Alberanteb4T

2022-11-24

inverse laplace transform - with symbolic variables:
$F\left(s\right)=\frac{2{s}^{2}+\left(a-6b\right)s+{a}^{2}-4ab}{\left({s}^{2}-{a}^{2}\right)\left(s-2b\right)}$
My steps:
$F\left(s\right)=\frac{2{s}^{2}+\left(a-6b\right)s+{a}^{2}-4ab}{\left(s+a\right)\left(s-a\right)\left(s-2b\right)}$
$=\frac{A}{s+a}+\frac{B}{s-a}+\frac{C}{s-2b}+K$
$K=0$
$A=F\left(s\right)\ast \left(s+a\right)$

### Answer & Explanation

inventorspotDkt

The routine way is to use the way we get in partial fractions and you correctly noted that. If we set:
$\frac{2{s}^{2}+\left(a-6b\right)s+{a}^{2}-4ab}{\left({s}^{2}-{a}^{2}\right)\left(s-2b\right)}=F=\frac{A}{s+a}+\frac{B}{s-a}+\frac{C}{s-2b}$
then by doing handy calculations we can find A,B and C. We have then:
$\frac{-Asa+2Aab+A{s}^{2}-2Asb+Bsa-2Bab+B{s}^{2}-2Bsb+C{s}^{2}-C{a}^{2}}{\left(-s+2b\right)\left(-{s}^{2}+{a}^{2}\right)}$
Now if we put s=−a in the numerators, we get $A=\frac{a+b}{a+2b}$ and this is what you already got. With the similar approach $s=+a$ for B and $s=2b$ for C, we get:

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