The magnitude of 2nd order low pass filter is given as $$|H(\omega)|^2= \frac{1}{(1-(\frac{\omega}{\omega_o})^2)^2+(\frac{2\zeta\omega}{\omega_o})^2}$$ Now in order to achieve maximally flat within pass band, we take the derivative of this equation and set it to zero to find the extremums. \begin{equation*} \frac{\partial(\frac{1}{|H(\omega)|^2})}{\partial\omega} = \frac{\partial[(1-\omega^2)^2+(2\zeta\omega)^2]}{\partial\omega} \end{equation*}

\begin{equation*} \begin{split} 2(1-\omega^2)(-2\omega)+8 \zeta^2\omega &= 0 \\ \omega(\omega^2-1)+2\zeta^2\omega &= 0 \\ \omega (\omega^2-1+2\zeta^2) &=0 \\ \end{split} \end{equation*}

Then we get $\omega =0$ or $\omega^2 = 1-2\zeta^2$

Now my question is, why do we set the second root to be zero to obtain no ripples in the passband so that we can derive: $$\zeta=1/\sqrt{2}$$ I understand that the slope of the magnitude is zero at these roots, but I couldn't really interpret the idea here. Any help is appreciated.


1 Answer 1


You have to check the next derivative, because the first one is zero anyway at $\omega=0$. The second derivative is


which can be made zero by choosing $\zeta=1/\sqrt{2}$.

  • $\begingroup$ Thank you so much! This has bothered me for a whole day! $\endgroup$
    – Yihan Hu
    Jul 16, 2020 at 18:49
  • 1
    $\begingroup$ Yes, please mark it accepted so it won’t come back unanswered to the feed ;) $\endgroup$
    – jojeck
    Jul 16, 2020 at 19:41

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