$ -plane mein half-plane ban jaati hai.

PICTURE — complex $s$ -plane: horizontal axis $σ$ , vertical axis $ω$ ; $σ = a$ par ek vertical line, daayein green shading (converges), baayein red (diverges). Imaginary $ω$ direction "free" hai.

Definition Hum kaunsi "convergence" maangte hain — absolute, sirf conditional nahi

Yahan do notions hain. Absolute convergence matlab $\int_{0}^{\infty} ∣ e^{- s t} f (t) ∣ d t = \int_{0}^{\infty} e^{- σ t} ∣ f (t) ∣ d t$ finite hai — sizes add up hoti hain. Conditional convergence matlab signed/oscillating integral $\int_{0}^{\infty} e^{- iω t} f (t) e^{- σ t} d t$ kisi limit par settle ho sakta hai spin $e^{- iω t}$ ke cancellation ki wajah se, tab bhi jab sizes add up na hoti hoon. Exact boundary $Re (s) = a$ par size-integral $\frac{M}{σ - a}$ blow up karta hai, toh absolute convergence wahan fail hoti hai. Ek cleverly oscillating $f$ kabhikabhi us boundary line par integral ko conditionally converge kara sakti hai — lekin Laplace transform absolute convergence ke zariye define hota hai, kyunki wohi guarantee karta hai ki algebra (differentiation-under-the-integral, sums aur integrals swap karna) baad mein legal hai. Isliye hum strict inequality $Re (s) > a$ par zor dete hain aur $Re (s) = a$ ko ROC ke bahar maante hain.

Step 8 — Degenerate case jiska koi transform nahi

KYA humne kiya: $t^{2}$ ko $σ t$ se race karaya.

KYUN yeh matter karta hai: chahe fade $σ$ kitna bhi bada karo, jab ek baar $t > σ$ ho jaata hai toh exponent $t^{2} - σ t$ positive hai aur chadh raha hai. Product explode karta hai, area infinite hai, ROC empty hai — $e^{t^{2}}$ ka koi Laplace transform nahi. Yeh poori theory ki boundary hai: exponential order koi technicality nahi, yeh entry ticket hai.

PICTURE — steepest orange fade bhi $e^{t^{2}}$ curve ko flatten nahi kar sakta; product phir bhi upar curl karta hai.

Step 9 — Doosri extreme: fast decay har jagah kaam karti hai

KYA humne kiya: check kiya ki kya hota hai jab $f$ super-fast decay karta hai.

KYUN yeh matter karta hai: yeh nail karta hai ki ROC rule $σ > a$ tab bhi hold karta hai jab $a < 0$ . Rapidly decaying signal ke liye wall $a$ bahut baayein khisar jaati hai, toh ROC practically poora plane cover kar leta hai jiske baare mein tum kabhi care karoge — tum backwards bhi fade kar sakte ho ( $σ < 0$ ) aur phir bhi finite area milega, kyunki signal already zero ki taraf collapse kar raha tha. Step 8 (wall at $+ \infty$ , ROC empty) ko is case se contrast karo (wall at $- \infty$ , ROC everywhere). Single rule $Re (s) > a$ dono extremes ko span karta hai.

PICTURE — blue $e^{- t^{2}}$ bump already zero se chipka hua hai; kisi bhi fade se multiply karo (yahan tak ki mild backwards-growing fade se bhi) phir bhi ek tiny finite area milta hai.

Ek-picture summary

Upar sab kuch compressed: ek signal ek exponential ceiling $M e^{a t}$ se cap hota hai; fade $e^{- σ t}$ us ceiling se ladhta hai; ladhai jeeti jaati hai (finite area, transform exist karta hai) exactly right half-plane $Re (s) > a$ mein, aur baayein hari jaati hai — jabki vertical $ω$ direction irrelevant hai. Wall $a$ kahin bhi baitha ho sakta hai: $+ \infty$ par (koi transform nahi, jaise $e^{t^{2}}$ ), ek finite value par (jaise $e^{a t}$ ), ya $- \infty$ par (har jagah converge karta hai, jaise $e^{- t^{2}}$ ).

