# Nyquist Stability Test

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 $$K=1$$.

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?

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

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:

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):

[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.