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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?

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  • $\begingroup$ Please don't write your last line as it is! It's false! Be specific and write, say, $\mathcal{L} (e^{-at})=...$ $\endgroup$ – Sean Roberson Dec 23 '19 at 3:05
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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.

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  • $\begingroup$ Thanks Ross! That's pretty amazing. If that's correct, then my solution is: $F(s) = \frac{-2 m k v R}{2 m R s^{2} + m k s + 2 k R}$ $F(s) = \frac{-2 m k v R}{2mR\left(s-\frac 1{4mR}\left(-mk-\sqrt{m^2k^2-16mkR^2}\right)\right)\left(s-\frac 1{4mR}\left(-mk+\sqrt{m^2k^2-16mkR^2}\right)\right)}$. Is that what you're saying? Because if so, from that I can easily apply a solution. Two questions: How can you assume $2mRs^{2}+mks+2kR=0$? I mean what if it doesn't equal zero? Also how did you do that massive quadratic solution? Is there some technique for those? Thanks. $\endgroup$ – mike Dec 23 '19 at 3:19
  • $\begingroup$ I didn't assume it equals $0$, I just used that to find the factorization. Yes, your $F(s)$ is what I was saying. Now the factors in the denominator are linear, so you can do partial fractions. That big expression just comes from plugging into the quadratic formula. You had a lot of constants in your equation, which is what makes it large. You can divide out $2mR$ for simplicity. The fact that the terms in the denominator are conjugate makes the partial fractions come out nicely. $\endgroup$ – Ross Millikan Dec 23 '19 at 4:49
  • $\begingroup$ Thanks Ross. Everything was perfect with that. $\endgroup$ – mike Dec 23 '19 at 4:55
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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 $$ F(s) = \frac{C}{(s-a)(s-b)} = \frac{C}{b-a}\left(\frac{1}{s-a}-\frac{1}{s-b}\right). $$

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