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

Short answer

The Nyquist plot crosses the negative real axis at -1.333, encircling -1 twice clockwise. No open-loop RHP poles (P=0), so Z=2. The closed-loop system is unstable with two RHP poles.

The open loop transfer function of closed loop system is G(s) = 2/[s(s+1)(2s+1)]. Using Nyquist Criterion, determine closed loop stability of this system.

G(s)=2s(s+1)(2s+1)G(s) = \frac{2}{s(s+1)(2s+1)}

Open-loop poles: 0,−1,−0.50, -1, -0.5, so P = 0. The contour is indented around s=0s = 0.

Frequency response

G(jω)=2ω1+ω21+4ω2∠(−90∘−tan⁡−1ω−tan⁡−12ω)G(j\omega) = \frac{2}{\omega\sqrt{1+\omega^2}\sqrt{1+4\omega^2}}\angle\left(-90^\circ - \tan^{-1}\omega - \tan^{-1}2\omega\right)
ω|G|∠GReIm
0⁺∞−90°−6−∞
0.119.5−107.0°−5.71−18.66
0.35.48−137.7°−4.05−3.69
0.52.53−161.6°−2.40−0.80
0.7071.333−180°−1.3330
10.632−198.4°−0.60+0.20
∞0−270°00
  • Low-frequency asymptote: Re → −K(T1+T2)=−2(1+2)=−6-K(T_1 + T_2) = -2(1 + 2) = -6.
  • Phase crossover: tan⁡−1ω+tan⁡−12ω=90∘⇒2ω2=1⇒ωpc=0.707\tan^{-1}\omega + \tan^{-1}2\omega = 90^\circ \Rightarrow 2\omega^2 = 1 \Rightarrow \omega_{pc} = 0.707 rad/s.
∣G∣=20.707×1.225×1.732=1.333|G| = \frac{2}{0.707 \times 1.225 \times 1.732} = 1.333
  • Indentation at the origin: maps to an infinite arc clockwise from +90° through 0° to −90°.
  • Infinite semicircle: maps to the origin.
               ..                          |
                ..                         |
                 ...                       |
                   ..                      |
                    ...                    |
                      ...                  |
                        ....               |
                           ......          |
--------------------------------*****@******------
                           ******          |
                        ****               |
                      ***                  |
                    ***                    |
                   **                      |
                 ***                       |
                **                         |
               **                          |
     -6                          -1.33     0
* w > 0   . w < 0 (mirror)   @ point (-1, j0)
crosses at -1.333 (w = 0.707); asymptote Re = -6

Nyquist criterion

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

The plot cuts the negative real axis at −1.333, to the left of −1. Together with its mirror image and the infinite clockwise arc, it encircles (−1,j0)(-1, j0) twice clockwise: N = 2.

Z=P+N=0+2=2Z = P + N = 0 + 2 = 2

The closed-loop system is unstable, with 2 RHP poles.

GM=11.333=0.75=−2.5 dBGM = \frac{1}{1.333} = 0.75 = -2.5\ \text{dB}

The gain must be reduced below 1.5 for stability. Check: 2s3+3s2+s+22s^3 + 3s^2 + s + 2 gives 3×1<2×23 \times 1 < 2 \times 2, so it is unstable. ✓

✓ Answer checked by Claude Opus 5.5 (high effort)

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