Convolution theorem — proof, applications: Question bank

The single golden rule to keep in your head while answering everything below:

True or false — justify

Every statement below is either subtly right or subtly wrong. Give the reason.

L^{- 1} {F (s) G (s)} = f (t) g (t)

, the ordinary pointwise product.

False. A product in the

s

-world becomes a convolution in the

t

-world, not a pointwise product. Test:

L {1} \cdot L {1} = 1/ s^{2}

, but

1 \cdot 1 = 1

has transform

1/ s \neq = 1/ s^{2}

; whereas

1 * 1 = t

has transform

1/ s^{2}

. ✓

Convolution is commutative, so

(f * g) (t) = (g * f) (t)

for all

t \geq 0

True. The substitution

u = t - τ

swaps which function carries

τ

and which carries

t - τ

, and flips the limits back, giving

g * f

. You may always put the easier-to-integrate factor "inside".

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

False. The theorem is about the convolution

f * g

, not the pointwise product

f \cdot g

. There is no simple formula for

L {f \cdot g}

in general — that direction requires a complex-plane (Bromwich) convolution, a much harder object.

The upper limit of the convolution integral can be replaced by

\infty

since Laplace transforms integrate to

\infty

False. The upper limit is

t

because for causal functions

g (t - τ) = 0

once

τ > t

(the argument

t - τ

would go negative). Extending to

\infty

would add fictitious contributions from "the future".

Convolution is associative:

(f * g) * h = f * (g * h)

True. In the

s

-world this is just

(F G) H = F (G H)

, ordinary multiplication of functions, which is associative. Transforming back gives associativity in the

t

-world for free.

f

and

g

are both continuous, then

f * g

is continuous.

True. Convolution is a smoothing (averaging) operation: it generally makes results smoother than the inputs, never rougher, so continuity is preserved and often improved.

δ * f = f

, where

δ

is the unit impulse.

True. The impulse is the identity element for convolution — sliding a spike across

f

just reads off

f

at that instant. In the

s

-world

L {δ} = 1

, and

1 \cdot F (s) = F (s)

. ✓

1 * 1 = 1

False.

1 * 1 = \int_{0}^{t} 1 \cdot 1 d τ = t

, not

1

. The constant

1

is not the identity for convolution — the impulse

δ

is. Confusing "multiply by 1" with "convolve with 1" is a classic slip.

The convolution theorem requires

f

and

g

to have the same Laplace transform.

False. They can be any two functions each possessing a Laplace transform; the theorem stitches together different

F (s)

and

G (s)

. The

sin t * sin t

example just happens to use two equal factors.

F (s) G (s)

is a rational function, convolution and partial fractions must give the same inverse.

True. They are two routes to the same unique inverse Laplace transform (which is unique among continuous functions). Convolution is often faster when the factors have clean inverses; partial fractions when they don't split naturally.

Spot the error

Each line contains a planted mistake. Find it and state the correct version.

(f * g) (t) = \int_{0}^{t} f (τ) g (τ) d τ

Both factors carry

τ

— that is a pointwise product, not convolution. One factor must be flipped and slid:

g (t - τ)

, so the arguments add to

t

(f * g) (t) = \int_{0}^{\infty} f (τ) g (t - τ) d τ

Wrong upper limit. For causal functions on

[0, \infty)

it must be

\int_{0}^{t}

, because

g (t - τ) = 0

when

τ > t

"Since

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

, we get

L^{- 1} {F (s) + G (s)} = (f * g) (t)

The theorem is about products

F (s) G (s)

, not sums. For a sum, linearity gives

L^{- 1} {F + G} = f (t) + g (t)

— the ordinary sum, no convolution.

"In the proof, changing from

(σ, τ)

(t, τ)

introduces a Jacobian factor of

s

The Jacobian is

1

, not

s

. Since

σ = t - τ

and

τ = τ

, the transformation is a shear with determinant

1

; area is preserved, so no extra factor appears.

"Because convolution smears things,

y (t) = t + \int_{0}^{t} y (τ) sin (t - τ) d τ

can't be solved by Laplace."

On the contrary — the integral is exactly

y * sin t

, so it Laplaces to

Y \cdot \frac{1}{s ^{2} + 1}

, turning the integral equation into simple algebra for

Y

L^{- 1} {1/ (s^{2} + 1)^{2}} = sin t \cdot sin t = sin^{2} t

Wrong operation — it is

sin t * sin t

, the convolution, which equals

\frac{1}{2} (sin t - t cos t)

, not

sin^{2} t

"The new integration region after

t = σ + τ

0 \leq t \leq τ

for each fixed

τ

Reversed. The constraints

σ = t - τ \geq 0

and

τ \geq 0

give

0 \leq τ \leq t

with

t

ranging

0

\infty

— the inner variable

τ

is bounded by

t

, not the other way round.

