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 _______________________________________________ Linux-audio-user mailing list Linux-audio-user@xxxxxxxxxxxxxxxxxxxx http://lists.linuxaudio.org/mailman/listinfo/linux-audio-user
