Laplace Convolution Theorem: Multiply Transforms

On this page: The Principle | Conditions | Failure Modes | EE Questions | Retrieval Practice | Practice Ground | Solve a Problem | Related Guides | FAQ

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The Principle

The move: replace the Laplace transform of a time-domain convolution by the product of the individual Laplace transforms.

The invariant: the transform-domain product represents the same time-domain convolution, but the convolution integral has become algebraic multiplication.

Pattern:

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

Legal route	Illegal route
$\mathcal{L}\{(f*g)(t)\}\Longrightarrow F(s)G(s)$	$\mathcal{L}\{f(t)g(t)\}\not\Longrightarrow F(s)G(s)$

The illegal route looks close because both expressions contain $f$ and $g$. The theorem is about convolution, not pointwise multiplication.

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Conditions of Applicability

Condition: $F(s)=\mathcal{L}\{f(t)\}$; $G(s)=\mathcal{L}\{g(t)\}$; convolution defined

This guide uses the ordinary one-sided Laplace-transform setting for differential equations, where

$$(f*g)(t)=\int_0^t f(\tau)g(t-\tau)\,d\tau.$$

Before applying, check: confirm that the expression inside the transform is a convolution of the same two functions whose transforms are $F(s)$ and $G(s)$.

If the condition is violated: the product $F(s)G(s)$ may represent a different time-domain object, or no licensed convolution theorem step at all.

• The theorem sends convolution in time to multiplication in the transform domain.
• It does not say $\mathcal{L}\{f(t)g(t)\}=F(s)G(s)$.
• If you start from a product $F(s)G(s)$ and return to time, the inverse move is a convolution.
• This guide builds on Laplace Transform Definition and usually appears before an inverse-transform step like Inverse Laplace Transform Relation.

Want the complete framework behind this guide? Read Masterful Learning.

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Common Failure Modes

Failure mode: treat ordinary multiplication $f(t)g(t)$ as if it were convolution -> the product $F(s)G(s)$ is attached to the wrong time-domain structure.

Debug: look for the convolution integral, or the notation $(f*g)(t)$, before multiplying transforms.

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Elaborative Encoding

Use these questions to build deep understanding. (See Elaborative Encoding for the full method.)

Within the Principle

• What does $(f*g)(t)$ mean in the one-sided Laplace setting, and where does the upper limit $t$ appear?
• Why does the transform-domain result use multiplication even though the time-domain expression is an integral?

For the Principle

• What check tells you whether a time-domain expression is eligible for the convolution theorem?
• How would the step change if you were given $F(s)G(s)$ and asked to return to a time-domain expression?

Between Principles

• How does this theorem connect Laplace Transform Definition to Inverse Laplace Transform Relation?

Generate an Example

• Write one legal convolution-transform step and one near miss where a student incorrectly transforms $f(t)g(t)$ as $F(s)G(s)$.

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Retrieval Practice

Answer from memory, then click to reveal and check. (See Retrieval Practice for the full method.)

State the move in one sentence: _____Transform a time-domain convolution into the product of the two individual Laplace transforms.

Write the canonical pattern: _____$\mathcal{L}\{(f*g)(t)\}=F(s)G(s)$

State the canonical condition: _____$F(s)=\mathcal{L}\{f(t)\};\, G(s)=\mathcal{L}\{g(t)\};\, \text{convolution defined}$

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Practice Ground

Use these exercises to build move-selection fluency. (See Self-Explanation for how to learn from worked examples.)

Procedure Walkthrough

Starting from $\mathcal{L}\{(e^t*\sin t)(t)\}$, apply the theorem and reach a product of transforms.

Step	Expression	Operation
0	$\mathcal{L}\{(e^t*\sin t)(t)\}$	Start with a stated convolution inside the Laplace transform.
1	$\mathcal{L}\{e^t\}\mathcal{L}\{\sin t\}$	Apply the convolution theorem.
2	$\frac{1}{s-1}\cdot \frac{1}{s^2+1}$	Substitute the known transforms.
3	$\frac{1}{(s-1)(s^2+1)}$	Write the product as one transform-domain expression.

