Thread Subject: bilinear transformation

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

"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

Thanks alot james. Ill give it a good try.

Regards,
Ana

hi there,

 %fp = 2*pi*2e03; For prewarping
 fs = 20e03;
 s = tf('s');
 num = 1;
 den = [1 1.414 1 ];
 
 [numd,dend] = bilinear(num,den,fs)
 
 
z= [1:0.1:20]';
m = filter(numd,dend,z)

freqz(m)

It seems that it is working now, if i understood well the command freqz is used to give the response of discrete time systems, like digital systems does this therefore mean that when obtianing the plot from the code above, I am getting the frequency response of a digital filter in this case the butterworth? Thanks alot for any help.

Regards,
Ana