Inverse Laplace transform — partial fractions, tables: Question bank

4.6.29 · D5 · Maths › Ordinary Differential Equations › Inverse Laplace transform — partial fractions, tables

Har item ek specific misconception ko point karta hai $F (s)$ ko atoms mein split karne aur table padhne ke baare mein. Agar koi group mushkil lage toh yeh prerequisites dobara dekhne layak hain: Partial Fraction Decomposition (Algebra), First and Second Shift Theorems, Hyperbolic functions sinh/cosh vs sin/cos.

Sach ya jhooth — justify karo

Sach ya jhooth:

L^{- 1}

linear hai, isliye

L^{- 1} {F (s) G (s)} = L^{- 1} {F} \cdot L^{- 1} {G}

Jhooth. Linearity sirf sums aur scalar multiples ke liye kaam karta hai;

s

-world mein products,

t

-world mein convolution ban jaate hain, product nahi — dekho Convolution Theorem.

Sach ya jhooth:

\frac{1}{s ^{2} + 4}

ek sine mein invert hota hai.

Sach, lekin constant fix karna padega:

\frac{1}{s ^{2} + 4} = \frac{1}{2} \cdot \frac{2}{s ^{2} + 4} \to \frac{1}{2} sin 2 t

, kyunki table numerator mein

ω = 2

hona chahiye.

Sach ya jhooth:

\frac{1}{s ^{2} - 4}

ek sine mein invert hota hai.

Jhooth. Sign minus hai, isliye poles real hain (

s = \pm 2

), jo

\frac{1}{2} sinh 2 t

deta hai — yeh ek hyperbolic, badhti hui function hai, oscillation nahi.

Sach ya jhooth: har proper rational

F (s)

ka ek unique inverse

f (t)

hota hai.

Sach piecewise-continuous functions of exponential order ke liye — Lerch's theorem forward transform ko one-to-one banata hai, aur yahi cheez lookup table ko justify karti hai.

Sach ya jhooth: agar

F (s) = \frac{2 s + 5}{s + 3}

hai toh seedha partial fractions pe ja sakte hain.

Jhooth. Top ki degree

=

bottom ki degree, isliye yeh improper hai; pehle long division karo (

= 2 - \frac{1}{s + 3}

), phir pieces ko invert karo.

Sach ya jhooth: ek factor

(s + 1)^{3}

exactly ek partial-fraction term contribute karta hai.

Jhooth. Yeh teen terms contribute karta hai, ek har power ke liye

(s + 1)^{1}, (s + 1)^{2}, (s + 1)^{3}

, kyunki iske upar ka numerator degree

< 3

ka koi bhi polynomial ho sakta hai.

Sach ya jhooth:

L^{- 1} {1}

ek ordinary function hai.

Jhooth.

F (s) = 1

ek proper rational nahi hai (constant top, degree 0 ka constant bottom), aur iska inverse Dirac delta

δ (t)

hai — dekho Heaviside Step & Dirac Delta.

Sach ya jhooth: completing the square sirf cosmetic hai; tum

\frac{1}{s ^{2} + 2 s + 5}

ko uske bina invert kar sakte ho.

Practice mein jhooth. Table mein linear term wale raw quadratic ka koi entry nahi hai;

(s + 1)^{2} + 4

ki tarah likhna hi shifted sine/cosine entries ko unlock karta hai.

Error dhundho

\frac{s}{s ^{2} + 9} \to sin 3 t

." Galti kahan hai?

Top pe

s

hona cosine ka matlab hai:

\frac{s}{s ^{2} + 9} \to cos 3 t

. Sirf akela

ω = 3

top pe sine deta hai.

\frac{s}{( s + 1 ) ^{2}} = \frac{A}{( s + 1 ) ^{2}}

, ek factor ek term, isliye answer

= A t e^{- t}

Double pole ko dono powers chahiye:

\frac{A}{s + 1} + \frac{B}{( s + 1 ) ^{2}}

. Pehla term drop karne se

e^{- t}

piece kho jaati hai; sahi answer

(1 - t) e^{- t}

hai.

(s - 1) (s + 2)

denominator mein hai, cover-up se

A

milta hai

s = 1

set karke poore fraction mein,

(s - 1)

sameta."

Pehle

(s - 1)

hatana padega (multiply through karo), phir

s = 1

set karo; warna zero se divide ho jaayega. Cover-up

(s - 1) F (s)

s = 1

pe evaluate karta hai.

(s + 1)^{2} + 4

, isliye

ω = 4

aur main

4 sin

likhta hoon."

Table

ω^{2}

use karta hai, isliye

ω = 4 = 2

; sine numerator aur argument dono

2

hone chahiye, jo

e^{- t} sin 2 t

deta hai.

\frac{2}{( s + 1 ) ^{2} + 4} \to e^{- t} cos 2 t

Constant numerator sine atom se match karta hai, aur yeh

ω = 2

ke barabar hona chahiye:

\frac{2}{( s + 1 ) ^{2} + 4} \to e^{- t} sin 2 t

. Cosine ke liye top pe

(s + 1)

chahiye.

