From the graph, we see that the first period is given by:

`f_1(t)=t*[u(t)-u(t-1)]` and that the period is `p=2`.

`Lap{f_1(t)}`

`= Lap{t*[(u(t)-u(t-1)]}`

`= Lap{t*u(t)}- Lap{t*u(t-1)}`

Now

`t*u(t-1)` `=(t-1)*u(t-1)+u(t-1)`

So

`Lap{t*u(t)}- Lap{t*u(t-1)}`

`= Lap{t*u(t)}-` ` Lap{(t-1)*u(t-1)+u(t-1)}`

`= Lap{t*u(t)}- ` `Lap{(t-1)*u(t-1)}-` ` Lap{u(t-1)}`

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

`=(1-e^(-s)-se^(-s))/(s^2)`

Hence, the Laplace transform of the periodic function, f(t) is given by:

`Lap{f(t)}` `=((1-e^(-s)-se^(-s))/s^2)xx1/(1-e^(-2s))`

`=(1-e^(-s)-se^(-s))/(s^2(1-e^(-2s))`