Inverse Laplace transform — partial fractions, tables: Visual walkthrough

4.6.29 · D2 · Maths › Ordinary Differential Equations › Inverse Laplace transform — partial fractions, tables

Woh fraction jise hum poore raaste chase karenge: $F (s) = \frac{3 s + 1}{( s - 1 ) ( s + 2 )} .$

Yahan $s$ sirf ek variable hai — ise "code-world coordinate" samjho, woh axis jis par hum kaam karte hain jab Laplace transform ek differential equation ko algebra mein badal deta hai. Hamara goal woh function $f (t)$ hai jiska transform exactly yahi $F (s)$ hai. Hum ise partial fractions aur ek lookup table se recover karte hain.

Step 1 — $F (s)$ aata kahan se hai? (round trip)

KYA. Algebra ko chhune se pehle, $F (s)$ ko map par locate karte hain. Ek problem real time $t$ mein ek differential equation ke roop mein shuru hoti hai. Laplace transform $L$ ise daayein $s$ -world mein le jaata hai jahan yeh seedha algebra ban jaata hai. Hum wahan solve karte hain aur $F (s)$ paate hain. Inverse $L^{- 1}$ woh return arrow hai jo humein wapas $f (t)$ par laata hai.

KYUN. Tum kisi cheez ko invert nahi kar sakte jab tak usse locate na kar sako. Loop dekhne se pata chalta hai kyun recognition (table lookup) kaafi hai: forward map one-to-one hai (Lerch's theorem), isliye har $F (s)$ ka exactly ek sachcha partner $f (t)$ hota hai. Koi ambiguity nahi — ek dictionary kaafi hai. Yahi machinery ODEs ko Laplace se solve karne ke peeche hai.

PICTURE. Do vertical lanes: $t$ -world (baayein, blue) aur $s$ -world (daayein, orange). Upar wala arrow $L$ daayein point karta hai; neecha wala arrow $L^{- 1}$ — jiske baare mein yeh poora page hai — baayein point karta hai.

Figure — Inverse Laplace transform — partial fractions, tables

Step 2 — $F (s)$ ko as-is invert kyun nahi kar sakte: pole map

KYA. Denominator $(s - 1) (s + 2)$ dekho. Values $s = 1$ aur $s = - 2$ ise zero bana dete hain — fraction wahan blow up ho jaata hai. Woh do spots poles kehlaate hain. Hamare fraction mein do poles ek expression mein glued hain.

KYUN. Lookup table sirf un fractions ko jaanta hai jinmein ek waqt mein single pole ho (ek $\frac{1}{s - a}$ ). Do-pole fraction table mein nahi hai — yeh ek compound hai, atom nahi. Toh sabse pehli rukawat yeh hai: yeh object itna complicated hai ki pehchana nahi ja sakta. Humein ise tod na hoga.

PICTURE. $s$ -axis (ek horizontal number line) jisme do poles marked hain: $s = 1$ par ek red dot aur $s = - 2$ par ek red dot. Har ek ke upar, ek vertical spike jo dikhata hai $∣ F (s) ∣ \to \infty$ . Caption likhta hai: "ek fraction, do blow-ups — table atom nahi."

$numerator P (s) 3 s + 1 / denominator Q (s) (s - 1) (s + 2) de g P = 1 < de g Q = 2.$

$P (s) = 3 s + 1$ — upar wala polynomial. Iska degree $1$ hai.
$Q (s) = (s - 1) (s + 2)$ — neecha wala. Iska degree $2$ hai.
Kyunki $de g P < de g Q$ , fraction proper hai — partial fractions ke liye precondition. (Agar nahi hota, toh pehle long-divide karte.)

Step 3 — Atoms mein split karo (partial fractions)

KYA. Hum propose karte hain ki hamar compound secretly do single-pole atoms ka sum hai, jisme unknown weights $A$ aur $B$ hain: $\frac{3 s + 1}{( s - 1 ) ( s + 2 )} = \frac{A}{s - 1} + \frac{B}{s + 2} .$

KYUN. $Q (s)$ ka har factor exactly ek pole own karta hai. Ek distinct linear factor $(s - a)$ ek atom $\frac{A}{s - a}$ contribute karta hai — ek shape jo table jaanta hai. Hum recipe guess kar rahe hain; weights $A, B$ woh hain jo humein abhi bhi find karne hain.

PICTURE. Baayein wala ek tall fraction split hota hai (ek scissors icon) daayein do separate chote fractions mein, har ek apne single pole ke upar draw kiya — compound apne atoms mein crack ho gaya.

