Reply by Fred Marshall●February 11, 20072007-02-11
<bob@coolgroups.com> wrote in message
news:1170997093.784442.53090@m58g2000cwm.googlegroups.com...
>I have a transfer function as follows:
>
> K*z^(-10)
> ----------------------
> 1 - z^(-1) + K*z^(-10)
>
> I'm trying to figure out for what values of K the system is stable.
> Is there an easy way to do this?
>
> Thanks.
>
> BTW, this isn't homework.
>
You can get the answer you need easily enough by using a root finder on the
denominator polynomial, manually assigning values to K and iteratively
finding a value for K that satisfies abs(roots(.))<1.
You can get fancier and put the test in a loop and search for K.
I was able to do the former in about 5 minutes.... that seems easy enough.
Fred
Reply by Tim Wescott●February 10, 20072007-02-10
bob@coolgroups.com wrote:
> I have a transfer function as follows:
>
> K*z^(-10)
> ----------------------
> 1 - z^(-1) + K*z^(-10)
>
> I'm trying to figure out for what values of K the system is stable.
> Is there an easy way to do this?
>
> Thanks.
>
> BTW, this isn't homework.
>
Turn it into a closed-loop problem, then do a Bode plot. It looks like
you can use
K*z^(-10)
H_ol(z) = -----------
1 - z^(-1)
Your gain margin with K = 1 will be the reciprocal of the allowable
maximum value of K.
If you need precision, use a zero-finder function along with a root
finder to narrow down the value of K.
--
Tim Wescott
Wescott Design Services
http://www.wescottdesign.com
Posting from Google? See http://cfaj.freeshell.org/google/
"Applied Control Theory for Embedded Systems" came out in April.
See details at http://www.wescottdesign.com/actfes/actfes.html
Reply by ●February 9, 20072007-02-09
On Feb 9, 5:58 pm, b...@coolgroups.com wrote:
> I have a transfer function as follows:
>
> K*z^(-10)
> ----------------------
> 1 - z^(-1) + K*z^(-10)
>
> I'm trying to figure out for what values of K the system is stable.
> Is there an easy way to do this?
>
> Thanks.
>
> BTW, this isn't homework.
This looks likes it's from a Root Locus problem. An open loop system
with unity negative feedback and gain k in the forward path. You need
a rootfinder - it's easy in Matlab.
F.
Reply by Randy Yates●February 9, 20072007-02-09
Randy Yates <yates@ieee.org> writes:
> [...]
PS: Andor, I consider this a rare scenario also. :)
--
% Randy Yates % "Remember the good old 1980's, when
%% Fuquay-Varina, NC % things were so uncomplicated?"
%%% 919-577-9882 % 'Ticket To The Moon'
%%%% <yates@ieee.org> % *Time*, Electric Light Orchestra
http://home.earthlink.net/~yatescr
Reply by Randy Yates●February 9, 20072007-02-09
"Andor" <andor.bariska@gmail.com> writes:
> On 9 Feb., 14:59, Randy Yates <y...@ieee.org> wrote:
>> "Andor" <andor.bari...@gmail.com> writes:
>> > On 9 Feb., 07:50, "Clay" <phys...@bellsouth.net> wrote:
>> >> On Feb 8, 11:58 pm, b...@coolgroups.com wrote:
>>
>> >> > I have a transfer function as follows:
>>
>> >> > K*z^(-10)
>> >> > ----------------------
>> >> > 1 - z^(-1) + K*z^(-10)
>>
>> >> > I'm trying to figure out for what values of K the system is stable.
>> >> > Is there an easy way to do this?
>>
>> >> > Thanks.
>>
>> >> > BTW, this isn't homework.
>>
>> >> The values of K that make the system stable are the ones that ensure
>> >> the 10 roots of the denominator polynomial are inside the unit circle.
>>
>> > stable <=> no poles on the unit circle!
>>
>> Really? So, for example, the right-handed sequence with z-transform
>>
>> X(z) = 1 / [(z - 1/3)(z - 2)]
>>
>> is stable? It doesn't have any poles on the unit circle.
>
> Allow me to let you answer yourself *):
>
> "
> Hey Randy!
