We observe that the Laplace inverse of this function will be periodic, with period T.

This is because of the part:

`1/(1-e^(-sT))`

We find the function for the first period [`f_1(t)`] by ignoring the `(1-e^((1-s)T))` part:

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

`=Lap^{:-1:}{(1)/(s-1)}-Lap^{:-1:}{(e^((1-s)T))/(s-1)}`

Now

`Lap^{:-1:}{(1)/(s-1)}` `=e^t*u(t)`

Considering the second fraction, we have:

`e^((1-s)T)=e^(T-sT)`,

which we can think of as:

`e^(T-sT)=e^Te^(-sT)`,

So

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

which is in the form:

`e^Txxe^(-as)G(s)`,

where `a = T`.

So we can use Rule (4) again:

`Lap^{:-1:}{(e^((1-s)T))/(s-1)}` `=e^T xx Lap^{:-1:}{e^(-sT) xx1/(s-1)}`

`G(s)=1/(s-1)` and so `g(t)=Lap^{:-1:}{(1)/(s-1)}=e^t`.

This gives `g(t-T)=e^(t-T)`

Using Rule (4):

`Lap^{:-1:}{e^(-sT) xx1/(s-1)}` `=e^(t-T)*u(t-T)`

`e^TxxLap^{:-1:}{e^(-sT) xx1/(s-1)}`

`=e^T xx e^(t-T)*u(t-T)`

`=e^t *u(t-T)`

So

`f_1(t) =e^t *u(t)-e^t *u(t-T)` `=e^t*[u(t)-u(t-T)]`

So the periodic function with `f(t)=f(t+T)` has the following graph:

t f(t) T 2T 3T 4T 5T 1
eT

Graph of `f(t)=e^t*[u(t)-u(t-T)]`, with `f(t)=f(t+T)`.