Here is the graph of our function.

π1tf(t)Open image in a new page

Graph of `f(t) = sin t * [u(t) − u(t − π)]`.

The function can be described using Unit Step Functions, since the signal is turned on at `t = 0` and turned off at `t=pi`, as follows:

`f(t) = sin t * [u(t) − u(t − π)]`

Now for the Laplace Transform:

`Lap{sin\ t * [u(t)-u(t-pi)]}` `=` `Lap{sin\ t * u(t)}- ` `Lap{sin\ t * u(t - pi)}`

Now, we need to express the second term all in terms of `(t - pi)`.

From trigonometry, we have:

`sin(t − π) = -sin\ t`

So we can write:

`Lap{sin\ t * u(t)}- ` `Lap{sin\ t * u(t - pi)}`

`= ` `Lap{sin\ t * u(t)}+ ` `Lap{sin(t - pi)* u(t - pi)}`

`=1/(s^2+1)+(e^(-pis))1/(s^2+1)`

`=(1+e^(-pis))/(s^2+1)`

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