Nyquist plot of K/[s(s²+2s+2)] and maximum K

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

Poles: $0,\ -1 \pm j1$. P = 0. The pole at the origin needs an indentation.

Frequency response

$$G(j\omega) = \frac{K}{-2\omega^2 + j\omega(2 - \omega^2)}$$ $$\text{Re} = \frac{-2K}{4\omega^2 + (2 - \omega^2)^2}, \qquad \text{Im} = \frac{-K(2 - \omega^2)}{\omega\left[4\omega^2 + (2 - \omega^2)^2\right]}$$

• Low-frequency asymptote: Re → −K/2.
• Real-axis crossing: Im = 0 when $2 - \omega^2 = 0$, so $\omega_{pc} = \sqrt2 = 1.414$ rad/s, where $G = K/(-4)$. The crossing is at −K/4.
• High frequency: reaches the origin along −270° (from the second quadrant).

Complete Nyquist plot

1. ω = 0⁺ → ∞: from $\infty\angle-90^\circ$ (along Re = −K/2), crossing at −K/4, to the origin.
2. ω = −∞ → 0⁻: mirror image.
3. Infinite semicircle: maps to the origin.
4. Indentation at s = 0: maps to an infinite arc clockwise from +90° through 0° to −90°.

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----@------------------------*************--------
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   -1                -0.5     -.25
* w > 0   . w < 0 (mirror)   @ point (-1, j0)
drawn for K = 1: crosses at -0.25 (w = 1.414);
asymptote Re = -0.5. For gain K, scale every point by K.

Maximum K for 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 $Z = 0$.

With P = 0, stability needs N = 0, i.e. the point −1 must lie to the left of the crossing:

$$\frac{K}{4} < 1 \;\Rightarrow\; K < 4$$

If $K > 4$ the crossing moves left of −1, giving N = 2 and Z = 2 (unstable). At $K = 4$ the plot passes through −1: marginal stability with oscillation at $\omega = \sqrt2$ rad/s.

Check (Routh): $s^3 + 2s^2 + 2s + K$ is stable if $2 \times 2 > K$, i.e. $K < 4$. ✓

Answer: K_max = 4. The system is stable for 0 < K < 4 and oscillates at 1.414 rad/s when K = 4.