Harald Bode Barberpole Phaser Questions

[Don Tillman]

   Date: Fri, 16 Jan 1998 12:55:47 -0800 (PST)
   From: Sean Costello <costello at costello.seanet.com>

   Don Tillman replied:
   >Wait a second...  What if it was a *completely* inaccurate quadrature
   >filter?  Well, that means you'd have both up-shift and down-shift
   >mixed together.  But that's just a simple ring modulator.  If you
   >replaced the frequency shifter with a simple ring modulator you'd
   >effectively have a rising barberpole mixed with a falling barberpole.
   >That might be nice effect.  Sure it's cooler to have rising and falling
   >barberpoles as a stereo pair, but if you're on a budget this might not
   >be a bad thing.  Hey Juergen, you're set up for this experiment right
   >now...

   Excitement prevails over the simplicity of the idea!  However, fellow
   Shallow Alto resident Harvey Devoe Thornburg replies with:

I prefer "Perfectly Pretentious Palo Alto" myself.  This *is* a really
nice town by the way.  If you've got Quicktime loaded on your computer
you can enjoy a remarkable virtual tour of Palo Alto and other nearby
places at this web site:
http://www.research.digital.com/PA/maps/qtvr-list-text-content.html

(The amazing thing is that the Palo Alto, Stanford and Menlo
Park shots were all taken within a 1.5 mile radius.)

If you don't have Quicktime loaded you can download a copy here for
free:
http://www.apple.com/quicktime/

   >Wouldn't this lead to amplitude modulation, because you are
   >essentially mixing some of the upshift with downshift, making 
   >part of the circuit act like a ring modulator? If you 
   >encountered even slight variations in the perceived signal
   >level this could be problematic (or useful depending on 
   >your goals).

Okay, the idea I mentioned above won't work, I spoke way too quickly,
but I think the reason why is worth noting...

When you have an ascending continuous phase shift (frequency shift up)
combined with a descending continuous phase shift (frequency shift
down) the resulting phase sits still.  It's effectively a node.  So
the single multiplier circuit is not going to be providing that
continuous phase shift that's so important to the barberpole process.
Unfortunately there's no free lunch.

  -- Don