TRANSFER FUNCTION
\(H(s)=\dfrac{s^2}{s^2+2\zeta\omega_ns+\omega_n^2}\)
To find the "relevant" poles, equate the denominator to zero and solve for s: -
\(s = \dfrac{-2\zeta\omega_n\pm\sqrt{4\zeta^2\omega_n^2-4\omega_n^2}}{2} = -\zeta\omega_n\pm\sqrt{\zeta^2\omega_n^2-\omega_n^2}\)
\(s=\omega_n(-\zeta\pm\sqrt{\zeta^2-1})\)
Peaking frequency and amplitude
\(H(j\omega)=\dfrac{1}{1-\dfrac{\omega_n^2}{\omega^2}-\dfrac{j2\zeta\omega_n}{\omega}}\)
The difference between high pass and low pass is that \(\omega_n\) and \(\omega\) have swapped places and the imaginary part is now negative.
\(|H(j\omega)|=\dfrac{1}{\sqrt{(1-\dfrac{\omega_n^2}{\omega^2})^2+\dfrac{4\zeta^2\omega_n^2}{\omega^2}}}\)
Differentiate the denominator and equate to zero to find the minima: -
\(0 = 4\omega_n^2 - 8\zeta^2\omega_n^2 - 4\dfrac{\omega_n^4}{\omega^2}\)
\(\omega^2 = \dfrac{4\omega_n^4}{4\omega_n^2 - 8\zeta^2\omega_n^2} = \dfrac{\omega_n^2}{1 - 2\zeta^2}\) hence, \(\omega = \omega_n\sqrt{\dfrac{1}{1-2\zeta^2}}\)
Note the similarity with the low pass filter's peaking frequency: \(\omega = \omega_n\sqrt{1-2\zeta^2}\)
To find the peak magnitude, replace the peaking frequency in the formula for \(|H(j\omega)|\): -
\(|H(j\omega)|=\dfrac{1}{\sqrt{(1-1+2\zeta^2)^2+4\zeta^2(1-2\zeta^2)}} = \dfrac{1}{\sqrt{4\zeta^4 + 4\zeta^2 - 8\zeta^4}}\)
\(=\dfrac{1}{2\zeta\sqrt{1-\zeta^2}}\) i.e. exactly the same as for the low-pass filter example.