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2^{nd} order low pass filter design  
BODE PLOT \(H(s)=\dfrac{\omega_n^2}{s^2+2\zeta\omega_ns+\omega_n^2}\)
To find the poles, equate the denominator to zero and solve for s: 
\(s = \dfrac{2\zeta\omega_n\pm\sqrt{4\zeta^2\omega_n^24\omega_n^2}}{2}\)
\(s = \zeta\omega_n\pm\sqrt{\zeta^2\omega_n^2\omega_n^2}\)
\(s=\omega_n(\zeta\pm\sqrt{\zeta^21})\) 


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Rearranging, \(\sqrt{\zeta^21}\Longrightarrow j\sqrt{1\zeta^2}\)
Hence poles are at \(\zeta\pm j\sqrt{1\zeta^2}\) and scaled by \(\omega_n\) 
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The bode plot and polezero diagram are two faces of the bigger 3D picture:  
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_____________________________ peaking freq and amplitude \(H(j\omega)=\dfrac{1}{1\dfrac{\omega^2}{\omega_n^2}+\dfrac{j2\zeta\omega}{\omega_n}}\)
and therefore \(H(j\omega)=\)
\(\dfrac{1}{\sqrt{(1\dfrac{\omega^2}{\omega_n^2})^2+\dfrac{4\zeta^2\omega^2}{\omega_n^2}}}\)
Differentiate \(\small\sqrt{\text{denominator}}\) and equate to zero to find the minima: 
\(4\dfrac{\omega^2}{\omega_n^2}=48\zeta^2\)
hence, \(\omega = \omega_n\sqrt{12\zeta^2}\)

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This is the peaking frequency.
To find the peak magnitude, replace the peaking frequency in the formula for \(H(j\omega)\): 
\(H(j\omega) = \dfrac{1}{1(\sqrt{12\zeta^2})^2+j2\zeta\sqrt{12\zeta^2}}\) \(=\dfrac{1}{2\zeta^2+j2\zeta\sqrt{12\zeta^2}}\) then convert to a magnitude value by squaring the terms: 
\(H(j\omega)= \dfrac{1}{\sqrt{(2\zeta^2)^2+4\zeta^2(12\zeta^2)}}\) = \(\dfrac{1}{\sqrt{4\zeta^4+4\zeta^2 8\zeta^4}}\) = \(=\dfrac{1}{2\zeta\sqrt{1\zeta^2}} \)
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Quality factor, amplitude and phase at \(\omega_n\)
\(H(j\omega)=\dfrac{1} {1\dfrac{\omega^2}{\omega_n^2}+\dfrac{j2\zeta\omega}{\omega_n}}\) and if \(\omega = \omega_n\), then \(H(j\omega)=\dfrac{1}{11+j2\zeta} = jQ\) \(=Q\angle{90}\)
still under construction
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