Question

Find the Laplace transforms of
f(t)={4,0<=t<=1
{3,t>1.

Answer 1

The Laplace transform is $F(s) = \int\limits_0^{\infty} e^{-st}f(t)\,dt.$

$F(s) = \begin{cases} \int\limits_0^{1} e^{-st}4\,dt, & 0\le t \le 1, \\ \int\limits_1^{\infty} e^{-st}3\,dt, & 1 < t \end{cases}\\$

$F(s) = \begin{cases} \dfrac{4}{s}\left(1 -{ e}^{-s}\right) , & 0\le t \le 1, \\[0.4cm] \dfrac{3}{s}\left(e^{-s} - \lim\limits_{t\to\infty}e^{-st} \right) , & 1 < t \end{cases}$

The last limit is finite only for s > 0 and transforms into $\dfrac{3}{s}e^{-s}.$