On May 15, 11:03*am, "v333k" <v3...@yahoo.com> wrote:
> >>On May 13, 1:24=A0pm, "v333k" <v3...@yahoo.com> wrote:
> >>> >On May 12, 8:28=3DA0am, "v333k" <v3...@yahoo.com> wrote:
> >>> >> Hi all,
> >>> >> I have a relatively basic question about applying a bilinear
> >>> >> transformation using MATLAB. =3DA0I am familiar with the bilinear
> >>> function,=3D
> >>> > but
> >>> >> the question states the following:
> >>> >> apply the bilinear transform to Ha(s) in order to obtain H(z) such
> >tha=
> >>t
> >>> w=3D
> >>> >1
> >>> >> =3D3D pi/4 radians.
>
> >>> >> Ha(s) =3D3D 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. =A0s =3D3D c (1-z^-1)/(1+z^-1) sometimes the "c" is
> written
> >a=
> >>s 2/
> >>> >T where "T" is the sampling period. =A0If 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. =A0The first part of the question
> >ask=
> >>ed
> >>> 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).
> =A0Also,
> >we
> >>> need to consider the frequency range 0<=3DOMEGA<=3D4 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 =3D approx 1.43 or 2.40
> >>radians/sec.
>
> >>For your bilinear transform, use
>
> >>s =3D c (1-z^-1)/(1+z^-1)
>
> >>here c=3DOMEGA*cot(omega/2) where OMEGA is the analog frequency that we
> >>are mapping to the digital frequency omega. For your case omega =3D
> 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
>
> Hi Clay,
> Thank you very much for the time you took to explain this to me.
> The only thing that I can't figure out (yet) is the last part, where you
> say "graph its phase and check the location of the -71 degrees spot. It
> should be right where the "c" equation puts it."
> When I graph the phase, I get 0.11 for 71 degrees. I know I am doing
> something wrong, but I can't seem to see my mistake. Here is the MATLAB
> code I am using:
>
> OMEGA_MAX = 4;
> num_samples = 1000;
> W = [0:1:num_samples]*OMEGA_MAX/num_samples;
>
> B = 20;
> A = [1 2 8 20 20];
>
> freqs(B,A,W);
>
> w1 = 0.25*pi; %pi/4
> c = 1.43*cot(w1/2);
> [b,a] = bilinear(B,A,c);
>
> figure;
> freqz(b,a); *% plot the frequency response
>
> Thanks once again.- Hide quoted text -
>
> - Show quoted text -

Since I don't use MatLab, I can't comment on the specifics of the
functions you are using. My last comment was basically was to note
that when you get your z-domain function via the bilinear transform
that the point of -71 degrees on the phase shift function now occurs
at pi/4. Essentially you need to evaluate your z domain eqn on the
unit circle and plot its phase vs frequency and the -71 spot should be
at the freq. of pi/4. This is a sanity check on you correctly doing
the bilinear transform. Oh yeah and make sure when you look at the
argument (phase) that you convert it to degrees if you are looking for
-71 degrees (-1.239 radians)

I hope this clarifies things,

Clay