Creating a 4p filter from 2 2p ones?
Guys! Id like to ask.
I'm loving the skf core filter. but i think thats only 2p.
I know this may be a very silly question.
But... if i cascade 2 of them des that mean i technically get a 4p filter?
sorry if this question is dumb.
Im learning 😅
Comments
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I've done something similar before where I'd crossfade between different filter amounts. I presume since a 2-pole is 12dB/Oct, if you add two of them together you'd get a reduction of 24dB/Oct. Just test it out with a frequency analyzer to see if the cutoff gets steeper.
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Just did a quick test, and yup they add up the same. Here's a quick photo.
The reason you don't see the B channel is because it's lined up perfectly with the A channel, thus showing the cutoff steepness is equal.
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Yes, it's true. A filter is a "linear" circuit, meaning that it contains nothing but addition, and multiplication by constants.
All bets are off, though, if a filter contains internal saturation or any other kind of waveshaping. Then it's "nonlinear".
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so yeah, you can cascade filters serially and the number of poles of filtering adds
however this does not make it a 4 pole filter as its usually understood. think svf and tkf are inherently 2 pole. usually ina 4 pole filter the feedback path goes through the entire structure. cascading two filters like you're talking about breaks that feedback path in two, so its not usually done, but if you decide it works for you there's absolutely no reason not to
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however this does not make it a 4 pole filter as its usually understood. think svf and tkf are inherently 2 pole. usually ina 4 pole filter the feedback path goes through the entire structure. cascading two filters like you're talking about breaks that feedback path in two, so its not usually done, but if you decide it works for you there's absolutely no reason not to
It does make a 4-pole. The caveat is what the resonance and the cutoff gain would be. For example, a 2-pole Butterworth (no resonance, maximally flat) has a Q of 1/sqrt(2), and -3 dB at the cutoff. If you chain two of them, you get -6 dB at the cutoff, so it's no longer a Butterworth. To get a 4-pole Butterworth by chaining 2-poles, you need Qs of 1/0.7654 and 1/1.8478. In general, the lp/hp cutoff gain is the product of the Qs of all the 2-pole stages, but you can get any given cutoff gain with different combinations of Qs, which would give different shapes. Vadim's 2nd-kind Butterworths (Butterworth2 in the core library) are an approach to this: they take a single resonance parameter from which they derive Qs for all the 2-pole stages and give a consistent resonant behavior.
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Also two 2-pole filters in series, making poor man's 4 pole filter, will have different phase shifts and so on, than real 4-pole filter.
But it will filter with steepness of 4-pole filter, no question....
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Some examples from the core library to clarify. This is the Butterworth LP4:
It's just two SVFs with the right values for 2R. An SVF with 2R = 1.4142 is a Butterworth LP2, but two of those in chain make a Linkwitz-Riley LP4, not a Butterworth.
This is the Butterworth2 LP4:
Again two SVFs, but the values for R are derived in a specific way from the resonance control.
As Chet said, the library SKF contains a saturator, which makes chaining it less easily predictable. You'd still get some kind of 4-pole, but as the Sallen-Key is kind of a 2-pole ladder, maybe you'd like one of the nonlinear versions (NL, NLA, NLB, D) of the ladder filter, which is 4-pole.
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Yah, you can put 10 filters in a row if you want to. But don't expect a sharp rolloff at the cutoff frequency. Typically the cutoff is 3db down, if you put 10 in series, the cutoff will be 30 db down. But it's still a matter of context. There is no rule saying what the filter has to sound like.
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