Consider the expression (s2+6s-9)/(s(s2-9)).
This can be expressed in partial fractions:
(s2+6s-9)/(s(s2-9))=(s2+6s-9)/(s(s-3)(s+3))=A/s+B/(s-3)+C/(s+3).
A, B and C are constants.
A(s2-9)+B(s2+3s)+C(s2-3s)≡s2+6s-9,
So A+B+C=1 (s2 coefficients)
3B-3C=6 (s coefficients)
-9A=-9 (constant)
Therefore A=1, C=-B, 6B=6, B=1, C=-1.
Partial fractions: 1/s+1/(s-3)-1/(s+3).
The complete expression is -8e-9s(1/s+1/(s-3)-1/(s+3)).
Now ℒ-1 can be applied to each term:
ℒ-1{-8e-9s/s}=-8u(t-9) where u is the unit-step function defined as:
u(t)=
⎛ 0 | t<0
⎝ 1 | t⩾0
In this case t-9⇒u is zero when t<9 and 1 when t⩾9. If t represents time, then this is known as time displacement or shift. The unit step (function=1) takes place after a delay of 9 time units. Before this lapse the function is zero.
ℒ-1{-8e-9s/(s-3)}=-8e3(t-9).u(t-9)
ℒ-1{8e-9s/(s+3)}=8e-3(t-9).u(t-9)
So, putting these together:
ℒ-1{-8e-9s(s2+6s-9)/(s(s2-9))}=ℒ-1{-8e-9s(1/s+1/(s-3)-1/(s+3))}=
-8u(t-9)-8e3(t-9).u(t-9)+8e-3(t-9).u(t-9)