Nyquist stability of 1/[s(1+2s)(1+s)] and gain margin

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

Nyquist contour

There is a pole at the origin, so the contour is indented around it by a small semicircle $s = \varepsilon e^{j\theta}$ with θ from −90° to +90°. Open-loop poles in the RHP: P = 0.

Mapping of each section

1. ω = 0⁺ to ∞ (positive $j\omega$-axis):

• Low-frequency asymptote: $\text{Re} \to -(2+1) = -3$. The plot starts at $\infty\angle -90^\circ$ along Re = −3.
• Phase crossover: $\tan^{-1}2\omega + \tan^{-1}\omega = 90^\circ \Rightarrow 2\omega^2 = 1 \Rightarrow \omega_{pc} = 0.707$ rad/s.
• $|G(j0.707)| = \dfrac{1}{0.707 \times \sqrt3 \times \sqrt{1.5}} = 0.667$, so it crosses at −0.667.

2. ω = −∞ to 0⁻: mirror image of the above.
3. Infinite semicircle: $G \to 0$, so it maps to the origin.
4. Small semicircle at origin: $G \approx 1/s = (1/\varepsilon)e^{-j\theta}$. As θ goes from −90° to +90°, $G$ sweeps an infinite-radius arc clockwise from +90° through 0° to −90° (right half of the G-plane), joining the ω = 0⁻ end to the ω = 0⁺ end.

                ..                     |
                 ..                    |
                  ..                   |
                   ..                  |
                    ...                |
                      ..               |
                       ....            |
                          .....        |
-----------------------------@**********----------
                          *****        |
                       ****            |
                      **               |
                    ***                |
                   **                  |
                  **                   |
                 **                    |
                **                     |
         -3                    -0.67   0
* w > 0   . w < 0 (mirror)   @ point (-1, j0)
crosses at -0.667 (w = 0.707); asymptote Re = -3

(The infinite clockwise arc on the right closes the curve.)

Gain margin

$$GM = \frac{1}{0.667} = 1.5 = 20\log1.5 = 3.52\ \text{dB}$$

The phase margin (for reference) is $11.4^\circ$ at $\omega_{gc} = 0.572$ rad/s.

Stability by Nyquist criterion

Nyquist criterion: $Z = P + N$, where $P$ = number of open-loop poles in the RHP, $N$ = net clockwise encirclements of $(-1, j0)$ by the Nyquist plot of $G(s)H(s)$, and $Z$ = number of closed-loop poles in the RHP. The closed loop is stable only if $Z = 0$.

The plot crosses the negative real axis at −0.667, to the right of −1, so the point $(-1, j0)$ is not encircled: N = 0.

$$Z = P + N = 0$$

The closed-loop system is stable. Check: $2s^3 + 3s^2 + s + 1$ has Routh first column $2, 3, 1/3, 1$, all positive. ✓

Answer: GM = 1.5 (3.52 dB) at $\omega_{pc}$ = 0.707 rad/s. N = 0, P = 0, Z = 0, so the system is stable.