Find the inverse laplace transform of $\displaystyle \frac{s}{a^2s^2+b^2}$.

Question

Find the inverse laplace transform of $\displaystyle \frac{s}{a^2s^2+b^2}$

My Thoughts: Take $\displaystyle s^*=\frac{s}{a}$ and $b^*=\frac{b}{a}$ and divide numerator and denominator by $a^2$.

Then we have, $\displaystyle \frac{1}{a}L^{-1}\left\{\frac{s^*}{s^{*2}+b^{*2}}\right\}$. This is equal to $\displaystyle \frac{1}{a}\cos\left(\frac{bt}{a}\right)$.

The given answer is $\displaystyle \frac{1}{a^2}\cos\left(\frac{bt}{a}\right)$.

I guess this has to do with the $s^*$, but in the definition of Laplace transform, we integrate with respect to $t$, so change in "$s$" shouldn't matter.

Is my line of reasoning correct ?

Answer 1

$\hat{h}(s) = \frac{s}{a^2s^2+b^2} = {1 \over a^2} {s \over s^2 + ({b \over a})^2 }$.

From this we get $h(t) = {1 \over a^2} \cos ({b \over a} t)$.

Answer 2

$$ L( \frac{1}{a}cos( \frac{bt}{a} ) ) = \frac{as^2}{a^2s^2 + b^2} $$

and hence,your reasoning must surely be wrong. Think about change of variable from $s$ to $s/a$ .