Re: [LAU] Wah update

From: Fons Adriaensen <fons@email-addr-hidden>
Date: Mon Jul 20 2009 - 12:51:07 EEST

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.

Ciao,

-- 
FA
Io lo dico sempre: l'Italia è troppo stretta e lunga.
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Received on Mon Jul 20 16:15:01 2009

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