Visual walkthrough — Convolution theorem — proof, applications

4.6.32 · D2 · Maths › Ordinary Differential Equations › Convolution theorem — proof, applications

Agar neeche koi bhi word naya lage, toh parent topic note aur Laplace Transform — definition and properties wahan dekhein — lekin aapko chahiye nahi hona chahiye; hum sab kuch yahan rebuild karte hain.

Step 0 — Teen characters (yeh padho warna kuch kaam nahi karega)

Step 1 se pehle, symbols se milo. Yeh skip karna har baad wali picture ko meaningless bana deta hai.

Yeh hain do functions aur unke do transforms jo hum fuse karenge:

$F(s)={\int }_0^{\infty }{e}^{-s\sigma }f(\sigma )d\sigma ,G(s)={\int }_0^{\infty }{e}^{-s\tau }g(\tau )d\tau .$

Notice karo maine inhe alag placeholder letters $\sigma$ aur $\tau$ diye hain purpose se. Step 2 batayega kyun.

Step 1 — Dono transforms ko areas ki tarah draw karo

KYA. $F(s)$ ko curve $e^{-s\sigma }f(\sigma )$ ke neeche area ki tarah picture karo, aur $G(s)$ ko $e^{-s\tau }g(\tau )$ ke neeche area ki tarah. Do alag 1-D pictures, do alag axes.

KYUN. Aap do cheezein "combine" nahi kar sakte jab tak unhe side by side na dekh lo aur notice na karo ki woh independent axes par rehti hain — ek $\sigma$ ke saath chalti hai, doosri $\tau$ ke saath. Independence Step 2 ka permission slip hai.

PICTURE. Left graph $\sigma$ sweep karta hai; right graph $\tau$ sweep karta hai. Abhi tak koi doosre ke baare mein nahi jaanta.

$\underset{\text{a number}}{\underbrace{F(s)}}={\int }_0^{\infty }\underset{\text{height at }\sigma }{\underbrace{e^{-s\sigma }f(\sigma )}}d\sigma ,\underset{\text{a number}}{\underbrace{G(s)}}={\int }_0^{\infty }\underset{\text{height at }\tau }{\underbrace{e^{-s\tau }g(\tau )}}d\tau .$

Har ek ek number hai: shaded area. Unka product $F(s)G(s)$ bas (area)×(area) hai.

Step 2 — Do areas multiply = ek plane par ek volume

KYA. Dono integrals multiply karo. "$\sigma$ par sum" times "$\tau$ par sum" ka product ek single "har pair $(\sigma ,\tau )$ par sum" hai: $F(s)G(s)={\int }_0^{\infty }{\int }_0^{\infty }{e}^{-s\sigma }f(\sigma )e^{-s\tau }g(\tau )d\sigma d\tau .$

KYUN. Yeh ordinary rule hai $\left({\sum }_i{a}_i\right)\left({\sum }_j{b}_j\right)={\sum }_{i,j}{a}_i{b}_j$, lekin integrals ke liye. Kyunki $\sigma$ aur $\tau$ kabhi interfere nahi karte (Step 1 ki independence), double sum valid hai. Exponentials combine hote hain: $e^{-s\sigma }{e}^{-s\tau }=e^{-s(\sigma +\tau )}$.

PICTURE. Ab hum ek flat quarter-plane par rehte hain: horizontal axis $\sigma \ge 0$, vertical axis $\tau \ge 0$. Har point $(\sigma ,\tau )$ ke upar ek chhota tower khada hai jiska height $e^{-s(\sigma +\tau )}f(\sigma )g(\tau )$ hai. Product un saare towers ka total volume hai.

