# Thread: [SOLVED] inverse of a laplace transform

1. ## [SOLVED] inverse of a laplace transform

Hey, I cannot figure out how to get these two probelms figured out. I need to find the inverse of the laplace transform.

Here is what ive tried: given H(s) find h(t). (take inverse laplace transform)
1) $H(s)=\dfrac{s^{2}-3s+9}{(s^{2}+16)^{2}}$

using partial fractions I split this up into:
$\dfrac{1}{s^{2}+16}+\dfrac{-3s-7}{(s^{2}+16)^{2}}$

I know:
$\dfrac{1}{s^{2}+16}=\dfrac{1}{4}sin(4t)$

which leaves me with:
$\dfrac{-3s-7}{(s^{2}+16)^{2}}$

From here i cant seem to split this up any further into a sin/cos/e^(x) or other simple function. Any help on the next step, or corrections to what I have already done would be great.

2) given K(s) find k(t) (take inverse of laplace transform).
$K(s)=\dfrac{2s}{s^2-6s+25}$

I cant factor the denominator so i tried writing the demoninator in various ways which didnt lead me anywhere. Without being able to split the denominator up I also cannot use partial fractions to help simplify this. Any ideas?
$K(s)=\dfrac{2s}{(s^2+16)+(-6s+9)}$, didn't seem to help.

2. $\frac{3s}{(s^{2}+16)^{2}}$

Gives

$\frac{3}{8}\cdot t \cdot \sin(4*t)$

and

$\frac{7}{(s^{2}+16)^{2}}$

Gives

$\frac{7}{128}\cdot (\sin(4*t) - 4 \cdot t \cdot \cos(4*t))$

That one might not be in your book. You may have to do it the hard way.

3. Nevermind, Im gonig to ask my professor at his office hours later today...Thanks again.