# How to derive 2nd order Butterworth condition for the damping coefficient mathematically?

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.

You have to check the next derivative, because the first one is zero anyway at $$\omega=0$$. The second derivative is
$$3\omega^2+2\zeta^2-1\Big|_{\omega=0}=2\zeta^2-1\tag{1}$$
which can be made zero by choosing $$\zeta=1/\sqrt{2}$$.