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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

    rude man

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    Both limits work, sort of.
    lim s → 0 sF(s) = lim t → ∞ f(t) = 25 which is correct. No problem here.

    lim s → ∞ sF(s) = lim t → 0 f(t) = ∞ which is also correct: f(0) is indeed infinite. But the latter limit just tells you f(0) = ∞ but not the "strength" of the infinity. In other words, the latter limit in the s domain can't tell if it's δ(t) or 10δ(t) or whatever.

    I would agree that using the final-value theorem for t = 0+ is invalid if the order of the numerator = order of the denominator.
     
  4. Dec 5, 2013 #3
    I don't see how you got those results though, because my calculator gives lim s → ∞ sF(s) = 25, whereas that's what you got for lim s → 0 sF(s). And for s → 0 sF(s) it gave me 0 as the answer.
     
  5. Dec 5, 2013 #4

    rude man

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    You got lim s → 0 sF(s) right & I got it wrong. Sorry. It's zero.

    As for lim s → ∞ of sF(s), it's obviously ∞. Look at sF(s) = 10s + (25s^2 + 45s)/(s^2 + 6s + 5). The first term is ∞. The second term does approach 25. But ∞ + 25 = ∞.
     
  6. Dec 5, 2013 #5
    I just talked to my professor about this and she said you don't multiply the 10 by s, only the fractional part.
     
  7. Dec 5, 2013 #6

    rude man

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    I have second thoughts.

    The initial value theorem just does not work if the order of the numerator >= order of denominator.

    In your case, sF(s) = 10 + (25s^2 + 45s)/(s^2 +6s + 5). If you ignore the 10s term you get f(0+) = 25, whereas the actual f(0+) = 10δ(t) + 25. So, put simply, the initial-value theorem gives an incorrect result if the 10s term is ignored.

    If the 10s term isn't ignored you get f(0+) = ∞ which is a lot closer to the truth than 25. The real f(0+) = 10δ(t) + 25 is a pulse of infinite height occurring at t = 0+ plus a constant of 25. Of course, as t increases the delta function disappears immediately after t = 0+ whereas the 25 dies down exponentially with time constants 1 and 5.
     
    Last edited: Dec 5, 2013
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