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.
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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