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From: "Ana " <filter_world@yahoo.com>
Newsgroups: comp.soft-sys.matlab
Subject: bilinear transformation
Date: Wed, 4 Nov 2009 22:33:03 +0000 (UTC)
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hi ,
If i understood correctly bilinear transformation is used in order to transform from s domain to the z domain. Now why then when using;
example;

 fp = 2e03;
 fs = 20e03;
 s = tf('s');
 num = 1;
 den = (s^2 + 1.414*s + 1); % butterworth 2nd order 
 [numd,dend] = bilinear(num,den,fs,fp)
it is telling me that function rdivide cannot be classified by tf, i guess numd and dend will give basically the transfer function in the zdomain.

Any ideas highly appreciated.Thanks beforhand

Ana