# Proof ;Q-factor of second order system

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:

Now how can i proceed?

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