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- Jan 26, 2012

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**Problem**: Let $\mathcal{L}\{g(t)\}$ denote the Laplace transform of $g(t)$. Show that

\[\mathcal{L}\left\{\left\lfloor \frac{t}{a} \right\rfloor\right\} = \frac{e^{-as}}{s(1-e^{-as})}\]

where $a>0$ and $\lfloor x\rfloor$ denotes the floor function (i.e. the greatest integer less than or equal to $x$).

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**Hint**:

The following formula will come in handy when taking the Laplace transform of the periodic function:

*If $f(t)$ is piecewise continuous and $p$-periodic, and $f_p(t)$ denotes one period of $f(t)$, then*

\[\mathcal{L}\{f(t)\} = \frac{1}{1-e^{-as}}\int_0^p e^{-st}f_p(t)\,dt = \frac{\mathcal{L}\{f_p(t)\}}{1-e^{-as}}\]

[HR][/HR]

Can't figure out the correct functions to use? It's alright...click the next spoiler to find out what I'm looking for!

\[\left\lfloor \frac{t}{a} \right\rfloor = \frac{t}{a} - f(t)\]

where $f(t)$ is the $a$-periodic sawtooth function pictured below.

\[u(t)=\begin{cases}1 & t\geq 0\\ 0 & t<0\end{cases}\] is the Heaveside step function.

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