Properties — linearity, first - second shift theorems, scaling

4.6.27 · Maths › Ordinary Differential Equations

1. Linearity

WHY ye sach hai (scratch se derive karo): Laplace transform ek integral hai, aur integration linear hoti hai.

L {a f + b g} = \int_{0}^{\infty} e^{- s t} (a f (t) + b g (t)) d t = a F (s) \int_{0}^{\infty} e^{- s t} f d t + b G (s) \int_{0}^{\infty} e^{- s t} g d t .

Ye step kyun? Constants bahar nikaalo aur sum ko split karo — dono legal hain kyunki integral operator linear hai. Ho gaya.

2. First Shift Theorem (s-shift / "frequency shift")

WHY / HOW (derive karo):

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

Ye step kyun? Exponentials ko combine karo: $e^{- s t} e^{a t} = e^{- (s - a) t}$ . Ab ye integral exactly $F$ ki definition hai lekin $s$ ki jagah $s - a$ hai. Isliye $F (s - a)$ .

3. Second Shift Theorem (t-shift / Heaviside delay)

WHY / HOW (derive karo):

L {f (t - a) u (t - a)} = \int_{0}^{\infty} e^{- s t} f (t - a) u (t - a) d t = \int_{a}^{\infty} e^{- s t} f (t - a) d t .

Ye step kyun? Step function $t = a$ ke neeche sab kuch khatam kar deta hai, isliye integral $a$ se shuru hota hai. Ab substitute karo $τ = t - a$ (so $t = τ + a$ , $d t = d τ$ , limits $0 \to \infty$ ):

= \int_{0}^{\infty} e^{- s (τ + a)} f (τ) d τ = e^{- a s} \int_{0}^{\infty} e^{- s τ} f (τ) d τ = e^{- a s} F (s) .

Ye step kyun? Substitution delayed $f$ ko original $f$ ke roop mein rewrite karta hai saath mein factor $e^{- a s}$ jo bahar nikal aata hai.

Figure — Properties — linearity, first - second shift theorems, scaling

4. Scaling (time scaling)

WHY / HOW (derive karo):

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

Substitute karo $τ = c t$ ( $t = τ / c$ , $d t = d τ / c$ , limits same rehte hain kyunki $c > 0$ ):

= \int_{0}^{\infty} e^{- s τ / c} f (τ) \frac{d τ}{c} = \frac{1}{c} \int_{0}^{\infty} e^{- (s / c) τ} f (τ) d τ = \frac{1}{c} F (\frac{s}{c}) .

Ye step kyun? Time ko $c$ se squeeze karna $s$ -axis ko $1/ c$ se stretch karta hai aur amplitude ko $1/ c$ se rescale karta hai — ek reciprocal trade-off, bilkul Fourier ki tarah.

Forecast-then-Verify drill

Recall Dekhne se pehle predict karo:

$L {e^{- t} t^{3}} = ?$ Forecast: $L {t^{3}} = \frac{3 !}{s ^{4}} = \frac{6}{s ^{4}}$ . First shift with $a = - 1$ ⇒ $s \to s + 1$ . Verify: $\frac{6}{( s + 1 ) ^{4}}$ . ✔

Common mistakes (Steel-manned)

Flashcards

Laplace transform ki linearity ka statement

L {a f + b g} = a F (s) + b G (s)

; sach hai kyunki integration linear hai.

First shift theorem

L {e^{a t} f (t)} = F (s - a)

— time mein

e^{a t}

se multiply ⇒

s

s - a

se replace karo.

Second shift theorem

L {f (t - a) u (t - a)} = e^{- a s} F (s)

—

a

se delay ⇒ transform ko

e^{- a s}

se multiply karo.

Scaling property

L {f (c t)} = \frac{1}{c} F (s / c)

for

c > 0

Second shift mein unit step kyun chahiye?

u (t - a)

integral ko

t = a

se shuru karata hai, jo substitution

τ = t - a

enable karta hai

e^{- a s}

factor out karne ke liye.

L {e^{3 t} cos 2 t}

\frac{s - 3}{( s - 3 ) ^{2} + 4}

First aur second shift mein key difference

First

s

-domain mein shift karta hai (

F (s - a)

); second

t

-domain mein ek delay tag karta hai (

e^{- a s} F (s)

L {(t - 2)^{2} u (t - 2)}

e^{- 2 s} \cdot \frac{2}{s ^{3}}

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

Socho ek machine hai jo wiggly time-pictures ko simple s-recipes mein badal deti hai. Linearity: agar aap do pictures add karo, machine bas unki recipes add kar deti hai. First shift: agar aap picture ko $e^{a t}$ se grow karoge, recipe sirf sideways slide ho jaati hai. Second shift: agar aap picture $a$ seconds late shuru karo, recipe wahi hoti hai lekin usmein ek sticker $e^{- a s}$ lag jaata hai jo kehta hai "main late aaya." Scaling: agar aap picture fast-forward karo, recipe doosri taraf stretch ho jaati hai aur thodi shrink ho jaati hai. Re-integrating ki zaroorat nahi — bas chaar tricks yaad rakho.

Connections

Laplace Transform — Definition and Existence
Laplace Transforms of Standard Functions (1, e^at, sin, cos, t^n)
Inverse Laplace Transform and Partial Fractions
Solving ODEs with Laplace Transforms
Unit Step and Dirac Delta Functions
Convolution Theorem