Laplace inverse of $\frac{{e}^{-s}(3{s}^{2}-s+2)}{(s-1)({s}^{2}+1)}$

Vrbljanovwu
2022-09-07
Answered

Laplace inverse of $\frac{{e}^{-s}(3{s}^{2}-s+2)}{(s-1)({s}^{2}+1)}$

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Colin Dougherty

Answered 2022-09-08
Author has **8** answers

I prefer the method of residues to partial fractions. If you recall, the residue of a simple pole ${s}_{0}$ of a function f(s) is

$$\underset{s\to {s}_{0}}{lim}[(s-{s}_{0})f(s)]$$

Here

$$f(s)=\frac{3{s}^{2}-s+2}{(s-1)({s}^{2}+1)}{e}^{st}$$

Therefore are poles at s=1 and $s=\pm i$. The sum of the residues at these poles are

$$2{e}^{t}+\frac{-1-i}{(-1+i)(2i)}{e}^{it}+\frac{-1+i}{(-1-i)(-2i)}{e}^{-it}=2{e}^{t}+\mathrm{cos}t$$

and that is the ILT.

$$\underset{s\to {s}_{0}}{lim}[(s-{s}_{0})f(s)]$$

Here

$$f(s)=\frac{3{s}^{2}-s+2}{(s-1)({s}^{2}+1)}{e}^{st}$$

Therefore are poles at s=1 and $s=\pm i$. The sum of the residues at these poles are

$$2{e}^{t}+\frac{-1-i}{(-1+i)(2i)}{e}^{it}+\frac{-1+i}{(-1-i)(-2i)}{e}^{-it}=2{e}^{t}+\mathrm{cos}t$$

and that is the ILT.

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