I read the paper "Bilateral Control of Teleoperators with Time Delay",which is written by Robert J. Anderson and Mark W. Spong. In that paper, they defined passivity of an n-port flow as
$\int^\infty_0 F^T (t) v(t) dt \geqslant 0$
$F$ means effort and $v$ means flow.
And they said if a two-port system is passive, the system is stable. They didn't said that directly, but they said an system can be unstable by being nonpassive. They said that without proof, so I want to know why an passive system is stable.
1 Answer
I didn't read the entire paper but only checked the statement you mentioned.
The two-port communication circuit for this system is nonpassive, and is the cause of the instability.
And later they write again
[...] it was shown that the instability [...] is due to a nonpassive communication block.
It looks to me like you mixed up sufficient and necessary conditions here. They write in the paper, as you already mentioned in your question, that nonpassive implies unstable.
This makes passivity a necessary condition for stability in this case.
They did not say that it is a sufficient condition, which would be how you understood it, that passive implies stable.
Wikipedia explains it a bit more compact than me.
If P is sufficient for Q, then knowing P to be true is adequate grounds to conclude that Q is true; however, knowing P to be false does not meet a minimal need to conclude that Q is false.
I hope I could explain it somewhat understandable, don't hesitate to comment if not.
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$\begingroup$ So that means if a system is stable, that is passive because passivity is a necessary condition. And and passive system is not always stable. I was confused about that before your answer. How can I proof "An stable system is passive"? Could you give me some informations or references to do that? $\endgroup$ Apr 8, 2018 at 5:59
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$\begingroup$ I'm sorry, I don't have references about that proof, but there are probably a lot of papers out there that include it. Maybe you also find something among the references of the paper that you mentioned in your question $\endgroup$ Apr 8, 2018 at 10:57