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Drills

Forward Step

Apply the convolution theorem once. Assume $F(s)=\mathcal{L}\{f(t)\}$, $G(s)=\mathcal{L}\{g(t)\}$, and the convolution is defined.

$$\mathcal{L}\{(f*g)(t)\}$$

Reveal

The convolution transforms to

$$F(s)G(s).$$

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Apply the theorem once. Assume the condition holds, $\mathcal{L}\{e^{2t}\}=\frac{1}{s-2}$, and $\mathcal{L}\{\cos t\}=\frac{s}{s^2+1}$.

$$\mathcal{L}\{(e^{2t}*\cos t)(t)\}$$

Reveal

Use the theorem on the convolution:

$$\mathcal{L}\{e^{2t}\}\mathcal{L}\{\cos t\} =\frac{1}{s-2}\cdot \frac{s}{s^2+1}.$$

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Reject or complete the step. Assume $F(s)=\mathcal{L}\{f(t)\}$ and $G(s)=\mathcal{L}\{g(t)\}$.

$$\mathcal{L}\{f(t)g(t)\}\Longrightarrow F(s)G(s)$$

Reveal

Reject the step. The expression is ordinary multiplication in time, not convolution. The theorem licenses

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

not $\mathcal{L}\{f(t)g(t)\}=F(s)G(s)$.

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Apply the theorem and substitute table transforms. Assume the convolution is defined.

$$\mathcal{L}\{(1*t)(t)\}$$

Reveal

Since $\mathcal{L}\{1\}=\frac{1}{s}$ and $\mathcal{L}\{t\}=\frac{1}{s^2}$,

$$\mathcal{L}\{(1*t)(t)\}=\frac{1}{s}\cdot\frac{1}{s^2}=\frac{1}{s^3}.$$

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Which inputs are eligible for the convolution theorem? Assume all named transforms exist when the input is structurally eligible.

A. $\mathcal{L}\{(f*g)(t)\}$
B. $\mathcal{L}\{f(t)g(t)\}$
C. $\mathcal{L}\left\{\int_0^t e^\tau\sin(t-\tau)\,d\tau\right\}$

Reveal

A and C are eligible. C is the convolution $(e^t*\sin t)(t)$ written as an integral. B is a pointwise product, not a convolution.

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Action Labels

What was done between these two steps? Assume the condition holds.

$$\mathcal{L}\{(f*g)(t)\} \quad \Longrightarrow \quad F(s)G(s)$$

Reveal

The Laplace Convolution Theorem was applied: the transform of a time-domain convolution became the product of the two individual transforms.

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What condition licenses this transition?

$$\mathcal{L}\{(e^t*\sin t)(t)\} \quad \Longrightarrow \quad \frac{1}{s-1}\cdot\frac{1}{s^2+1}$$

Reveal

The condition is that $\frac{1}{s-1}=\mathcal{L}\{e^t\}$, $\frac{1}{s^2+1}=\mathcal{L}\{\sin t\}$, and the convolution $(e^t*\sin t)(t)$ is defined.

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A student claims this transition uses the convolution theorem. What is wrong?

$$\mathcal{L}\{t\sin t\} \quad \Longrightarrow \quad \mathcal{L}\{t\}\mathcal{L}\{\sin t\}$$

Reveal

The input is the product $t\sin t$, not the convolution $(t*\sin t)(t)$. The convolution theorem does not split ordinary products.

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Name the move in this chain. Assume the condition holds.

$$\int_0^t (t-\tau)e^\tau\,d\tau \quad \Longrightarrow \quad \mathcal{L}\left\{\int_0^t (t-\tau)e^\tau\,d\tau\right\} \quad \Longrightarrow \quad \frac{1}{s^2}\cdot\frac{1}{s-1}$$

Reveal

The final transition uses the Laplace Convolution Theorem after recognizing the integral as $(t*e^t)(t)$.

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Transition Identification

Where does the convolution theorem enter this chain?

$$(1*\sin t)(t)=\int_0^t \sin(t-\tau)\,d\tau \quad \Longrightarrow \quad \mathcal{L}\{(1*\sin t)(t)\} \quad \Longrightarrow \quad \frac{1}{s}\cdot\frac{1}{s^2+1}$$

Reveal

It enters in the second transition, where the transform of the convolution becomes the product of the individual transforms.

