We can see from the graph that

`f_1(t)=a/bt*[u(t)-u(t-b)]`

and that the period is `p = b`.

So we have

`Lap{f_1(t)}`

`= Lap{a/bt*[u(t)-u(t-b)]}`

`=a/b Lap{t*u(t)-t*u(t-b)}`

(We next subtract, then add a "`b`" term in the middle, to achieve the required form.)

`=a/b Lap{t*u(t)-` `{:(t-b+b)*u(t-b)}`

`=a/b Lap{t*u(t)-` `(t-b)*u(t-b) -` `{: b*u(t-b)}`

(We now find the Laplace Transform of the individual pieces.)

`=a/b[ Lap{t*u(t)}-` ` Lap{(t-b)*u(t-b)}-` `{: Lap{b*u(t-b)}{:]`

`=a/b(1/s^2-(e^(-bs))/s^2-(be^(-bs))/s)`

`=(a(1-e^(-bs)-bse^(-bs)))/(bs^2)`

So the Laplace Transform of the periodic function is given by

`Lap{f(t)}` `=(a(1-e^(-bs)-bse^(-bs))) / (bs^2(1-e^(-bs))`

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