# Thread: la place inverse transform

1. ## la place inverse transform

hey
how do we do inverse la place transforms for $s/(s-9)(s+8)$
and $9(s+3)/((s+3)^2+81)((s+3)^2+36)$?

thanks in advance

2. we would use partial faction decomp.

this gives us

$\frac{s}{(s-9)(s+8)}=\frac{\frac{9}{17}}{s-9}+\frac{\frac{8}{17}}{s+8}$ the inverse transform of these are

$\frac{9}{17}e^{9t}+\frac{8}{17}e^{-8t}$

3. thanks for the first one mate but do u know how u would approach the second problem? that one is more important to me

4. Originally Posted by ramzouzy
thanks for the first one mate but do u know how u would approach the second problem? that one is more important to me
You bet...

Once again we to use partial fracion decomp so

$\frac{9(s+3)}{[(s+3)^2+81][(s+3)^2+36]}=\frac{-1}{5}\frac{(s+3)}{[(s+3)^2+9^2]}+\frac{1}{5}\frac{(s+3)}{[(s+3)^2+6^2]}$

S axis translation theorem backwards we get...

$\mathit{L}(e^{at}f(t)=F(s-a)$

$\frac{-1}{5}e^{-3t}\cos(9t)+\frac{1}{5}e^{-3t}\cos(6t)$

5. You have to know the ROC, to find out the inverse transform. (but I may be wrong).