Hassan H. answered • 08/22/13

Math Tutor (All Levels)

Hello Sun,

You will simply want to "split up" the expression for F(s), and then use known (and oftentimes tabulated) results on the Laplace transform to finish the computation. So, write

F(s) = G(s)H(s)

where H(s) = 1/(s^{2}+1), and note that the inverse Laplace transform of H(s) is

h(t) = £^{-1}{H(s)} = sin t.

Now, if we denote the inverse Laplace transform of G(s) by

g(t) = £^{-1}{G(s)},

we can immediately apply the "convolution theorem" to obtain the inverse Laplace transform, f(t), of F(s):

f(t) = £^{-1}{F(s)} = £^{-1}{G(s)H(s)} = ∫_{0}^{t} h(t-τ)g(τ) dτ = ∫_{0}^{t} sin(t-τ)g(τ) dτ.

Hope this clears things up so that you can do the rest of your homework problems confidently on your own.

Regards,

Hassan H.