The system transfer function for an unstable discrete time system is given by H(z) = 0.8970 + 2.6911z-1 + 2.6911z-2 + 0.8970z-3 / (1.0 + 2.3822z-1 + 3.0000z-2 + 0.7941z-3) H(z) is unstable because one or more of its poles is outside the unit circle. Note that nothing is mentioned regarding the region on convergence - assume it is free to be chosen. Also assume there are no restrictions on the corresonding sequence, such as causality. No, this is not a homework problem. -- % Randy Yates % "My Shangri-la has gone away, fading like %% Fuquay-Varina, NC % the Beatles on 'Hey Jude'" %%% 919-577-9882 % %%%% <yates@ieee.org> % 'Shangri-La', *A New World Record*, ELO http://home.earthlink.net/~yatescr
Is this statement correct?
Started by ●September 10, 2004
Reply by ●September 10, 20042004-09-10
Randy Yates <yates@ieee.org> writes:> The system transfer function for an unstable discrete time system is given by > > H(z) = 0.8970 + 2.6911z-1 + 2.6911z-2 + 0.8970z-3 / (1.0 + 2.3822z-1 + 3.0000z-2 + 0.7941z-3)That should have been H(z) = (0.8970 + 2.6911z-1 + 2.6911z-2 + 0.8970z-3) / (1.0 + 2.3822z-1 + 3.0000z-2 + 0.7941z-3) (parens around the numerator) -- % Randy Yates % "And all that I can do %% Fuquay-Varina, NC % is say I'm sorry, %%% 919-577-9882 % that's the way it goes..." %%%% <yates@ieee.org> % Getting To The Point', *Balance of Power*, ELO http://home.earthlink.net/~yatescr
Reply by ●September 10, 20042004-09-10
Randy Yates wrote:> Randy Yates <yates@ieee.org> writes: > > >>The system transfer function for an unstable discrete time system is given by >> >> H(z) = 0.8970 + 2.6911z-1 + 2.6911z-2 + 0.8970z-3 / (1.0 + 2.3822z-1 + 3.0000z-2 + 0.7941z-3) > > > That should have been > > H(z) = (0.8970 + 2.6911z-1 + 2.6911z-2 + 0.8970z-3) / (1.0 + 2.3822z-1 + 3.0000z-2 + 0.7941z-3) > > (parens around the numerator)A z-plane pole outside the unit circle implies an s-plane pole in the right half plane. In turn, that implies an s with a positive real part. Then exp(st) = exp((a+jb)t) = exp(at)*cis(bt), which grows without limit for positive a. What was the question again? Jerry -- Engineering is the art of making what you want from things you can get. �����������������������������������������������������������������������
Reply by ●September 10, 20042004-09-10
Jerry Avins wrote:> Randy Yates wrote: > >> Randy Yates <yates@ieee.org> writes: >> >> >>> The system transfer function for an unstable discrete time system is >>> given by >>> >>> H(z) = 0.8970 + 2.6911z-1 + 2.6911z-2 + 0.8970z-3 / (1.0 + 2.3822z-1 >>> + 3.0000z-2 + 0.7941z-3) >> >> >> >> That should have been >> >> H(z) = (0.8970 + 2.6911z-1 + 2.6911z-2 + 0.8970z-3) / (1.0 + >> 2.3822z-1 + 3.0000z-2 + 0.7941z-3) >> >> (parens around the numerator) > > > A z-plane pole outside the unit circle implies an s-plane pole in the > right half plane. In turn, that implies an s with a positive real part. > Then exp(st) = exp((a+jb)t) = exp(at)*cis(bt), which grows without limit > for positive a. What was the question again? > > JerrySo, Jerry -- do you always have to translate to the s domain, or are you just doing it for rhetorical reasons? -- Tim Wescott Wescott Design Services http://www.wescottdesign.com
Reply by ●September 10, 20042004-09-10
Randy Yates wrote:> The system transfer function for an unstable discrete time system is given by > > H(z) = 0.8970 + 2.6911z-1 + 2.6911z-2 + 0.8970z-3 / (1.0 + 2.3822z-1 + 3.0000z-2 + 0.7941z-3) > > H(z) is unstable because one or more of its poles is outside the unit circle. > > Note that nothing is mentioned regarding the region on convergence - > assume it is free to be chosen. Also assume there are no restrictions > on the corresonding sequence, such as causality. > > No, this is not a homework problem.Ordinarily H(z) would describe an unstable system. In fact we know it describes an unstable system because you say so. But. If we are taking the z transform from k = -infinity to infinity and H(z) describes a system (ignoring rounding) who's impulse response is h(k) = 0.897 + 0.3734 * 2.8908^k * u(-k) - 0.4155^k * (0.1786 * cos(2.308 * k) - 0.2950 * sin(2.308 * k)) * u(k) (where u(k) is the unit step) then the system is stable (albeit completely unrealizable). So unless you specify that you are taking the one-sided z transform of the system (which is a restriction on causality, after all) you cannot say that an "unstable" pole of H(z) indicates that the described system is unstable. -- Tim Wescott Wescott Design Services http://www.wescottdesign.com
