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Consider the periodic function $F(t), t \geq 0 .$ Of period $w=2$ such
that $F(t)=\left\{\begin{array}{ll}{3,0 \leq t \leq 1} \\ {0,1 \leq t \leq 2} & {\text { Find } L F(t)}\end{array}\right.$

$L[F(t)]=\frac{1}{1-e^{-2 s}} \int_{0}^{2} e^{-s t} F(t) d t$

$=\int_{0}^{1} e^{-s t}(3) d t+\int_{1}^{2} e^{-s t}(0) d(t)$

$=3 \int_{0}^{1} e^{-s t} d t+0$

$=3 \frac{1}{-s}\left[e^{-s t}\right]_{0}^{1}$

$=\frac{-3}{s}\left(e^{-s}-1\right)$

$\mathcal {L}[F(t)]=\frac{1}{1-e^{-2 s}} \int_{0}^{2} e^{-s t} F(t) d(t)$

$\mathcal{L}[F(t)]=\frac{1}{1-e^{-2 S}}-\frac{3}{s}\left(e^{-s}-1\right)$

$=\frac{3}{s\left(1+e^{-s}\right)}$

Find the Inverse Laplace transform $F(t)=L^{-1}[f(s)]$ of the
function $f(s)=\frac{s-2}{s^{2}-4 s+5} e^{-3 s}$

$\mathcal{L}^{-1}\left[\frac{s-2}{s^{2}-4 s+5}\right]=L^{-1}\left[\frac{s-2}{s^{2}-4 s+4+1}\right]$

$=L^{-1}\left[\frac{s-2}{(s-2)^{2}+1}\right]$

$s \rightarrow s+2$

$=L^{-1}\left[\frac{s-2}{(s-2)^{2}+1}\right]$

$=e^{2 t} L^{-1}\left[\frac{s}{s^{2}+1}\right]=e^{2 t} \cos t$

\$L^{-1}\left[\frac{s-2}{s^{2}-4s+5} e^{-3 s}\right]$

$t \rightarrow t-3$

$=e^{2(t-3)} \cos (t-3) u_{3}(t)$

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