To get this into a useful form, we need to multiply numerator and denominator by `(1-e^(-sT))`.

(We will use the basic algebraic identity, `(a+b)(a-b)=a^2 - b^2`.)

This gives

`(1-e^(-sT))/(s(1+e^(-sT)))xx(1-e^(-sT))/(1-e^(-sT))`

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

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

The `(1-e^(-2sT))` part indicates that the inverse function will be periodic, with period `2T`.

So

`f_1(t)` `=Lap^{:-1:}{(1-2e^(-sT)+e^(-2sT))/s}`

`=Lap^{:-1:}{1/s-(2e^(-sT))/s+(e^(-2sT))/s}`

`=u(t)-2u(t-T)+u(t-2T)`

`=[u(t)-u(t-T)]+` `(-1)[u(t-T)-u(t-2T)]`

So `f(t)` will repeat this pattern every `t = 2T`.

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