[sdiy] 2 Pole SVF manipulations
Richie Burnett
rburnett at richieburnett.co.uk
Wed Feb 27 22:59:57 CET 2013
From: "Justin Owen" <juzowen at gmail.com>
> http://www.sdiy.org/juz/svf_poles.pdf
>
> Part 1 shows a simplified block/signal-flow diagram of the 2 Pole SVF that
> so many people have worked on over the years.
>
> I was wondering if anyone could answer (or speculate) what would happen at
> Outputs A through E for variations 2, 3 and 4 with regards to Filter Type
> (HP, LP etc), Phase and Attenuation/Slope?
The short answer is that it depends where the poles are.
I also think you might be confusing the state-variable filter with the moog
cascade type of filter. These are different topologies. If diagram 1 is
supposed to represent the basic 2nd-order state-variable filter then both
poles are located at DC. They are integrators. It is only the gain term of
these two blocks that changes when you modulate the cutoff frequency. There
should also be a weighted feedback path from point B to the input that
controls the damping. So, in summary the most basic SVF is two cascaded
integrators with feedback being a weighted mix of the two integrator
outputs. Output A is HP, output B is BP and output C is LP. They're all
2nd-order responses.
What you have drawn in diagram 4 represents the moog cascade. This filter
only has a single feedback path from the final output, exactly as you have
drawn it. The four poles are normally all located at the same cutoff
frequency. All 4 poles move in unison when you modulate the cutoff
frequency. In short, for diagram 4, output B is a first-order LP output
(-6dB/oct,) output C is 2nd-order LP output (-12dB/oct,) output D is
3rd-order LP (-18dB/oct) and output E is 4th-order LP (-24dB/oct) but this
only holds true when there is no feedback (zero resonance.) Once you apply
feedback around the loop things get more complicated! You conceptually roll
up the whole thing into a ball and call it ALL a 4th order closed-loop
system. All of the outputs mathematically become 4-pole filter responses.
Although, the application of the feedback causes zeroes in the closed-loop
transfer function that cancel out the effects of some of the poles for some
of the outputs. Here's a simple example of this effect: If you increase
the global feedback, output B developes a big resonant peak (this behaviour
requires at least a 2nd order system) but it's roll-off response ultimately
tends to only -6dB/oct slope in the stop-band, giving the illusion that it
may be only 1st-order.
I'm not going to analyse the other circuits as proper explanations get
mathematical and complicated really quickly. Suffice to say that diagram 3
operates similarly to diagram 4, but diagram 2 will always only give a
1st-order roll-off, with no resonance even with feedback.
There is also a far better tool for doing this kind of investigation... --->
Simulation! <---
Grab a copy of PSpice and enter each of the poles directly as a Laplace
block. You can then drop markers on the schematic to your hearts content
and graph all of the resulting output responses.
-Richie,
PS. Also try graphing the responses from mixtures of several outputs.
That's how the Oberheim Expander managed to offer all those different filter
options. They're all just mixtures of the outputs ABCDE in your diagram 4.
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