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Easy Limit Question (Final Value Theorem)

  1. Feb 14, 2010 #1
    1. The problem statement, all variables and given/known data

    I am given finction in the Laplace domain

    [tex]X(s) = \frac{3s+7}{s^2(s+9)}[/tex]

    and I am asked to find:

    [tex]\lim_{t\rightarrow\infty}x(t)[/tex]

    I solved this by partial fraction expansion and transformed it to the time domain, took the limit and the result was an infinite limit.

    I feel like I could have used the Final Value Theorem which says that [itex]\lim_{t\rightarrow\infty}x(t) = \lim_{s\rightarrow 0}X(s)[/itex] and made this easier. Does anyone see how? As it stands, I cannot evaluate the limit as s-->0 because of the denominator. But if I could get it into an 'indeterminate form' I could use L'Hopital's rule.

    Any thoughts?
     
    Last edited: Feb 14, 2010
  2. jcsd
  3. Feb 14, 2010 #2

    vela

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    I think you have a typo in the function. In any case, what's the problem? Both methods gave you the same result: x(t) diverges as [itex]t\rightarrow\infty[/itex].
     
  4. Feb 14, 2010 #3
    Hi vela. I don't you read what I wrote correctly. Perhaps take another look. :wink:

    [typo fixed]


    The whole point of the thread is that I did not use FVT because I do not know how to get X(s) into a form that allows me too. So how would I know that both methods yield the same result?

    I appreciate your help :smile:

    Casey
     
  5. Feb 14, 2010 #4

    vela

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    You can't use the final value theorem because the limit of sX(s) doesn't exist as s approaches 0. The top approaches 7 while the bottom goes to 0. There's no way to get around it diverging.

    I'm not sure why you're expecting to be able to produce a finite answer via the FVT. You already found that x(t) diverges as t goes to infinity by explicitly finding x(t) and taking its limit.
     
  6. Feb 14, 2010 #5
    I am not expecting to produce a finite answer. I thought I might be able to confirm my answer using FVT. But I think that I agree that there isn't anyway to re-write X(s) such that L'Hopital's rule could be used.

    Do you know if FVT only works for finite-valued limits?

    Thanks again :smile:
     
  7. Feb 14, 2010 #6

    vela

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    Well, in a sense you did. Applying the FVT says the limit diverges, which is consistent with what you found by taking the limit of x(t).

    To say that the limits are equal assumes that the limits exist, but I'd guess that if one limit blows up, the other one will too. I've seen various statements of the theorem. Most seem to say it applies if the poles lie in the LHS of the complex plane, which isn't true for your particular X(s). The other statements are sloppier, just stating the result without the necessary givens.
     
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