Intuition The one core idea
The convolution theorem says that when you multiply two functions in the Laplace (s ) world, the matching operation back in the time (t ) world is a special "blend-as-you-slide" integral called convolution . Everything on the parent page is just machinery for reading, writing, and trusting that one sentence.
This page assumes you have seen nothing . Before we can even read L { f ∗ g } = F ( s ) G ( s ) , we must earn every squiggle in it: what a function is, what an integral pictures, what that big ∫ 0 ∞ with e − s t actually does , what a dummy variable is, and why "arguments that add to t " is the whole trick. We build them in an order where each one leans only on the ones before it.
area under a curve = integral
improper integral to infinity
the exponential e to the minus s t
shift and flip g of t minus tau
solve ODEs and integral equations
Read it top to bottom: functions and integrals feed everything; the exponential builds the Laplace transform ; shifting/flipping builds convolution ; the two streams meet at the theorem .
A function f is a rule: give it an input number t , it returns exactly one output number, written f ( t ) ("f of t "). In our chapter t is time , always t ≥ 0 (we start the clock at 0 ).
The picture is a graph : the horizontal axis is the input t , the vertical axis is the output f ( t ) . A curve is just "for each t , how tall is f ".
Intuition Why the topic needs this
Convolution, Laplace, integral equations — every object we manipulate is a function of time. If "f ( t ) " is fuzzy, nothing after it can be sharp. Two functions we'll meet constantly: f ( t ) = t (a straight ramp) and g ( t ) = sin t (a wave). Keep those two pictures in your head.
Definition Definite integral
∫ a b f ( t ) d t means: chop the interval from t = a to t = b into ultra-thin vertical strips of width d t , each of height f ( t ) , and add up all their areas . The result is the signed area between the curve and the horizontal axis (area below the axis counts as negative).
The stretched-S symbol ∫ is literally an old-fashioned "S" for Sum . The d t is not decoration — it tells you which variable you are sweeping and reminds you each strip has width "d t ".
Intuition Why the topic needs this
Both the Laplace transform and convolution are defined as integrals . The whole proof on the parent page is just "rearrange a double integral". If you see ∫ as "total accumulated amount = sum of tiny pieces", every later step reads as common sense.
Recall Signed area, all cases
f ( t ) > 0 on [ a , b ] ::: integral is positive (area above axis).
f ( t ) < 0 on part ::: that part subtracts (area below axis).
a = b ::: integral is 0 (no width, nothing to add).
f ( t ) = 0 everywhere ::: integral is 0 .
Definition Dummy (bound) variable
In ∫ 0 t f ( τ ) d τ the letter τ ("tau") is a dummy variable : it only exists inside the integral to label the sliver we are currently adding. Once the sum is done, τ is gone. You could rename it u , x , anything — the answer is identical.
Picture τ as a pointer sweeping from the lower limit to the upper limit, tapping each strip once.
Common mistake Confusing the dummy
τ with the real limit t
In ∫ 0 t f ( τ ) g ( t − τ ) d τ the τ is swept from 0 up to t , but t itself is a fixed number for that whole integral — it's the "now" moment. Why the confusion: they look similar. The fix: τ = "which past moment", t = "the present moment we're evaluating at". They are different roles even though the top limit is t .
Intuition Why the topic needs this
Convolution has τ inside and t outside. The proof of the theorem deliberately uses two different dummy letters σ and τ so two separate integrals can be merged without clashing. Dummy-variable fluency is exactly what lets that merge feel legal.
Definition Integral to infinity
∫ 0 ∞ f ( t ) d t means: take ∫ 0 b f ( t ) d t and let the top edge b run off to infinity. If that area settles down to a finite number, we say the integral converges and equals that number; if it grows without bound, it diverges .
Intuition Why the topic needs this
The Laplace transform (next section) integrates all the way to ∞ . For it to give a finite answer we need something that squashes the far-away part of the curve down to nothing. That squasher is the exponential — which is exactly why it appears.
Definition The exponential decay
e is a fixed number ≈ 2.718 . Raising it to a negative power gives a curve e − s t that starts at 1 (when t = 0 ) and slides down toward 0 as t grows. Bigger s ⇒ faster slide. Here s is a new independent variable (think of it as a "tuning knob"), and t is still time.
this function and not another?
We need a weight that (1) is 1 at the start so it doesn't distort t = 0 , (2) shrinks fast enough to force ∫ 0 ∞ to converge, and (3) turns products into sums of exponents : e − s σ ⋅ e − s τ = e − s ( σ + τ ) . That last property — multiplying exponentials just adds the exponents — is the single algebraic fact that makes the convolution proof work. No other elementary function does all three.