>
> This is one of those very rare occasions when you're wrong! Perhaps
> you
> just read the words a little too fast and thought Andor was saying
> something
> else, but this is very true, i.e., IN GENERAL, if the ROC includes the
> unit circle,
> the system is stable. I think for whatever reason most of us have been
> taught in the
> digital domain to consider causal sequences only, and in that case, a
> stable
> system must have the poles inside the unit circle since a causal
> sequence
> always has a ROC outside the outermost pole.
> "
>
> Regards,
> Andor
>
> *) From:
> http://groups.google.com/group/comp.dsp/msg/395804ea55745999
Hi Andor,
I agree with my words to Robert in that rare scenario, but they don't
apply here to myself. That's because I'm not saying the same thing
Robert said.
You stated that a system is stable iff there are no poles on the unit
circle. This is, in general, wrong. What is true is that a stable
system must have a z-transform in which the unit circle is included in
the region of convergence. However, that is not what you stated, i.e.,
you stated nothing about the ROC, instead basing your conditions
solely on the location of the poles. In other words, since your
statement lacked any constraints on the ROC, we must assume it is true
no matter what ROC we choose, which, by the example I gave, it is not.
--
% Randy Yates % "She has an IQ of 1001, she has a jumpsuit
%% Fuquay-Varina, NC % on, and she's also a telephone."
%%% 919-577-9882 %
%%%% <yates@ieee.org> % 'Yours Truly, 2095', *Time*, ELO
http://home.earthlink.net/~yatescr
Reply by Oli Charlesworth●February 9, 20072007-02-09
On Feb 9, 3:03 pm, "Andor" <andor.bari...@gmail.com> wrote:
> On 9 Feb., 14:59, Randy Yates <y...@ieee.org> wrote:
>
>
>
> > "Andor" <andor.bari...@gmail.com> writes:
> > > On 9 Feb., 07:50, "Clay" <phys...@bellsouth.net> wrote:
> > >> On Feb 8, 11:58 pm, b...@coolgroups.com wrote:
>
> > >> > I have a transfer function as follows:
>
> > >> > K*z^(-10)
> > >> > ----------------------
> > >> > 1 - z^(-1) + K*z^(-10)
>
> > >> > I'm trying to figure out for what values of K the system is stable.
> > >> > Is there an easy way to do this?
>
> > >> > Thanks.
>
> > >> > BTW, this isn't homework.
>
> > >> The values of K that make the system stable are the ones that ensure
> > >> the 10 roots of the denominator polynomial are inside the unit circle.
>
> > > stable <=> no poles on the unit circle!
>
> > Really? So, for example, the right-handed sequence with z-transform
>
> > X(z) = 1 / [(z - 1/3)(z - 2)]
>
> > is stable? It doesn't have any poles on the unit circle.
>
> Allow me to let you answer yourself *):
>
> "
> Hey Randy!
>
> This is one of those very rare occasions when you're wrong! Perhaps
> you
> just read the words a little too fast and thought Andor was saying
> something
> else, but this is very true, i.e., IN GENERAL, if the ROC includes the
> unit circle,
> the system is stable. I think for whatever reason most of us have been
> taught in the
> digital domain to consider causal sequences only, and in that case, a
> stable
> system must have the poles inside the unit circle since a causal
> sequence
> always has a ROC outside the outermost pole.
> "
stable <=> ROC includes the unit circle
covers:
stable => no poles on the unit circle
but not:
stable <= no poles on the unit circle
--
Oli
Reply by Andor●February 9, 20072007-02-09
On 9 Feb., 14:59, Randy Yates <y...@ieee.org> wrote:
> "Andor" <andor.bari...@gmail.com> writes:
> > On 9 Feb., 07:50, "Clay" <phys...@bellsouth.net> wrote:
> >> On Feb 8, 11:58 pm, b...@coolgroups.com wrote:
>
> >> > I have a transfer function as follows:
>
> >> > K*z^(-10)
> >> > ----------------------
> >> > 1 - z^(-1) + K*z^(-10)
>
> >> > I'm trying to figure out for what values of K the system is stable.
> >> > Is there an easy way to do this?
>
> >> > Thanks.
>
> >> > BTW, this isn't homework.
>
> >> The values of K that make the system stable are the ones that ensure
> >> the 10 roots of the denominator polynomial are inside the unit circle.
