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Homework Help: Damping ratio from a transfer function

  1. Jul 6, 2010 #1
    I have a transfer function for system.

    23.23*s + 1.421
    ------------------------------------- = tf
    s^2 + 25.88*s + 1.421

    Since the numerator has a non-zero coefficient for "s" I am wary about equating

    25.88 = 2 * zeta * omega [the stuff we usually do for calculating the damping ratio].

    Can someone shed any light on this?


  2. jcsd
  3. Jul 8, 2010 #2
    Just a refresher. Can someone answer the above question?
  4. Jul 8, 2010 #3


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    Hello sgsawant,

    It's been awhile since I've looked at stuff like this. But it appears that your system is some sort of lowpass + bandpass combination.

    I believe your "25.88 = 2 * zeta * omega" is still valid though.

    Have you considered the fact that

    [tex] \frac{23.23s + 1.421}{s^2 + 25.88s + 1.421} = \frac{23.23s}{s^2 + 25.88s + 1.421} \ + \ \frac{1.421}{s^2 + 25.88s + 1.421}[/tex]
  5. Jul 8, 2010 #4
    Thanks for your reply. But the cases where 2 * zeta * omega is valid - for those that I have seen - have omega squared in the numerator and also as the constant in the quadratic of s in the denominator. Thus omega can be usually simply calculated (either by taking the root of the numerator or that of the constant term in the denominator).

    What method would you advice in this case to calculate omega - and then eventually zeta?


  6. Jul 9, 2010 #5


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    Yes, that's the case for a second order lowpass system (or a second order lowpass section of a more complex system).

    The transfer function of a second order lowpass system H(s) takes the form

    [tex] H_{LP}(s) = A_0 \frac{\omega_0^2}{s^2 + 2 \zeta \omega_0 s + \omega_0^2} [/tex]

    Note that at low frequencies (frequencies can be seen by replacing s with ), H() approaches A0. But at high frequencies, H() approaches 0; meaning only the low frequencies get through.

    But there are other types of second order systems. A second order highpass system takes the form

    [tex] H_{HP}(s) = A_0 \frac{s^2}{s^2 + 2 \zeta \omega_0 s + \omega_0^2}. [/tex]

    Here, only the high frequencies make it through.

    Similarly for a bandpass we have:

    [tex] H_{BP}(s) = A_0 \frac{2 \zeta \omega_0 s}{s^2 + 2 \zeta \omega_0 s + \omega_0^2}. [/tex]

    And bandstop,

    [tex] H_{BS}(s) = A_0 \frac{s^2 + \omega_0^2}{s^2 + 2 \zeta \omega_0 s + \omega_0^2}. [/tex]

    Note that the denominator is the same in all types.

    The numerator (i.e. zeros) of your transfer function doesn't seem to quite fit with any of the standard second order systems. But if I remember correctly, both the the resonant (natural) frequency and the damping factor are functions of the pole locations, not the zeros.
    Last edited: Jul 9, 2010
  7. Jul 9, 2010 #6
    Thanks mate! That was quite informative. The poles argument really made it clear.
  8. Mar 27, 2011 #7
    That is the damping offered by the base movement.. you do not have to worry about that.. 2*zeta*omega holds good
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