ODE6.3

Find the Laplace transform of $f(t) = \begin{cases} 1 & 0 < t \le 1 \\ -1 & 1 < t \le 2 \end{cases}\,$

Use this property:

$L\{g(t)\} = \frac{\int_0^P e^{-st} g(t)\,dt}{1-e^{-Ps}}\,$ , with period P.

$L\{f(t)\} = \frac{\int_0^2 e^{-st} f(t)\,dt}{1-e^{-2s}}\,$

$= \frac{\int_0^1 e^{-st}(1)\,dt + \int_1^2e^{-st}(-1)\,dt}{1-e^{-2s}}\,$

$= \frac{\frac{-e^{-st}}{s}\bigg|_{t=0}^1+\frac{e^{-st}}{s}\bigg|_{t=1}^2}{1-e^{-2s}}\,$

$= \frac{\frac{1}{s}(1-2e^{-s}+e^{-2s})}{1-e^{-2s}}\,$

$= \frac{\frac{1}{s}(1-e^{-s})^2}{(1+e^{-s})(1-e^{-s})}\,$

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

$= \frac{1}{s}\left(\frac{e^{s/2} - e^{-s/2}}{e^{s/2}+e^{-s/2}}\right)\,$

$= \frac{1}{s} \tanh \frac{s}{2}\,$