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Subject: Re: bilinear transformation
Date: Wed, 4 Nov 2009 22:42:02 +0000 (UTC)
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"Ana " <filter_world@yahoo.com> wrote in message <hcsvev$gbb$1@fred.mathworks.com>...
> 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

Hi Ana,

I think it's just the syntax you're using is incorrect.
Try:

fp = 2e03;
 fs = 20e03;
 s = tf('s');
 num = 1;
 den = [2 1.414 1]; % butterworth 2nd order
 [numd,dend] = bilinear(num,den,fs,fp)

Hope that helps,
James