Let \(\displaystyle{x}^{{\ast}}{\left({s}\right)}={\left({\frac{{{1}}}{{{\left({s}+\mu_{{1}}+\mu_{{2}}\right)}{\left({s}+\hat{{\lambda}}_{{2}}\right)}{\left({s}+\lambda_{{1}}+\lambda_{{2}}\right)}}}}\right)}\) be the laplace transform in question, where

Leia Sullivan 2022-04-13 Answered
Let
x(s)=(1(s+μ1+μ2)(s+λ^2)(s+λ1+λ2))
be the laplace transform in question, where μ1,μ2,λ1,λ2,λ^2 are just real parameters.
How can i get its inverse by means of convolution, x(t)=L1(x(s))
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Answers (2)

abangan85s0
Answered 2022-04-14 Author has 16 answers

You right, first thing here is a convolution
x(t)=L-1(1(s+μ1+μ2)(s+λ^2)(s+λ1+λ2))=L-1(1s+μ1+μ2)*L-1(1s+λ^2)*L-1(1s+λ1+λ2)=f(t)*g(t)*w(t)
where f(t)=L1(1s+μ1+μ2) , g(t)=L1(1s+λ^2) and w(t)=L1(1s+λ1+λ2)
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(t)=(f(t)g(t))w(t)=f(t)(g(t)w(t))=(g(t)f(t))w(t)=f(t)(w(t)g(t))=(f(t)w(t))g(t))
Now use the formula (fg)(t)=0tf(τ)g(tτ)dτ and finish the transform.

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cab65699m
Answered 2022-04-15 Author has 14 answers
By the residue theorem:
1(xa)(xb)(xc)=1(ab)(ac)(xa)+1(ba)(bc)(xb)+1(ca)(cb)(xc)
and L1(1xc)=ecs
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