How to find damping ratio of a 4th order system?

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

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):

.

open loop transfer function:

the expansion of open loop system fraction:

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?

• How the damping ratio is defined for 4'th order system? Commented Jan 17, 2018 at 17:34
• This transfer function does not exhibit any peak in gain. take the derivative and see why. Commented Jan 17, 2018 at 18:42
• 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.
– Chu
Commented Jan 17, 2018 at 18:55

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.

• I didnt find out my response Commented Jan 17, 2018 at 17:47
• @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.
– jonk
Commented Jan 17, 2018 at 17:55
• @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.)
– jonk
Commented Jan 17, 2018 at 18:02
• @jonk I would find it interesting too. Commented Jan 17, 2018 at 18:06

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.

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