$\frac{A}{s - 1}$ — woh atom jo $s = 1$ par pole carry karta hai. Weight $A$ abhi unknown hai.
$\frac{B}{s + 2}$ — woh atom jo $s = - 2$ par pole carry karta hai. Weight $B$ abhi unknown hai.

Step 4 — Cover-up trick se har atom ko weigh karo

KYA. $A$ find karne ke liye, poori identity ko $(s - 1)$ se multiply karo aur phir $s = 1$ set karo: $A = [(s - 1) F (s)]_{s = 1} = \frac{3 s + 1}{s + 2}_{s = 1} = \frac{3 ( 1 ) + 1}{1 + 2} = \frac{4}{3} .$ $B$ ke liye, $(s + 2)$ se multiply karo aur $s = - 2$ set karo: $B = \frac{3 s + 1}{s - 1}_{s = - 2} = \frac{3 ( - 2 ) + 1}{- 2 - 1} = \frac{- 5}{- 3} = \frac{5}{3} .$

KYUN. $(s - 1)$ se multiply karna $A$ ke apne term ke neeche $(s - 1)$ ko cancel kar deta hai, toh $A$ akela khada rehta hai. Har doosra term abhi bhi ek factor $(s - 1)$ carry karta hai, jo $s = 1$ par $0$ ban jaata hai — woh neighbours vanish ho jaate hain. Toh pole par evaluate karna exactly ek weight ko isolate karta hai. Koi linear system solve nahi karna.

PICTURE. Atom $\frac{A}{s - 1}$ highlighted; ek "cover" (gray patch) original fraction mein $(s - 1)$ factor ko hide karta hai; ek arrow $s = 1$ par axis par neeche drop karta hai aur $4/3$ read karta hai. Doosra term fade hota dikhta hai.

$(s - 1) F (s)$ — us pole ko remove karta hai jise hum target kar rahe hain.
$_{s = 1}$ — woh substitution jo har rival term ko kill kar deti hai.
$A = \frac{4}{3}, B = \frac{5}{3}$ — do residues, har atom ki "matra."

Toh: $F (s) = \frac{4/3}{s - 1} + \frac{5/3}{s + 2} .$

Step 5 — Table mein har atom look up karo

KYA. Woh ek table fact jo humein chahiye: $L^{- 1} {\frac{1}{s - a}} = e^{a t} .$ Ise har atom par apply karo, constant weight ko linearity se through pull karte hue: $L^{- 1} {\frac{4/3}{s - 1}} = \frac{4}{3} e^{1 \cdot t}, L^{- 1} {\frac{5/3}{s + 2}} = \frac{5}{3} e^{- 2 t} .$

KYUN. $s = a$ par ek single pole ek pure exponential $e^{a t}$ ka transform hai — yeh seedha forward integral $\int_{0}^{\infty} e^{- s t} e^{a t} d t = \frac{1}{s - a}$ se aata hai ( $s > a$ ke liye valid). Toh pole ki location exponential ki growth rate ban jaati hai. $s = 1$ par pole → badhta $e^{t}$ ; $s = - 2$ par pole → ghatata $e^{- 2 t}$ .

PICTURE. $s$ -axis par har pole ek arrow se apne time-graph se mapped hai: $s = 1$ pole → ek badhti blue curve $\frac{4}{3} e^{t}$ ; $s = - 2$ pole → ek girati orange curve $\frac{5}{3} e^{- 2 t}$ . Pole position ↔ exponential rate visually.

$a = 1$ (origin ke daayein) → positive exponent → term grow karta hai.
$a = - 2$ (origin ke baayein) → negative exponent → term decay karta hai.
Weights $\frac{4}{3}, \frac{5}{3}$ sirf har curve ko vertically scale karte hain.

Step 6 — Pieces ko wapas glue karo (linearity)

KYA. Do inverted atoms ko add karo: $f (t) = \frac{4}{3} e^{t} + \frac{5}{3} e^{- 2 t} .$

KYUN. Kyunki $L^{- 1}$ linear hai, ek sum ka inverse pieces ke inverses ka sum hai. Hum split karke conquer karte hain, atom-by-atom invert karte hain, aur ab re-assemble karte hain. Yeh sum hi $f (t)$ hai: woh unique time-function jiska Laplace transform hamaara original $F (s)$ tha.

PICTURE. Blue growing curve aur orange decaying curve pointwise add hokar ek green total curve $f (t)$ banti hai — shuruaat mein decaying term matter karta hai, lekin large $t$ ke liye growing $e^{t}$ dominate karta hai.

