Questions: Convolution Theorem

Question 1

You need to find ℒ⁻¹[1/(s(s+3))]. Recognizing F(s) = 1/s and G(s) = 1/(s+3), which approach correctly uses the convolution theorem?

• A. Multiply the individual inverses: f(t)·g(t) = 1·e^{-3t} = e^{-3t}
• B. Compute the convolution integral: ∫₀ᵗ e^{-3(t-τ)} dτ = (1 - e^{-3t})/3
• C. Add the individual inverses: f(t) + g(t) = 1 + e^{-3t}
• D. The theorem only applies when F(s) and G(s) have the same denominator

Answer: (B)

The convolution theorem states ℒ⁻¹[F(s)G(s)] = (f * g)(t) = ∫₀ᵗ f(τ)g(t-τ) dτ, not f(t)·g(t). With f(t) = 1 and g(t) = e^{-3t}, the convolution is ∫₀ᵗ 1·e^{-3(t-τ)} dτ = e^{-3t}∫₀ᵗ e^{3τ} dτ = e^{-3t}·[e^{3τ}/3]₀ᵗ = (1 - e^{-3t})/3. Option A is the most tempting error: the inverse Laplace transform distributes over addition but NOT over multiplication. You can verify the answer via partial fractions: 1/(s(s+3)) = (1/3)(1/s - 1/(s+3)), giving inverse (1 - e^{-3t})/3.

Question 2

In the convolution integral y(t) = ∫₀ᵗ h(t-τ)g(τ) dτ describing a driven system, what is the physical meaning of h(t-τ)?

• A. The current value of the forcing function at time t
• B. The system's response to a unit impulse applied at time τ, evaluated at current time t
• C. The transfer function evaluated at the frequency corresponding to time t
• D. The average of the forcing function over the interval [0, t]

Answer: (B)

h(t-τ) is the impulse response of the system — how the system responds to a unit spike applied at time τ, observed at the current time t. The convolution integral accumulates these responses: each past input g(τ) contributes an impulse response h(t-τ) scaled by g(τ), and all contributions are summed. This reveals that the system's output at time t is a superposition of decaying responses to every past input — the system has 'memory' of all previous forcing.

Question 3

If ℒ[f] = F(s) and ℒ[g] = G(s), then ℒ⁻¹[F(s)G(s)] = f(t)g(t).

• T. True
• F. False

Answer: False

This is the central misconception of the convolution theorem. The inverse Laplace transform distributes over addition: ℒ⁻¹[F(s) + G(s)] = f(t) + g(t). But it does NOT distribute over multiplication. The correct result is ℒ⁻¹[F(s)G(s)] = (f * g)(t) = ∫₀ᵗ f(τ)g(t-τ) dτ. Multiplication in the s-domain corresponds to convolution in the t-domain, not pointwise multiplication.

Question 4

Convolution is commutative: (f * g)(t) = (g * f)(t) for all t ≥ 0.

• T. True
• F. False

Answer: True

Since F(s)G(s) = G(s)F(s) in the s-domain, and the convolution theorem identifies ℒ⁻¹[F(s)G(s)] = (f * g)(t), commutativity follows from the commutativity of multiplication. It can also be verified directly by substituting u = t - τ in the integral: ∫₀ᵗ f(τ)g(t-τ) dτ = ∫₀ᵗ f(t-u)g(u) du = (g * f)(t).

Question 5

Explain why the convolution theorem is useful for inverting a Laplace transform of the form Y(s) = F(s)G(s), and what the alternative approach would be.

Model answer: When Y(s) factors as F(s)G(s) and both individual inverses f(t) and g(t) are known, the convolution theorem gives the inverse directly as the convolution integral ∫₀ᵗ f(τ)g(t-τ) dτ, without needing to algebraically combine F(s)G(s) into a form amenable to a table lookup. The alternative is partial fraction decomposition, which works when F(s)G(s) can be decomposed into simpler rational terms — but convolution is preferable when the forcing function is complicated or when the factored form already reveals physical structure (impulse response and input).

Both approaches give the same answer, as the worked example in the Explainer confirms. The deeper value of convolution is conceptual: it shows that the output is a superposition of the system's impulse response over all past inputs, revealing causality and memory in the system's behavior. This interpretation is invisible when you just apply partial fractions and look up table entries.