Inverse Laplace transform with unit step function

Click For Summary

Discussion Overview

The discussion revolves around the process of taking the Laplace transform of a function that results in a unit step function. Participants are exploring the steps involved in the transformation, particularly focusing on the use of partial fractions and the notation used in expressing the functions and their transforms.

Discussion Character

  • Technical explanation
  • Mathematical reasoning
  • Debate/contested

Main Points Raised

  • One participant expresses confusion about the transition from a polynomial fraction to a sum of simpler terms, specifically questioning how the expression changes from \frac{e^{-s}}{s(s-1)} to \frac{e^{-s}}{s} - \frac{e^{-s}}{s-1}.
  • Another participant suggests that the original poster is unclear in their notation, emphasizing the importance of distinguishing between the function and its Laplace transform.
  • A later reply asks if the original poster understands how to decompose a polynomial fraction into simpler terms, implying that this step is crucial for solving the problem.

Areas of Agreement / Disagreement

There appears to be disagreement regarding the clarity of the original poster's explanation and notation. Some participants are attempting to clarify the process, while the original poster maintains that they are discussing the Laplace transform rather than the inverse transform.

Contextual Notes

There are unresolved issues regarding the notation used for the functions and their transforms, as well as the steps involved in the partial fraction decomposition. The discussion does not reach a consensus on these points.

Who May Find This Useful

Readers interested in the mathematical processes involved in Laplace transforms, particularly those who may be struggling with notation and partial fraction decomposition.

shaqywacky
Messages
4
Reaction score
0
Hello again.

First off, I wasn't sure how to say this in the title but I'm not taking the inverse Laplace transform of a unit step function. I'm taking the Laplace transform of something that comes out to the unit step function.

I have this question, which is a similar version of the question I am trying to solve. This is also the solution but I have no idea what happened.

laplace_unit.png

(sorry, I don't know why that is so small, click to make it bigger)

So the first part of that image is just the inverse Laplace transform I am trying to solve. The first step I saw was to do partial fractions. When I do that I get:
\frac{e}{s+1} + \frac{1}{s}
I'm omitting the inverse Laplace transform here because I don't know how to do it in latex.

But as you can see, the solution maunual got:
\frac{e^{-s}}{s} - \frac{e^{-s}}{s+1}

Clearly there is something I do not understand about this. I can get the correct answer after this, I just don't understand this step.

Any help would be greatly appreciated.

Thank you.
 
Last edited:
Physics news on Phys.org
Hi !
You question seems confused.
You must not use the same variable (s) for the function and for the Laplace transform of the function.
Please write clearly :
"The Laplace transform of f(x) is F(s)"
or
"The inverse Laplace transform of F(s) is f(x)"
Then, write clearly the expression of the known function : is it f(x) or F(s) ?
 
I don't think I did. I never took any inverse Laplace transforms in my post. I'm talking about inside the Laplace transform. I didn't know how to use the Laplace transform in Latex, so I omitted it and made a note of it but I don't think I was being very clear. I'm just omitting the notation of the Laplace transform. So the equations I posted (besides the ones in the picture) are what's inside of the Laplace transform I'm trying to solve.

To try to clarify, I just don't understand the first step that is in the picture. So I don't see how they went from
\frac{e^{-s}}{s(s-1)}[/size]

to

\frac{e^{-s}}{s} - \frac{e^{-s}}{s-1}[/size]

Thanks.
 
Hi !

Do you know how to write a polynomial fraction as a sum of simple terms ?
Have a look to the attachment :
 

Attachments

  • obviously.JPG
    obviously.JPG
    7.2 KB · Views: 614

Similar threads

  • · Replies 1 ·
Replies
1
Views
1K
  • · Replies 1 ·
Replies
1
Views
3K
  • · Replies 17 ·
Replies
17
Views
4K
  • · Replies 7 ·
Replies
7
Views
7K
  • · Replies 5 ·
Replies
5
Views
5K
  • · Replies 2 ·
Replies
2
Views
3K
  • · Replies 1 ·
Replies
1
Views
4K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 5 ·
Replies
5
Views
3K