Solve using Convolution theorem (Inverse Laplace)

Question

Using Convolution theorem, find: $$L^{-1} [\frac{s^2}{(s^2+1)^3}]$$

Note: This was an exam question and was worth "3" marks only, so it should not be so "long" I suppose.

So my initial attempt was: $$L^{-1} [\frac{s^2}{(s^2+1)^3}]=\cos(t)*\cos(t)*\sin(t)$$ But this yielded a much longer problem!

Is there a faster way that I am not seeing?

Thanks in advance.

Answer 1

We can write it as

$$\mathcal{L}^{-1}\left( \frac{s^2}{(s^2+1)^3}\right) = \mathcal{L}^{-1}\left(\left(\frac{s}{\left(s^2+1\right)}\right)\left(\frac{s}{\left(s^2+1\right)^2}\right)\right)$$

We have

$$\mathcal{L}^{-1}\left(\frac{s}{\left(s^2+1\right)}\right) = \cos(t)$$

$$\mathcal{L}^{-1}\left(\frac{s}{\left(s^2+1\right)^2}\right) = \dfrac{1}{2}t \sin (t)$$

Now we just need to convolve

$$\cos(t) * \dfrac{1}{2}t \sin (t)$$

Using the definition of convolution, we have

$$\int_0^t \frac{1}{2} (v \sin (v)) \cos (t-v) \, dv = \frac{1}{8} \left(t^2 \sin (t)+\sin (t)-t \cos (t)\right)$$