Calculate the inverse Laplace transform of a right-sided sequence with transfer function

$G\left(s\right)=\frac{7s-6}{({s}^{2}-s-6)}\mathrm{.}$

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The characteristic equation of G(s) is given by s² − s − 6 = 0, which has two roots at s = 3 and −2. Using the partial fraction expansion, the Laplace transform G(s) is expressed as

$G\left(s\right)=\frac{7s-6}{(s+2)(s-3)}\equiv \frac{{k}_{1}}{(s+2)}+\frac{{k}_{2}}{(s-3)}\mathrm{.}$

The coefficients of the partial fractions ${k}_{1}$k1 and ${k}_{2}$k2 are given by

${k}_{1}={\left[\right(s+2\left)\frac{(7s-6)}{(s+2)(s-3)}\right]}_{s=-2}={\left[\frac{(7s-6)}{(s-3)}\right]}_{s=2}=4$

and

${k}_{2}={\left[\right(s-3\left)\frac{(7s-6)}{(s+2)(s-3)}\right]}_{s=3}={\left[\frac{(7s-6)}{(s+2)}\right]}_{s=3}=3.$

The partial fraction expansion of the Laplace transform G(s) is therefore given by

$G\left(s\right)=\frac{4}{(s+2)}+\frac{3}{(s-3)}\mathrm{.}$

$g\left(t\right)=(4{\mathrm{e}}^{-2t}+3{\mathrm{e}}^{3t})u\left(t\right)\mathrm{.}$