Step 7 — Edge cases jinpar kabhi trip mat karna

KYA / KYUN / PICTURE. Usi machine ke teen degenerate variations, har ek itna draw kiya gaya hai ki tum trap ko dekhte hi pehchaan sako.

Case A — repeated pole. Agar instead $Q (s) = (s + 1)^{2}$ ho, toh do poles ek jagah collapse ho jaate hain. Ek atom kaafi nahi hota: ek factor $(s - a)^{k}$ ko $k$ terms chahiye hote hain, $\frac{s}{( s + 1 ) ^{2}} = \frac{A}{s + 1} + \frac{B}{( s + 1 ) ^{2}},$ kyunki upar ka numerator degree $< 2$ ka koi bhi polynomial ho sakta hai. $s = A (s + 1) + B$ solve karne par $A = 1, B = - 1$ milta hai, aur $L^{- 1} {1/ (s + 1)^{2}} = t e^{- t}$ ke saath $f (t) = (1 - t) e^{- t}$ milta hai. Picture: do red poles ek double pole mein slide karke aapas mein aa jaate hain; answer mein ek $t$ -factor aa jaata hai.

Case B — complex poles (irreducible quadratic). Agar $Q (s) = s^{2} + 2 s + 5$ ke real roots nahi hain, toh poles axis se complex plane mein chale jaate hain. Square complete karo: $s^{2} + 2 s + 5 = (s + 1)^{2} + 4$ , toh $a = - 1, ω = 4 = 2$ . First shift theorem ke zariye, $L^{- 1} {\frac{s + 3}{( s + 1 ) ^{2} + 4}} = e^{- t} cos 2 t + e^{- t} sin 2 t .$ Real poles ne exponentials diye; ek complex pair $e^{a t}$ envelope ke andar oscillation $sin / cos$ deta hai. Picture: poles axis se uth jaate hain; time-graph ek decaying wiggle ban jaata hai.

Case C — real symmetric poles: $sinh / cosh$ , $sin$ nahi. $\frac{1}{s ^{2} - 4}$ ke liye poles real hain $s = \pm 2$ par (denominator $s^{2} - 4$ , ek minus). Split karo: $\frac{1}{s ^{2} - 4} = \frac{1}{4} (\frac{1}{s - 2} - \frac{1}{s + 2}) \Rightarrow f (t) = \frac{1}{4} (e^{2 t} - e^{- 2 t}) = \frac{1}{2} sinh 2 t .$ $s^{2} - ω^{2}$ mein minus sign ka matlab real poles → hyperbolic $sinh / cosh$ ; $s^{2} + ω^{2}$ mein plus ka matlab imaginary poles → ordinary $sin / cos$ . Picture: denominator mein sign answer ko oscillation aur hyperbolic growth ke beech flip karta hai.

Ek-picture summary

Ek diagram poori walkthrough compress karta hai: do poles wala ek compound fraction → scissors → do atoms → table arrows → do time-curves → $f (t)$ mein unhe glue karta ek plus-sign. $s$ -world mein pole positions $t$ -world mein exponential rates set karti hain.

Recall Feynman retelling (hidden)

$s$ -world mein ek messy fraction ek compound hai jo atoms se bana hai jinhe hum poora nahi padh sakte. Hum dhundh te hain kahan woh blow up karta hai — woh poles hain. Har pole ek atom hai. Hum compound ko uske atoms mein split karte hain (partial fractions) aur har atom ko uske apne factor ko cover karke aur uske pole par value read karke weigh karte hain. Phir hum dictionary kholte hain: $s = a$ par ek pole ka matlab hai time-function mein ek piece $e^{a t}$ hai — zero ke daayein woh grow karta hai, baayein woh decay karta hai. Agar do poles merge ho jayein, humein ek extra term chahiye aur ek $t$ andar aa jaata hai. Agar poles real axis se off complex space mein ud jayein, toh pieces waves $sin / cos$ ban jaate hain jo ek exponential mein wrapped hote hain. Agar denominator ek minus carry kare, real twins $sinh / cosh$ dete hain instead. Aakhir mein hum saare time-pieces ko ek plus sign se wapas glue karte hain — kyunki transform ko undo karna linear hai — aur woh sum hi saccha $f (t)$ hai. Split, weigh, look up, glue. Ghar.

Prerequisites & neighbours: Laplace Transform — definition and existence · Solving ODEs with Laplace Transforms · Partial Fraction Decomposition (Algebra) · First and Second Shift Theorems · Convolution Theorem · Heaviside Step & Dirac Delta · Hyperbolic functions sinh/cosh vs sin/cos.