According to argument principle, if a contour encircles a number of poles and zeros of a transfer function, the number of origin encirclements can be deduced by : N = Z - P

In which N can be both positive and negative which means it can either be counter clockwise or clockwise.

But in Nyquist theorem, most of the time we are only counting the number of counter clockwise encirclements of point (-1,0) with no mentions of possible clockwise encirclements. Why is this?

  • $\begingroup$ Can you add a reference? To my knowledge, I've never encountered such a statement. $\endgroup$ – useless-machine Jul 6 '20 at 8:47

After looking through the book "Feedback control of Dynamic Systems", I should say the clockwise encirclements are as much mentioned as the counter clockwise encirclements. However, the importance of both differs whether the open loop transfer function has unstable poles. So as you might know, N the number of clockwise encirclements in the nyquist plot, equals Z the number of unstable poles in the closed loop transfer function minus P the number of open loop unstable poles. P is in these examples assumed to be known. it is desired to get Z = 0. Therefore, if the open loop has RHP poles (P>0), The nyquist plot should have P counter clockwise encirclements. However, if the open loop has no poles in RHP, there should be no encirclements of the point (-1,0) in the nyquist plot. Every encirclement (which can be assumed to be clockwise) will show that the closed loop TF is unstable. If P > 0 and you count clockwise encirclements, the closed loop has more poles in RHP than the open loop has. So in short, counting counter clockwise encirclements to achieve stability, if there is a clockwise encirclement (you dont have to count), the system is unstable anyway.

I hope this helped

  • $\begingroup$ So if a nyquist plot has both the same number of counterclockwise encirclement and clockwise encirclement of -1, if it hadn't had any rhp poles in the open loop transfer function, then the system is stable? $\endgroup$ – Farhood ET Jul 7 '20 at 21:19
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    $\begingroup$ if the system is open loop stable, the nyquist plot would not encircle the point (-1, 0) counter-clockwise. Every counter-clockwise encirclement indicates an open loop unstable pole. As such, the net number of encirclements would be your value for N. since that is 0, your closed loop transfer function is as unstable as the open loop is. $\endgroup$ – Petrus1904 Jul 8 '20 at 10:36
  • $\begingroup$ Aha. So the case never actually happens that an open loop system is stable, and has a net zero of counter clock wise and clockwise encirclements (while both are not zero), right? $\endgroup$ – Farhood ET Jul 9 '20 at 8:44

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