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.
transfer function
Started by ●February 9, 2007
Reply by ●February 9, 20072007-02-09
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. Now you just gotta factor that beast! Clay
Reply by ●February 9, 20072007-02-09
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. Regards, Andor
Reply by ●February 9, 20072007-02-09
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(). Alternatively, if you need to do the maths explicitly, see http:// en.wikipedia.org/wiki/Companion_matrix, for instance. -- Oli
Reply by ●February 9, 20072007-02-09
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.> Alternatively, if you need to do the maths explicitly, see http:// > en.wikipedia.org/wiki/Companion_matrix, for instance.Regards, Andor
Reply by ●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
Reply by ●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 ●February 9, 20072007-02-09
Oli Charlesworth wrote: And I fixed the link:> ... if you need to do the maths explicitly, see > http://en.wikipedia.org/wiki/Companion_matrix, for instance.Jerry -- Engineering is the art of making what you want from things you can get. ¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯
Reply by ●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
Reply by ●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
