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For example in this 4th order transfer function how the damping ratio would be calculated? enter image description here


in fact I` m encountered with this problem: If I didnt realize the concept of problem plz guide me:

problem: calculate gain magnitude (k) if damper ratio of closed loop system is 0.7 (it means zita=0.7):

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open loop transfer function:

enter image description here


the expansion of open loop system fraction:

enter image description here

and now how much is zita? quadratic faraction isn t standard form because there s a 's' in numerator, am I right?


how much is k due to the zita has a magnitude of 0.7? enter image description here

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    \$\begingroup\$ How the damping ratio is defined for 4'th order system? \$\endgroup\$
    – Eugene Sh.
    Jan 17, 2018 at 17:34
  • \$\begingroup\$ This transfer function does not exhibit any peak in gain. take the derivative and see why. \$\endgroup\$ Jan 17, 2018 at 18:42
  • \$\begingroup\$ In this case, it probably means the damping coefficient of the bracketed 2nd order term, since this is under-damped. But strictly it's undefined for a TF higher than 2nd order. \$\endgroup\$
    – Chu
    Jan 17, 2018 at 18:55

2 Answers 2

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The damping ratio is a parameter, usually denoted by ζ (zeta),1 that characterizes the frequency response of a second order ordinary differential equation.

The quote above is taken from Wikipedia: Damping ratio. In other words it relates to a 2nd order transfer function and not a 4th order system. Having said that, if it is possible to reduce the denominator to two multiplying equations each of the form: -

\$s^2 +2s\zeta \omega_n + \omega_n^2\$ (where \$\zeta\$ is damping ratio and \$\omega_n\$ is natural resonant frequency)

Then it would have some meaning.

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  • \$\begingroup\$ I didnt find out my response \$\endgroup\$
    – muhamad
    Jan 17, 2018 at 17:47
  • \$\begingroup\$ @somebody If you haven't yet got it, the concept "damping factor" only applies to quadratics. Suppose you have a quadratic: the names of each constant are: quadratic coefficient, linear coefficient and constant coefficient, in that order. The linear coefficient has a certain meaning. Now, suppose someone gave you a 4th order equation. The linear coefficient would NOT have the same meaning, anymore. The whole idea is in the trash, so to speak. There is no understood and well-defined meaning for "damping ratio" on a 4th order system. So you need to define your meaning. \$\endgroup\$
    – jonk
    Jan 17, 2018 at 17:55
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    \$\begingroup\$ @somebody The idea of a damping factor in 2nd order is as an index of its tendency towards oscillation. If you can work out an index that achieves this goal for 4th order systems, then I think you'd already have your answer anyway. But feel free to discuss it, as a comprehensive study of the set of all possible 4th order filters in this context would actually be interesting to read (to me, anyway.) \$\endgroup\$
    – jonk
    Jan 17, 2018 at 18:02
  • \$\begingroup\$ @jonk I would find it interesting too. \$\endgroup\$
    – Andy aka
    Jan 17, 2018 at 18:06
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In this case, it probably means the \$\small \zeta\$ of the bracketed 2nd order term, since this is under-damped. But strictly it's undefined for a TF higher than 2nd order.

Also, this appears to be an open loop TF, so the question may be looking for the 'equivalent' 2nd order closed loop TF where the \$\small \zeta\$ value is given by the ROT: \$\small\zeta\approx 0.01\times PM\$, and \$\small PM\$ is the phase margin in degrees.

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  • \$\begingroup\$ how is the manner of calculating its phase margine? \$\endgroup\$
    – muhamad
    Jan 17, 2018 at 19:37
  • \$\begingroup\$ From the open loop Bode plot, it's (180 - phase angle) at the frequency where the gain is 0dB \$\endgroup\$
    – Chu
    Jan 18, 2018 at 7:12

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