>
> > stable <=> no poles on the unit circle!
>
> Really? So, for example, the right-handed sequence with z-transform
>
> X(z) = 1 / [(z - 1/3)(z - 2)]
>
> is stable? It doesn't have any poles on the unit circle.
Allow me to let you answer yourself *):
"
Hey Randy!
This is one of those very rare occasions when you're wrong! Perhaps
you
just read the words a little too fast and thought Andor was saying
something
else, but this is very true, i.e., IN GENERAL, if the ROC includes the
unit circle,
the system is stable. I think for whatever reason most of us have been
taught in the
digital domain to consider causal sequences only, and in that case, a
stable
system must have the poles inside the unit circle since a causal
sequence
always has a ROC outside the outermost pole.
"
Regards,
Andor
*) From:
http://groups.google.com/group/comp.dsp/msg/395804ea55745999
Jerry
--
Engineering is the art of making what you want from things you can get.
¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯
Reply by Oli Charlesworth●February 9, 20072007-02-09
On Feb 9, 1:35 pm, "Andor" <andor.bari...@gmail.com> wrote:
> On 9 Feb., 14:14, "Oli Charlesworth" <c...@olifilth.co.uk> wrote:
>
>
>
> > On Feb 9, 10:41 am, "Andor" <andor.bari...@gmail.com> wrote:
>
> > > On 9 Feb., 07:50, "Clay" <phys...@bellsouth.net> wrote:
>
> > > > On Feb 8, 11:58 pm, b...@coolgroups.com wrote:
>
> > > > > I have a transfer function as follows:
>
> > > > > K*z^(-10)
> > > > > ----------------------
> > > > > 1 - z^(-1) + K*z^(-10)
>
> > > > > I'm trying to figure out for what values of K the system is stable.
> > > > > Is there an easy way to do this?
>
> > > > > Thanks.
>
> > > > > BTW, this isn't homework.
>
> > > > The values of K that make the system stable are the ones that ensure
> > > > the 10 roots of the denominator polynomial are inside the unit circle.
>
> > > stable <=> no poles on the unit circle!
>
> > > > Now you just gotta factor that beast!
>
> > > Yup, looks like a bummer.
>
> > With something like MATLAB, this shouldn't be too much of a problem;
> > in MATLAB the particular command you'll need is roots().
>
> Roots() finds the zeros of a polynomial using numeric methods. This
> command will be of no help to determine all values of K such that no
> pole in the above transfer function lies on the unit circle.
Ah yes, I missed the requirement "to find which values of K make it
stable" requirement in the OP's post!
--
Oli
Reply by Randy Yates●February 9, 20072007-02-09
"Andor" <andor.bariska@gmail.com> writes:
> On 9 Feb., 07:50, "Clay" <phys...@bellsouth.net> wrote:
>> On Feb 8, 11:58 pm, b...@coolgroups.com wrote:
>>
>> > I have a transfer function as follows:
>>
>> > K*z^(-10)
>> > ----------------------
>> > 1 - z^(-1) + K*z^(-10)
>>
>> > I'm trying to figure out for what values of K the system is stable.
>> > Is there an easy way to do this?
>>
>> > Thanks.
>>
>> > BTW, this isn't homework.
>>
>> The values of K that make the system stable are the ones that ensure
>> the 10 roots of the denominator polynomial are inside the unit circle.
>
> stable <=> no poles on the unit circle!
Really? So, for example, the right-handed sequence with z-transform
X(z) = 1 / [(z - 1/3)(z - 2)]
is stable? It doesn't have any poles on the unit circle. This
is from Example 10.6 in \cite{signalsandsystems}.
Clay's statement is true assuming a right-handed sequence.
--Randy
@BOOK{signalsandsystems,
title = "{Signals and Systems}",
author = "{Alan~V.~Oppenheim, Alan~S.~Willsky, with Ian~T.~Young}",
publisher = "Prentice Hall",
year = "1983"}
--
% Randy Yates % "Watching all the days go by...
%% Fuquay-Varina, NC % Who are you and who am I?"
%%% 919-577-9882 % 'Mission (A World Record)',
%%%% <yates@ieee.org> % *A New World Record*, ELO
http://home.earthlink.net/~yatescr