I didn't see it right away but here is what I have
s^4+4=(s^2+2s+2)(s^2-2s+2)=((s+1)^2+1)((s-1)^2+1)
this will also change the numerator so what I ended up getting overall is
\frac{(s+1)-\frac{1}{2}s}{(s+1)^2+1}-\frac{(s-1)-\frac{1}{2}s}{(s+1)^2+1}
Now just trying to figure out...
If I break it out to be
[itex](s^2+2)^2-4s^2[\itex]
which is just the [itex] e^{-at}sin(bt) [\itex] unfortunately I have the [itex] s^2 [\itex] terms I need to get rid of. I think I can work it out if I break it down into complex numbers from here but not sure if this is the right move.
I split the denominator into
(s^2+2j)(s^2-2j)
How can I split the above into partial fractions since I have the imaginary numbers in both.
Also, how do I use the Latex in writing these equations, I am trying to use my regular Latex syntax but for some reason it has no effect here...
Find the inverse Laplace transform
$F(s)=\frac{4}{s^4+4}$
I tried factoring out the solution, but run into the problem with the imaginary numbers and am still stuck with the s^2+2j, which I have to factor out once more, and that's where the problem gets even messier. What do I do?