Why questions

Answer with the mechanism, not just a restatement.

Why does a product in the

s

-world correspond to an integral in the

t

-world at all?

Because

F (s) G (s)

is secretly a double integral over the quarter-plane, and grouping it by the diagonals

σ + τ = t

collects all

(σ, τ)

pairs summing to a given

t

— that "sum along a diagonal" is precisely the convolution integral.

Why do we deliberately use different dummy variables

σ

and

τ

when writing

F (s) G (s)

as integrals?

So the two integrals live on independent axes and can be merged into one honest double integral. Reusing the same letter would falsely tie the two integration variables together.

Why is the substitution

t = σ + τ

(rather than, say,

t = σ τ

) the "right" move in the proof?

We are steering toward the standard Laplace form

\int_{0}^{\infty} e^{- s t} (\dots) d t

. Since the exponent is

e^{- s (σ + τ)}

, choosing

t = σ + τ

makes it

e^{- s t}

exactly — the exponent tells us what the new time variable must be.

Why does convolution "help" a Volterra integral equation but a pointwise-product integral wouldn't?

A Volterra kernel appears as

\int_{0}^{t} y (τ) k (t - τ) d τ

— arguments adding to

t

, i.e. a convolution — so it Laplaces to the product

Y (s) K (s)

. A genuine pointwise product

\int y (τ) k (τ) d τ

has no such clean transform.

Why can you swap which factor gets the

t - τ

argument when computing a convolution by hand?

Commutativity (

f * g = g * f

) lets you flip the roles freely, so you park the harder function at

g (t - τ)

or the easier one — whichever integral is simpler — without changing the answer.

Why does convolving a function with the unit step

u (t)

give its running integral

\int_{0}^{t} f (τ) d τ

Because

u (t - τ) = 1

for all

τ \leq t

and

0

beyond, so

f * u = \int_{0}^{t} f (τ) \cdot 1 d τ

— the step acts as a "keep everything up to now" mask, which is integration. In the

s

-world this is

F (s) \cdot \frac{1}{s}

, the known integration rule.

Edge cases

Boundary and degenerate inputs — the scenarios the formula must survive.

What is

(f * g) (0)

for any bounded

f, g

Zero. The integral

\int_{0}^{0} (\dots) d τ

collapses to an empty interval, so every convolution starts at

0

— a useful sanity check on any answer.

What is

f * 0

(convolution with the zero function)?

The zero function. The integrand is identically zero, so the integral is

0

for all

t

. In the

s

-world this is

F (s) \cdot 0 = 0

. ✓

c

(a constant) an identity for convolution, i.e. does

c * f = f

for some

c

No constant works —

c * f = c \int_{0}^{t} f (τ) d τ

is an integral of

f

, not

f

itself. The identity is the impulse

δ

, which is not an ordinary function at all.

f

grows like

e^{2 t}

and

g

e^{3 t}

, does

f * g

still exist?

Yes as an integral for each finite

t

(a finite interval

[0, t]

always integrates a continuous function). The transform

F (s) G (s)

then exists only for

Re (s) > 5

, the sum of the two abscissas of convergence.

What does convolving a spiky function with a wide smooth one do to the spike?

It smears (blurs) the spike out over the width of the smooth function — convolution is a weighted local average, so sharp features get softened. This is exactly the "glowing sand grain" picture from the parent note.

g (t - τ)

is a delayed step turning on only at

t = a

(i.e.

g (t) = u (t - a)

), how does the lower effective limit change?

Contributions start only once

t - τ \geq a

, i.e.

τ \leq t - a

, so the effective integral becomes

\int_{0}^{t - a} f (τ) d τ

and is exactly

0

for

t < a

— the delay simply postpones all output. This is the

s

-world shift

e^{- a s} / s

Recall One-line summary to carry away

Every trap on this page is a disguise of the same two ideas: (1) product in $s$ ↔ convolution (not product) in $t$ , and (2) the integrand's arguments must add to $t$ , with the integral running $0$ to $t$ because functions are causal. Check any convolution claim against these two, and you will rarely be fooled.

Connections

Parent: full proof & worked examples
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