hey
how do we do inverse la place transforms for $\displaystyle s/(s-9)(s+8)$
and $\displaystyle 9(s+3)/((s+3)^2+81)((s+3)^2+36)$?
thanks in advance
hey
how do we do inverse la place transforms for $\displaystyle s/(s-9)(s+8)$
and $\displaystyle 9(s+3)/((s+3)^2+81)((s+3)^2+36)$?
thanks in advance
You bet...
Once again we to use partial fracion decomp so
$\displaystyle \frac{9(s+3)}{[(s+3)^2+81][(s+3)^2+36]}=\frac{-1}{5}\frac{(s+3)}{[(s+3)^2+9^2]}+\frac{1}{5}\frac{(s+3)}{[(s+3)^2+6^2]}$
S axis translation theorem backwards we get...
$\displaystyle \mathit{L}(e^{at}f(t)=F(s-a)$
$\displaystyle \frac{-1}{5}e^{-3t}\cos(9t)+\frac{1}{5}e^{-3t}\cos(6t)$