Damping Ration of Second Order System

Discussion in 'Homework Help' started by Hanck, Sep 4, 2009.

  1. Hanck

    Thread Starter Member

    Nov 25, 2008
    18
    0
    I know the prototype for second order circuit is

    T = w^2 / [s^2 + 2*zita*w*s + w^2]

    but what about the system with zeros instead of a constant gain?

    for exam T = [s+1]/[s^2+ s + 1] ?

    does the frequency and damping ratio follow the prototype?
    (i.e w = 1, damping ratio zita = 1/2) If yes, what about the frequency on numerator?

    Thanks in advance.
     
  2. steveb

    Senior Member

    Jul 3, 2008
    2,433
    469
    In my opinion, you can say that the natural frequency and the damping ratio are meaningful. It's important to keep in mind that the second order system, with a zero, typically behaves differently from the pure second order system, but there are also some similarities. The natural frequency will be the peak resonance frequency, when the system is underdamped, and this relation still holds up. Even if you place a zero exactly at the natural frequency (in an underdamped system), you can't cancel out the resonance because the poles are a complex conjugate pair (i.e with imaginary parts), while the zero must be a real number (in a physically realizable system). Also, I believe that the damping ratio is still an indicator of whether the system is underdamped, overdamped or critically damped.

    Note that, if the system is critically damped, or overdamped, then it is possible to have a "pole-zero-cancellation" and the system becomes a first order system.
     
    Last edited: Sep 4, 2009
  3. Hanck

    Thread Starter Member

    Nov 25, 2008
    18
    0
    Thanks steveb, that is informative :)
     
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