Your browser does not support inline frames or is currently configured not to display inline frames. The magnitude of the voltage transfer function is derived from the potential divider equation: -   $$\dfrac{V_{OUT}}{V_{IN}} = \dfrac {j\omega L}{R + j\omega L +\frac{1}{j\omega C}} = \dfrac{-\omega^2 LC}{j\omega RC - \omega^2 LC +1} = \dfrac{-\omega^2}{-\omega^2 + j\omega\frac{RC}{LC}+\frac{1}{LC}}$$   For a 2nd order filter $$\frac{1}{LC} = \omega_n^2$$   and $$\dfrac{R}{\omega_n L} = \dfrac{1}{Q}= 2\zeta$$, we find: -   $$\dfrac{V_{OUT}}{V_{IN}} = \dfrac{1}{1-\dfrac{\omega_n^2}{\omega^2}-j2\zeta\dfrac{\omega_n}{\omega}}$$    this yields the TF's phase angle: $$-\arctan\left[\dfrac{2\zeta\dfrac{\omega_n}{\omega}}{1-\dfrac{\omega_n^2}{\omega^2}}\right]$$   Multiplying numerator and denominator by the conjugate of the denominator we get: -   $$\dfrac{V_{OUT}}{V_{IN}} =\dfrac{1-\dfrac{\omega_n^2}{\omega^2}+j2\zeta\dfrac{\omega_n}{\omega}}{(1-\dfrac{\omega_n^2}{\omega^2})^2+4\zeta^2\dfrac{\omega_n^2}{\omega^2}}$$   The magnitude is found by taking the square root of the squared numerator terms: -   $$|H(j\omega)| = \dfrac{\sqrt{(1-\dfrac{\omega_n^2}{\omega^2})^2+4\zeta^2\dfrac{\omega_n^2}{\omega^2}}}{(1-\dfrac{\omega_n^2}{\omega^2})^2+4\zeta^2\dfrac{\omega_n^2}{\omega^2}} = \dfrac{1}{\sqrt{(1-\dfrac{\omega_n^2}{\omega^2})^2+4\zeta^2\dfrac{\omega_n^2}{\omega^2}}}$$   $$|H(j\omega)| = \dfrac{1}{\sqrt{1+\dfrac{\omega_n^2}{\omega^2}\cdot(4\zeta^2-2)+\dfrac{\omega_n^4}{\omega^4}}}$$   As a sanity check, if $$\omega$$ is set to the natural frequency we get: -   $$|H(j\omega)| = \dfrac{1}{\sqrt{1+(4\zeta^2-2) +1}} = \dfrac{1}{2\zeta} = Q$$   as expected.

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