new here plus a question...

[Magnus Danielson]

From: Don Tillman <don at till.com>
Subject: Re: new here plus a question...
Date: Sun, 21 Jan 2001 02:24:41 -0800 (PST)

>    Date: Thu, 18 Jan 2001 19:46:11 +0100 (MET)
>    From: Bram de Jong <Bram.DeJong at rug.ac.be>
> 
>    On Thu, 18 Jan 2001, Happy Hairy Harry wrote: > The "tone controls"
>    in the Duncan application are typical of > guitar amps.  If you
>    model them, you will have a first-order > approximation of how they
>    "sound".
> 
>    Aha! But, my question really was: how do I do this :)
> 
> Write out the equations for the current into each node of the circuit,
> and solve those equations simultaneously.  If you'd like more details,
> it's basically a full semester course in electrical engineering.
> You'd need to get some textbooks on "linear circuit analysis".

Agree.

>    > But most of us think that there is still something missing from
>    > the model. Maybe parasitic components of the circuit, or
>    > non-linear processes that are or are NOT obvious.
> 
> I'll suggest that the parasitics and nonlinearities are not important
> to the tone stack.  The problem is more fundamental.
> 
>    > The other thread was "Why does the Moog Ladder filter have some
>    > quality that does not allow it to be modeled exactly" This is
>    > probably way beyond what you are trying to get to right now.
> 
>    Hmmm. Isn't this allso because the straight s-lane to z-plane
>    transform introduces delay-free feedback loops? (that is,
>    additional to the 'unlinear' part)

The z-plane has nothing to do with the Moog ladder. z-transform and
thus the z-plane is of interest only if you sample, the Moog ladder
has nothing to do with that.

> The Moog Ladder is a separate case because it's nonlinear in a really
> interesting way.  You're just interested in simulating a Fender tone
> stack, right?  (Or the whole amp?)
> 
> (Of course I've got to wonder why anybody would go to the trouble of
> simulation a simple circuit made from $10 worth of parts...)
> 
> I think the main issue is that translating from the s-plane to the
> z-plane is a very rude approximation.  z-poles are inherently
> different than s-poles, they won't sound the same as the original, and
> I'll bet you would have no problem telling the difference between the
> frequency response plots visually.

z-plane and z-poles has IMHO nothing to do with it. If they do, please
explain.

Cheers,
Magnus