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I have an old digital synthesizer that has a Butterworth-like 2nd order filter with resonance. However, with high frequency tones it appears there is no steep cutoff near Nyquist (I am sure it is the Nyquist as the tones fold beyond that point), as is usual with the bilinear-transformed Butterworth filter. This poses few questions.

  1. Is the attenuation caused by aliasing or something different, and what is it called?
  2. Are there digital alternatives to the Butterworth or bilinear transform that do not suffer from the attenuation near Nyquist, or is this a usual feature?
  3. What filter could an old resource-strained synthesizer have used to achieve a response without the attenuation?

Clarification: By Butterworth-like response it is meant that the response appears similar at lower frequencies, I have not confirmed that there is no ripple or anything like that. Nonetheless, a 2nd order filter of some sort.

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  • $\begingroup$ I think it depends on how the analog transfer function was mapped to the z domain. If you use the bilinear transform, that's impossible. But maybe if you use impulse response matching or pole matching. $\endgroup$
    – Ben
    Mar 12 at 12:01
  • $\begingroup$ What do you mean "with resonance"? $\endgroup$ Mar 12 at 12:38
  • $\begingroup$ @a concerned citizen Synthesizer filters usually include resonance control to feed the filter back to itself, creating a boosted response at the cutoff. See old question: dsp.stackexchange.com/questions/72686/… for Butterworth with resonance (and answer if you can ;)). $\endgroup$
    – Tony
    Mar 12 at 12:52
  • $\begingroup$ @Tony A Butterworth filter has flat passband and strictly monotonically decreasing stopband. Any deviation from this response means it's no longer a Butterworth. There's no such thing as a Butterworth with peaking, or resonance. Also, when using the ping operator, @, use <TAB> to cycle between names and leave them as they are selected, there should be no spaces. $\endgroup$ Mar 12 at 13:44
  • $\begingroup$ @aconcernedcitizen It's a Butterworth filter with added resonance control, as given by dsp.stackexchange.com/questions/19262/…. When the resonance is zero, then it becomes a Butterworth filter. $\endgroup$
    – Tony
    Mar 12 at 14:10
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Is the attenuation caused by aliasing or something different, and what is it called?

The two standard methods of mapping an analog filter to a digital one are either the bilinear transform or the impulse invariant transform. Neither does a particularly good job at preserving the transfer function at higher frequencies. The bilinear transform will map whatever happens at "infinite" analog frequency to the digital Nyquist frequency. For any analog lowpass filter (Butterworth or not), this will indeed create the steep roll off at Nyquist.

Are there digital alternatives to the Butterworth or bilinear transform that do not suffer from the attenuation near Nyquist, or is this a usual feature?

There are plenty of alternatives. Impulse invariant is one of them, although it to tends overshoot by 6dB at Nyquist. One heuristic method is to start with bilinear, and then move the zeros from z = -1 towards z = 0 until you have something you "like" by whatever criteria is important for your application.

What filter could an old resource-strained synthesizer have used to achieve a response without the attenuation?

A biquad is a biquad is a biqaud. Computational cost is not a function of the specific filter coefficients. There is really no difference between a Butterworth or any other second order filter.

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  • $\begingroup$ By moving the zero it suffices to $z\rightarrow(z-a)$ on the polynomial for zeros, or is the intention to have more zeros? $\endgroup$
    – Tony
    Mar 13 at 10:23
  • $\begingroup$ It seems like a state variable filter is a good candidate as well. In which case they wouldn’t be calculating coefficients in the same sense as the more proper biquad topologies. $\endgroup$
    – Dan Szabo
    Mar 13 at 13:53
  • $\begingroup$ @DanSzabo It worked great! wiki.analog.com/resources/tools-software/sigmastudio/toolbox/… For 4-pole is the idea to combine two of these or something different? $\endgroup$
    – Tony
    Mar 14 at 12:39
  • $\begingroup$ @Tony glad to hear it. If you cascaded two in series you would have a pretty easy to control 4-pole filter. How exactly you control is up to you, since 4-pole filters don’t have a ‘resonance’ parameter per se. If you keep the Q the same, your magnitude response will be the square of the 2-pole version, so super easy to picture. However, if you wanted say a Butterworth response, the Qs would be different. $\endgroup$
    – Dan Szabo
    Mar 14 at 15:29

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