Amended the previous question to be more concise

I have the following open-loop plant:


These are the poles of the plant, and as you can see, there are no unstable poles:


If I have a positive feedback, I know for sure it's unstable, since one of the poles of $$L_i(s)=\frac{1}{1-G}$$ is 2.6443.

However, this is the Nyquist plot from MATLAB when using


enter image description here

I see no encirclement about -1. So what gives?

By the way, here's the negative feedback (which is asymptotically stable) Nyquist plot. Again, I don't see encirclement.

enter image description here

  • \$\begingroup\$ @AJN This small denominator is just a numerical artifice and is a red-herring. See the modified question with a much simpler transfer function. \$\endgroup\$
    – JZYL
    Aug 21, 2021 at 15:44
  • \$\begingroup\$ What happens when you change the x and y axis limits of the diagram? Does the contour appear to change the direction from which it enters the diagram. In the current scale, the contour appears to come from Ist quadrant. Please check if the contour is encircling via an infinitely large circle which is not obvious in the present x and y axis limits. \$\endgroup\$
    – AJN
    Aug 21, 2021 at 16:12
  • \$\begingroup\$ @AJN Not as far as I can see. There's no sign it's encircling back as s->0 \$\endgroup\$
    – JZYL
    Aug 21, 2021 at 18:15

1 Answer 1


I think I found my answer through: https://lpsa.swarthmore.edu/Nyquist/NyquistStability.html

Because there are poles on the imaginary axes (in this case, two origin poles), there is a full 360 deg clockwise encirclement at infinity in the Nyquist plot, which is not apparent from a numerical generation of the Nyquist plot.

Thanks @AJN for the hint.


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