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Sketching Transfer function in time domain

  1. Oct 12, 2015 #1

    Maylis

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    Gold Member

    1. The problem statement, all variables and given/known data
    $$ y(s) = \frac {s}{(s+1)(2s+1)} u(s) $$
    Where ##u(s)## is the step function ##\frac {1}{s}##

    Find the output at t=0 and t= infinity

    2. Relevant equations


    3. The attempt at a solution
    My question is kind of basic, so I know the final and initial value theorem

    $$ \lim_{s \to 0} sY(s) = \lim_{t \to \infty} y(t) $$

    But should I include the step function, or leave it out. Meaning, should I evaluate

    $$ \lim_{s \to 0} \frac {s^{2}}{(s+1)(2s+1)} \frac {1}{s}$$
    or rather,
    $$ \lim_{s \to 0} \frac {s^{2}}{(s+1)(2s+1)} $$

    The reason I am hesitating on this is because in the textbook example problem, they do not mention what the input function is, and proceed to solve without the step function. Then I solved a homework problem where they asked to match the transfer function output with a step function input, and at that time I did not even realize, so I was only doing limits of Y(s), not sY(s), and got them all right. So now my head is all jumbled up and I just want to get this thing cleared up!
     
  2. jcsd
  3. Oct 13, 2015 #2

    rude man

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    You need to include U(s) = 1/s.
    If you used the initial & final value theorems without the U(s) input you'd be solving the problem with the input = kδ(t), the delta function aka impulse input, where k = 1 Volt-sec. If you're not familiar with the delta function, be careful with it. Unlike u(t) which is dimensionless, δ(t) has dimension 1/t.
     
  4. Oct 13, 2015 #3

    Maylis

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    Makes sense why I got them all right then, the s and 1/s cancel anyways
     
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