Our function is `f(t)=e^(t-a)`. This is an exponential function (see Graphs of Exponential Functions).

When `t = a`, the graph has value `e^(a-a)= e^0= 1`.

t f(t) a b 1
eba

Graph of `f(t)=e^(t-a)*{u(t-a)-u(t-b)}`.

The function has the form:

`f(t)=e^(t-a)*{u(t-a)-u(t-b)}`

We will use the Time Displacement Theorem:

`Lap{u(t-a)*g(t-a)}=e^(-as)G(s)`

Now, in this example, `G(s)=` `Lap{e^t}=1/(s-1)`

`Lap{e^(t-a)*[u(t-a)-u(t-b)]}`

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

We now make use of a trick, by noting `(t-a) = (b-a ) + (t-b)` and re-writing `e^(t-a)` as `e^(b-a)e^(t-b)`:

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

[We have introduced eb−a, a constant, for convenience.]

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

[Each part is now in the form `u(t − c) · g(t − c)`, so we can apply the Time Displacement Theorem.]

`=e^(-as)xx1/(s-1)` `-e^(b-a)xxe^(-bs)xx1/(s-1)`

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

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

Easy to understand math videos:
MathTutorDVD.com