--- title: "Nyquist plot of K/[s(s²+2s+2)] and maximum K" subject: "Control Systems" chapter: 6 key: "semester_3/control_systems:nyquist-k-s-s2-2s-2" source: https://www.bctengineer.com/q/semester_3/control_systems/nyquist-k-s-s2-2s-2 site: BCT Engineering --- # Nyquist plot of K/[s(s²+2s+2)] and maximum K Control Systems · Chapter 6: Frequency Response Techniques - Asked in: 2081 Bhadra, 8+2 marks - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_3/control_systems/nyquist-k-s-s2-2s-2 - Whole chapter: https://www.bctengineer.com/md/semester_3/control_systems/chapter-6 ## Question Construct complete Nyquist plot for a unity feedback control system whose open loop transfer function is: G(s) = K/[s(s²+2s+2)]. Find maximum value of K for which the system is stable as per Nyquist Criterion. ## Short answer The Nyquist plot crosses the real axis at –K/4. For P=0, N must be zero, so –K/4 must lie to the right of –1, giving K < 4. Maximum stable gain is K_max = 4 (marginal, oscillation at ω = √2 rad/s). ## Answer $$ G(s) = \frac{K}{s(s^2 + 2s + 2)} $$ Poles: $0,\ -1 \pm j1$. **P = 0**. The pole at the origin needs an indentation. #### Frequency response $$ G(j\omega) = \frac{K}{-2\omega^2 + j\omega(2 - \omega^2)} $$ $$ \text{Re} = \frac{-2K}{4\omega^2 + (2 - \omega^2)^2}, \qquad \text{Im} = \frac{-K(2 - \omega^2)}{\omega\left[4\omega^2 + (2 - \omega^2)^2\right]} $$ | ω | \|G\|/K | ∠G | Re/K | Im/K | |---|---|---|---|---| | 0⁺ | ∞ | −90° | −0.5 | −∞ | | 0.5 | 0.99 | −119.7° | −0.49 | −0.86 | | 1 | 0.447 | −153.4° | −0.40 | −0.20 | | 1.414 | 0.25 | −180° | −0.25 | 0 | | 2 | 0.112 | −206.6° | −0.10 | +0.05 | | ∞ | 0 | −270° | 0 | 0 | - **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°. ```text . | . | . | . | .. | .. | ... | ... | ----@------------------------*************-------- *** | *** | ** | ** | * | * | * | * | -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. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. For G(s) = K/[s(s^2+2s+2)], what is P and what special treatment does the contour need? A. P = 0; indent around the pole at the origin B. P = 1; indent around the pole at the origin C. P = 2; no indentation is needed D. P = 0; no indentation is needed Answer: A. Poles are 0 and -1 +/- j1, none in the RHP so P = 0, but the pole at s = 0 lies on the jw axis and must be bypassed by a small semicircle. (Key point: Open-loop poles and P) 2. Where does the Nyquist plot cut the negative real axis and at what frequency? A. At -K/2 at w = 1 rad/s B. At -K/4 at w = sqrt(2) rad/s C. At -K/4 at w = 2 rad/s D. At -K/2 at w = sqrt(2) rad/s Answer: B. Im = 0 when 2 - w^2 = 0, so w = sqrt(2), and G = K/(-4), giving -K/4. (-K/2 is the low-frequency asymptote.) (Key point: Real-axis crossing) 3. With P = 0, what condition on the crossing point gives the maximum K for stability? A. K/4 < 1, so K < 4 B. K/2 < 1, so K < 2 C. K/4 > 1, so K > 4 D. K < 8 Answer: A. The crossing at -K/4 must stay to the right of -1 so that N = 0; this gives K < 4, matching the Routh result. (Key point: Maximum K from the criterion) 4. If K > 4, what happens according to the Nyquist criterion? A. N = 0 and Z = 0, still stable B. N = 1 and Z = 1 C. N = 2 and Z = 2, unstable D. N = -2 and Z = -2, stable Answer: C. The crossing moves left of -1, so the plot encircles -1 twice clockwise: Z = P + N = 0 + 2 = 2 RHP closed-loop poles. 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