be the laplace transform in question, where

How can i get its inverse by means of convolution,

Leia Sullivan
2022-04-13
Answered

Let

${x}^{\ast}\left(s\right)=\left(\frac{1}{(s+{\mu}_{1}+{\mu}_{2})(s+{\hat{\lambda}}_{2})(s+{\lambda}_{1}+{\lambda}_{2})}\right)$

be the laplace transform in question, where$\mu}_{1},{\mu}_{2},{\lambda}_{1},{\lambda}_{2},{\hat{\lambda}}_{2$ are just real parameters.

How can i get its inverse by means of convolution,$x\left(t\right)={\mathcal{L}}^{-1}\left({x}^{\ast}\left(s\right)\right)$

be the laplace transform in question, where

How can i get its inverse by means of convolution,

You can still ask an expert for help

abangan85s0

Answered 2022-04-14
Author has **16** answers

You right, first thing here is a convolution

$x\left(t\right)={\mathcal{L}}^{-1}\left(\frac{1}{(s+{\mu}_{1}+{\mu}_{2})(s+{\hat{\lambda}}_{2})(s+{\lambda}_{1}+{\lambda}_{2})}\right)={\mathcal{L}}^{-1}\left(\frac{1}{s+{\mu}_{1}+{\mu}_{2}}\right)*{\mathcal{L}}^{-1}\left(\frac{1}{s+{\hat{\lambda}}_{2}}\right)*{\mathcal{L}}^{-1}\left(\frac{1}{s+{\lambda}_{1}+{\lambda}_{2}}\right)=f\left(t\right)*g\left(t\right)*w\left(t\right)$

where $f\left(t\right)={\mathcal{L}}^{-1}\left(\frac{1}{s+{\mu}_{1}+{\mu}_{2}}\right)$ , $g\left(t\right)={\mathcal{L}}^{-1}\left(\frac{1}{s+{\hat{\lambda}}_{2}}\right)$ and $w\left(t\right)={\mathcal{L}}^{-1}\left(\frac{1}{s+{\lambda}_{1}+{\lambda}_{2}}\right)$

Just to note, since convolution is associative and commutative, you can evaluate it in any order (which is why no brackets were needed above),

$x\left(t\right)=(f\left(t\right)\cdot g\left(t\right))\cdot w\left(t\right)=f\left(t\right)\cdot (g\left(t\right)\cdot w\left(t\right))=(g\left(t\right)\cdot f\left(t\right))\cdot w\left(t\right)=f\left(t\right)\cdot (w\left(t\right)\cdot g\left(t\right))=(f\left(t\right)\cdot w\left(t\right))\cdot g\left(t\right))$

Now use the formula $(f\cdot g)\left(t\right)={\int}_{0}^{t}f\left(\tau \right)g(t-\tau )d\tau$ and finish the transform.

cab65699m

Answered 2022-04-15
Author has **14** answers

By the residue theorem:

$\frac{1}{(x-a)(x-b)(x-c)}=\frac{1}{(a-b)(a-c)(x-a)}+\frac{1}{(b-a)(b-c)(x-b)}+\frac{1}{(c-a)(c-b)(x-c)}$

and$\mathcal{L}}^{-1}\left(\frac{1}{x-c}\right)={e}^{cs$

and

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