Polar and Nyquist plot of K/[s(s+1)(0.1s+1)]

The polar plot is the locus of $G(j\omega)$ in the complex plane as $\omega$ goes from 0 to $\infty$. The Nyquist plot adds the mirror image for $-\infty<\omega<0$ and the map of the small indentation around the pole at the origin, so that the whole Nyquist contour is mapped.

Polar plot

$$G(j\omega)=\frac{K}{j\omega(1+j\omega)(1+j0.1\omega)}$$ $$|G(j\omega)|=\frac{K}{\omega\sqrt{1+\omega^2}\sqrt{1+0.01\omega^2}},\qquad \angle G(j\omega)=-90^\circ-\tan^{-1}\omega-\tan^{-1}0.1\omega$$

• Low-frequency asymptote: as $\omega\to0$, $\text{Re}\,G \to -K(T_1+T_2) = -K(1+0.1) = -1.1K$. The plot starts at infinity parallel to the negative imaginary axis, along the line $\text{Re}=-1.1K$.
• Negative real axis crossing: set phase $=-180^\circ$: $\tan^{-1}\omega+\tan^{-1}0.1\omega=90^\circ \Rightarrow 0.1\omega^2=1 \Rightarrow \omega_{pc}=\sqrt{10}=3.162$ rad/s.

$$\begin{aligned} |G(j\omega_{pc})| &= \frac{K}{3.162\times\sqrt{11}\times\sqrt{1.1}} = \frac{K}{11} \end{aligned}$$

So the plot crosses the negative real axis at $-K/11$ and ends at the origin, tangent to the positive imaginary axis ($-270^\circ$).

            Im
             |
   -K/11     |
 ----o-------+--------> Re
    / \_____/|
   |  w=3.16 |
   |         |
   |  w ->0+ (to -inf j,
   |   asymptote Re=-1.1K)

Nyquist plot

1. Draw the polar plot for $\omega = 0^+ \to \infty$ (third quadrant, crosses at $-K/11$, ends at origin).
2. Draw its mirror image about the real axis for $\omega=-\infty \to 0^-$.
3. The infinite semicircle of the Nyquist contour ($s=Re^{j\theta}$, $R\to\infty$) maps to the origin.
4. The small semicircle around the pole at $s=0$ ($s=\epsilon e^{j\theta}$, $\theta: -90^\circ\to+90^\circ$) maps to $\frac{K}{\epsilon}e^{-j\theta}$, i.e. an infinite-radius semicircle traversed clockwise from $\omega=0^-$ to $\omega=0^+$ through the positive real axis.

The resulting closed curve is the Nyquist plot. It cuts the negative real axis only at $-K/11$.

Range of K for stability

Nyquist criterion: $Z = P + N$, where $P$ = open-loop RHP poles, $N$ = clockwise encirclements of $-1+j0$.

• $P = 0$ (poles at $0,-1,-10$; the origin pole is excluded by the indentation).
• For stability, $Z=0$, so $N=0$: the point $-1$ must not be enclosed, i.e. it must lie to the left of the crossing point.

$$\frac{K}{11} < 1 \;\Rightarrow\; K < 11$$

If $K>11$, the curve encircles $-1$ twice clockwise ($N=2$, $Z=2$): unstable. At $K=11$ the system oscillates at $\omega=3.162$ rad/s.

Check by Routh: $0.1s^3+1.1s^2+s+K=0$ gives $1.1\times1 > 0.1K \Rightarrow K<11$.

Answer: The closed-loop system is stable for $0 < K < 11$.