Nyquist range of K for K/[s(s+1)(s+2)]

Frequency response

$$G(j\omega)=\frac{K}{j\omega(1+j\omega)(2+j\omega)}=\frac{K}{-3\omega^2+j\omega(2-\omega^2)}$$ $$|G|=\frac{K}{\omega\sqrt{1+\omega^2}\sqrt{4+\omega^2}},\qquad \angle G=-90^\circ-\tan^{-1}\omega-\tan^{-1}\frac{\omega}{2}$$

Key points of the plot

• Low-frequency asymptote: $\text{Re}\,G=\dfrac{-3K\omega^2}{9\omega^4+\omega^2(2-\omega^2)^2}\to-\dfrac{3K}{4}$ as $\omega\to0$.
• Negative real-axis crossing: imaginary part of denominator zero: $2-\omega^2=0 \Rightarrow \omega_{pc}=\sqrt2=1.414$ rad/s.

$$G(j\sqrt2)=\frac{K}{-3(2)}=-\frac{K}{6}$$

• As $\omega\to\infty$, $G\to0$ at $-270^\circ$.

Nyquist plot

            Im
   w<0 mirror|
     ___     |
    /   \    |
 --o-----+---+------- Re
 -K/6 \_/    |
   w=1.414   |
  w->0+: to -inf j along Re=-0.75K
  pole at s=0 -> infinite arc,
  clockwise from w=0- to w=0+

1. $\omega=0^+\to\infty$: polar plot above.
2. $\omega=-\infty\to0^-$: mirror image about the real axis.
3. Small indentation at $s=0$ maps to an infinite clockwise semicircle.
4. Infinite $s$-semicircle maps to the origin.

Applying the criterion

$Z=P+N$ with $P=0$ (poles at $0,-1,-2$).

• If $K/6<1$: $-1$ lies outside the plot, $N=0$, $Z=0$: stable.
• If $K/6>1$: $-1$ is encircled twice clockwise, $N=2$, $Z=2$: unstable.
• If $K=6$: the plot passes through $-1$: marginally stable, oscillation at $\omega=\sqrt2=1.414$ rad/s.

$$\frac{K}{6}<1 \Rightarrow K<6$$

Check by Routh: $s^3+3s^2+2s+K=0$ requires $3\times2>K$, i.e. $K<6$.

Answer: the closed-loop system is stable for $0<K<6$; at $K=6$ it oscillates at 1.414 rad/s.