Reply by Bernhard Holzmayer●September 13, 20042004-09-13
Randy Yates wrote:
> Bernhard,
>
> I'm not sure I follow your analogy completely. It is my belief
> that the filter can be completely stable if the proper ROC is
> chosen.
>
> --Randy
>
In an introduction to the explanation of Lyapunov stability,
I found this fundamental thought (transl. from German):
If the total enery of a mechanical (or electrical) system is
continuously dissipating, the system (linear or non-linear)
approaches an equilibrium condition.
Based on that, I'd expect that a completely stable filter will
end with zero output, when let alone with no excitation.
Despite any earlier excitation.
If the filter requires, that the ROC is never overridden (maybe by
any spurious/transient signal which had occurred in the past),
I would not call it stable.
In contrary, I use such a filter to create an oscillator.
Just by relying on spurious effects which cause the filter to leave
the ROC, it is started.
It then oscillates at a certain frequency which is defined by
carefully selecting the position of the poles.
If your filter becomes stable, under the condition that a certain
ROC is kept, then the filter contains such a ROC-keeper.
Your transfer function currently doesn't.
If you would include it, this would certainly modify your transfer
function so that the poles after that modification would be inside
the unit circle.
Bernhard
Reply by Randy Yates●September 12, 20042004-09-12
Tim Wescott <tim@wescottnospamdesign.com> writes:
> [...]
> I am going to go bury my head in some dirt. You and Randy are right,
> I was wrong when I said I was wrong -- for |d| < 1 you either have a
> causal stable system or a non-causal non-stable one, and for |d| > 1
> you either have a non-causal stable system or a causal non-stable one.
>
> For |d| = 1 you still have a metastable system, no matter what you do,
> so we can thank the Gods of Mathematics for some -- er -- stability in
> our lives.
Tim, I have many days like the one you had, if it helps...
--
% Randy Yates % "With time with what you've learned,
%% Fuquay-Varina, NC % they'll kiss the ground you walk
%%% 919-577-9882 % upon."
%%%% <yates@ieee.org> % '21st Century Man', *Time*, ELO
http://home.earthlink.net/~yatescr
Reply by Tim Wescott●September 12, 20042004-09-12
Martin Eisenberg wrote:
> Tim Wescott wrote:
>
>
>>If you take the z transform of a signal h(k) = d^k * u(k) for
>>any |d|<1, then it will converge nicely and indicate a stable
>>system with a z transform of z/(z-d).
>>
>>If you have a system with |d|>1 then the system will have a
>>growing response, but if you have a _signal_ equal to d^k *
>>u(-k) then the signal will be quite nicely finite.
>>
>>If you stupidly apply half-understood theory, then you'll say
>>that the z transform of the above signal is z/(z-d). If you
>>actually do the math, however, you'll get a result of sum{k=0 to
>>infinity}((z/d)^k) = -d/(z-d), which is indistinguishable from a
>>causal signal with a decaying exponential.
>
>
> You're confusing me. How are z/(z-d) and -e/(z-e)
> (0 < |d| < 1, e = 1/d) indistinguishable?
>
I am going to go bury my head in some dirt. You and Randy are right, I
was wrong when I said I was wrong -- for |d| < 1 you either have a
causal stable system or a non-causal non-stable one, and for |d| > 1 you
either have a non-causal stable system or a causal non-stable one.
For |d| = 1 you still have a metastable system, no matter what you do,
so we can thank the Gods of Mathematics for some -- er -- stability in
our lives.
--
Tim Wescott
Wescott Design Services
http://www.wescottdesign.com
Reply by Jerry Avins●September 12, 20042004-09-12
Tim Wescott wrote:
> Jerry Avins wrote:
>
>> Tim Wescott wrote:
>>
> - snip -
>
>>
>> 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
>
>
> Answer A: Practical application? Eew.
>
> Answer B: To maintain mental flexibility, and/or to demonstrate the
> limitations of the tools that we're using.
>
> Answer C: Because it's common to approximate a realizable, high-delay
> filter with a filter that's referenced to time zero, and why shouldn't
> you extend this to a filter that's infinite in extent as well?
The output of an unstable filter grows with only saturation as a limit
limit as more and more samples are run through it. How can reordering
the sample's indices alter that?
Jerry
--
Engineering is the art of making what you want from things you can get.
�����������������������������������������������������������������������
Reply by Martin Eisenberg●September 12, 20042004-09-12
Tim Wescott wrote:
> If you take the z transform of a signal h(k) = d^k * u(k) for
> any |d|<1, then it will converge nicely and indicate a stable
> system with a z transform of z/(z-d).
>
> If you have a system with |d|>1 then the system will have a
> growing response, but if you have a _signal_ equal to d^k *
> u(-k) then the signal will be quite nicely finite.
>
> If you stupidly apply half-understood theory, then you'll say
> that the z transform of the above signal is z/(z-d). If you
> actually do the math, however, you'll get a result of sum{k=0 to
> infinity}((z/d)^k) = -d/(z-d), which is indistinguishable from a
> causal signal with a decaying exponential.
You're confusing me. How are z/(z-d) and -e/(z-e)
(0 < |d| < 1, e = 1/d) indistinguishable?
--
Quidquid latine dictum sit, altum viditur.
