new here plus a question...

[Don Tillman]

   Date: Sun, 21 Jan 2001 19:02:06 +0100
   From: Magnus Danielson <cfmd at swipnet.se>

   From: Don Tillman <don at till.com>
   Date: Sun, 21 Jan 2001 02:24:41 -0800 (PST)

   > 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.

We're talking about a digital implementation of a given linear analog
filter circuit, right?

In the analog universe we have poles and zeroes in the s-plane, and
the resulting frequency and phase responses that we know so well.  

In the digital universe, the poles and zeroes are in the z-plane, and
the effect of each of these poles and zeroes can be seen in the
resulting frequency and phase responses, but these effects are
different from what you have in the analog universe because the
digital filter mechanism is fundamentally different.

Further, in translating the positions of the analog poles and zeroes
to their z-plane equivalents (the bilinear transform) things get moved
around.

(That's a gruesome distillation of Oppenheim and Schafer's "Digital
Signal Processing", chapters 2 and 5.)

  -- Don

-- 
Don Tillman
Palo Alto, California, USA
don at till.com
http://www.till.com