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What is missing from this worked chain?

$$\mathcal{L}\{(f*g)(t)\} \quad \Longrightarrow \quad F(s)G(s)$$

Reveal

The chain is missing the condition check. It should state that $F(s)=\mathcal{L}\{f(t)\}$, $G(s)=\mathcal{L}\{g(t)\}$, and the convolution is defined before multiplying the transforms.

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Backward Step

Return the product to the time domain as a convolution. Assume $\mathcal{L}\{e^t\}=\frac{1}{s-1}$ and $\mathcal{L}\{\sin t\}=\frac{1}{s^2+1}$.

$$\mathcal{L}^{-1}\left\{\frac{1}{(s-1)(s^2+1)}\right\}$$

Reveal

Factor the product into two known transforms:

$$\frac{1}{(s-1)(s^2+1)}=\frac{1}{s-1}\cdot\frac{1}{s^2+1}.$$

So the inverse transform is the convolution

$$(e^t*\sin t)(t).$$

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Solve a Problem

Apply what you’ve learned with Problem Solving.

Problem: Starting from $\frac{1}{s^2(s+1)}$, return to a time-domain convolution using Laplace Convolution Theorem.

Full solution

Step	Expression	Move
0	$\frac{1}{s^2(s+1)}$	Start from a transform-domain product candidate.
1	$\frac{1}{s^2}\cdot\frac{1}{s+1}$	Factor the expression into two known transform factors.
2	$\mathcal{L}\{t\}\mathcal{L}\{e^{-t}\}$	Identify each factor as a Laplace transform.
3	$\mathcal{L}\{(t*e^{-t})(t)\}$	Use the convolution theorem backward to identify the product as a convolution transform.
4	$\mathcal{L}^{-1}\left\{\frac{1}{s^2(s+1)}\right\}=(t*e^{-t})(t)$	Return to the time-domain convolution.

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Related Guides

• Differential Equations Subdomain - Return to the transform-and-boundary-methods lane.
• Laplace Transform Definition - Review what $F(s)$ and $G(s)$ mean before applying the theorem.
• Inverse Laplace Transform Relation - See how transform-domain products return to time-domain functions.
• Laplace Derivative Transform - Compare convolution-transform moves with derivative-transform moves in IVP work.
• Principle Structures - Keep the name, equation, condition, and legal move separate while studying.

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FAQ

What is Laplace Convolution Theorem?

Laplace Convolution Theorem is the rule $\mathcal{L}\{(f*g)(t)\}=F(s)G(s)$. It says that convolution in the time domain becomes multiplication in the Laplace-transform domain.

When is Laplace Convolution Theorem valid?

It is valid when $F(s)=\mathcal{L}\{f(t)\}$, $G(s)=\mathcal{L}\{g(t)\}$, and the convolution $(f*g)(t)$ is defined. In ordinary one-sided Laplace work, that convolution is $\int_0^t f(\tau)g(t-\tau)\,d\tau$.

Is convolution the same as multiplying functions?

No. Multiplication gives $f(t)g(t)$ at the same time value. Convolution combines shifted values through an integral, so $\mathcal{L}\{f(t)g(t)\}$ is not generally $F(s)G(s)$.

Why is the theorem useful in differential equations?

It lets you recognize some transform-domain products as time-domain convolutions, especially after algebraic solution steps produce products of known transforms. That makes it a bridge between transform-table algebra and time-domain solution forms.

Can I use the theorem backward?

Yes, when the conditions are met. If $F(s)$ and $G(s)$ are the transforms of $f(t)$ and $g(t)$, then $\mathcal{L}^{-1}\{F(s)G(s)\}=(f*g)(t)$.

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How This Fits in Unisium

Within the differential equations subdomain, Unisium treats this as a condition-first transform move: identify convolution structure, verify the two matching transforms, and only then multiply in the $s$ domain. The Unisium Study System pairs that habit with retrieval practice, self-explanation, and compact problem-solving chains so the theorem becomes a selectable move rather than a table entry memorized in isolation.

Ready to practice differential equations with structure? Check access and join the Unisium waitlist or explore the complete framework in Masterful Learning.

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