Reply by Tim Wescott●September 11, 20042004-09-11
David Gelbart wrote:
> In article <10k3hoq934rd25c@corp.supernews.com>,
> Tim Wescott <tim@wescottnospamdesign.com> wrote:
>
>>Jerry Avins wrote:
>>
>>
>>>Tim Wescott wrote:
>>>
>>
>>- snip -
>>
>>>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
>>
>>Answer A: Practical application? Eew.
>
>
> In an application where we have all the input data already
> buffered before we start our digital signal processing, I don't
> see why causality should be a requirement.
>
> Let's say we have a causal stable filter h(n) = a^n * u(n),
> for |a| < 1. If we have a finite-duration, previously recorded input
> signal x(n), we can approximate h(n) with the first K
> samples of h(n), choosing K according to (e.g.) when h(n)
> shrinks below the smallest representable value of a number
> on our computing system. Then a linear convolution
> of x(n) with the truncated h(n) is straightforward.
>
> If we have an anticausal stable filter h(n) = a^n * u(-n),
> for |a| > 1, then I think the same reasoning applies.
>
> If you prefer a biquad IIR filter sort of implementation,
> I think you could implement the first filter by running
> the input signal through the system y(n) = x(n) + a * y(n-1),
> with n increasing from the index of the first nonzero sample in
> x(n). I think you could implement the second filter using the
> system y(n) = x(n) + a * y(n+1), with n decreasing from the
> index of the last nonzero sample in x.
>
>
Yea, well, that's what answer C was all about. Answer A was just to
poke fun at theorists.
--
Tim Wescott
Wescott Design Services
http://www.wescottdesign.com
Reply by David Gelbart●September 11, 20042004-09-11
In article <10k3hoq934rd25c@corp.supernews.com>,
Tim Wescott <tim@wescottnospamdesign.com> wrote:
>Jerry Avins wrote:
>
>> Tim Wescott wrote:
>>
>- snip -
>>
>> 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
>
>Answer A: Practical application? Eew.
In an application where we have all the input data already
buffered before we start our digital signal processing, I don't
see why causality should be a requirement.
Let's say we have a causal stable filter h(n) = a^n * u(n),
for |a| < 1. If we have a finite-duration, previously recorded input
signal x(n), we can approximate h(n) with the first K
samples of h(n), choosing K according to (e.g.) when h(n)
shrinks below the smallest representable value of a number
on our computing system. Then a linear convolution
of x(n) with the truncated h(n) is straightforward.
If we have an anticausal stable filter h(n) = a^n * u(-n),
for |a| > 1, then I think the same reasoning applies.
If you prefer a biquad IIR filter sort of implementation,
I think you could implement the first filter by running
the input signal through the system y(n) = x(n) + a * y(n-1),
with n increasing from the index of the first nonzero sample in
x(n). I think you could implement the second filter using the
system y(n) = x(n) + a * y(n+1), with n decreasing from the
index of the last nonzero sample in x.
Reply by Eric Jacobsen●September 11, 20042004-09-11
On Fri, 10 Sep 2004 12:19:31 -0400, Jerry Avins <jya@ieee.org> wrote:
>Randy Yates wrote:
>
> ...
>
>> Thanks for the (indirect) complement, Jerry. That was kind of you.
>
>Not kind. Well deserved.
>
>> But the truth is, I feel very stupid most of the time. Maybe that's
>> how smart people feel? In any case, it's not pleasant, and my desire
>> is to remove those nasty ignorances.
>
>You just inspired a wonderful sig. See below.
>
>Jerry
>--
> It takes great intellect to fathom the depths of one's own ignorance.
>�����������������������������������������������������������������������
Ahg! A conundrum in a paradox! How can I ever claim to be stupid
again?!
Eric Jacobsen
Minister of Algorithms, Intel Corp.
My opinions may not be Intel's opinions.
http://www.ericjacobsen.org
Reply by Jerry Avins●September 10, 20042004-09-10
Tim Wescott wrote:
> Jerry Avins wrote:
>
>> Tim Wescott wrote:
>>
> - snip -
>
>>
>> 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
>
>
> Answer A: Practical application? Eew.
>
> Answer B: To maintain mental flexibility, and/or to demonstrate the
> limitations of the tools that we're using.
>
> Answer C: Because it's common to approximate a realizable, high-delay
> filter with a filter that's referenced to time zero, and why shouldn't
> you extend this to a filter that's infinite in extent as well?
I don't claim you shouldn't. I don't know why you would want to. :-)
I like Answer B best. I'll buy it.
Jerry
P.S. I'm a queer duck. I understand why participants and those who bet
on them care about the outcomes of sporting games. I don't understand
why others do. I like to watch a good game. Competence and occasional
excellence can excite me. Outcomes much less so.
--
Engineering is the art of making what you want from things you can get.
�����������������������������������������������������������������������
Reply by Jerry Avins●September 10, 20042004-09-10
Randy Yates wrote:
...
> Thanks for the (indirect) complement, Jerry. That was kind of you.
Not kind. Well deserved.
> But the truth is, I feel very stupid most of the time. Maybe that's
> how smart people feel? In any case, it's not pleasant, and my desire
> is to remove those nasty ignorances.
You just inspired a wonderful sig. See below.
Jerry
--
It takes great intellect to fathom the depths of one's own ignorance.
�����������������������������������������������������������������������