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Second order system

  1. Apr 23, 2017 #1
    < Mentor Note -- thread moved to HH from the technical physics forums, so no HH Template is shown >

    how we can find contact time of second order system ? 2y**+4y*+8y=8x I want to find damping coefficient ... howz possible
     
    Last edited by a moderator: Apr 27, 2017
  2. jcsd
  3. Apr 23, 2017 #2

    Hesch

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    ( 2s2 + 4s + 8 ) * y(s) = 8 * x(s)

    Finding y(s)/x(s) the characteristic equation of the transfer function will be:

    2s2 + 4s + 8 = 0 →

    s2 + 2s + 4 = 0

    which can be formulated

    s2 + 2ξωns + ωn2 = 0

    So ωn = 2 , the damping coefficient, ξ = 0.5
     
  4. Apr 23, 2017 #3
    actually I got this question from a book of M.handa .I solved my answer are same but the answer given in the book is damping coefficient is 1 ..... can you tell me about time constant . how we can find it
     
  5. Apr 23, 2017 #4

    Hesch

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    The equation

    2s2 + 4s + 8 = 0

    has two complex roots: s = -1 ± j√3.

    If the characteristic equation were to have a damping ratio = 1, it should have two real roots at the same location, for example.
    s1 = -1.2 , s2 = -1.2.
    In this case the characteristic equation could be written:
    s2 + 2.4s + 1.44 = 0
    and the transfer function would have a double time constant = 1/1.2 sec.

    If the characteristic equation were
    s2 + 7s + 10 = 0
    the two roots would be s1 = -2 , s2 = -5 , and the transfer function would have two time constants:
    τ1 = 0.5 sec. , τ2 = 0.2 sec.

    When a root is complex, we cannot speak of a "time constant". The time is complex. The system will oscillate within some damping coefficient.

    2000px-2nd_Order_Damping_Ratios.svg.png
     
    Last edited: Apr 23, 2017
  6. Apr 24, 2017 #5
    thankyou :)
     
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