On May 13, 1:24*pm, "v333k" <v3...@yahoo.com> wrote:
> >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 bilinear
> function,=
> > but
> >> the question states the following:
> >> apply the bilinear transform to Ha(s) in order to obtain H(z) such that
> 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 info
> out thinking that it is irrelevant. *The first part of the question asked
> to plot the Ha(s) phase response in degrees and find the frequency OMEGA1
> when the phase is at -71 degrees).
> The second part was asking the original question in this thread, and the
> part I left out was w1 is the digital counterpart of OMEGA1). *Also, we
> need to consider the frequency range 0<=OMEGA<=4 rad/sec.
>
> Thanks in advance (again).
>
> v333k- Hide quoted text -
>
> - Show quoted text -

Now what you have added makes sense - thanks.


On your part A of the problem, I hope you found two spots where the
phase is -71 degrees. For these two spots OMEGA = approx 1.43 or 2.40
radians/sec.


For your bilinear transform, use

s = c (1-z^-1)/(1+z^-1)


here c=OMEGA*cot(omega/2) where OMEGA is the analog frequency that we
are mapping to the digital frequency omega. For your case omega = pi/4
and OMEGA is equal to 1.43 or the 2.40 radians/sec number.

These two values give a "c" equal to approx 3.45 or 5.78. You will
need to do this in full precision to get it to work well, but when you
get your bilinear transformed function, graph its phase and check the
location of the -71 degree spots. It should be right where the "c"
equation puts it.

IHTH,

Clay