Recall Feynman retelling — plain words mein poora walkthrough

Ek signal ko ek aisi aawaz ki tarah imagine karo jo time ke saath tez ho sakti hai. Tum ek magic headphone pehnte ho jisme ek "fade knob" $σ$ hai — ek plain real number: jitna bada ghuma, utni tezi se har aawaz fade hogi. Phir tum saari faded aawaz ko ek single loudness number mein add karte ho — lekin "forever add karna" ka matlab hai: kisi stopping-time $T$ tak add karo, phir $T$ ko aage dhakelo aur dekho running total settle hota hai ya nahi. Agar original aawaz bahut wildly nahi grow karti (woh kisi exponential ceiling $M e^{a t}$ ke neeche rehti hai), toh ek fade speed kaafi hai total settle karaane ke liye: koi bhi $σ$ jo aawaz ke khud ke growth rate $a$ se bada ho woh kaam karta hai. Knob zyada gently ghuma ( $σ < a$ ) toh aawaz fade ko outrun kar degi — total kabhi settle nahi hoga. Ek doosra knob $ω$ bhi hai jo aawaz ko sirf spin karata hai, kabhi louder nahi, toh woh kabhi affect nahi karta ki total finite hai ya nahi — isliye answer ek right half-plane hai, sirf half-line nahi, aur hum insist karte hain ki sizes add up honi chahiye (absolute convergence), isliye razor's-edge line $Re (s) = a$ bahar rakhi jaati hai. Do extremes ise pin karte hain: $e^{t^{2}}$ jaisi insanely fast growing aawaz kisi bhi fade se pakdi nahi ja sakti (koi transform nahi), jabki $e^{- t^{2}}$ jaisi already collapsing aawaz har fade speed ke liye kaam karti hai, yahan tak ki negative ones ke liye bhi.

Recall Rapid self-check

$L {e^{3 t}}$ ki ROC wall? ::: $Re (s) > 3$ . $s = σ + iω$ ka kaunsa part convergence control karta hai? ::: sirf $σ = Re (s)$ . $e^{- iω t}$ ROC ko affect kyun nahi karta? ::: uski magnitude hamesha $1$ hoti hai; woh sirf spin karta hai. $\int_{0}^{\infty}$ ka matlab kya hai? ::: $lim_{T \to \infty} \int_{0}^{T}$ — running area ka limit. Strict $Re (s) > a$ kyun, $\geq$ kyun nahi? ::: hum absolute convergence maangte hain, jo boundary $Re (s) = a$ par fail hoti hai. $e^{t^{2}}$ ka koi transform kyun nahi? ::: $t^{2}$ har $σ t$ ko hara deta hai, toh product sabhi $σ$ ke liye diverge karta hai. $e^{- t^{2}}$ ki ROC kya hai? ::: sab $s$ — uski growth rate $a = - \infty$ hai, toh $Re (s) > a$ hamesha true hai. $L {e^{a t}}$ ki value? ::: $\frac{1}{s - a}$ , valid for $Re (s) > a$ .

Connections

Improper Integrals — woh $\int_{0}^{\infty} = lim_{T \to \infty} \int_{0}^{T}$ machinery jiska convergence hi ROC hai.
Exponential Order and Growth Rates — ceiling $M e^{a t}$ jo wall $a$ define karti hai.
Laplace Transform of Derivatives — kyun $\frac{d}{d t} e^{- s t} = - s e^{- s t}$ property asli prize hai.
Inverse Laplace Transform — $F (s)$ ko undo karna; ROC uniqueness restore karta hai.
Solving ODEs with Laplace Transforms — payoff.
Fourier Transform — wahi picture vertical line $Re (s) = 0$ ke saath padhi jaaye.

#flashcard