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

Signal Processing Asked by Yihan Hu on October 24, 2021

The magnitude of 2nd order low pass filter is given as
$$|H(omega)|^2= frac{1}{(1-(frac{omega}{omega_o})^2)^2+(frac{2zetaomega}{omega_o})^2}$$
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})}{partialomega} = frac{partial[(1-omega^2)^2+(2zetaomega)^2]}{partialomega}
end{equation*}

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

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

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.

One Answer

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

$$3omega^2+2zeta^2-1Big|_{omega=0}=2zeta^2-1tag{1}$$

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

Answered by Matt L. on October 24, 2021

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