$F(s)G(s)={\iint }_{\sigma ,\tau \ge 0}\underset{\text{depends only on }\sigma +\tau }{\underbrace{e^{-s(\sigma +\tau )}}}\underset{\text{tower height}}{\underbrace{f(\sigma )g(\tau )}}d\sigma d\tau .$

Annotated exponent dekho: yeh sirf sum $\sigma +\tau$ ki parwah karta hai, $\sigma$ aur $\tau$ alag-alag ki nahi. Yahi observation poore theorem ka seed hai.

Step 3 — Sum ka naam do: $t=\sigma +\tau$ introduce karo

KYA. Exponent $e^{-s(\sigma +\tau )}$ bheekh maang raha hai ki use $e^{-st}$ likha jaaye. Toh ek naya variable define karo: $t=\sigma +\tau .$ $\tau$ ko doosra variable rakho. Tab automatically $\sigma =t-\tau$ ho jaata hai.

KYUN. Hum ${\int }_0^{\infty }{e}^{-st}(\text{something})dt$ ki shape dhoondh rahe hain, kyunki woh shape, Step 0 ki definition se, "something ka Laplace transform" hai. Agar hum apna double integral us mein bend kar sakein, toh bracket $F(s)G(s)$ ka ${\mathcal{L}}^{-1}$ hona chahiye — jo exactly wahi hai jo hum dhoondhna chahte hain.

PICTURE. Lines "$\sigma +\tau =\text{constant}$" diagonals hain jo top-left se bottom-right ki taraf jaati hain. Har diagonal $t$ ki ek fixed value hai. Coloured bands = fixed $t$, baahir ki taraf sweep karte hue.

$\underset{\text{new: how far out the diagonal is}}{\underbrace{t}}=\sigma +\tau ,\underset{\text{old}}{\underbrace{\sigma }}=t-\tau ,\underset{\text{kept}}{\underbrace{\tau }}=\tau .$

Ab plane par har point (diagonal $t$ kaun si, us par kitna aage $\tau$) se address hota hai instead of $(\sigma ,\tau )$.

Step 4 — Kya area stretch hota hai? Jacobian check

KYA. Jab hum points relabel karte hain, chhote area elements badh ya ghad sakte hain. Jacobian $J$ woh number hai jis se $d\sigma d\tau$, $dtd\tau$ ban jaata hai. Hume ise compute karna hai warna volume galat aayega.

KYUN. Yeh skip karna sabse common tarika hai jisse log theorem "prove" karte hain aur ek stray factor aa jaata hai. Humara change hai $\sigma =t-\tau ,\tau =\tau$ — ek shear (ek slanting slide), aur shears area preserve karte hain, toh hum expect karte hain $J=1$. Confirm karte hain, assume nahi karte.

$J=\left|\det \left(\begin{array}{cc}\frac{\partial \sigma }{\partial t}& \frac{\partial \sigma }{\partial \tau }\\ \frac{\partial \tau }{\partial t}& \frac{\partial \tau }{\partial \tau }\end{array}\right)\right|=\left|\det \left(\begin{array}{cc}1& -1\\ 0& 1\end{array}\right)\right|=|1\cdot 1-(-1)\cdot 0|=1.$

• $\partial \sigma /\partial t=1$: diagonal ko $1$ baahir push karo, $\sigma$ $1$ se badhta hai.
• $\partial \sigma /\partial \tau =-1$: diagonal ke saath slide karo, $\sigma$ ghadta hai jab $\tau$ badhta hai (woh trade off karte hain, sum fixed).
• Bottom row: $\tau$ bas apna aap hai.

PICTURE. Ek chhota square grid slant hokar same area ke parallelograms mein badal jaata hai — yahi $J=1$ dikhta hai. Koi squishing nahi, koi stretching nahi.

Toh $d\sigma d\tau =1\cdot dtd\tau$. Volume unchanged hai; sirf hamara description badla.

