Laplace transform of a rectangular pulse: For a pulse of constant magnitude E applied from t = 0 to t = a, determine L{v(t)}.

Introduction / Context:In signals and systems, the Laplace transform of time-limited rectangular pulses is a staple result used in control, communications, and circuit analysis. Recognizing how to build a pulse from step functions lets us derive the transform quickly and correctly.

Given Data / Assumptions:

• The signal is a rectangular pulse of constant magnitude E starting at t = 0 and ending at t = a.
• Use unilateral Laplace transform with Re{s} large enough for convergence.
• Unit step u(t) indicates switching actions.

Concept / Approach:

A finite-duration pulse can be written as the difference of two steps: E[u(t) − u(t − a)]. The Laplace transform of u(t) is 1/s (for Re{s} > 0), and the time shift property gives L{u(t − a)} = e^{-a s}/s. Linearity then yields the pulse transform.

Step-by-Step Solution:

Verification / Alternative check:

Direct integration also works: V(s) = ∫₀^a E e^{-s t} dt = E(1 − e^{-a s})/s, which matches the step-based result exactly.

Why Other Options Are Wrong:

• E(1 − e^{-a s}) lacks the essential 1/s factor.
• E/(s + a) is the transform of E e^{-a t} u(t), not a finite pulse.
• (E a)/s corresponds to the transform of a constant of area Ea, not a finite \u201cburst\u201d in time.
• E e^{-a s}/s is the transform of a delayed step of magnitude E, not a pulse between 0 and a.

Common Pitfalls:

• Forgetting to subtract the delayed step, which removes the signal after t = a.
• Dropping the 1/s factor that always accompanies step transforms.

Final Answer:

(E / s) * (1 - e^{-a s})