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Inverse Laplace with given limits

  1. Jun 12, 2012 #1
    1. The problem statement, all variables and given/known data
    Problem reads: find inverse Laplace transform of f(t) of F(s)=(2s+3)/(s(s2+7s+10) What is the value of the function f(t) at t=0 and t=∞?


    2. Relevant equations
    Inverse laplace transform


    3. The attempt at a solution
    I solved F(t) down to F(t)= [itex].3/s[/itex]+[itex]0.166/s+2[/itex]-[itex]0.465/s+5[/itex]

    thus lead me to the inverse of f(t)=0.3 +0.1666e2t-0.465e5t

    Correct me on the math if i am wrong or whether i am on the wrong track up to this point. So, I’m guessing that I put 1 in for t which would be f(1)=-67.486 and for f(∞) I’m getting it as "undef" can someone shed light as to whether I’m on the right track?
     
    Last edited: Jun 12, 2012
  2. jcsd
  3. Jun 12, 2012 #2
    I think you should expect decaying exponentials, so check the signs on those puppies. Also, you mentioned both t=0 and t=1, did you do what you meant to do?
     
  4. Jun 12, 2012 #3

    vela

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    That should be F(s). Use parentheses. What you wrote means
    $$F(s) = \frac{0.3}{s}+\frac{0.166}{s} +2+\frac{0.465}{s}+5$$
     
  5. Jun 12, 2012 #4
    I apologize I have been using the "latex reference" but am still getting use to it. WhatI meant to say was that F(s)=[itex]0.3/\left(s\right)[/itex]+[itex]0.166/\left(s+2\right)[/itex] -[itex]0.465/\left(s+5\right)[/itex]. Now looking at this am I correct in assuming that when the problem asks for f(t) at t=0 and t=∞. So, @t=0 f(0)=-66.1498 and f(∞)=undefined??
     
    Last edited: Jun 12, 2012
  6. Jun 12, 2012 #5

    vela

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    Did you check your work like algebrat suggested? You should get a finite answer for t=∞. Your answer for f(0) doesn't look correct either.
     
  7. Jun 12, 2012 #6
    Yeah I have been working Laplace transforms for a while now and am still making stupid mistakes. I think I caught it: f(t)=0.3+0.166e-2t-0.465e-5t; thus f(0)=0.01 and f(∞)=0.3
     
  8. Jun 12, 2012 #7

    vela

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    It might have been intended for you to use the Laplace transform limit theorems. They say that
    \begin{align*}
    \lim_{t \to 0^+} f(t) &= \lim_{s \to \infty} sF(s) \\
    \lim_{t \to \infty} f(t) &= \lim_{s \to 0} sF(s)
    \end{align*} Try those and see if you get the same answers.

    (Or maybe not, now that I've reread the original post.)
     
    Last edited: Jun 12, 2012
  9. Jun 12, 2012 #8
    Im would not think I would have to use the limit theorem, mainly because it has not been in the course work I have been studying. I will look into it though and thank you very much for your help.
     
  10. Jun 13, 2012 #9

    Ray Vickson

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    You should get out of the bad habit of extreme roundoff, which you have performed above. Your F(s) is NOT
    [tex] \frac{0.3}{s} +\frac{0.166}{s+2} - \frac{0.465}{s+5}.[/tex] It is
    [tex] F(s) = \frac{3}{10 s} + \frac{1}{6(s+2)} - \frac{7}{15(s+5)}
    \doteq \frac{0.3}{s} + \frac{0.1666667}{s+2} - \frac{0.4666667}{s+5}.[/tex] In fact, there is no reason at all to convert to decimal numbers; using rationals is just as easy.

    RGV
     
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