How to get partial fractions for this equation? (To go from Laplace to Time Domain)

Question

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 simple partial fractions with respect to $s$, 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?

What would the simplest partial fractions form of this equation be?

Answer 1

It is a quadratic in $s$, so you can factor it using the quadratic formula. $$2 m R s^{2} + m k s + 2 k R=0 \implies s=\frac 1{4mR}\left(-mk\pm\sqrt{m^2k^2-16mkR^2}\right)\\ \\ \text{ } \\\text{so} \\ \text { }\\ 2 m R s^{2} + m k s + 2 k R=2mR\left(s-\frac 1{4mR}\left(-mk-\sqrt{m^2k^2-16mkR^2}\right)\right)\left(s-\frac 1{4mR}\left(-mk+\sqrt{m^2k^2-16mkR^2}\right)\right)$$ and you can use partial fractions.

Answer 2

First, use the quadratic formula to find the roots of the denominator and write $F(s)$ in the following form: $$ F(s) = \frac{C}{(s-a)(s-b)}. $$ If $a=b$, you are done. Otherwise, the partial fraction expansion will be $$ F(s) = \frac{C}{(s-a)(s-b)} = \frac{C}{b-a}\left(\frac{1}{s-a}-\frac{1}{s-b}\right). $$