1
\$\begingroup\$

If we have a 4th order system,such as the one in snapshot,

enter image description here

how we can find its undamped natural frequency and damping ratio? Because on wikipedia ,i found case of 2nd order system and couldn't find case/scenario of fourth order system

apparently another question on EE SE has same subject

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

But the issue with this question is that it has no answer that is direct/simple and to the point as can be seen from the wording of those two answers ,they are containing a bit level of uncertainty, and the fact that OP hasn't accepted anyone of those two answers yet?

\$\endgroup\$
14
  • \$\begingroup\$ You can find the 2nd order ‘equivalent’ from the phase margin and gain cross-over frequency. \$\endgroup\$ – Chu Dec 29 '19 at 17:47
  • 3
    \$\begingroup\$ Does this answer your question? How to find damping ratio of a 4th order system? \$\endgroup\$ – user103380 Dec 29 '19 at 17:50
  • 1
    \$\begingroup\$ The op hasn’t accepted an answer because he hasn’t been back to this site since the day after he raised the question. Zeta is a 2nd order thing so break your equation into two 2nd order equations that are multiplied together and solve for zeta on both but separately. \$\endgroup\$ – Andy aka Dec 29 '19 at 19:17
  • 1
    \$\begingroup\$ There is no such thing as "zeta for a 4th order system". Zeta is a 2nd order thing. \$\endgroup\$ – Andy aka Jan 5 '20 at 16:50
  • 1
    \$\begingroup\$ @Andyaka Once one understands any of this, your statement is so obviously correct. And now that you suggest it, I am wondering if (a) engr is Man; or, b) engr is just copying Man, like a bot. Either way, I don't have the time or inclination to add an answer until engr engages me in a discussion about the approach I suggested. (Which is something I've never tried, but as I imagine it I believe may work "acceptably." If not acceptably, then certainly "interestingly." When I get some time, I'm definitely going to play with the idea and see how it compares to a non-trivial 4th order evaluation.) \$\endgroup\$ – jonk Jan 5 '20 at 20:31
1
\$\begingroup\$

Two possible methods of several:

  1. A 2nd order CLTF model of a higher order system (viz. the \$\small\zeta\$ and \$ \omega _n\$ values), can be derived from the open-loop gain cross-over frequency, and phase margin, thus:

$$\small\left(\frac{\omega _c}{\omega _n}\right)^2 \approx \sqrt{(4\zeta ^4 +1)}\:-2\zeta^2 $$

and

$$\small\zeta \approx 0.01\times PM $$

where \$\small PM\$ is the phase margin in degrees, and \$\small \omega _c\$ is the frequency at which the open loop gain crosses \$\small 0\:dB\$ (i.e. the frequency at which the phase margin is defined).

This assumes that \$ \omega _c\$ (and consequently the phase margin) exists.

  1. Dominant poles: the dominant poles are located at \$\small s=0\$ and \$\small s=-5\$, since the other two poles are at least three times further from the origin than the \$\small (s+5)\$ pole. Hence ignoring the less dominant poles gives the OLTF:

$$\small G_c (s)=\frac{K(s+7)}{s(s+5)} $$

with CLTF:

$$\small G(s)=\frac{K(s+7)}{s(s+5)+K(s+7)}=\frac{K(s+7)}{s^2 +(K+5)s+7K} $$

\$\endgroup\$
2
  • \$\begingroup\$ What do you mean? Which are less dominant poles?those at -15 and -12? \$\endgroup\$ – Man Jan 4 '20 at 16:11
  • \$\begingroup\$ Yes, further from origin = less dominant. Check by finding the residues at the poles for a simple 2nd order unit step response like: \$\frac{10}{s(s+1)(s+10)}\$ compared with: \$\frac{1}{s(s+1)}\$. \$\endgroup\$ – Chu Jan 4 '20 at 16:37

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.