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Initial and final value theorems

  1. Dec 4, 2013 #1
    1. The problem statement, all variables and given/known data
    Find f(t) for the function F(s)=(10s^2+85s+95)/(s^2+6s+5) and apply the initial and final value theorems to each transform pair


    2. Relevant equations
    Initial value theorem: f(0)=lim s->∞ s(F(s))
    Final value theorem: f(∞) = lim s->0 s(F(s))


    3. The attempt at a solution
    After dividing due to improper fraction:
    F(s)= 10 + (25s+45)/(s^2+6s+5)

    F(s)= 10+5/(s+1)+20/(s+5)
    f(t)= 10δ(t)+[5e^(-t)+20e^(-5t)]u(t)

    Where I'm confused is how I would apply the value theorems since there's an impulse function. When my professor did a similar problem and applied the theorems, I couldn't follow what she did, but the answer solution to this problem says the value theorems can't be applied to the function because the function is improper and the corresponding f(t) function contains an impulse.
    How was my professor able to do it if it supposedly can't be done? Can someone please clarify this for me?
     
  2. jcsd
  3. Dec 5, 2013 #2

    ShayanJ

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    Which transform are you talking about? Fourier,Laplace or what?
     
  4. Dec 5, 2013 #3
    Laplace
     
  5. Dec 5, 2013 #4

    ShayanJ

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    Dirac delta is frequently explained as being zero everywhere except at one point which becomes infinite there.So [itex] \lim_{t\rightarrow 0}\delta(t)=0 [/itex]because although t is very near to 0,but it isn't equal to 0!

    EDIT:Looks like I'm wrong...because [itex] \lim_{s\rightarrow\infty}sF(s)=\infty[/itex]!
    So it seems we should have [itex] \lim_{t\rightarrow 0}\delta(t)=\infty[/itex]!!!
     
  6. Dec 5, 2013 #5
    How did you get infinity for lim s → ∞ sF(s)? This says it's 25:

    http://www.wolframalpha.com/input/?i=lim+as+x-%3Einfinity+%2825x^2%2B45x%29%2F%28x^2%2B6x%2B5%29
     
  7. Dec 5, 2013 #6

    ShayanJ

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    You forgot the 10s part!
     
  8. Dec 5, 2013 #7
    Are you sure the limit applies to the 10 as well? When my professor did this type of problem she only applied the limit to the fractional part. I'm so confused because my professor, the solutions manual, and you are all saying different things so I have no idea what is correct.
     
  9. Dec 5, 2013 #8

    ShayanJ

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    Its simple.The theorem urges us to find [itex] \lim_{s \rightarrow \infty} sF(s) [/itex] and whether you like it or not,10s is part of sF(s)!
    Its helpful to ask one of your classmates about the explanations that your teacher were giving while doing that.
    Also you can tell us what your solution manual and your teacher say,not only the results,but also the explanations!
     
  10. Dec 5, 2013 #9
    Well I won't have time for that, my final is in 10 hours lol. This is what the solution manual says:

    20ab6lv.jpg
     
  11. Dec 5, 2013 #10

    ShayanJ

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    Well,I think that's right.Because I was right about the limit of dirac delta being zero as the argument goes to zero.www.Wolframalpha.com confirms that! and because sF(s) becomes infinite as s goes to infinity,you can't apply the theorem.
     
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