Inverse laplace transform of $e^{-\frac{\pi}{2}s}\cdot\frac{s}{(s^2+1)^2}$

Question

I want to find the inverse la place transformation of: $ e^{-\frac{\pi}{2}s}\cdot\frac{s}{(s^2+1)^2}$. I know that the inverse la place transformation of $\frac{s}{(s^2+1)^2}$ is $f(t) =\frac{1}{2}t \sin t$. I have difficult, however, to exactly find what the exponential changes. I know that it shifts the function $f(t)$ by $\frac{\pi}{2}$ to the right. Would it make that the inverse la place transform equal to: $ g(t) = H_{\frac{\pi}{2}}(t)\cdot(\frac{1}{2}(t-\frac{\pi}{2})\sin (t-\frac{\pi}{2}))$ with $H$ being the Heaviside function?

I found that the answer should be $\frac{1}{2}H_{\frac{\pi}{2}}(t)\cdot((t-\frac{\pi}{2})\sin (t)-\cos(t))$. What did I do wrong?

Thanks for reading,

K. Kamal

Answer 1

Well, solving a more general problem:

$$\text{f}_{\space\alpha,\beta}\left(t\right):=\mathscr{L}_\text{s}^{-1}\left[\exp\left(\alpha\cdot\text{s}\right)\cdot\frac{\text{s}}{\left(\beta+\text{s}^2\right)^2}\right]_{\left(t\right)}\tag1$$

Using the 'convolution' property of the Laplace transform, we can write:

$$\text{f}_{\space\alpha,\beta}\left(t\right)=\int_0^t\mathscr{L}_\text{s}^{-1}\left[\exp\left(\alpha\cdot\text{s}\right)\right]_{\left(t-\tau\right)}\cdot\mathscr{L}_\text{s}^{-1}\left[\frac{\text{s}}{\left(\beta+\text{s}^2\right)^2}\right]_{\left(\tau\right)}\space\text{d}\tau\tag2$$

Using the 'delayed impulse' in the table of the Laplace transforms, we can write:

$$\text{f}_{\space\alpha,\beta}\left(t\right)=\int_0^t\delta\left(t-\tau+\alpha\right)\cdot\mathscr{L}_\text{s}^{-1}\left[\frac{\text{s}}{\left(\beta+\text{s}^2\right)^2}\right]_{\left(\tau\right)}\space\text{d}\tau\tag3$$

And:

$$\mathscr{L}_\text{s}^{-1}\left[\frac{\text{s}}{\left(\beta+\text{s}^2\right)^2}\right]_{\left(\tau\right)}=\frac{\tau\cdot\sin\left(\tau\cdot\sqrt{\beta}\right)}{2\cdot\sqrt{\beta}}\tag4$$

So, we get:

$$\text{f}_{\space\alpha,\beta}\left(t\right)=\int_0^t\delta\left(t-\tau+\alpha\right)\cdot\frac{\tau\cdot\sin\left(\tau\cdot\sqrt{\beta}\right)}{2\cdot\sqrt{\beta}}\space\text{d}\tau=$$ $$\frac{1}{2\cdot\sqrt{\beta}}\cdot\int_0^t\delta\left(t-\tau+\alpha\right)\cdot\tau\cdot\sin\left(\tau\cdot\sqrt{\beta}\right)\space\text{d}\tau\tag5$$