Using the Table of Laplace Transforms, we have:

`Lap^{:-1:}{1/(s^2+9)e^(-pis//2)}`

`=1/3Lap^{:-1:}{3/(s^2+9)e^(-pis//2)}`

`=1/3 sin 3(t-pi/2)*u(t-pi/2)`

`=1/3 cos 3t*u(t-pi/2)`

NOTE:

`sin 3(t-pi/2)`

`=sin(3t-(3pi)/2)`

`=sin 3t\ cos ((3pi)/2)` `-cos 3t\ sin ((3pi)/2)`

`=cos 3t`

So the Inverse Laplace transform is given by:

`g(t)=1/3cos 3t*u(t-pi/2)`

The graph of the function (showing that the switch is turned on at `t=pi/2 ~~ 1.5708`) is as follows:

0.5ππ1.5π0.10.20.30.4-0.1-0.2-0.3-0.4tg(t)Open image in a new page

Graph of `g(t)=1/3cos 3t*u(t-pi/2)`.

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