Recall Behaviour of
e − s t in every case (t ≥ 0 )
t = 0 ::: value is e 0 = 1 .
s > 0 , t large ::: value → 0 (decays — good for convergence).
s = 0 ::: value is 1 for all t (no shrinking).
s < 0 ::: value grows , integral may diverge — Laplace needs s large enough.
Definition Laplace transform
L { f } ( s ) = F ( s ) = ∫ 0 ∞ e − s t f ( t ) d t
Read it as: "take the time-function f ( t ) , multiply it by the shrinking weight e − s t , and add up (integrate) over all time." The result is no longer a function of t — it's a new function F ( s ) living in the "s -world". The script L is just the name of this whole machine.
Intuition Why the topic needs this
The convolution theorem is a statement about L : it compares what happens in the t -world versus the s -world. Without knowing F ( s ) means "the reweighted-and-summed version of f ", the equation L { f ∗ g } = F ( s ) G ( s ) is just symbols. Full details of this machine live in Laplace Transform — definition and properties .
Before convolution can make sense, we must understand what plugging t − τ into a function does to its picture .
Definition Reflected-and-shifted argument
g ( − τ ) is g flipped left-to-right (a mirror through τ = 0 ).
g ( t − τ ) is that flipped copy slid right by t .
So as t increases, the flipped g slides across the fixed f .
Intuition Why "arguments add to
t " is the whole trick
Inside f ( τ ) g ( t − τ ) the two inputs sum: τ + ( t − τ ) = t . Picture it as pairing "what f did at time τ " with "how g responds after the remaining time t − τ ". Every pair whose times add to t contributes to the value at t . That pairing-by-a-fixed-total is precisely the FLIP-and-SLIDE mnemonic from the parent page.
Definition Convolution (now fully readable)
( f ∗ g ) ( t ) = ∫ 0 t f ( τ ) g ( t − τ ) d τ
Every symbol is now defined: ∫ 0 t (§2, §4) sums slivers as the dummy τ (§3) sweeps from 0 to t ; f ( τ ) is f at the swept time (§1); g ( t − τ ) is g flipped and slid (§7); the arguments add to t (§7).
Common mistake Why the top limit is
t , not ∞
For causal functions (g ( x ) = 0 when x < 0 , because nothing happens before the clock starts), g ( t − τ ) = 0 as soon as τ > t . So slivers past τ = t add exactly 0 — the integral naturally stops at t . This is the causality point the parent page insists on.
This convolution then powers the applications: solving ODEs (see Solving Linear ODEs with Laplace Transforms ) and Volterra Integral Equations , and it reappears (with different limits) in the Fourier Transform Convolution Theorem and in Transfer Functions and Impulse Response .
1 ∗ 1
Take f ( t ) = 1 and g ( t ) = 1 (constant functions). Then
( 1 ∗ 1 ) ( t ) = ∫ 0 t 1 ⋅ 1 d τ = [ τ ] 0 t = t .
Check via the theorem: L { 1 } = 1/ s , so the product in s -world is s 1 ⋅ s 1 = s 2 1 = L { t } . The t -world blend ( 1 ∗ 1 ) = t matches. ✓ This is the smallest possible demonstration that product in s = convolution in t .
Test yourself — cover the right side of each ::: and see if you can answer instantly.
What does f ( t ) mean in one phrase? A rule turning input number t into exactly one output number; its graph is height-vs-t .
What does ∫ a b f ( t ) d t picture? The signed area under the curve between t = a and t = b = sum of thin strips.
What is a dummy variable like τ ? A name that lives only inside the integral, labelling the current strip; renaming it changes nothing.
Why does the Laplace integral need e − s t ? It's a weight that is 1 at t = 0 , shrinks to force convergence over ∫ 0 ∞ , and turns products of exponentials into sums of exponents.
What is F ( s ) ? The Laplace transform of f : ∫ 0 ∞ e − s t f ( t ) d t , a function in the s -world (not t ).
What are L { t } and L { sin t } ? 1/ s 2 and 1/ ( s 2 + 1 ) .
What does g ( t − τ ) do to the graph of g ? Flips it left-right, then slides it right by t .
Why do the convolution arguments add to t ? τ + ( t − τ ) = t ; each pair of times summing to t contributes to the value at t .
Why is the convolution upper limit t ? Causality — g ( t − τ ) = 0 for τ > t , so those slivers add zero.
What is ( 1 ∗ 1 ) ( t ) ? t .