2nd order HIGH pass filter design





To find the "relevant" poles, equate the denominator to zero and solve for s: -


\(s = \dfrac{-2\zeta\omega_n\pm\sqrt{4\zeta^2\omega_n^2-4\omega_n^2}}{2} = -\zeta\omega_n\pm\sqrt{\zeta^2\omega_n^2-\omega_n^2}\)




Peaking frequency and amplitude


The difference between high pass and low pass is that \(\omega_n\) and \(\omega\) have swapped places and the imaginary part is now negative.


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.


Quality factor, amplitude and phase at \(\omega_n\) 

\(H(j\omega)=\dfrac{1} {1-\dfrac{\omega^2}{\omega_n^2}-\dfrac{j2\zeta\omega}{\omega_n}}\)    and if \(\omega = \omega_n\), then 

\(H(j\omega)=\dfrac{1}{1-1-j2\zeta} = +jQ\)  \(=Q\angle{+90}\)




still under construction

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