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




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


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



`=[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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