Question

Find the Laplace transform of the signal x(t)=e^t  sin⁡2t for t≤0.

(a) \(\frac{2}{(s-1)^2+2^2}\)

(b) \(-\frac{2}{(s-1)^2+2^2}\)

(c) \(\frac{2}{(s+1)^2+2^2}\)

(d) \(-\frac{2}{(s+1)^2+2^2}\)

This question was addressed to me in a job interview.

The doubt is from The Laplace Transform topic in division Laplace Transform and System Design of Signals and Systems

Answer 1

Right answer is (b) \(-\frac{2}{(s-1)^2+2^2}\)

The explanation is: Given x(t) = e^t  sin⁡2t for t≤0

∴    x(t) = e^t  sin⁡2t u(-t)

L{x(t)} = X(s) = \(\int_{-∞}^∞ x(t) e^{-st} \,dt = \int_{-∞}^∞ e^t  \,sin⁡2t \,u(-t) e^{-st} \,dt\)

= \(\int_{-∞}^0 \left(e^{j2t} – e^{-j2t}{2j}\right) = \frac{1}{2j} \int_{-∞}^0 [e^{(1-s+j2)t} – e^{(1-s-j2)t}]dt\)

= \(\frac{1}{2j} \left(\frac{1}{1-s+j2}-\frac{1}{1-s-j2}\right)\)

=\(-\frac{2}{(s-1)^2+2^2}\).