Convolution theorem — proof, applications

WHAT is convolution?

WHY this particular form? Because Laplace transforms turn integrals/derivatives into algebra, and the product $F(s)G(s)$ "wants" exactly this overlapping integral — we'll see why in the proof.

Step 1 — Write both transforms as integrals. $F(s)G(s)=\left(\int_0^\infty e^{-s\sigma}f(\sigma)\,d\sigma\right)\left(\int_0^\infty e^{-s\tau}g(\tau)\,d\tau\right)$ Why this step? We use different dummy variables $\sigma,\tau$ so we can merge into one double integral without confusion.

Step 2 — Combine into a double integral. $F(s)G(s)=\int_0^\infty\!\!\int_0^\infty e^{-s(\sigma+\tau)}f(\sigma)g(\tau)\,d\sigma\,d\tau$ Why this step? A product of integrals over independent variables is one double integral over the quarter-plane $\sigma,\tau \ge 0$ .

Step 3 — Change variables to make the exponent look like $e^{-st}$ . Let $t=\sigma+\tau$ (so the exponent becomes $e^{-st}$ ). Keep $\tau$ as the other variable, so $\sigma = t-\tau$ . Why this step? We are aiming for the standard form $\int_0^\infty e^{-st}(\cdots)\,dt$ , which would identify the bracket as a Laplace transform.

The Jacobian of $(\sigma,\tau)\to(t,\tau)$ is $1$ (since $\sigma=t-\tau,\ \tau=\tau$ ). New region: $\tau\ge0$ and $\sigma=t-\tau\ge0\Rightarrow 0\le\tau\le t$ , with $t$ from $0$ to $\infty$ .

$F(s)G(s)=\int_{t=0}^\infty\int_{\tau=0}^{t} e^{-st} f(t-\tau)g(\tau)\,d\tau\,dt$

Step 4 — Recognise the inner integral. $=\int_0^\infty e^{-st}\underbrace{\left[\int_0^t f(t-\tau)g(\tau)\,d\tau\right]}_{=(f*g)(t)}dt=\mathcal{L}\{f*g\}.$ Why this step? The inner integral is exactly the convolution (using $g*f=f*g$ ). Done. $\blacksquare$

APPLICATIONS

App 1 — Inverse Laplace without partial fractions

Worked example

$\mathcal{L}^{-1}\left\{\dfrac{1}{s^2(s^2+1)}\right\}$ Split as a product: $F(s)=\dfrac1{s^2}\Rightarrow f(t)=t$ , and $G(s)=\dfrac1{s^2+1}\Rightarrow g(t)=\sin t$ . Why this step? Each factor has a known simple inverse; convolution stitches them. $\mathcal{L}^{-1}=\int_0^t \tau\,\sin(t-\tau)\,d\tau$ Why this step? Apply the theorem: $(f*g)(t)=\int_0^t f(\tau)g(t-\tau)d\tau$ . Integrate by parts (let $u=\tau$ , $dv=\sin(t-\tau)d\tau\Rightarrow v=\cos(t-\tau)$ ): $=\big[\tau\cos(t-\tau)\big]_0^t-\int_0^t\cos(t-\tau)d\tau = t-\big[\sin(t-\tau)\big]_0^t\cdot(-1)$ Carefully: $\int_0^t \cos(t-\tau)d\tau=\sin(t-\tau)\big|... = \sin t$ . Final: $\boxed{\;t-\sin t\;}$ Check: $\mathcal L\{t-\sin t\}=\frac1{s^2}-\frac1{s^2+1}=\frac{(s^2+1)-s^2}{s^2(s^2+1)}=\frac1{s^2(s^2+1)}.\ \checkmark$

App 2 — Solving ODEs / integral equations

Worked example Solve

$y'' + y = \sin t,\quad y(0)=0,\ y'(0)=0$ Take Laplace: $s^2Y - sy(0)-y'(0) + Y = \dfrac{1}{s^2+1}$ $\Rightarrow (s^2+1)Y=\frac1{s^2+1}\Rightarrow Y=\frac{1}{(s^2+1)^2}$ Why convolution helps? $Y=\dfrac1{s^2+1}\cdot\dfrac1{s^2+1}$ is a product of two known inverses ( $\sin t$ each): $y(t)=\sin t * \sin t=\int_0^t \sin\tau\,\sin(t-\tau)\,d\tau$ Use product-to-sum $\sin A\sin B=\tfrac12[\cos(A-B)-\cos(A+B)]$ with $A=\tau,B=t-\tau$ : $=\frac12\int_0^t[\cos(2\tau-t)-\cos t]\,d\tau=\frac12\left[\frac{\sin(2\tau-t)}{2}-\tau\cos t\right]_0^t$ $=\frac12\left[\frac{\sin t - \sin(-t)}{2}-t\cos t\right]=\frac12\left[\sin t - t\cos t\right]=\boxed{\tfrac12(\sin t - t\cos t)}$

