>On May 15, 11:03=A0am, "v333k" <v3...@yahoo.com> wrote:
>> >>On May 13, 1:24=3DA0pm, "v333k" <v3...@yahoo.com> wrote:
>> >>> >On May 12, 8:28=3D3DA0am, "v333k" <v3...@yahoo.com> wrote:
>> >>> >> Hi all,
>> >>> >> I have a relatively basic question about applying a bilinear
>> >>> >> transformation using MATLAB. =3D3DA0I am familiar with th
bilinea=
>r
>> >>> function,=3D3D
>> >>> > but
>> >>> >> the question states the following:
>> >>> >> apply the bilinear transform to Ha(s) in order to obtain H(z
such
>> >tha=3D
>> >>t
>> >>> w=3D3D
>> >>> >1
>> >>> >> =3D3D3D pi/4 radians.
>>
>> >>> >> Ha(s) =3D3D3D 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 th
bilinear
>> >>> >transform. =3DA0s =3D3D3D c (1-z^-1)/(1+z^-1) sometimes the "c
is
>> written
>> >a=3D
>> >>s 2/
>> >>> >T where "T" is the sampling period. =3DA0If the Laplace expressio
i=
>s
>> >>> >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. =3DA0The first part of th
questi=
>on
>> >ask=3D
>> >>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).
>> =3DA0Also,
>> >we
>> >>> need to consider the frequency range 0<=3D3DOMEGA<=3D3D4 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 =3D3D approx 1.43 o
2.=
>40
>> >>radians/sec.
>>
>> >>For your bilinear transform, use
>>
>> >>s =3D3D c (1-z^-1)/(1+z^-1)
>>
>> >>here c=3D3DOMEGA*cot(omega/2) where OMEGA is the analog frequenc
that =
>we
>> >>are mapping to the digital frequency omega. For your case omeg
=3D3D
>> 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 whe
you
>> >>get your bilinear transformed function, graph its phase and chec
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, wher
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 th
MATLAB
>> code I am using:
>>
>> OMEGA_MAX =3D 4;
>> num_samples =3D 1000;
>> W =3D [0:1:num_samples]*OMEGA_MAX/num_samples;
>>
>> B =3D 20;
>> A =3D [1 2 8 20 20];
>>
>> freqs(B,A,W);
>>
>> w1 =3D 0.25*pi; %pi/4
>> c =3D 1.43*cot(w1/2);
>> [b,a] =3D bilinear(B,A,c);
>>
>> figure;
>> freqz(b,a); =A0% 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
>

Hi Clay,
I think I understand it now.
Normally, as all books and examples, the equation for bilinear transform
is: s = 2/T(1-z^1)/(1+z^1)
However, the question I am working with is actually diverting from this
default parameter, and by mapping w1 (pi/4) to OMEGA1 which is found at -71
degrees of the phase shift.
So, what you were saying is the following:
eq1. s = 2/T(1-z^1)/(1+z^1)
eq2. W = 2/T *tan(w/2)
let c = 2/T
s = c * (1-z^1)/(1+z^1)
and from eq2, c = W/tan(w/2) = W * cot(w/2)

applying the bilinear transform:
I will obtain the digital coefficients of H(z) and when checking the phase
response plot at -71 degrees, I should get pi/4 which is the mapped
frequency.

This is great!
I really appreciate your time and effort!

Thanks a lot.
VK