>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) suc
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 mappin
between
>> >Laplace and z-domains where 0 -> 0 and infinity -> pi and then
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 writte
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" i
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 som
info
>> out thinking that it is irrelevant. =A0The first part of the questio
ask=
>ed
>> to plot the Ha(s) phase response in degrees and find the frequenc
OMEGA1
>> when the phase is at -71 degrees).
>> The second part was asking the original question in this thread, an
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
>
>

I 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 yo
say "graph its phase and check the location of the -71 degrees spot. I
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 doin
something wrong, but I can't seem to see my mistake. Here is the MATLA
code I am using:

OMEGA_MAX = 4;
num_samples = 1000;
W = [0:1:num_samples]*OMEGA_MAX/num_samples;

B = 105;
A = [1 9.992 44.9534 104.9207 104.99118];

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.