Question

Find the Laplace transform of e^-at  sin⁡ωt u(t).

(a) \(\frac{s+a}{(s+a)^2-ω^2}\)

(b) \(\frac{ω}{(s+a)^2-ω^2}\)

(c) \(\frac{s+a}{(s+a)^2+ω^2}\)

(d) \(\frac{ω}{(s+a)^2+ω^2}\)

The question was asked during an internship interview.

Question is taken from The Laplace Transform in section Laplace Transform and System Design of Signals and Systems

Answer 1

Right option is (d) \(\frac{ω}{(s+a)^2+ω^2}\)

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

L{x(t)} = X(s) = \(L{e^{-at} \frac{e^{jωt}-e^{-jωt}}{2j} \,u(t)} = \frac{1}{2j} L[e^{-(a-jω)t} \,u(t)] – \frac{1}{2j} L[e^{-(a+jω)t} \,u(t)]\)

X(s) = \(\frac{1}{2j} [\frac{1}{s+(a-jω)} – \frac{1}{s+(a+jω)}] = \frac{1}{2j} [\frac{2jω}{(s+a)^2+ω^2}] = \frac{ω}{(s+a)^2+ω^2}\)

e^^-at  sin⁡ωt u(t) \(\underleftrightarrow{LT} \frac{ω}{(s+a)^2+ω^2}\);ROC Re(s)>-a.