Transforms of standard functions — proofs — Hinglish notes

4.6.26 · Maths › Ordinary Differential Equations

Wo ek definition jisse sab kuch aata hai

HOW to read it: $s$ ek parameter hai, $t$ integration variable hai. Integrate karne ke baad, $t$ gayab ho jaata hai aur hum sirf $s$ ka function lekar bachte hain.

1. $f (t) = 1$ ka Transform

Derivation (scratch se). $L {1} = \int_{0}^{\infty} e^{- s t} \cdot 1 d t = [\frac{e ^{- s t}}{- s}]_{0}^{\infty} .$

Yeh step kyun? $e^{- s t}$ ka antiderivative w.r.t. $t$ hai $e^{- s t} / (- s)$ (chain rule ulta lagao).

$t \to \infty$ pe: $e^{- s t} \to 0$ bas tab jab $s > 0$ (warna yeh blow up ho jaata hai — yahan se condition $s > 0$ paida hoti hai). $t = 0$ pe: $e^{0} = 1$ . To $= 0 - \frac{1}{- s} = \frac{1}{s} .$

2. $f (t) = t^{n}$ ka Transform (aur raaste mein $e^{a t}$ bhi)

Pehle $L {e^{a t}}$ derive karo (yeh sabse aasaan hai aur sabse zyada reusable bhi): $L {e^{a t}} = \int_{0}^{\infty} e^{- s t} e^{a t} d t = \int_{0}^{\infty} e^{- (s - a) t} d t = [\frac{e ^{- (s - a) t}}{- ( s - a )}]_{0}^{\infty} = \frac{1}{s - a},$

Kyun $s > a$ ? Humein $s - a > 0$ chahiye taaki exponential infinity pe decay kare. Yeh bas $L {1}$ ka proof hai jisme $s \to s - a$ ho gaya.

$L {t^{n}}$ ko recursion se derive karo (integration by parts). Maano $I_{n} = \int_{0}^{\infty} e^{- s t} t^{n} d t$ . $\int u d v = uv - \int v d u$ use karo jahan $u = t^{n}, d v = e^{- s t} d t$ : $I_{n} = [t^{n} \frac{e ^{- s t}}{- s}]_{0}^{\infty} + \frac{n}{s} \int_{0}^{\infty} e^{- s t} t^{n - 1} d t .$

Yeh choice kyun? $t^{n}$ ko differentiate karna uski power ghataata hai, ek recursion banata hai.

Boundary term zero ho jaata hai: $\infty$ pe, $e^{- s t}$ kisi bhi polynomial ko beat kar deta hai ( $s > 0$ ); $0$ pe, $n \geq 1$ ke liye $t^{n} = 0$ . Isliye $I_{n} = \frac{n}{s} I_{n - 1} .$ $I_{0} = L {1} = 1/ s$ se unroll karo: $I_{n} = \frac{n}{s} \cdot \frac{n - 1}{s} \dots \frac{1}{s} \cdot \frac{1}{s} = \frac{n !}{s ^{n + 1}} .$

3. $sin a t$ aur $cos a t$ ke Transforms

Derivation (slick complex tarika). $e^{ia t} = cos a t + i sin a t$ use karo. Step 2 se $a \to ia$ karte hain: $L {e^{ia t}} = \frac{1}{s - ia} = \frac{s + ia}{( s - ia ) ( s + ia )} = \frac{s + ia}{s ^{2} + a ^{2}} .$

Yeh step kyun? Upar aur neeche conjugate $s + ia$ se multiply karo taaki denominator real ho jaaye.

