Nyquist stability of 10/[s(s²+2s+4)] and GM

Frequency response

$$G(j\omega)=\frac{10}{j\omega\left[(4-\omega^2)+j2\omega\right]} = \frac{10}{-2\omega^2 + j\omega(4-\omega^2)}$$ $$|G| = \frac{10}{\omega\sqrt{(4-\omega^2)^2+4\omega^2}},\qquad \angle G = -90^\circ-\tan^{-1}\frac{2\omega}{4-\omega^2}$$

• Low-frequency asymptote: $\text{Re}\,G = \dfrac{-20\omega^2}{4\omega^4+\omega^2(4-\omega^2)^2} \to -\dfrac{20}{16} = -1.25$ as $\omega\to0$.

Phase crossover frequency

The imaginary part of the denominator is zero when $4-\omega^2=0$:

$$\omega_{pc}=2\ \text{rad/s},\qquad G(j2)=\frac{10}{-2(4)}=-1.25$$

So the plot cuts the negative real axis at $-1.25$.

Nyquist plot

              Im
     w<0 mirror  |
       ____      |
      /    \     |
 ----o--x---+----+----- Re
   -1.25 -1 \__/ |
      w=2        |
   w->0+ : to -inf j
   (asymptote Re=-1.25)
  small semicircle at s=0 maps
  to infinite arc, clockwise

• $\omega: 0^+\to\infty$: starts at $-1.25-j\infty$, passes through $-1.25$ at $\omega=2$, ends at the origin at $-270^\circ$.
• $\omega: -\infty\to0^-$: mirror image.
• Indentation around the pole at $s=0$ maps to an infinite semicircle clockwise (through the positive real axis).
• Infinite semicircle of the $s$-plane maps to the origin.

Nyquist criterion

$Z = P + N$

• $P=0$ (open-loop poles: $0$, $-1\pm j1.732$; none in the RHP).
• The crossing at $-1.25$ lies to the left of $-1$, so the point $-1+j0$ is encircled twice clockwise: $N=2$.
• $Z = 0+2 = 2$ closed-loop poles in the RHP.

The closed-loop system is unstable. (Check by Routh: $s^3+2s^2+4s+10$: $2\times4=8<10$, two sign changes.)

Gain margin

$$GM = \frac{1}{|G(j\omega_{pc})|}=\frac{1}{1.25}=0.8,\qquad GM_{dB}=20\log0.8=-1.94\ \text{dB}$$

The negative GM confirms instability; the gain must be reduced below $10\times0.8=8$ for stability.

Answer: $\omega_{pc}=2$ rad/s, GM = 0.8 (−1.94 dB); N = 2, Z = 2, so the closed-loop system is unstable.