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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?

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    \$\begingroup\$ 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. \$\endgroup\$
    – Andy aka
    Commented Feb 15, 2018 at 13:13
  • \$\begingroup\$ 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. \$\endgroup\$
    – LvW
    Commented Feb 15, 2018 at 15:37
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    \$\begingroup\$ Only for a second-order bandpass the pole quality factor Qp equals the expression (wo/dwo). \$\endgroup\$
    – LvW
    Commented Feb 15, 2018 at 16:31
  • \$\begingroup\$ Oh ......so can anyone please provide me the general formula to calculate Q-factor of any type of transfer function \$\endgroup\$
    – Suresh
    Commented Feb 16, 2018 at 5:09
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    \$\begingroup\$ 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: electronics.stackexchange.com/questions/233654/… \$\endgroup\$
    – Andy aka
    Commented Feb 20, 2018 at 12:37

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