Visual walkthrough — Convolution theorem — proof, applications

If any word below is new, the parent topic note and Laplace Transform — definition and properties are the places to look first — but you shouldn't need them; we rebuild everything here.

Step 0 — The three characters (read this or nothing else works)

Before Step 1, meet the symbols. Skipping this makes every later picture meaningless.

Here are the two functions and their two transforms we'll fuse:

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

Notice I gave them different placeholder letters $\sigma$ and $\tau$ on purpose. Step 2 will show why.

Step 1 — Draw the two transforms as areas

WHAT. Picture $F(s)$ as the area under the curve $e^{-s\sigma }f(\sigma )$, and $G(s)$ as the area under $e^{-s\tau }g(\tau )$. Two separate 1-D pictures, two separate axes.

WHY. You can't "combine" two things until you can see them side by side and notice they live on independent axes — one runs along $\sigma$, the other along $\tau$. Independence is the permission slip for Step 2.

PICTURE. Left graph sweeps $\sigma$; right graph sweeps $\tau$. Neither knows about the other yet.

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

Each is one number: the shaded area. Their product $F(s)G(s)$ is just (area)×(area).

Step 2 — Two areas multiplied = one volume over a plane

WHAT. Multiply the two integrals. A product of "sum over $\sigma$" times "sum over $\tau$" is a single "sum over every pair $(\sigma ,\tau )$": $F(s)G(s)={\int }_0^{\infty }{\int }_0^{\infty }{e}^{-s\sigma }f(\sigma )e^{-s\tau }g(\tau )d\sigma d\tau .$

WHY. This is the ordinary rule $\left({\sum }_i{a}_i\right)\left({\sum }_j{b}_j\right)={\sum }_{i,j}{a}_i{b}_j$, but for integrals. Because $\sigma$ and $\tau$ never interfere (Step 1's independence), the double sum is legal. The exponentials combine: $e^{-s\sigma }{e}^{-s\tau }=e^{-s(\sigma +\tau )}$.

PICTURE. We now live on a flat quarter-plane: horizontal axis $\sigma \ge 0$, vertical axis $\tau \ge 0$. Above each point $(\sigma ,\tau )$ sits a little tower of height $e^{-s(\sigma +\tau )}f(\sigma )g(\tau )$. The product is the total volume of all those towers.

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

Look at the annotated exponent: it only cares about the sum $\sigma +\tau$, not about $\sigma$ and $\tau$ separately. That observation is the seed of the whole theorem.

Step 3 — Name the sum: introduce $t=\sigma +\tau$

WHAT. The exponent $e^{-s(\sigma +\tau )}$ is begging to be written $e^{-st}$. So define a new variable $t=\sigma +\tau .$ Keep $\tau$ as the second variable. Then automatically $\sigma =t-\tau$.

WHY. We are hunting for the shape ${\int }_0^{\infty }{e}^{-st}(\text{something})dt$, because that shape is, by Step 0's definition, "the Laplace transform of something." If we can bend our double integral into it, the bracket must be ${\mathcal{L}}^{-1}$ of $F(s)G(s)$ — which is exactly what we want to find.

PICTURE. The lines "$\sigma +\tau =\text{constant}$" are the diagonals running top-left to bottom-right. Each diagonal is one fixed value of $t$. Coloured bands = fixed $t$, sweeping outward.

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

Now every point on the plane is addressed by (which diagonal $t$, how far along it $\tau$) instead of $(\sigma ,\tau )$.

Step 4 — Does the area stretch? The Jacobian check

WHAT. When we relabel points, tiny area elements can grow or shrink. The Jacobian $J$ is the number by which $d\sigma d\tau$ scales to become $dtd\tau$. We must compute it or the volume comes out wrong.

WHY. Skipping this is the single most common way people "prove" the theorem and get a stray factor. Our change is $\sigma =t-\tau ,\tau =\tau$ — a shear (a slanting slide), and shears preserve area, so we expect $J=1$. Let's confirm, not assume.

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

• $\partial \sigma /\partial t=1$: push the diagonal out by $1$, $\sigma$ grows by $1$.
• $\partial \sigma /\partial \tau =-1$: slide along the diagonal, $\sigma$ drops as $\tau$ rises (they trade off, sum fixed).
• Bottom row: $\tau$ is just itself.

PICTURE. A little square grid gets slanted into parallelograms of the same area — that's what $J=1$ looks like. No squishing, no stretching.

So $d\sigma d\tau =1\cdot dtd\tau$. The volume is unchanged; only our description changed.

Step 5 — Fix the new limits (this is where causality lives)

WHAT. Under the old picture $\sigma \ge 0$ and $\tau \ge 0$. Translate those into the new coordinates $(t,\tau )$:

• $\tau \ge 0$ stays $\tau \ge 0$.
• $\sigma \ge 0$ becomes $t-\tau \ge 0$, i.e. $\tau \le t$.
• Together: $0\le \tau \le t$, and $t$ itself runs from $0$ to $\infty$.

WHY. The upper limit $t$ (not $\infty$!) is not a choice — it is forced by $\sigma =t-\tau \ge 0$. This is exactly the "causality" the parent note warns about: you can't have used more time than has passed. On the diagonal at level $t$, $\tau$ can only slide between the two axes, from $0$ up to $t$.

PICTURE. The whole quarter-plane, re-sliced into diagonal segments. Each segment (fixed $t$) starts on the $\tau$-axis and ends on the $\sigma$-axis; its length in $\tau$ is $0$ to $t$.

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

Term by term: $e^{-st}$ is constant along one diagonal (that's why we chose $t$); $f(t-\tau )g(\tau )$ is the tower height re-expressed; the inner integral marches along a diagonal, the outer integral marches across diagonals.

Step 6 — Recognise the convolution, and finish

WHAT. Pull $e^{-st}$ out of the inner integral (it doesn't depend on $\tau$) and stare at what's left: $F(s)G(s)={\int }_0^{\infty }{e}^{-st}\underset{\text{call this }h(t)}{\underbrace{\left[{\int }_0^{t}f(t-\tau )g(\tau )d\tau \right]}}dt.$

WHY. The bracket $h(t)$ is a function of $t$ alone (the $\tau$ got integrated away). And ${\int }_0^{\infty }{e}^{-st}h(t)dt$ is literally the Laplace transform of $h$ (Step 0's definition). So $F(s)G(s)=\mathcal{L}\{h\}$.

Now identify $h$. Compare with the parent's definition $(g\ast f)(t)={\int }_0^{t}g(\tau )f(t-\tau )d\tau$. Our bracket is exactly that — and since convolution is commutative ($g\ast f=f\ast g$), we may write it either way: $h(t)=(f\ast g)(t).$

PICTURE. One diagonal, extracted: it collects the product $f(t-\tau )g(\tau )$ as $\tau$ slides from $0$ to $t$. Flip $g$, slide it across $f$, multiply, add — that summed diagonal is the value $(f\ast g)(t)$.

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

Step 7 — Degenerate & edge cases (never leave a gap)

The one-picture summary

Everything above, compressed: two 1-D areas → one flat quarter-plane of towers → re-sliced along diagonals $\sigma +\tau =t$ (area-preserving shear) → each diagonal is one convolution value → summed with weight $e^{-st}$ = a 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