Polar plot of 10/[s(s+1)²] with GM and PM

$$G(s) = \frac{10}{s(s+1)^2}, \qquad G(j\omega) = \frac{10}{\omega(1+\omega^2)}\angle\left(-90^\circ - 2\tan^{-1}\omega\right)$$

Key points of the polar plot

• Low-frequency asymptote: $\text{Re}\,G \to -K\sum T = -10(1+1) = -20$. The plot starts at $\infty\angle -90^\circ$ along the line Re = −20.
• Real-axis crossing: $\angle G = -180^\circ \Rightarrow 2\tan^{-1}\omega = 90^\circ \Rightarrow \omega = 1$, where $|G| = 10/(1 \times 2) = 5$. The plot cuts the negative real axis at −5.
• High frequency: the plot reaches the origin along $-270^\circ$ (from the second quadrant, tangent to the +j axis).

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      -20                       -5        0
* w > 0   . w < 0 (mirror)   @ point (-1, j0)
crosses real axis at -5 (w = 1); asymptote Re = -20

(iv) Phase crossover frequency (pcf)

$\omega_{pc} = 1$ rad/s.

(i) Gain margin

$$GM = \frac{1}{|G(j\omega_{pc})|} = \frac{1}{5} = 0.2 = 20\log 0.2 = -13.98\ \text{dB}$$

(iii) Gain crossover frequency (gcf)

$$|G| = 1 \Rightarrow \omega(1 + \omega^2) = 10 \Rightarrow \omega^3 + \omega - 10 = 0 \Rightarrow \omega_{gc} = 2\ \text{rad/s}$$

(Check: $8 + 2 = 10$.)

(ii) Phase margin

$$\begin{aligned} \angle G(j2) &= -90^\circ - 2\tan^{-1}2 = -90^\circ - 126.87^\circ = -216.87^\circ \\ PM &= 180^\circ + \angle G(j\omega_{gc}) = -36.87^\circ \end{aligned}$$

(v) Stability

GM < 0 dB and PM < 0. The polar plot crosses the real axis at −5, to the left of −1, so the $(-1, j0)$ point is enclosed. With P = 0, the complete Nyquist plot encircles −1 twice clockwise, so Z = 2. The closed-loop system is unstable, with two RHP poles.

Check with Routh: $s^3 + 2s^2 + s + 10$ gives $2 \times 1 < 10$, so it is unstable. For stability $K$ must be below 2, i.e. the gain must be reduced by a factor of 5.

Answer: GM = −13.98 dB, PM = −36.87°, gcf = 2 rad/s, pcf = 1 rad/s. The system is unstable.