I'm having difficulties calculating a simple Laplace inverse :

$\frac{S-4}{{S}^{2}-2S-11}$

Haylee Bowen
2022-04-07
Answered

I'm having difficulties calculating a simple Laplace inverse :

$\frac{S-4}{{S}^{2}-2S-11}$

You can still ask an expert for help

enchantsyseq

Answered 2022-04-08
Author has **19** answers

When Laplace transforms are rational functions, the typical line of attack is to put it into the form of partial fractions. Note that ${S}^{2}-2S-11={(S-1)}^{2}-12$ has roots $1\pm \sqrt{12}=1\pm 2\sqrt{3}$ , so:

$\frac{S-4}{{S}^{2}-2S-11}=\frac{S-4}{(S-1-2\sqrt{3})(S-1+2\sqrt{3})}$

So, we try to write it in the form$\frac{A}{S-1-2\sqrt{3}}+\frac{B}{S-1+2\sqrt{3}}$ , with A,B constants. One way of solving for these constants is the cover-up method. We apply this here:

$A=\frac{S-4}{S-1+2\sqrt{3}}{\mid}_{S=1+2\sqrt{3}}=\frac{-3+2\sqrt{3}}{4\sqrt{3}}=\frac{2-\sqrt{3}}{4}$

$B=\frac{S-4}{S-1-2\sqrt{3}}{\mid}_{S=1-2\sqrt{3}}=\frac{-3-2\sqrt{3}}{-4\sqrt{3}}=\frac{2+\sqrt{3}}{4}$

Thus, our function's Laplace transform can be written as$\frac{1}{4}(\frac{2-\sqrt{3}}{S-1+2\sqrt{3}}+\frac{2+\sqrt{3}}{S-1-2\sqrt{3}})$

We now recall that$e}^{at$ has Laplace transform $\frac{1}{s-a}$ , and this allows us to recover our function as:

$f\left(t\right)=\frac{1}{4}((2-\sqrt{3}){e}^{(1-2\sqrt{3})t}+(2+\sqrt{3}){e}^{(1+2\sqrt{3})t})$

So, we try to write it in the form

Thus, our function's Laplace transform can be written as

We now recall that

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