Using a translation theorem to determine that
$ℒ{f(t-a)u(t-a)}$ is equivalent to $F(s) e^{-as}$,
I determined that $ℒ\sin(2t)u(t-\pi)$ is equal to
$ℒ\sin(2t) * e^{-(\pi)t}$
and found the Laplace transform to be
$2/(s^2+4) * e^{-(\pi)t}$
is this correct?
$\endgroup$ 31 Answer
$\begingroup$Your answer is correct, but only because $\sin(x)=\sin(x+2\pi)$.
${\cal L}(\sin(2t)u_\pi(t)) = e^{-\pi s} \cdot {\cal L}(\sin(2(t+\pi)))= e^{-pi s}\cdot{\cal L}(\sin(2t + 2\pi)) = e^{-\pi s}\cdot{\cal L}(\sin(2t)) = e^{-\pi s}\cdot {2\over s^2+4}$
$\endgroup$ 4