Step 5 — Nayi limits fix karo (yahan causality rehti hai)

KYA. Purani picture mein $\sigma \ge 0$ aur $\tau \ge 0$ tha. Inhe naye coordinates $(t,\tau )$ mein translate karo:

• $\tau \ge 0$ waisa hi rehta hai $\tau \ge 0$.
• $\sigma \ge 0$ ban jaata hai $t-\tau \ge 0$, matlab $\tau \le t$.
• Milake: $0\le \tau \le t$, aur $t$ khud $0$ se $\infty$ tak chalta hai.

KYUN. Upper limit $t$ (naa ki $\infty$!) ek choice nahi hai — yeh $\sigma =t-\tau \ge 0$ se force hota hai. Yahi "causality" hai jiske baare mein parent note warn karta hai: aap itna time use nahi kar sakte jo ab tak guzra hi nahi. Level $t$ wali diagonal par, $\tau$ sirf do axes ke beech slide kar sakta hai, $0$ se $t$ tak.

PICTURE. Poora quarter-plane, diagonal segments mein re-slice kiya gaya. Har segment (fixed $t$) $\tau$-axis par shuru hota hai aur $\sigma$-axis par khatam; $\tau$ mein iska length $0$ se $t$ tak hai.

$F(s)G(s)={\int }_{t=0}^{\infty }{\int }_{\tau =0}^t{e}^{-st}f(t-\tau )g(\tau )\underset{\text{along the diagonal}}{\underbrace{d\tau }}\underset{\text{across diagonals}}{\underbrace{dt}}.$

Term by term: $e^{-st}$ ek diagonal par constant hai (isliye humne $t$ choose kiya); $f(t-\tau )g(\tau )$ tower height hai re-expressed; inner integral ek diagonal ke saath chalta hai, outer integral diagonals ke aare chalta hai.

Step 6 — Convolution pehchano, aur finish karo

KYA. $e^{-st}$ ko inner integral se bahar nikalo (yeh $\tau$ par depend nahi karta) aur dekho jo bacha: $F(s)G(s)={\int }_0^{\infty }{e}^{-st}\underset{\text{call this }h(t)}{\underbrace{\left[{\int }_0^{t}f(t-\tau )g(\tau )d\tau \right]}}dt.$

KYUN. Bracket $h(t)$ sirf $t$ ka function hai ($\tau$ integrate ho gaya). Aur ${\int }_0^{\infty }{e}^{-st}h(t)dt$ literally $h$ ka Laplace transform hai (Step 0 ki definition). Toh $F(s)G(s)=\mathcal{L}\{h\}$.

Ab $h$ identify karo. Parent ki definition $(g\ast f)(t)={\int }_0^{t}g(\tau )f(t-\tau )d\tau$ se compare karo. Hamara bracket exactly wahi hai — aur kyunki convolution commutative hai ($g\ast f=f\ast g$), hum ise kisi bhi taraf likh sakte hain: $h(t)=(f\ast g)(t).$

PICTURE. Ek diagonal, extracted: yeh $f(t-\tau )g(\tau )$ ka product collect karta hai jab $\tau$ $0$ se $t$ tak slide karta hai. $g$ ko flip karo, $f$ ke upar slide karo, multiply karo, add karo — woh summed diagonal hi $(f\ast g)(t)$ ki value hai.

$\boxed{\mathcal{L}\{f\ast g\}=F(s)G(s)}■$

Step 7 — Degenerate & edge cases (koi gap mat chhodna)

Ek-picture summary

Upar sab kuch, compressed: do 1-D areas → ek flat quarter-plane of towers → diagonals $\sigma +\tau =t$ ke saath re-slice kiya gaya (area-preserving shear) → har diagonal ek convolution value hai → $e^{-st}$ weight ke saath sum kiya gaya = ek Laplace transform.

Connections

• Parent: Convolution Theorem
• Laplace Transform — definition and properties
• Inverse Laplace Transform — partial fractions
• Solving Linear ODEs with Laplace Transforms
• Volterra Integral Equations
• Fourier Transform Convolution Theorem
• Transfer Functions and Impulse Response

Concept Map