[sdiy] Oberheim Xpander Multimode VCF Question

[Steven Cook]

Thanks for this, Magnus. I see what you mean about my use of '-24dB/oct' - 
perhaps I should have said 4 pole. I think I've made another mistake as 
well: (+) and (-) are swapped in my example, so it should have been:

input - pole1*3 + pole2*6 - pole3*4

I still don't really understand the math(s), though. Is it possible to work 
out by educated guesswork, like by noting that the allpass given above 
consists of a 24dB highpass minus a 24dB lowpass and a 6dB lowpass? I'm 
thinking, get rid of the 6dB lowpass contribution and end up with:

input - pole1*4 + pole2*6 - pole3*4

Or am I barking up the wrong tree?

Steven Cook.

http://www.spcplugins.com/
+44(0)1271 867288


--------------------------------------------------
From: "Magnus Danielson" <magnus at rubidium.dyndns.org>
Sent: Thursday, March 11, 2010 11:04 AM
To: "Steven Cook" <stevenpaulcook at tiscali.co.uk>
Cc: <synth-diy at dropmix.xs4all.nl>
Subject: Re: [sdiy] Oberheim Xpander Multimode VCF Question

> Steven Cook wrote:
>> Hi,
>>
>> Is it possible to get a -24dB/oct allpass response out of an Oberheim 
>> Xpander type of 4-pole filter arrangement? I've got the following for 
>> an -18dB allpass, from a post by David G. Dixon...
>>
>> input + pole1*3 - pole2*6 + pole3*4
>>
>> Which seems to work fine (in software).
>
> I'm in a hurry, so you will not get the complete answer now... from me at 
> least.
>
> First of all, you can't talk about -24 dB/Oct slope for an all-pass 
> filter. While I am happy that you say -24 dB/Oct for low-pass, it is 
> meaningless for all-pass as they are 0 dB/Oct. This only once again brings 
> out the point... the slope does not tell how many poles and zeros there is 
> in a filter, just how many there at least needs to be.
>
> Now, the actual question is, can you create a 4-pole/4-zero all-pass 
> filter from the Xpander module? The quick answer is yes.
>
> What I don't have time to help you with is how to combine the outputs from 
> the different taps, but you are on the right path when you combine input 
> with the taps.
>
> For non-complex poles, you have (s+a)/(s-a) for each stage. For complex 
> poles (s+a-jb)(s+a+jb)/(s-a-jb)/(s-a+jb). Notice how the real term a shift 
> sign from poles to zero. Any all-pass filter is then just a combination of 
> those using various sets of a and b. You should be able to pull it out 
> from there with some paper-and-pen work, which is the thing I don't have 
> time for right now.
>
> If you still have problem, let me know. I might be able to do the homework 
> for you later.
>
> Cheers,
> Magnus