Region of convergence for the Laplace transform of exp(a t) with a > 0 In continuous-time process control, the unilateral Laplace transform of f(t) = exp(a t) exists only for certain values of the complex variable s. For which condition on s is the transform defined, and why is this restriction necessary?

Introduction / Context:
Laplace transforms are a cornerstone of control systems and process dynamics. They convert time-domain signals into the complex s-domain, enabling straightforward algebraic manipulation of differential equations. However, the transform only exists where the defining integral converges. This question tests your understanding of the region of convergence for the classic exponential input f(t) = exp(a t) with a > 0.

Given Data / Assumptions:

• Unilateral Laplace transform definition: F(s) = ∫₀^∞ f(t) * exp(−s t) dt.
• Function of interest: f(t) = exp(a t), with a > 0.
• We consider real s for convergence insight (extension to complex s uses Re{s}).

Concept / Approach:
Substituting f(t) = exp(a t) gives F(s) = ∫₀^∞ exp((a − s) t) dt. An improper integral of the form ∫ exp(α t) dt from 0 to ∞ converges only when α < 0. Therefore, we require a − s < 0, i.e., s > a. In complex analysis terms, the region of convergence is Re{s} > a. Only within this half-plane does the integral produce a finite result.

Step-by-Step Solution:

Verification / Alternative check:
Differentiate F(s) w.r.t. s or evaluate a numerical approximation for s slightly greater/less than a; you will see divergence as s approaches a from above, confirming the boundary Re{s} = a.

Why Other Options Are Wrong:

Common Pitfalls:
Confusing existence of a closed-form result with convergence; F(s) = 1/(s − a) is valid only where the integral converges. Also, forgetting that for general complex s the condition becomes Re{s} > a, not simply s > a on the real line.

Final Answer:
the Laplace transform integral of exp(a t) has finite value only when s > a