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

In summary: 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
  • #1
1s1
20
0

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?
 
Physics news on Phys.org
  • #2
1s1 said:

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.
 
  • #3
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.
 
Last edited:
  • #4
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.
 
  • #5
1s1 said:
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.
 
  • #6
^What is this "calculus 101?" It sounds horrible.
Use the integration rule
$$\mathcal{L}^{-1} \{ \mathrm{F}(s) \} = t \, \mathcal{L}^{-1} \left\{ \int_s^\infty \! \mathrm{F}(u) \, \mathrm{d}u \right\} \\
\text{so} \\
\mathcal{L}^{-1} \left\{ \frac{s}{(s^2+1)^2} \right\}=t \, \mathcal{L}^{-1} \left\{\int_s^\infty \! \frac{u}{(u^2+1)^2} \, \mathrm{d}u \right\}$$
 
  • #7
lurflurf said:
^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 :)
 
  • #8
That formula is in most of the engineering books as far as I know. The form may be slightly different such as
$$\mathcal{L}\left\{\frac{\mathrm{f}(t)}{t}\right\}=\int_s^\infty \! \mathrm{F}(u) \mathrm{d}u\\
\text{where}\\
\mathrm{F}(s)=\mathcal{L}\{ \mathrm{f}(t) \}=\int_0^\infty \! \mathrm{f}(t)e^{-s \, t} \, \mathrm{d}t$$
 
  • Like
Likes 1 person

What is the inverse Laplace transform of s/(s^2+1)^2?

The inverse Laplace transform of s/(s^2+1)^2 is (1/2)(1-cos t).

What is the domain and range of the inverse Laplace transform of s/(s^2+1)^2?

The domain of the inverse Laplace transform of s/(s^2+1)^2 is t≥0 and the range is 0≤y≤1.

What is the significance of the inverse Laplace transform of s/(s^2+1)^2?

The inverse Laplace transform of s/(s^2+1)^2 represents the solution to a differential equation with a double pole at s=±i. It is also used in signal processing and control theory to analyze and design systems.

How do you calculate the inverse Laplace transform of s/(s^2+1)^2?

To calculate the inverse Laplace transform of s/(s^2+1)^2, you can use partial fraction decomposition and the table of Laplace transforms to find the inverse transform as (1/2)(1-cos t).

Are there any real-world applications of the inverse Laplace transform of s/(s^2+1)^2?

Yes, the inverse Laplace transform of s/(s^2+1)^2 has various applications in engineering, physics, and mathematics. It can be used to analyze the response of electrical circuits, solve differential equations in control systems, and model vibrations in mechanical systems. It is also used in the study of oscillations and wave phenomena.

Similar threads

  • Calculus and Beyond Homework Help
Replies
10
Views
1K
  • Calculus and Beyond Homework Help
Replies
4
Views
1K
  • Calculus and Beyond Homework Help
Replies
1
Views
653
  • Calculus and Beyond Homework Help
Replies
7
Views
710
  • Calculus and Beyond Homework Help
Replies
2
Views
942
  • Calculus and Beyond Homework Help
Replies
8
Views
2K
  • Calculus and Beyond Homework Help
Replies
3
Views
1K
  • Calculus and Beyond Homework Help
Replies
9
Views
1K
  • Calculus and Beyond Homework Help
Replies
1
Views
1K
  • Calculus and Beyond Homework Help
Replies
5
Views
1K
Back
Top