Step 1 — Dono transforms ko integrals ke roop mein likho.F(s)G(s)=(∫0∞e−sσf(σ)dσ)(∫0∞e−sτg(τ)dτ)Yeh step kyun? Hum alag dummy variablesσ,τ use karte hain taaki ek double integral mein merge kar sakein bina confusion ke.
Step 2 — Ek double integral mein combine karo.F(s)G(s)=∫0∞∫0∞e−s(σ+τ)f(σ)g(τ)dσdτYeh step kyun? Independent variables ke upar integrals ka product, quarter-plane σ,τ≥0 ke upar ek double integral hota hai.
Step 3 — Variables change karo taaki exponent e−st jaisa dikhe.t=σ+τ let karo (toh exponent e−st ban jaata hai). τ ko doosra variable rakho, toh σ=t−τ.
Yeh step kyun? Hum standard form ∫0∞e−st(⋯)dt ki taraf ja rahe hain, jisme bracket ek Laplace transform ki tarah identify hoga.
(σ,τ)→(t,τ) ka Jacobian 1 hai (kyunki σ=t−τ,τ=τ). Naya region: τ≥0 aur σ=t−τ≥0⇒0≤τ≤t, aur t0 se ∞ tak.
F(s)G(s)=∫t=0∞∫τ=0te−stf(t−τ)g(τ)dτdt
Step 4 — Inner integral ko pehchano.=∫0∞e−st=(f∗g)(t)[∫0tf(t−τ)g(τ)dτ]dt=L{f∗g}.Yeh step kyun? Inner integral exactly convolution hai (using g∗f=f∗g). Ho gaya. ■
Recall Feynman: ek 12-saal ke bachche ko explain karo
Do machines imagine karo. Machine f alag-alag times pe sand giraaती hai; machine g kehti hai ki har grain land hone ke baad kitni der tak glow karta hai. Abhi (t) brightness, saare pehle gire grains ki glow ka sum hai — har grain us hisaab se dim hoti hai ki woh kitni der pehle land hui thi. "Time τ pe kya giraya × woh t−τ ke baad abhi kitna glow karta hai" — yeh jodna hi convolution hai. Magic theorem kehta hai: ek special math language (Laplace) mein, yeh complicated jodna sirf do numbers multiply karna ban jaata hai.
