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Jul 9, 2019 at 21:17 vote accept felipeek
Apr 6, 2019 at 10:24 history edited Matt L.
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Apr 5, 2019 at 11:20 answer added Matt L. timeline score: 4
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Apr 6, 2017 at 20:42 answer added Dan Boschen timeline score: 7
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S Mar 7, 2017 at 17:48 history suggested Tendero CC BY-SA 3.0
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Dec 8, 2016 at 14:20 comment added rrogers @fibonatic I see, the -1 refers to the closed loop stability and the question is referring to standard transfer function stability. In any case: both answers point out the fact that, basically, the examples above never actually reach 0,0 with any output left.
Dec 7, 2016 at 14:28 comment added fibonatic @rrogers Usually you do use this, but this is based on the fact that if you let $s$ follow some closed curve, then $G(s)$ will have $Z$ (equal to the number op zeros of $G(s)$ inside the area enclosed by the curve) anti-clockwise encirclements of the origin and $P$ (equal to the number op poles of $G(s)$ inside the area enclosed by the curve) clockwise encirclement of the origin.
Dec 7, 2016 at 11:59 answer added LJSilver timeline score: 0
Dec 7, 2016 at 0:38 answer added fibonatic timeline score: 0
Dec 6, 2016 at 18:02 comment added rrogers I don't use Nyquist plots so correct me if I am wrong; but aren't you supposed to be count encirclements around -1 ?
Dec 5, 2016 at 20:41 history tweeted twitter.com/StackSignals/status/805874687788154880
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Dec 5, 2016 at 5:25 history asked felipeek CC BY-SA 3.0