Talk:Marginal stability

Latest comment: 1 year ago by 77.137.73.60 in topic What about a simple pole at the origin?

Significance

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To be expanded: What is the signifcance of being Marginally Stable? We have a definition but it doesn't tell us why we'd care. What is the advantage? RJFJR 03:29, 14 October 2005 (UTC)Reply

Nonpositive vs. zero

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Nonpositive is something different than zero. A (CT,LTI) system with one zero eigenvalue/pole and all other in the left half plane is still marginally stable. Just undid this change Zeebrommer (talk) 19:21, 11 August 2014 (UTC)Reply

Additions 11 August 2014

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On the rare occasion of encountering an article that I felt I had something to add to, I contributed to Wikipedia for the first time. I hope I did it about right, feel free to revert/edit anything I added! Zeebrommer (talk) 21:29, 11 August 2014 (UTC)Reply

Misleading phrase: "no final steady-state output"

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In the section "System response" it reads "A bounded offset or oscillations in the output will persist indefinitely, and so there will in general be no final steady-state output." It looks to me that this paragraph says a non-constant values implies no steady-state. If that's case, it's WRONG. Steady-state need not be constant; for example in AC electric circuits with sinusoidal sources, there still exists steady-state (and that's exactly why it's valid to use phasors!; if we wanted the transient response, phasors wouldn't be useful.)

If you don't believe me, you can read the section 10.2 of the textbook Engineering Circuit Analysis (8th edition) by W. Hayt, J. Kemmerly and S. Durbin, which reads "Unfortunately, steady state carries the connotation of 'not changing with time' in the minds of many students. This is true for dc forcing functions, but the sinusoidal steady-state response is definitely changing with time."

So, my suggestion is to change the phrase "no final steady-state output" to "no final value." Otherwise I may change it to avoid possible misinterpretations from future readers.

What about a simple pole at the origin?

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Consider H(s)=1/s, hence h(t)=u(t). The system is marginally stable. However, it does not satisfy the condition: "one or more poles have zero real part and non-zero imaginary part" 77.137.73.60 (talk) 22:10, 16 February 2023 (UTC)Reply