Why Is the Inverse Laplace Transform of This Function 2e^{-t} - e^{-2t}?

In summary, the conversation discusses finding the inverse Laplace transform of a function using tables and the use of partial fractions. The answer is given as 2e^{-t} - e^{-2t}. One person suggests using the Heaviside formula while another suggests using partial fractions to solve the problem. The conversation also touches on working backwards with the correct answer and the possibility of using an alternative method for higher order polynomials.
  • #1
cyberdeathreaper
46
0
I am asked to find the inverse laplace transform of the following function:

[tex]
\frac{ \left( s+3 \right) }{ \left( s+1 \right) \left( s+2 \right) }
[/tex]

Using tables, can anyone help me understand why the answer is:

[tex]
2e^{-t} - e^{-2t}
[/tex]

I'm completely loss on this one, and yet the book assumes this is easily determined. Any ideas?

Note: I already realize that the bottom can be rewritten using partial fractions, but it seems to me that the book assumes that isn't even necessary - which makes sense, since it doesn't seem to get me anywhere anyway.
 
Last edited:
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  • #2
cyberdeathreaper said:
I am asked to find the inverse laplace transform of the following function:

[tex]
\frac{ \left( s+3 \right) }{ \left( s+1 \right) \left( s+2 \right) }
[/tex]

Using tables, can anyone help me understand why the answer is:

[tex]
2e^{-t} - e^{-2t}
[/tex]

I'm completely loss on this one, and yet the book assumes this is easily determined. Any ideas?

Note: I already realize that the bottom can be rewritten using partial fractions, but it seems to me that the book assumes that isn't even necessary - which makes sense, since it doesn't seem to get me anywhere anyway.

Look's like it might be this Heaviside formula for the ratio of polynomials. Have you encountered this before?

http://www.plmsc.psu.edu/~www/matsc597/fourier/laplace/node10.html
 
Last edited by a moderator:
  • #3
you have to do partial fractions that's all. [tex]\frac{\left( s+3 \right) }{ \left( s+1 \right) \left( s+2 \right) }= \frac{A}{s+1} + \frac{B}{s+2} [/tex]

s+3= A(s+2) + B(s+1)= (A+B)s + (2A+B)
A=2 B=-1

so you get
[tex] \frac{s+3}{(s+1)(s+2)} = \frac{2}{s+1} - \frac{-1}{s+2} [/tex]

from there you can see pretty simply from [tex]e^{at}= \frac {1}{s-a}[/tex] that you get the textbooks given answer.
 
  • #4
Good grief...

That's totally correct Gale17 - thanks. I was trying to use partial fractions on just this part:

[tex]
\frac{ 1 }{ \left( s+1 \right) \left( s+2 \right) }
[/tex]

and then mutliply the answer by s+3. Obviously that was not getting me anywhere, because I still had an s term on the top to deal with.

Thanks again! :biggrin:

PS: OlderDan, thanks for the suggestion though. I don't think we have touched on that approach specifically yet, but based on your link, it does seem feasible.
 
  • #5
an interesting way to use partial fractions... :wink: also, when you have the right answer, try working backwards with it.

you're welcome! (i like feeling useful anyways... sides, i have my own diff eq final coming up.. woo... )
 
  • #6
cyberdeathreaper said:
PS: OlderDan, thanks for the suggestion though. I don't think we have touched on that approach specifically yet, but based on your link, it does seem feasible.

The partial fraction approach is definitely the way to go. I thought you were looking for an alternative. That other thing might come in handy for higher order polynomials.
 

What is an Inverse Laplace Transform?

An inverse Laplace transform is a mathematical operation that is used to convert a function in the complex frequency domain back into its original form in the time domain. It is the opposite of the Laplace transform, which is used to convert a function from the time domain to the frequency domain.

What is the formula for the Inverse Laplace Transform?

The formula for the inverse Laplace transform is given by: f(t) = (1/2πi) ∫σ-i∞σ+i∞F(s)estds, where f(t) is the original function in the time domain, F(s) is the function in the complex frequency domain, and σ is a constant known as the abscissa of convergence.

What are the applications of the Inverse Laplace Transform?

The inverse Laplace transform has many applications in various fields such as engineering, physics, and mathematics. It is used to solve differential equations, analyze electrical circuits, model dynamic systems, and study the behavior of signals and systems.

What are the properties of the Inverse Laplace Transform?

Some of the important properties of the inverse Laplace transform include linearity, time shifting, scaling, differentiation, integration, and convolution. These properties make it easier to manipulate functions in the time domain using the inverse Laplace transform.

What are some common techniques to find the Inverse Laplace Transform?

Some commonly used techniques to find the inverse Laplace transform include partial fraction decomposition, convolution, and the use of tables or Laplace transform pairs. Additionally, the use of complex analysis and contour integration can also be helpful in finding the inverse Laplace transform of some functions.

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