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 &amp;= 0 \\
    \omega(\omega^2-1)+2\zeta^2\omega &amp;= 0 \\
    \omega (\omega^2-1+2\zeta^2) &amp;=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 $\zeta=1/\sqrt{2}$.