Reply by ●September 10, 20042004-09-10
Randy Yates wrote:> The system transfer function for an unstable discrete time system > is given by > > H(z) = 0.8970 + 2.6911z-1 + 2.6911z-2 + 0.8970z-3 / (1.0 + > 2.3822z-1 + 3.0000z-2 + 0.7941z-3) > > H(z) is unstable because one or more of its poles is outside the > unit circle. > > Note that nothing is mentioned regarding the region on convergence > - assume it is free to be chosen. Also assume there are no > restrictions on the corresonding sequence, such as causality. >A sort of handwaving approach: Put a billard ball inside a bowl. It will come to a rest at the deepest point. If you nudge it a bit, it will move around, then come to a rest at the same point as before. It's a stable system. Now put the bowl upside down and place the billard ball on top. Every slight touch or nudge on the ball will initiate a movement which makes the ball fall off the bowl. It will never come back to the original point. It's instable. As long as the system is not disturbed, it doesn't matter if it's stable or not. Once it is disturbed, it does. With discrete systems you need an additional aspect. Imagine a rim on top of the inversed bowl: Although the system is instable, and the ball would fall down, as soon as the excitation were big enough, it behaves stable as long as the excitation is not too big. The rim describes the radius of convergence. Though the filter you describe is generally instable, it might be of use depending of the application. What you need is some sort of mechanism which makes sure that either the ball doesn't "fall over the rim" or which "kicks it back again once it falls over the rim". Once I happened to work with such an instable filter because of rounding errors which caused the poles to be outside the unit circle. I didn't realize this, the filter worked perfectly. Until I happened to have one curious input signal, which made the filter go crazy. That was rather difficult to debug. Nevertheless, such a filter could be useful if you want to observe the convergence radius. As long as the condition is met, it works as low pass filter. As soon, as the convergence radius is overridden, by its feed-forward character it starts ringingwhich might cause an alert. Bernhard
Reply by ●September 10, 20042004-09-10
In article <10k2dkfn0o0b77d@corp.supernews.com>, Tim Wescott <tim@wescottnospamdesign.com> wrote:>Randy Yates wrote: > >> The system transfer function for an unstable discrete time system is >> given by >> >> H(z) = 0.8970 + 2.6911z-1 + 2.6911z-2 + 0.8970z-3 / (1.0 + 2.3822z-1 >> + 3.0000z-2 + 0.7941z-3) >> >> H(z) is unstable because one or more of its poles is outside the unit >> circle. >> >> Note that nothing is mentioned regarding the region on convergence - >> assume it is free to be chosen. Also assume there are no restrictions >> on the corresonding sequence, such as causality. >> >> No, this is not a homework problem.I am a little shaky on this without having the book handy to refer to, but my impression is that if you do not place restrictions on the region of convergence and do not require causality, then the only thing that will make BIBO stability impossible is if there are poles on the unit circle. See Chapter 5 of Discrete-Time Signal Processing (2nd edition) by Oppenheim and Schafer. A subtle thing about this is that there can be different discrete-time sequences corresponding to the same H(z), but at most one of the sequences can be stable (again, check the book to be sure, since I am a little shaky). Google turns up the following, from some class notes: "A discrete-time LTI system is stable if and only if the transfer function has a region of convergence that includes the unit circle." http://ptolemy.eecs.berkeley.edu/eecs20/lectures/chapter12part1student.pdf If you don't mind sharing, I am curious how your question arose.