Ab real aur imaginary parts match karo. Kyunki $L$ linear hai, $L {e^{ia t}} = L {cos a t} + i L {sin a t}$ : $real L {cos a t} = \frac{s}{s ^{2} + a ^{2}}, imag L {sin a t} = \frac{a}{s ^{2} + a ^{2}} .$

4. $sinh a t$ aur $cosh a t$ ke Transforms

Derivation. Hyperbolic functions ko exponentials mein likho aur linearity + step 2 use karo: $L {cosh a t} = L {\frac{e ^{a t} + e ^{- a t}}{2}} = \frac{1}{2} (\frac{1}{s - a} + \frac{1}{s + a}) = \frac{s}{s ^{2} - a ^{2}} .$ $L {sinh a t} = L {\frac{e ^{a t} - e ^{- a t}}{2}} = \frac{1}{2} (\frac{1}{s - a} - \frac{1}{s + a}) = \frac{a}{s ^{2} - a ^{2}} .$

Kyun $s > ∣ a ∣$ ? Humein $e^{a t}$ aur $e^{- a t}$ dono pieces converge karni chahiye, matlab $s > a$ aur $s > - a$ .

Figure — Transforms of standard functions — proofs

Forecast-then-Verify

Recall Padhne se pehle predict karo!

Q: Bina compute kiye, $L {cos a t}$ guess karo, given ki tumhe pata hai $L {sin a t} = \frac{a}{s ^{2} + a ^{2}}$ . Predict: numerator $s$ hona chahiye (cos→even, $t = 0$ pe value $1$ ), denominator same. Verify: $\frac{s}{s ^{2} + a ^{2}}$ ✓. Note the trick: $a \to ia$ replace karne se $sin / cos$ , $sinh / cosh$ mein badal jaate hain, matlab $s^{2} + a^{2} \to s^{2} - a^{2}$ .

Common mistakes (steel-manned)

Active-recall flashcards

Laplace transform of

f (t)

ka defining integral kya hai?

\int_{0}^{\infty} e^{- s t} f (t) d t

s

ke liye region of convergence mein.

L {1} = ?

aur uski condition

1/ s

, valid for

s > 0

L {e^{a t}} = ?

\frac{1}{s - a}

for

s > a

L {1}

ke integral ko

s > 0

kyun chahiye?

Taaki

e^{- s t} \to 0

jab

t \to \infty

; warna area diverge karta hai.

L {t^{n}} = ?

aur yeh derive kaise hota hai?

n! / s^{n + 1}

; recursion

I_{n} = \frac{n}{s} I_{n - 1}

se, integration by parts ke zariye,

I_{0} = 1/ s

se start karke.

L {sin a t}

aur

L {cos a t}

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

aur

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

Sin/cos transforms

L {e^{ia t}}

se kaise milte hain?

\frac{1}{s - ia} = \frac{s + ia}{s ^{2} + a ^{2}}

compute karo, phir real part (cos) aur imaginary part (sin) lo.

L {cosh a t}

aur

L {sinh a t}

\frac{s}{s ^{2} - a ^{2}}

aur

\frac{a}{s ^{2} - a ^{2}}

, for

s > ∣ a ∣

Trig denominator hyperbolic mein kaise banta hai?

a \to ia

replace karo:

s^{2} + a^{2} \to s^{2} - a^{2}

Recall Feynman: 12-saal ke bacche ko samjhao

Socho tum ek function ko wazan kar rahe ho. Har moment $t$ pe jo weight daala jaata hai wo hai $e^{- s t}$ — shuruaat ke paas bhaari, baad mein fade hota hua. Tum "function × weight" ko poore time pe add karte ho taaki $s$ ki har choice ke liye ek number mile. Simple functions ke liye — jaise flat line, badhta exponential, ya hilta-dulta sine — yeh add-up saaf fractions deta hai. Hum har ek ko carefully add-up karke prove karte hain (ek integral), aur note karte hain ki $s$ kitna bada hona chahiye taaki total infinity pe na bhaag jaaye.

Connections

Laplace Transform — Definition and Existence
Linearity and First Shifting Theorem
Transform of Derivatives — solving ODEs
Inverse Laplace Transform — Partial Fractions
Integration by Parts · Euler's Formula $e^{i\theta}$ · Improper Integrals