--- title: "Nyquist stability of 2/[s(s+1)(2s+1)]" subject: "Control Systems" chapter: 6 key: "semester_3/control_systems:nyquist-2-s-s1-2s1" source: https://www.bctengineer.com/q/semester_3/control_systems/nyquist-2-s-s1-2s1 site: BCT Engineering --- # Nyquist stability of 2/[s(s+1)(2s+1)] Control Systems · Chapter 6: Frequency Response Techniques - Asked in: 2076 Asoj, 8 marks - Video: [#215 Problem on Nyquist plot in control systems || EC Academy](https://www.youtube.com/watch?v=5PEVdDxZWjA) — EC Academy - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_3/control_systems/nyquist-2-s-s1-2s1 - Whole chapter: https://www.bctengineer.com/md/semester_3/control_systems/chapter-6 ## Question 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. ## 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. ## Answer $$ G(s) = \frac{2}{s(s+1)(2s+1)} $$ Open-loop poles: $0, -1, -0.5$, so **P = 0**. The contour is indented around $s = 0$. #### Frequency response $$ G(j\omega) = \frac{2}{\omega\sqrt{1+\omega^2}\sqrt{1+4\omega^2}}\angle\left(-90^\circ - \tan^{-1}\omega - \tan^{-1}2\omega\right) $$ | ω | \|G\| | ∠G | Re | Im | |---|---|---|---|---| | 0⁺ | ∞ | −90° | −6 | −∞ | | 0.1 | 19.5 | −107.0° | −5.71 | −18.66 | | 0.3 | 5.48 | −137.7° | −4.05 | −3.69 | | 0.5 | 2.53 | −161.6° | −2.40 | −0.80 | | 0.707 | 1.333 | −180° | −1.333 | 0 | | 1 | 0.632 | −198.4° | −0.60 | +0.20 | | ∞ | 0 | −270° | 0 | 0 | - **Low-frequency asymptote:** Re → $-K(T_1 + T_2) = -2(1 + 2) = -6$. - **Phase crossover:** $\tan^{-1}\omega + \tan^{-1}2\omega = 90^\circ \Rightarrow 2\omega^2 = 1 \Rightarrow \omega_{pc} = 0.707$ rad/s. $$ |G| = \frac{2}{0.707 \times 1.225 \times 1.732} = 1.333 $$ - **Indentation at the origin:** maps to an infinite arc clockwise from +90° through 0° to −90°. - **Infinite semicircle:** maps to the origin. ```text .. | .. | ... | .. | ... | ... | .... | ...... | --------------------------------*****@******------ ****** | **** | *** | *** | ** | *** | ** | ** | -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:** $Z = P + N$, 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 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 $(-1, j0)$ **twice clockwise**: N = 2. $$ Z = P + N = 0 + 2 = 2 $$ The **closed-loop system is unstable**, with 2 RHP poles. $$ GM = \frac{1}{1.333} = 0.75 = -2.5\ \text{dB} $$ The gain must be reduced below 1.5 for stability. Check: $2s^3 + 3s^2 + s + 2$ gives $3 \times 1 < 2 \times 2$, so it is unstable. ✓ ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. For G(s) = 2/[s(s+1)(2s+1)], what is P? A. P = 3 B. P = 1 C. P = 2 D. P = 0, with an indentation around s = 0 Answer: D. Poles at 0, -1 and -0.5 are not in the open RHP, so P = 0; the origin pole needs a small indentation. (Key point: Poles and P) 2. How is the phase crossover found and what are its values? A. tan^-1 w + tan^-1 2w = 90 deg gives w = 0.707 rad/s and |G| = 1.333 B. tan^-1 w = 90 deg gives w = 1, |G| = 0.632 C. w = 2 rad/s, |G| = 0.5 D. w = 0.5 rad/s, |G| = 2.53 Answer: A. The total phase is -180 deg when the two lags sum to 90 deg, so 2w^2 = 1, w = 0.707 and |G| = 2/(0.707 x 1.225 x 1.732) = 1.333. (Key point: Phase crossover) 3. The plot crosses at -1.333. What follows? A. N = 0, Z = 0, stable B. N = 2 clockwise, Z = 2, unstable C. N = 1 anticlockwise, Z = -1 D. Marginally stable, Z = 0 Answer: B. -1.333 is left of -1 so (-1, j0) is encircled twice clockwise; Z = P + N = 2. (Key point: Encirclements) 4. What is the GM and what must be done to stabilise? A. GM = 2.5 dB; the gain can be increased B. GM = infinity C. GM = -2.5 dB (0.75); the gain must be reduced below 1.5 D. GM = 1.333 dB; the gain must be doubled Answer: C. GM = 1/1.333 = 0.75 = -2.5 dB, so loop gain 2 must fall below 2 x 0.75 = 1.5. 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