Re: [LAU] Wah update

On Mon, Jul 20, 2009 at 11:51:07AM +0200, Fons Adriaensen wrote:
> On Sun, Jul 19, 2009 at 08:10:47PM -0700, Ken Restivo wrote:
>
> > Just a quick update on the wah research.
> >
> > A friend owns a Dunlop "Jimi Hendrix Wah", which says it is the "Original Thomas Design", by which I assume they mean to claim it's the same design as the Thomas Organ Wah, formerly Vox.
> >
> > This website's describes the frequency response as a lowpass with a resonant peak:
> > http://www.geofex.com/Article_Folders/wahpedl/wahped.htm
> >
> > So here is what JAPA says it does (and I believe JAPA more than some random website):
> >
> > When fully closed, it's a bandpass, with a VERY high Q!
> > http://restivo.org/misc/lowend-jimi.png
> >
> > But, wait, when I open it up, suddenly it becomes more like a highpass, but with a lot of resonance:
> > http://restivo.org/misc/midrange-jimi.png
> >
> > When it's fully opened, it's definitely a highpass, but with a helluva peak:
> > http://restivo.org/misc/high-jimi.png
> >
> > So, not only is the opposite of what that article says, but it's also kind of non-linear. I'll poke around the various LADSPA plugins and see if I can find something nearly like this.
> >
> > Another guitar-player friend has a different wah (IIRC, either a "Cry Baby", or a Morley), and I'll see if I can run his through this and see what it comes up looking like.
>
>
> AFAICS this is a resonant (which is not the same as bandpass) filter.
> If the response near Fs/2 bcomes flat, that does not mean it is a
> highpass.
>
> Remember that any digital filter is 'mirrored' to the other side
> of Fs/2. Also the magnitude of the response must be continuous or
> zero at all points (for finite order).
>
> The result of all this is that at Fs/2 the response must be either
> zero or have a zero derivative, i.e. be horizontal.
>
> In a high order filter you can make the 'roundoff' region near
> Fs/2 very small, but it's always there, unless the response is
> zero at that frequency.
>
> You can probably get this type of response using the MOOG VCF
> by taking the output at a different point in the algorithm.
>
> The MOOG VCF is 4th order, this is overkill as the analog
> circuit is very likely to be just 2nd order.
>

Thanks. Alas, that seems like a very concise explanation, but I don't have the mathematical background to implement that.

If someone feels like modifying the Moog VCF to make it a Vox/Thomas Wah, I'd be eternally grateful. But it's pretty clear I don't have the skills to take this over the finish line.

-ken
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Received on Tue Jul 21 04:15:04 2009