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

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

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

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

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


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

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

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

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

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


$$s \rightarrow s+2$$


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

\$$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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