# How to convert from Laplace Domain to Time Domain?

I want to convert the following equation from Laplace domain to continuous time domain:

$$F(s) = \frac{-2 m k v R}{2 m R s^{2} + m k s + 2 k R}$$

m, k, v, R are all constants.

If I can factor or put this into partial fractions, I can use the table here to convert: https://lpsa.swarthmore.edu/LaplaceZTable/LaplaceZFuncTable.html

eg.

$$\frac{1}{s+a} = e^{-at}$$

But I don't know how to break it down into that form. Is it doable? If so, how?

I have learned how to convert Laplace into the z-domain but I have found some problems with that. In particular, I need continuous time equations to set up the [n-1] and [n-2] etc. samples for the initial run or I won't get useful outputs.

• I think your question is really how to do partial fraction expansions -- the mathematics site would probably be best for that. – Dan Boschen Dec 23 '19 at 1:44
• Is $Fk(s)$ the Laplace transform of a signal, or is it a transfer function in the Laplace domain? – TimWescott Dec 23 '19 at 2:37
• I'm not sure which you'd call it. I don't think it's a transform of a signal. It's just an equation for force using the Laplace domain because it worked well in the derivatives, etc. Ie. It is of the nature that if $F(s) = 1/(s+3)$, then $F(t) = e^{-3t}$ as I have done that sort of conversion with simpler force equations and it worked. This is just too complicated for me to simply match up like that. – mike Dec 23 '19 at 2:44
• Re-reading your question, I see that what you're stuck on is just the partial fraction expansion -- I agree with @DanBoschen that you need learn how to do that. Partial fraction expansion is a topic that should be covered in the typical University course in Calculus, so if you kept your book it's in there somewhere. It should also be covered on the Web, someplace. You'll probably want to plug the numbers into your expression and then do the partial fraction expansion, although with a great deal of work, and the acceptance of complex arithmetic, you could make a general expression. – TimWescott Dec 23 '19 at 3:30
• lol. I never took calculus in unversity. :) But I posted it to math here: math.stackexchange.com/questions/3485053/… and someone gave me a solution. I don't actually need to do partial fraction expansion if that answer is correct. The format it's in as $\frac{1}{(s+a)(s+b)}$ matches one of the conversion cheats from here: lpsa.swarthmore.edu/LaplaceZTable/LaplaceZFuncTable.html – mike Dec 23 '19 at 3:35

## 1 Answer

$$F(s) = \frac{-2 m k v R}{2mR\left(s-\frac 1{4mR}\left(-mk-\sqrt{m^2k^2-16mkR^2}\right)\right)\left(s-\frac 1{4mR}\left(-mk+\sqrt{m^2k^2-16mkR^2}\right)\right)}$$

$$F(s) = \frac{-2 m k v R}{2mR}*\frac{1}{\left(s-\frac 1{4mR}\left(-mk-\sqrt{m^2k^2-16mkR^2}\right)\right)\left(s-\frac 1{4mR}\left(-mk+\sqrt{m^2k^2-16mkR^2}\right)\right)}$$

$$F(s) = - k v *\frac{1}{\left(s-\frac 1{4mR}\left(-mk-\sqrt{m^2k^2-16mkR^2}\right)\right)\left(s-\frac 1{4mR}\left(-mk+\sqrt{m^2k^2-16mkR^2}\right)\right)}$$

Then I can use:

$$\frac{1}{(s+a)(s+b)} = \frac{e^{-at}-e^{-bt}}{b-a}$$

And that should do it!