Question

Find the inverse transform of:
F(s)=[2s+14]/[s^2+8s+32]=[2s+14]/[(s+4)^2+4^2]

Answer 1

Not sure. Looks complicated.

Answer 2

You have already done the first step. We so far have \(F(s)=\frac{2s+14}{(s+4)^2+4^2}\). We're looking for its inverse laplace transform. We can rewrite the expression as:
\[{2s+14 \over (s+4)^2+4^2}={2s+8 \over (s+4)^2+4^2}+{6 \over (s+4)^2+4^2}=2{s+4 \over (s+4)^2+4^2}+{6 \over 4}{4 \over (s+4)^2+4^2}.\] Let f(t) be the inverse, then \(f(t)=2\cos(4t)e^{-4t}+\frac{3}{2}\sin(4t)e^{-4t}\). You also can take \(e^{-4t}\) as a common factor if you like.