[sdiy] Re: moog high pass

[Magnus Danielson]

From: Karl Dalen <karldalen at yahoo.se>
Subject: Re: [sdiy] Re: moog high pass
Date: Wed, 30 Jul 2003 03:50:08 +0200 (CEST)

>  --- Magnus Danielson <cfmd at swipnet.se> skrev: > From: Karl Dalen
> <karldalen at yahoo.se>
> > Subject: Re: [sdiy] Re: moog high pass
> > Date: Wed, 30 Jul 2003 02:56:53 +0200 (CEST)
> > 
> > > >Ethan Duni <eduni at ucsd.edu> skrev: > Except for the inversion, isn't
> > > >that exactly what you'd expect out of a highpass filter with a cutoff
> > > > of 0? 
> > > 
> > > Yes, but his suggestion with comparing with inverted "unprocessed"
> > > input signals will leave a inverted input signal at the output when Cf
> > > is closing down to zero. Imagin this topology with high Q. 
> > 
> > Actually, what you gain is a fiddeling of the zeros in a sufficiently
> > close  fashion as one wants.
> > 
> > Summing with the input signal isn't a bad idea,
> 
> It still dont give me real HP filtering, it fades between a 
> inverted input signal and a LP filtered signal and that is not
> a real HP filter. (holding my thumbs real hard, hoping im not
> entierly wrong into this :-)..!!).

Nope, it is real HP filtering.

For all filters with n poles you have n zeros. For a lowpass filter all those
zeros is placed in the neighborhood of infinity (we assume they are at infinity
but they are normally just sufficiently far away from the passband for us to
be fairly correct in that assumption).

The s-plane responce can be viewed as being

          n
         --
         ||  s - s
         i=1      z,i     Z(s)
H(s) = G ------------ = G ----
          n               P(s)
         --
         ||  s - s
         i=1      p,i

where s is the LaPlaceian variable (two-dimensional, so solutions is located in
a 2D-plane), G is the pure gain, s(z,i) is the i:th zero location, s(p,i) is
the i:th pole location.

Now, if we view the zeros and poles as their polynomials Z(s) and P(s) then
subtracting the input signal with the output of the filter, then we would have

                       Z(s)   P(s) - Z(s)   Z'(s)
H (s) = 1 - H(s) = 1 - ---- = ----------- = -----
 1                     P(s)      P(s)       P(s)

where Z'(s) = P(s) - Z(s).

So, doing such an operation only changes the zeros, but not the poles. When we
tune up the resonance we move around the poles, but not necessarilly the zeros.

Anyway, for a lowpass filter we have all zeroes in infinity, i.e. in s - s(z,i)
is the s(z,i) term heavier than the se term, so we can forget about the s term
and we effectively get all multiplies together, and it is just a DC gain which
we can move away, so often we write it as Z(s) = 1 (please note that this is
a specific transformation of the values that one does, but it is convenient).

For a highpass filter, we put all the zeroes usually in DC, i.e. 0Hz, so then
s(z,i) is naturally zero (since s - s(z,i) = 0 has a solution for s = 0 for all
s(z,i)) and the polynomial naturally gets the form of Z(s)=s^n. However,
nothing prohibits the zeroes to be differently located and you still get a
highpass responce in the end.

If you want to make it properly, you need to cancel out the lower order of the
polynomial coefficients. One of these is however dynamic due to the Q-value,
but by propperly sum in the feedback term of the Q-curcuit you hand handle that
as well.

So "Real" HP can only be when you have only the Z(n) = s^n, but the way you
acheive it has nothing to do with it.

Have you now figured out why I have spent years pushing this thing infront of
me and saying "another day"?

> >easpecially if you
> > consider the alternative, which is to dump multiple, if not all, of the
> > diffrential voltages over the caps in the ladder and then combine these
> >in a matrix  form  into the state-variable signals of a 4-pole
> >state-variable filter. It's  possible but messy.
> 
> Exactly, that is implementing the OBI Xpander filter
> solution to a Ladder design as you say messy but should work.
> Infact i had some ideas of going this way! 

In filter theory it is a classic thing, and the state-variable filters have
the benefit that you easilly can build up the zeroes by dimensioning your
summing from the polynomial coefficients. The Moog-ladder (or any other ladder)
is however not a state-variable filter, but you can form similar outputs.

However, even if one does all the tricks correctly from a linear point of view,
it is not sure that it will give the right "feel" from the non-linear
responces.

> > You should really invert the lowpass signal then. You get a difference
> > signal
> > either way, the only thing is the sign of the difference signal.
> 
> Yes, and as far as i understand that wont give me a real 
> HP filter either, as you say it only changes sign. Still as i can
> understand the input signal goes straigt trough when Cf is closed
> no matter if the LP signal or the input signal is inverted!

Yes, and if you have a highpass filter with the cutoff set low, all of the
signal will pass through (this is why it is called passband) essentially
unmodified. Ever played a MS-20?

So what is wrong?

In my point of view, it will not give a "correct" HP-filter, but probably a
sufficiently good one to be usefull musically. The trick has been used before
and with success, so there is nothing terribly wrong with it.

Cheers,
Magnus