"Kyunki

\frac{1}{s - a} \to e^{a t}

hai, isliye

\frac{1}{s + 2} \to e^{2 t}

hai."

Yahan

a = - 2

hai, isliye yeh

e^{- 2 t}

hai; pole ka sign dhyan se padho — exponent denominator ki root hai.

\frac{5}{s ^{2} + 2 s + 5}

mein top pe

s

kahi nahi hai, isliye answer ek pure sine hai bina kisi exponential ke."

Completing the square se

(s + 1)^{2} + 4

milta hai; shift

s \to s + 1

e^{- t}

envelope force karta hai: answer

\frac{5}{2} e^{- t} sin 2 t

hai, akela sine nahi.

Why questions

Kyun First Shift Theorem ek shifted denominator ko damped oscillation mein badal deta hai?

\int e^{- s t} f d t

ke andar

s

s - a

se replace karne par ek extra

e^{a t}

factor aa jaata hai, isliye completed-square quadratic

(s - a)^{2} + ω^{2}

sine/cosine ko multiply karta hua ek

e^{a t}

envelope carry karta hai.

Kyun hum partial fractions ki taklif uthate hain seedhe inverse integral ki jagah?

Table sirf ek pole ek waqt jaanta hai; partial fractions ek compound

F (s)

ko un single-pole atoms mein split karta hai taaki linearity invert karke reassemble kar sake — recognition, integration se behtar hai.

Kyun cover-up method sirf simple (non-repeated) poles ke liye kaam karta hai?

(s - a)

se multiply karke

s = a

set karna us residue ko tabhi isolate karta hai jab koi doosra term mein abhi bhi

(s - a)

na ho; ek repeated factor surviving

(s - a)

powers chhodata hai, isliye lower coefficients ke liye matching/differentiation chahiye.

Kyun table se invert karne se pehle

F (s)

proper hona zaroori hai?

Har table entry

s \to \infty

par zero ho jaati hai; ek improper fraction nahi hoti, isliye uska polynomial part

δ (t)

aur uske derivatives se correspond karta hai, jo ordinary-function table se bahar hain.

Kyun

s^{2} + ω^{2}

oscillation deta hai jabki

s^{2} - ω^{2}

hyperbolics deta hai?

Sign decide karta hai ki poles imaginary hain (

\pm iω

, oscillating

e^{\pm iω t}

) ya real (

\pm ω

, growing/decaying

e^{\pm ω t}

) — dekho Hyperbolic functions sinh/cosh vs sin/cos.

Kyun

L^{- 1}

ki uniqueness zaroori hai — kya hum guess nahi kar sakte?

Uniqueness ke bina, do alag

f (t)

ek hi

F (s)

share kar sakte the aur table ka "answer yeh hai..." meaningless ho jaata; Lerch's theorem guarantee karta hai ki return ticket exactly ek function pe land kare.

Edge cases

L^{- 1} {1/ s}

kya hai,

1/ (s - a) \to e^{a t}

a \to 0

limit?

Yeh constant

1

hai (

= e^{0 \cdot t}

) — degenerate exponential ek flat line hai, jo table ki pehli row se consistent hai.

\frac{ω}{s ^{2} + ω ^{2}}

ω \to 0

par kya hoga?

Numerator vanish ho jaata hai aur pole

0

pe double pole mein collapse ho jaata hai; sine

sin ω t

0

ki taraf flatten ho jaata hai, jo

ω t

ke negligible hone se match karta hai — koi oscillation nahi bachti.

L^{- 1} {1/ (s^{2} + ω^{2})^{2}}

conceptually kya hai — kya yeh abhi bhi table lookup hai?

Koi single atom ek squared quadratic ke liye fit nahi hota; ya toh Convolution Theorem use karo (

\frac{1}{ω} sin ω t

ka self-convolution) ya

s

mein differentiation, kyunki basic table first-power denominators pe ruk jaata hai.

Kya

L^{- 1} {e^{- 2 s} / s}

sirf shift

1/ s \to 1

hai?

Nahi — ek

e^{- a s}

factor ek time shift hai (Second Shift), jo Heaviside step

u (t - 2)

deta hai,

s

-shift envelope nahi; dekho First and Second Shift Theorems.

Kya purely real repeated pole

(s - a)^{2}

kabhi oscillation produce karta hai?

Kabhi nahi — real poles sirf

e^{a t}

aur

t e^{a t}

dete hain; oscillation ke liye complex pole chahiye (ek irreducible quadratic jo apne vertex ke relative negative discriminant rakhta ho).

Recall Ek-line self-test

Agar tum kisi bhi $F (s)$ ke liye yeh bata sako ki har atom kyun oscillatory / exponential / polynomial / delta hai table chhoone se pehle, toh yeh topic tumhara hai. $\frac{s}{( s + 1 ) ^{2} + 9}$ ke liye mera reason? ::: Shifted quadratic $\Rightarrow$ $e^{- t}$ envelope; numerator mein shifted $(s + 1)$ andar chupi hui hai, isliye cosine-plus-correction: $e^{- t} (cos 3 t - \frac{1}{3} sin 3 t)$ .