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Currently I'm learning about Nyquist Stability Test and I'm few days stuck on one thing and I don't understand that so I wanted look here for help.

Given is transfer function $K\frac{2s-11}{s(s^2+s+1}$. We take that $K=1$. enter image description here

Line represens the Nyquist contour for K=1.

Than in my book it's written that that function is stabil when

$\triangle arg(1+L(Iw))=(max(grad(z_L),grad(n_L)-N_{-}(n_L)+N_{+}(n_L) )\pi$

than they see from this picture above $\triangle arg(1+L(Iw))=-\pi$

and

$(max(grad(z_L),grad(n_L)-N_{-}(n_L)+N_{+}(n_L) )\pi=(max(1,3)-2+0)\pi=\pi$

Not stabil.

What I don't undestand is - $\triangle arg(1+L(Iw))=-\pi$? What this could be? What angle? How do I see that from picture?

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  • $\begingroup$ Could you check your open-loop transfer function? I'm not sure if it's correct. $\endgroup$ – Matt L. Nov 26 '18 at 22:03
  • $\begingroup$ what exactly @MattL. $\endgroup$ – Alena Nov 26 '18 at 22:19
  • $\begingroup$ I get a different Nyquist plot for the transfer function in your question. $\endgroup$ – Matt L. Nov 26 '18 at 22:28
  • $\begingroup$ can you put photo of it? @MattL. $\endgroup$ – Alena Nov 26 '18 at 23:24
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The open-loop transfer function

$$G(s)=\frac{2s-11}{s(s^2+s+1)}\tag{1}$$

given in your question has the following Nyquist plot:

enter image description here

which is different from the one in your question.

Note that in any case it's important to know how the trace is closed, i.e., how the points $\omega=0^-$ and $\omega=0^+$ are connected. In this case they are connected by a semi-circle with infinite radius in the left half-plane (the dashed curve in the figure below):

enter image description here

[generated with WolframAlpha]. This is a consequence of the pole of $G(s)$ at $s=0$.

The Nyquist criterion is actually simpler than you made it look in your question. The number of clockwise encirclements ($N$) of the point $-1+j0$ equals the number of right half-plane zeros of $1+G(s)$ ($Z$) minus the number of right half-plane poles ($P$):

$$N=Z-P\tag{2}$$

Since $G(s)$ has no poles in the right half-plane ($P=0$), the number of clockwise encirclements equals the number of zeros in the right half-plane. From the Nyquist plot we see that $N=1$, so there is one zero in the right half-plane, which corresponds to one pole of the closed-loop transfer function in the right half-plane. Consequently, the system is not stable.

This is also easily verified by directly computing the poles of the closed-loop transfer function

$$Q(s)=\frac{G(s)}{1+G(s)}=\frac{2s-11}{s^3+s^2+3s-11}\tag{3}$$

roots([1,1,3,-11])
ans =

  -1.2833 + 2.3182i
  -1.2833 - 2.3182i
   1.5667 + 0.0000i

which shows that there is one pole with a positive real part, as predicted from the Nyquist plot.

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