Running a bode plot and AC analysis both show cutoff frequencies as desired. I have a two stage band pass Butterworth filter with cutoff frequencies of 0.05 and 40 Hz. If I'm not mistaken they should do the same thing. Also, \$w_n=w_p\$, causes an infinite response (undamped system - oscillator). Hello all, I've been running into an inconsistency between the bode plot and AC Sweep. The meaning of \$w_n\$ for the Butterworth response is the same as for the first-order case, that is, \$w_n\$ represents the -3 dB frequency, also called cuttoff frequency. The magnitude curve is sais to be maximally flat (no peak). In filter theory, that special value for \$\zeta=0.707\$ corresponds to a Butterworth response. Note on figure below: When varying the damping ratio \$\zeta\$, the peak follows a specific curve. Where \$\omega_n\$ is the natural frequency (also called corner frequency when considering assymptotes), the peak Peaks in the frequency response can only exist in systems with conjugate complex poles.įor an underdamped (\$\zeta 0.5\$) second-order system, the peak appears specifically for \$\zeta<1/\sqrt$$
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