Nyquist plot of 2.2/[s(s+1)(s²+2s+2)]

$$G(s) = \frac{2.2}{s(s+1)(s^2 + 2s + 2)}$$

Poles: $0, -1, -1 \pm j1$, so P = 0. The pole at the origin is indented.

Frequency response

$$\angle G = -90^\circ - \tan^{-1}\omega - \tan^{-1}\frac{2\omega}{2 - \omega^2}, \qquad |G| = \frac{2.2}{\omega\sqrt{1 + \omega^2}\sqrt{(2 - \omega^2)^2 + 4\omega^2}}$$

• Low-frequency asymptote: $G = \dfrac{1.1}{s(1+s)(1 + s + 0.5s^2)}$, so Re → $-1.1(1 + 1) = -2.2$.
• Phase crossover: $s^4 + 3s^3 + 4s^2 + 2s$ at $s = j\omega$ has imaginary part $-3\omega^3 + 2\omega = 0$, so $\omega_{pc} = \sqrt{2/3} = 0.816$ rad/s. The real part is $\omega^4 - 4\omega^2 = 0.444 - 2.667 = -2.222$, so $G = 2.2/(-2.222) = -0.990$.
• The plot cuts the negative real axis at −0.99, then passes through the second quadrant (crossing the +j axis at ω = 2, where ∠G = −270°) and reaches the origin along −360°.

              ..                          |
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                  ...                     |
                    ...                   |
                      .....     **********|
--------------------------**@****--------**-------
                      *****     ..........|
                    ***                   |
                  ***                     |
                 **                       |
                **                        |
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               *                          |
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         -2.2             -0.99           0
* w > 0   . w < 0 (mirror)   @ point (-1, j0)
crosses at -0.99 (w = 0.816), just right of -1

Stability

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 crossing at −0.99 lies just to the right of −1, so the point $(-1, j0)$ is not encircled: N = 0 and $Z = 0 + 0 = 0$.

The closed-loop system is stable, but only barely:

$$GM = \frac{1}{0.99} = 1.0101 = 0.087\ \text{dB}, \qquad PM \approx 0.5^\circ \;(\omega_{gc} = 0.811\ \text{rad/s})$$

The closed-loop poles are at $-0.0026 \pm j0.814$ and $-1.497 \pm j1.039$. The dominant pair is almost on the $j\omega$-axis, so the step response will be a very lightly damped oscillation at about 0.81 rad/s. The gain limit (Routh) is $K < 2.222$, so a 1% gain increase would make the system unstable.

Answer: N = 0, Z = 0, so the system is stable (marginally: GM ≈ 0.09 dB, PM ≈ 0.5°).