Inverse Laplace Transform of s/(s^2+1)^2)

[1s1]

Homework Statement

$\mathcal{L}^{-1}\Big\{\frac{s}{(s^2+1)^2}\Big\}$

I'm trying to figure out how to find the inverse Laplace transform of this expression. Is this something you just look up in a table or is there a way to find it directly, maybe by Convolution?

 

[LCKurtz]

Homework Statement

$\mathcal{L}^{-1}\Big\{\frac{s}{(s^2+1)^2}\Big\}$

I'm trying to figure out how to find the inverse Laplace transform of this expression. Is this something you just look up in a table or is there a way to find it directly, maybe by Convolution?

Yes, by convolution. Write the transform as the product of these transforms$$
\frac{1}{s^2+1}\frac{s}{s^2+1}$$both of which are easy to inverse.

 

[vela]

You might be able to find it using a table or using a combination of table entries. You could also use convolution. Give it a shot and show us what you come up with.

Here's one way you might get it if your table tells you how integration or differentiation affects Laplace transforms. You can integrate $\frac{s}{(s^2+1)^2}$ pretty easily, and the result of the integration should be in your table.

 

[1s1]

Thanks for the help!


$$\mathcal{L}^{-1}\Big\{\frac{s}{s^2+1}\Big\} = cos(t)$$
$$\mathcal{L}^{-1}\Big\{\frac{1}{s^2+1}\Big\} = sin(t)$$
$f(t)=cos(t)$ and $g(t)=sin(t)$

So you can use $\mathcal{L}^{-1}\Big\{F(s)G(s)\Big\} = f*g$

and do the convolution $cos(t)*sin(t)$

Plugging this into the definition of a convolution, you get: $\int_{0}^{t} cos(\tau)sin(t-\tau )d\tau$

That's where things get dicey for me ... I guess my trig identities and integral calculus are pretty rusty. I'm not sure what to do with the $sin(t-\tau )d\tau$

I guess this problem has evolved into an integration problem. Any suggestions on how to solve this integral? I've been looking at the sum and difference formulas for sine and cosine, but that doesn't seem to get me anywhere.

 

[Ray Vickson]

Thanks for the help!


$$\mathcal{L}^{-1}\Big\{\frac{s}{s^2+1}\Big\} = cos(t)$$
$$\mathcal{L}^{-1}\Big\{\frac{1}{s^2+1}\Big\} = sin(t)$$
$f(t)=cos(t)$ and $g(t)=sin(t)$

So you can use $\mathcal{L}^{-1}\Big\{F(s)G(s)\Big\} = f*g$

and do the convolution $cos(t)*sin(t)$

Plugging this into the definition of a convolution, you get: $\int_{0}^{t} cos(\tau)sin(t-\tau )d\tau$

That's where things get dicey for me ... I guess my trig identities and integral calculus are pretty rusty. I'm not sure what to do with the $sin(t-\tau )d\tau$

I guess this problem has evolved into an integration problem. Any suggestions on how to solve this integral? I've been looking at the sum and difference formulas for sine and cosine, but that doesn't seem to get me anywhere.

Expand $\sin(t-\tau)$ using the trig addition formula; then you just need to be able to integrate $\sin^2(\tau)$ and $\sin(\tau) \cos(\tau)$, and those are both exercises in calculus 101.

 

[lurflurf]

^What is this "calculus 101?" It sounds horrible.
Use the integration rule
$$\mathcal{L}^{-1} \{ \mathrm{F}(s) \} = t \, \mathcal{L}^{-1} \left\{ \int_s^\infty \! \mathrm{F}(u) \, \mathrm{d}u \right\} \\
\text{so} \\
\mathcal{L}^{-1} \left\{ \frac{s}{(s^2+1)^2} \right\}=t \, \mathcal{L}^{-1} \left\{\int_s^\infty \! \frac{u}{(u^2+1)^2} \, \mathrm{d}u \right\}$$

 

[1s1]

^What is this "calculus 101?" It sounds horrible.
Use the integration rule
$$\mathcal{L}^{-1} \{ \mathrm{F}(s) \} = t \, \mathcal{L}^{-1} \left\{ \int_s^\infty \! \mathrm{F}(u) \, \mathrm{d}u \right\} \\
\text{so} \\
\mathcal{L}^{-1} \left\{ \frac{s}{(s^2+1)^2} \right\}=t \, \mathcal{L}^{-1} \left\{\int_s^\infty \! \frac{u}{(u^2+1)^2} \, \mathrm{d}u \right\}$$

This integration rule seems really great, but I can't seem to find it in my undergraduate advanced engineering mathematics textbook. Either I don't know what I'm looking for or I need to upgrade textbooks :)

 

[lurflurf]

That formula is in most of the engineering books as far as I know. The form may be slightly different such as
$$\mathcal{L}\left\{\frac{\mathrm{f}(t)}{t}\right\}=\int_s^\infty \! \mathrm{F}(u) \mathrm{d}u\\
\text{where}\\
\mathrm{F}(s)=\mathcal{L}\{ \mathrm{f}(t) \}=\int_0^\infty \! \mathrm{f}(t)e^{-s \, t} \, \mathrm{d}t$$