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We've been requested to find the resonant frequencies of a 3rd order plant in class and were presented to the following method to do so: Example included

My question is, when using the $\ \omega_r $ formula, to find both of the resonant freq's do I need to place $\ \omega_{n1} $ with $\ ζ_{1} $ to get $\ \omega_{r1} $ (and $\ -\omega_{n2} $ with $\ ζ_{2} $ to get $\ \omega_{r2} $), or I am supposed to use the same $\ ζ_{2} $ (of the Pole) for both, like this:

$$ \omega_{r1} = f(\omega_{n1}, ζ_{2}) $$

$$ \omega_{r2} = f(\omega_{n2}, ζ_2) $$

I hope the question is clear enough, thanks!

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  • $\begingroup$ Do you need to use this formula to find $\omega_r$ ? You can find the magnitude in terms of $j \omega$ and check where the derivative of the magnitude is zero (for $ω=ω_r$). $\endgroup$ – Teo Protoulis Feb 17 '20 at 15:59
  • $\begingroup$ Definitely, though this method can save lots of time if used correctly :) $\endgroup$ – Yarden2y Feb 17 '20 at 19:40
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The resonance is caused by the poles, therefore $\omega_r = \omega_{n_2} \sqrt{1-2\zeta_2^2} \approx 2.9462$.

The zeros do not cause a resonance, but an antiresonance. Thus, the antiresonance frequency is located at $\omega_{ar} = \omega_{n_1} \sqrt{1-2\zeta_1^2}\approx 0.9592$.

You can verify this by drawing the Bode DiagramBode Diagram Note that there is a small deviation between the values obtained using the formula and the Bode Diagram created with MATLAB. The small error is caused by the poles and zeros being close to each other, while the used formula only holds for a pure second order system. The further away the poles are from each other, the smaller the effect and the better the formula will estimate the (anti)resonance frequency.

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    $\begingroup$ Thank you very much for you time and the great answer! $\endgroup$ – Yarden2y Feb 22 '20 at 21:35

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