Reply by ●September 10, 20042004-09-10
Tim Wescott wrote:> Jerry Avins wrote: > >> Randy Yates wrote: >> >>> Randy Yates <yates@ieee.org> writes: >>> >>> >>>> The system transfer function for an unstable discrete time system is >>>> given by >>>> >>>> H(z) = 0.8970 + 2.6911z-1 + 2.6911z-2 + 0.8970z-3 / (1.0 + >>>> 2.3822z-1 + 3.0000z-2 + 0.7941z-3) >>> >>> >>> >>> >>> That should have been >>> >>> H(z) = (0.8970 + 2.6911z-1 + 2.6911z-2 + 0.8970z-3) / (1.0 + >>> 2.3822z-1 + 3.0000z-2 + 0.7941z-3) >>> >>> (parens around the numerator) >> >> >> >> A z-plane pole outside the unit circle implies an s-plane pole in the >> right half plane. In turn, that implies an s with a positive real part. >> Then exp(st) = exp((a+jb)t) = exp(at)*cis(bt), which grows without limit >> for positive a. What was the question again? >> >> Jerry > > > So, Jerry -- do you always have to translate to the s domain, or are you > just doing it for rhetorical reasons?Somewhere, I still have my spirule. (I know where my slide rules are.) I avoid math whenever I can. I go as far back to basics as is easy for me when my aim is to learn or to prove something. Randy is a smart cookie. When I think I know something that he doesn't, I seek the firmest ground available to me. Jerry -- Engineering is the art of making what you want from things you can get. �����������������������������������������������������������������������
Reply by ●September 10, 20042004-09-10
Tim Wescott wrote:> Randy Yates wrote: > >> The system transfer function for an unstable discrete time system is >> given by >> >> H(z) = 0.8970 + 2.6911z-1 + 2.6911z-2 + 0.8970z-3 / (1.0 + 2.3822z-1 >> + 3.0000z-2 + 0.7941z-3) >> >> H(z) is unstable because one or more of its poles is outside the unit >> circle. >> >> Note that nothing is mentioned regarding the region on convergence - >> assume it is free to be chosen. Also assume there are no restrictions >> on the corresonding sequence, such as causality. >> >> No, this is not a homework problem. > > > Ordinarily H(z) would describe an unstable system. In fact we know it > describes an unstable system because you say so. > > But. If we are taking the z transform from k = -infinity to infinity > and H(z) describes a system (ignoring rounding) who's impulse response is > > h(k) = 0.897 + 0.3734 * 2.8908^k * u(-k) - 0.4155^k * (0.1786 * > cos(2.308 * k) - 0.2950 * sin(2.308 * k)) * u(k) > > (where u(k) is the unit step) then the system is stable (albeit > completely unrealizable). So unless you specify that you are taking the > one-sided z transform of the system (which is a restriction on > causality, after all) you cannot say that an "unstable" pole of H(z) > indicates that the described system is unstable.Tim, Without a lot of work I don't care to do, I can't follow your path. A question, though: why might I care about something nobody could build? Jerry -- Engineering is the art of making what you want from things you can get. �����������������������������������������������������������������������
Reply by ●September 10, 20042004-09-10
Tim, Thanks for responding. This is the two-sided z-transform, i.e., H(z) = \sum_{-\infty}^{+\infty} h[n]*z^{-n} --Randy Tim Wescott <tim@wescottnospamdesign.com> writes:> Randy Yates wrote: > > > The system transfer function for an unstable discrete time system is given by > > H(z) = 0.8970 + 2.6911z-1 + 2.6911z-2 + 0.8970z-3 / (1.0 + > > 2.3822z-1 + 3.0000z-2 + 0.7941z-3) > > > H(z) is unstable because one or more of its poles is outside the > > unit circle. > > > Note that nothing is mentioned regarding the region on convergence - > > > assume it is free to be chosen. Also assume there are no restrictions > > on the corresonding sequence, such as causality. > > No, this is not a homework problem. > > > Ordinarily H(z) would describe an unstable system. In fact we know it > describes an unstable system because you say so. > > > But. If we are taking the z transform from k = -infinity to infinity > and H(z) describes a system (ignoring rounding) who's impulse response > is > > > h(k) = 0.897 + 0.3734 * 2.8908^k * u(-k) - 0.4155^k * (0.1786 * > cos(2.308 * k) - 0.2950 * sin(2.308 * k)) * u(k) > > > (where u(k) is the unit step) then the system is stable (albeit > completely unrealizable). So unless you specify that you are taking > the one-sided z transform of the system (which is a restriction on > causality, after all) you cannot say that an "unstable" pole of H(z) > indicates that the described system is unstable. > > > -- > > Tim Wescott > Wescott Design Services > http://www.wescottdesign.com-- Randy Yates Sony Ericsson Mobile Communications Research Triangle Park, NC, USA randy.yates@sonyericsson.com, 919-472-1124