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I was under the impression that the differences between a Chebyshev, Butterworth, and Bessel filter was that they represented the cases of an underdamped, critically damped, and overdamped systems, respectively. However, after revisiting the transfer function for a second-order Sallen-Key filter, it seems I am mistaken as a Butterworth filter is defined with \$\mathrm{Q} = \frac{1}{\sqrt{2}}\$ or 0.707 placing it in the underdamped category. For a critically-damped filter, Q must be equal to 0.5. Why then doesn't an ideal Butterworth filter seem to exhibit ringing as can be seen with a Chebyshev filter?

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  • \$\begingroup\$ You must not mix the terms "ringing" (overshoot; defined in the TIME domain only) and "peaking resp. ripple" (defined in the FREQUENCY domain only). \$\endgroup\$ – LvW May 2 '18 at 18:41
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A Butterworth filter has a flat pass band (spectral response): -

enter image description here

But it will still ring when a transient comes along (time response): -

enter image description here

Only a critically damped or over-damped filter will not produce overshoot (ringing).


EDIED to show that a 2nd order Bessel filter does produce 0.433 % overshoot: - enter image description here

Calculator source

The resistor value is manipulated to give a zeta of precisely 0.866 with Fd of 0.5*Fn (aka Bessel).

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  • \$\begingroup\$ Not only a critically damped filter ... any overdamped filter won't overshoot either - such as a Bessel. \$\endgroup\$ – Brian Drummond Jan 17 '17 at 12:01
  • \$\begingroup\$ @BrianDrummond very true I shall make amends. \$\endgroup\$ – Andy aka Jan 17 '17 at 12:02
  • \$\begingroup\$ Brian Drummond - it is not quite correct that a Bessel step response "won`t overshoot". It does! However, just a little. This applies to all 2nd-order functions with Qp>0.5. \$\endgroup\$ – LvW Jan 17 '17 at 12:27
  • \$\begingroup\$ Bessel doesn't overshoot. Some approximations to Bessel, for instance those found in A.B.Williams, do very slightly. \$\endgroup\$ – Neil_UK Jan 17 '17 at 12:33
  • \$\begingroup\$ Bessel overshoot is 0.43% - see the formula in my answer. \$\endgroup\$ – LvW Jan 17 '17 at 13:28
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The step response of a second order lowpass filter does not depend on the selected filter topolgy. It is rather a function of the pole location - which can be expressed using the pole quality factor Qp. The following expression for the overshoot (in percent) can be used:

overshoot (gamma)=100*exp[-pi/SQRT(4Qp²-1)].

As you can see, for Qp=0.5 the exponent approaches "minus infinity" and gamma is zero. As an example, the Bessel response (Qp=0.577) gives an overshoot of only 0.43 %

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The Butterworthiness is a principle to make maximally flat frequency response at the center frequency of the passband or at 0 Hz if the filter is LPF. This principle leads some ringing - not much, but some. Its mathematics.

Sallen-Keyness is the principle how the component values have been picked from the vast set of possiblities that in theory imply the same filtering effect.

The inventors, Sallen and Key picked the combination that allows maximally loose component tolerances for a given error in frequency response.

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Yes, a Butterworth should ring, but who knows what should happen when you implement it with 10% or worse caps? Those poles can move around substantially.

One of my old electronics labs as a student involved tweaking out a Butterworth using a spectrum analyzer. It was not easy.

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