Question

Find the Laplace transform of e^-at u(t) and its ROC.

(a) \(\frac{1}{s-a}\), Re{s}>-a

(b) \(\frac{1}{s}\), Re{s}>a

(c) \(\frac{1}{s×a}\), Re{s}>a

(d) \(\frac{1}{s+a}\), Re{s}>-a

The question was asked in an interview for job.

My question comes from The Laplace Transform topic in section Laplace Transform and System Design of Signals and Systems

Answer 1

Right choice is (d) \(\frac{1}{s+a}\), Re{s}>-a

Easiest explanation: Laplace transform, L{x(t)} = X(s) = \(\int_{-∞}^∞ x(t) e^{-st} \,dt\)

L{e^-at) u(t)} = \(\int_{-∞}^∞ e^{-at} u(t) e^{-st} \,dt = \int_0^∞ e^{-at} e^{-st} \,dt =  \frac{1}{s+a}\)  when (s+a)>0

(σ+a)>0

σ>-a

ROC is Re{s}>-a.