Question

Can anyone find inverse Laplace of this (s^2-pi^2)/(s^2+pi^2)^2

Answer 1

\[\frac{(s^{2}-\pi ^{2})}{(s ^{2}+\pi ^{2})^{2}}\]

Answer 2

\[\frac{ (s- \pi) (s+ \pi)}{ (s+ \pi) (s+ \pi) }\] now you can do it ;)

Answer 3

The denominator is not quite right?

Answer 4

thanks itsmylife

Answer 5

i think its done , you gotta separate terms now thats all ;) your welcome @shafqat_uet

Answer 6

holda thers a mistake

Answer 7

in denomirator you made a mistake

Answer 8

\[\frac{ (s - \pi)(s + \pi) }{ (s^2 + \pi^2)(s^2+ \pi^2) } \] this looks pretty easier but m stuck now :(

Answer 9

\[\frac{(s^{2}-\pi ^{2})}{(s ^{2}+\pi ^{2})^{2}}\]\[=\frac{(s^{2})}{(s ^{2}+\pi ^{2})^{2}} - \frac{\pi^2}{{(s ^{2}+\pi ^{2})^{2}}}\] \[=(\frac{s^2}{(s ^{2}+\pi ^{2})})(\frac{1}{(s ^{2}+\pi ^{2})}) - \frac{\pi^2}{{(s ^{2}+\pi ^{2})}}\times \frac{1}{{(s ^{2}+\pi ^{2})}}\]

Answer 10

@itsmylife good try but I have also tried like this. Is it possible using derivative or integral of the laplace

Answer 11

no i guess wat @Callisto has done is quite easier , you gotta take inverse now

Answer 12

Sorry, last step:
\[=(\frac{s}{(s ^{2}+\pi ^{2})})(\frac{s}{(s ^{2}+\pi ^{2})}) - \frac{\pi}{{(s ^{2}+\pi ^{2})}}\times \frac{\pi}{{(s ^{2}+\pi ^{2})}}\]