Having difficulty finding the inverse laplace transform

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Homework Help Overview

The discussion revolves around finding the inverse Laplace transform of the function f(s) = (5se^{-3s} - e^{-3s}) / (s^2 - 4s + 17). Participants are exploring methods to manipulate the function's denominator and address the challenges posed by its structure.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants discuss rewriting the denominator in a different form and express concerns about the presence of the constant 13. There are attempts to factor out terms and relate the expression to known transforms. Some participants suggest using properties of exponential shifts in the Laplace domain.

Discussion Status

The conversation is ongoing, with various suggestions being made. Some participants are questioning their understanding of the material and expressing difficulties with the problem setup. There is no explicit consensus, but multiple approaches are being explored, including completing the square and recognizing the nature of the roots.

Contextual Notes

Participants note that they are new to Laplace transforms and are facing challenges with a problem that is intended as a challenge exercise. There is an acknowledgment that the roots of the polynomial in the denominator are complex, which complicates the problem further.

Theelectricchild
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Having difficulty finding the inverse laplace transform!

Hello everyone, I am really stuck on finding the inverse Laplace transform for this:

[tex]f(s)=\frac{5se^{-3s} - e^{-3s}}{s^{2}-4s+17}[/tex]

Heres my reasoning: I feel that I should rewrite the denominator in some kind of form such as (s-2)^2 + 13, and note the similarity with some of the problems I've been doing before, however its that 13 that is bothering me! Its not something you can take the squareroot of--- and also in addtion, I tried factoring out e^-3s on top and splitting this into two equations, but its this denominator that I absolutely despise.

Any help with finding the right method would be greatly appreciated thank you!
 
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I've forgotten just about every transform formula I knew (and I'm not in the mood rederiving them), but you may write [tex]13=(\sqrt{13})^{2}[/tex]

if that helps..
 
hmm i still am having difficulty---
 
Well, what about rewriting:
[tex]e^{-3s}=e^{-6}e^{-3(s-2)}[/tex]

Then you would get one expression on the form:
[tex]k\frac{e^{-3w}}{w^{2}+a^{2}},k=-e^{-6},w=s-2,a=\sqrt{13}[/tex]

Is this a familiar transform in w?
 
The exponentials are due to shifts in the t-domain:

L{f(t-T)}=e-sTF(s).

Just find the inverse transform of the rational functions of s, and then let the t-domain functions be delayed by the appropriate amount, given by the coefficient in the exponents of the exp functions.
 
interesting let me see if i can get anywhere with that...
 
Uh oh, you see, I cannot split the denominator into two linear factors like the book seems to be doing with most of the problems---- and because I cant, i don't know how to proceed like the examples do... I am sorry if I am a bit slow but we just started Laplace and this is a challenge problem id like to know to prepare.
 
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Well I cannot solve it but thanks for your help anyway...
 
Theelectricchild said:
Uh oh, you see, I cannot split the denominator into two linear factors like the book seems to be doing with most of the problems---- and because I cant, i don't know how to proceed like the examples do... I am sorry if I am a bit slow but we just started Laplace and this is a challenge problem id like to know to prepare.

You could factor the denominator into 2 linear factors, but the roots of that polynomial are complex. It would be better instead to complete the square in the denominator. The solution will be t-shifted, exponentially damped sines and cosines.
 

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