# Find the inverse Laplace transform

1. Sep 6, 2011

### paczan85

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?

2. Sep 6, 2011

### micromass

Factor the denominator first and split into partial fractions...

3. Sep 6, 2011

### paczan85

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.

Thanks

4. Sep 6, 2011

### micromass

You should be able to factor it without using complex numbers. That is, you should be able to find a,b,c and d such that

$$s^4+4=(s^2+as+b)(s^2+cs+d)$$

Enclose them in $.... [/ itex] brackets (without the spaces). So instead of writing s^2, you write [itex] s^2 [/ itex]. The equivalent of  ...  is $$... [/ tex] 5. Sep 6, 2011 ### paczan85 If I break it out to be [itex](s^2+2)^2-4s^2[\itex] which is just the [itex] e^{-at}sin(bt) [\itex] unfortunately I have the [itex] s^2 [\itex] terms I need to get rid of. I think I can work it out if I break it down into complex numbers from here but not sure if this is the right move. 6. Sep 6, 2011 ### paczan85 And looks like my Latex syntax is still not working 7. Sep 6, 2011 ### micromass Have you tried factoring it the way I suggested?? Also, you have the wrong / in your tex brackets. 8. Sep 6, 2011 ### paczan85 I didn't see it right away but here is what I have [itex] s^4+4=(s^2+2s+2)(s^2-2s+2)=((s+1)^2+1)((s-1)^2+1)$ this will also change the numerator so what I ended up getting overall is $\frac{(s+1)-\frac{1}{2}s}{(s+1)^2+1}-\frac{(s-1)-\frac{1}{2}s}{(s+1)^2+1}$ Now just trying to figure out the right functions to transform this back into the t domain, any suggestions? 9. Sep 6, 2011 ### micromass That's good, so working it out a bit further. We get that we need to take the inverse Laplace transform of [tex]\frac{1}{2}\frac{s}{(s+1)^2+1}$$

$$\frac{1}{(s+1)^2+1}$$

$$-\frac{1}{2}\frac{s}{(s-1)^2+1}$$

$$\frac{1}{(s-1)^2+1}$$

This shouldn't be too hard. It should be of the form $e^{-at}\sin(\omega t)u(t)$ or $e^{-at}\cos(\omega t)u(t)$