Question

Hi, need help with an inverse Laplace transform:
my output Y(s)=4/(s^2+6s+3)
I think I need to use e^at(sinh(bt)=>b/((s-a)^2-b^2) but can't manage the manipulation... Any pointers???

Answer 1

ok so if we are to use e^at(sinh(bt)), then we know that the denominator of this must also be equal to the denom. of Y(s) so we'll have:
s^2+6s+3=(s-a)^2-b^2
s^2-2as+a^2-b^2=s^2+6s+3
so we'll have the equations:
-2a=6
a^2-b^2=3
solving this gives us a=-3 and b=sqrt(6) however our transform is not in the form we want it to be so we rewrite it as:
Y(s)=((4sqrt(6)/sqrt(6)))/((s+3)^2-(sqrt(6))^2)=(4/sqrt(6))(sqrt(6)/((s+3)^2-(sqrt(6))^2)
and so our inverse will just be:
(4/sqrt(6))e^(-3t)(sinh((sqrt(6))t))