[sdiy] Designing 4-pole filters with identical 2-pole stages - why not?

[Tom Wiltshire]

Hi Neil,

On 21 Dec 2015, at 22:54, Neil Johnson <neil.johnson71 at gmail.com> wrote:

>> The usual way to do this is to set the Q of the first stage to 0.541, and the Q of the second to 1.307. Multiplying one by the other gives an overall Q of 0.707, which is our Butterworth response.
> 
> Hmm.... in my experience the "usual way" is to specify what you want,
> then grab a filter cookbook like Williams and Taylor or the venerable
> Zverev, read out the coefficients from the appropriate table,
> frequency and impedance scale, and there's your answer.  Not much to
> it.
> 
> Or am I missing something?

The fun bit! The fun bit is where I accidentally discovered that you can get very close (within 1dB passband peaking) to a 4-pole Butterworth response with two identical stages if you set the resonance to 0.841. Since my original assumption (that it was only the overall q that mattered) was wrong, this number doesn't have any magic quality, and you can probably get closer by reducing the q a bit more.

My intention was just to try and simplify a schematic a bit more by reducing the number of individual component values. Building two stages with Qs of 0.541 and 1.307 tends to lead to a lot of odd values, and even if not odd values, usually eight separate values. I was looking for ways to simplify that. I'd tried equal-R and equal-C filters, and couldn't get the response I needed, so I decided to try making the two *stages* identical instead. And it works, or close enough for rock'n'roll.

Tom