Inverse Laplace transform of $\frac{1}{(s^2 + 6s + 45)^2}$

Question

Inverse Laplace transform of $\frac{1}{(s^2 + 6s + 45)^2}$. So we can see the determinant of the denominator is negative which means we have two conjugate roots. I then checked the book which gives a formula for such situation and tried to expand the fraction like that and tried to find the undetermined constants, but found the constants are all zero except one, which gives me the original form of the fraction.

Answer 1

When you are looking to do inverse Lapace transforms you are looking to "factor" the denominator like so...

$s^2 + 6s + 45 = (s + 3)^2 + 36$

Not a true factoring, but complete the square.

Now I suggest you work backward and look at:

$\mathcal L \{\cos at\} = \frac {s}{s^2 + a^2}\\ \mathcal L \{e^{bt}\cos at\} = \frac {(s-b)}{(s-b)^2 + a^2}\\ \mathcal L \{\sin at\} = \frac {a}{s + a^2}\\ \mathcal L \{t f(t)\} = -\frac {d}{ds}\mathcal L \{f(t)\}$

Now you have to figure out how to get from

$\frac {1}{((s+3)^2 + 36)^2}$ to some combination of the above.

It is going to be something with the structure: $Ae^{-3t}\cos 6t + Be^{-3t}\sin 6t + Cte^{-3t}\cos 6t + Dte^{-3t}\sin 6t$

Answer 2

Rewrite $$ \frac 1{(s^2 + 6s + 45)^2} = \frac 1{((s + 3)^2 + 6^2)^2} = \frac 1{((s + 3) - 6i)^2((s + 3) + 6i)^2} = \\ \frac i{864} \frac 1{(s + 3 + 6i)} \\ \quad + \frac{-1}{144} \frac 1{(s + 3 + 6i)^2} \\ \quad - \frac{i}{864} \frac{1}{s + 3 - 6i}\\ \quad - \frac{-1}{144} \frac 1{(s + 3 - 6i)^2} $$