App 3 — Volterra integral equations

COMMON MISTAKES

Recall Feynman: explain to a 12-year-old

Imagine two machines. Machine $f$ drops sand at different times; machine $g$ says how long each grain keeps glowing after it lands. The brightness right now ( $t$ ) is the sum of glows from all grains dropped earlier, each grain dimmed by how much time has passed since it landed. Adding up "what was dropped at time $\tau$ × how much it still glows after $t-\tau$ " — that adding-up is convolution. The magic theorem says: in a special math language (Laplace), this complicated adding-up becomes just multiply two numbers.

Active Recall

Define the convolution

(f*g)(t)

\int_0^t f(\tau)g(t-\tau)\,d\tau

, arguments add to

t

State the convolution theorem.

\mathcal{L}\{f*g\}=F(s)G(s)

, equivalently

\mathcal{L}^{-1}\{F(s)G(s)\}=f*g

Is convolution commutative? Prove the idea.

Yes; substitute

u=t-\tau

to swap

f

and

g

Why does product in

s

-world become convolution, not product, in

t

-world?

Because

F(s)G(s)

is a double integral over

\sigma+\tau=t

diagonals (Jacobian 1), giving

\int_0^t f(t-\tau)g(\tau)d\tau

What is

\mathcal{L}^{-1}\{1/(s^2(s^2+1))\}

t-\sin t

What is

\sin t * \sin t

\tfrac12(\sin t - t\cos t)

Why is the convolution upper limit

t

Causality:

g(t-\tau)=0

for

\tau>t

, so contribution vanishes.

How do convolutions help solve Volterra integral equations?

An integral

\int_0^t y(\tau)k(t-\tau)d\tau

Laplaces to

Y(s)K(s)

, turning the equation into algebra.

Connections

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

Hinglish (regional understanding)

Intuition Hinglish mein samjho

Dekho, Convolution theorem ka core idea bilkul simple hai: Laplace world (jaha sab kuch $s$ ke terms me hai) me agar do functions ka product $F(s)G(s)$ aata hai, to time world me uska inverse koi simple multiplication nahi hota — wo ek special integral hota hai jise convolution kehte hain: $(f*g)(t)=\int_0^t f(\tau)g(t-\tau)d\tau$ . Yaad rakhne ka trick: dono arguments add hoke $t$ ban jaate hain ( $\tau + (t-\tau)=t$ ).

Proof ka soul ye hai: $F(s)G(s)$ ko do alag dummy variables $\sigma,\tau$ ke saath ek double integral bana do, fir $t=\sigma+\tau$ substitute karo (Jacobian 1 hota hai), aur jadoo se andar wala integral exactly convolution ban jaata hai. Geometrically, hum quarter-plane ke diagonals $\sigma+\tau=t$ pe sum kar rahe hain — yahi convolution hai.

Iska faayda kya? Do badi jagah. Pehla: inverse Laplace bina partial fractions ke nikal lo — jaise $1/(s^2(s^2+1))$ ko $1/s^2 \cdot 1/(s^2+1)$ likh ke $t * \sin t = t - \sin t$ . Dusra: ODEs aur Volterra integral equations (jaha integral me $y(\tau)k(t-\tau)$ aata hai) ko algebra me convert karke easily solve karo.

Sabse common galti: students product in $s$ ko product in $t$ samajh lete hain — galat. Product in $s$ = convolution in $t$ . Aur upper limit hamesha $t$ hoti hai, $\infty$ nahi, kyunki causal functions me $g(t-\tau)=0$ jab $\tau>t$ . Bas isko rato mat, "FLIP-and-SLIDE" picture imagine karo — ek function ulta karke doosre ke upar slide karte ho, overlap ka area hi convolution value hai.

Convolution theorem — proof, applications

WHAT is convolution?

THE CONVOLUTION THEOREM

Derivation from scratch (no formula dump)

APPLICATIONS

App 1 — Inverse Laplace without partial fractions

App 2 — Solving ODEs / integral equations

App 3 — Volterra integral equations

COMMON MISTAKES

Active Recall

Connections

Concept Map

Hinglish (regional understanding)

Go deeper — visual, from zero

Test yourself — Ordinary Differential Equations

Connections