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I want to proof Quality Factor of 2nd order system is \$ \frac{1}{2 \times \zeta} \$ ;
that is
$$if \quad H(s)= \frac{k \times \omega_n^2}{s^2 + 2 \times \zeta \times \omega_n \times s + \omega_n^2} \quad then \quad Q= \frac{1}{2 \times \zeta} $$

My Approach: enter image description here

Now how can i proceed?

Suresh
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    The peaking frequency is \$\omega_n\sqrt{1-2\zeta^2}\$ for a 2nd order low pass filter. This isn't the so-called "resonant frequency". For a BP filter the peaking frequency IS the resonant frequency. – Andy aka Feb 15 '18 at 13:13
  • You have started with the wrong definition for Q. For a 2nd order system the "Q-factor" is defined by the pole position (Quality Qp of the pole). The definition is Qp=1/2cos(alpha) - with alpha being the angle between the negativ-real axis of the s-plane and the pointer between the origin and the pole position. Note that the cos function contains the real part of the pole ("sigma") - and this gives the relation between Qp and the damping ratio. – LvW Feb 15 '18 at 15:37
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    Only for a second-order bandpass the pole quality factor Qp equals the expression (wo/dwo). – LvW Feb 15 '18 at 16:31
  • Oh ......so can anyone please provide me the general formula to calculate Q-factor of any type of transfer function – Suresh Feb 16 '18 at 05:09
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    Q factor only applies to 2nd order filters or filters that can be broken down to 2nd order filters. You might start with Q = the amplitude response at the natural resonant frequency of a 2nd order low pass filter. See the bottom picture of my answer here: https://electronics.stackexchange.com/questions/233654/intuition-for-resonant-natural-and-oscillatory-frequencies-of-rlc-circuits/233852#233852 – Andy aka Feb 20 '18 at 12:37

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