>On May 12, 8:28=A0am, "v333k" <v3...@yahoo.com> wrote:
>> Hi all,
>> I have a relatively basic question about applying a bilinear
>> transformation using MATLAB. =A0I am familiar with the bilinea
function,=
> but
>> the question states the following:
>> apply the bilinear transform to Ha(s) in order to obtain H(z) such tha
w=
>1
>> =3D pi/4 radians.
>>
>> Ha(s) =3D 20/(s^4 + 2s^3 + 8s^2 + 20s + 20)
>>
>> my question/confusion is where does the w1 play a role in this?
>>
>> thanks in advance!
>
>Perhaps this question has some context around it that you are not
>posting here. In general the bilinear transform does a mapping between
>Laplace and z-domains where 0 -> 0 and infinity -> pi and then a third
>point is adjustable in the sense of where you do the mapping. I.e.
>this is the "c" or whatever stadard variable you see in the bilinear
>transform. s =3D c (1-z^-1)/(1+z^-1) sometimes the "c" is written as 2/
>T where "T" is the sampling period. If the Laplace expression is
>normalized to 1 rad/sec, then "c" is simply cot(pi*f) where "f" is the
>normalized digital frequency that gets mapped to 1 rad/sec.
>
>Clay
>

Hi Clay,
thanks for your reply.
Since I am not fully understanding the question, maybe I left some inf
out thinking that it is irrelevant. The first part of the question aske
to plot the Ha(s) phase response in degrees and find the frequency OMEGA
when the phase is at -71 degrees).
The second part was asking the original question in this thread, and th
part I left out was w1 is the digital counterpart of OMEGA1). Also, w
need to consider the frequency range 0<=OMEGA<=4 rad/sec.

Thanks in advance (again).

v333k