--- title: "Nyquist stability of 1/[s(1+2s)(1+s)] and gain margin" subject: "Control Systems" chapter: 6 key: "semester_3/control_systems:nyquist-1-s-1-2s-1" source: https://www.bctengineer.com/q/semester_3/control_systems/nyquist-1-s-1-2s-1 site: BCT Engineering --- # Nyquist stability of 1/[s(1+2s)(1+s)] and gain margin Control Systems · Chapter 6: Frequency Response Techniques - Asked in: 2078 Bhadra, 6+2 marks; 2075 Chaitra, 10 marks; 2081 Bhadra, 6+2+2 marks (asked 3 times) - 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-1-s-1-2s-1 - Whole chapter: https://www.bctengineer.com/md/semester_3/control_systems/chapter-6 ## Question A unity feedback system has open loop transfer function G(s) = 1/[s(1+2s)(1+s)]. Sketch Nyquist plot for the system and therefore obtain the gain margin. Check stability by Nyquist criterion. ## Short answer The Nyquist plot crosses the real axis at -0.667, yielding gain margin 1.5 (3.52 dB) at phase crossover 0.707 rad/s. With P=0 and no encirclements of -1, the system is stable. ## Answer $$ G(s) = \frac{1}{s(1+2s)(1+s)}, \qquad G(j\omega) = \frac{1}{\omega\sqrt{1+4\omega^2}\sqrt{1+\omega^2}}\angle\left(-90^\circ - \tan^{-1}2\omega - \tan^{-1}\omega\right) $$ #### Nyquist contour There is a pole at the origin, so the contour is indented around it by a small semicircle $s = \varepsilon e^{j\theta}$ with θ from −90° to +90°. Open-loop poles in the RHP: **P = 0**. #### Mapping of each section 1. **ω = 0⁺ to ∞ (positive $j\omega$-axis):** | ω | \|G\| | ∠G | Re | Im | |---|---|---|---|---| | 0.1 | 9.76 | −107.0° | −2.86 | −9.33 | | 0.3 | 2.74 | −137.7° | −2.02 | −1.84 | | 0.5 | 1.26 | −161.6° | −1.20 | −0.40 | | 0.707 | 0.667 | −180° | −0.667 | 0 | | 1 | 0.316 | −198.4° | −0.30 | 0.10 | | ∞ | 0 | −270° | 0 | 0 | - Low-frequency asymptote: $\text{Re} \to -(2+1) = -3$. The plot starts at $\infty\angle -90^\circ$ along Re = −3. - Phase crossover: $\tan^{-1}2\omega + \tan^{-1}\omega = 90^\circ \Rightarrow 2\omega^2 = 1 \Rightarrow \omega_{pc} = 0.707$ rad/s. - $|G(j0.707)| = \dfrac{1}{0.707 \times \sqrt3 \times \sqrt{1.5}} = 0.667$, so it crosses at **−0.667**. 2. **ω = −∞ to 0⁻:** mirror image of the above. 3. **Infinite semicircle:** $G \to 0$, so it maps to the origin. 4. **Small semicircle at origin:** $G \approx 1/s = (1/\varepsilon)e^{-j\theta}$. As θ goes from −90° to +90°, $G$ sweeps an **infinite-radius arc clockwise** from +90° through 0° to −90° (right half of the G-plane), joining the ω = 0⁻ end to the ω = 0⁺ end. ```text .. | .. | .. | .. | ... | .. | .... | ..... | -----------------------------@**********---------- ***** | **** | ** | *** | ** | ** | ** | ** | -3 -0.67 0 * w > 0 . w < 0 (mirror) @ point (-1, j0) crosses at -0.667 (w = 0.707); asymptote Re = -3 ``` (The infinite clockwise arc on the right closes the curve.) #### Gain margin $$ GM = \frac{1}{0.667} = 1.5 = 20\log1.5 = 3.52\ \text{dB} $$ The phase margin (for reference) is $11.4^\circ$ at $\omega_{gc} = 0.572$ rad/s. #### Stability by 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 crosses the negative real axis at −0.667, to the right of −1, so the point $(-1, j0)$ is **not encircled**: N = 0. $$ Z = P + N = 0 $$ The **closed-loop system is stable**. Check: $2s^3 + 3s^2 + s + 1$ has Routh first column $2, 3, 1/3, 1$, all positive. ✓ **Answer:** GM = **1.5 (3.52 dB)** at $\omega_{pc}$ = 0.707 rad/s. N = 0, P = 0, Z = 0, so the system is **stable**. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. G(s) has a pole at the origin. How is the Nyquist contour handled there? A. Indented with a small semicircle, which maps to an infinite-radius clockwise arc through the right half of the G-plane B. The pole is ignored C. Indented to the left so P = 1 D. The contour is a straight line through the origin Answer: A. Near s = 0, G is about 1/s, so the small semicircle maps to an infinite arc swept clockwise from +90 to -90 degrees, closing the curve. (Key point: Indentation at the origin) 2. How is the phase crossover frequency found and what is it? A. tan^-1(2w) + tan^-1(w) = 90 degrees gives 2w^2 = 1, w = 0.707 rad/s B. |G| = 1 gives w = 0.572 rad/s C. tan^-1(2w) = tan^-1(w) gives w = 0 D. 2w^2 = 3 gives w = 1.22 rad/s Answer: A. The total phase is -90 - tan^-1(2w) - tan^-1(w) = -180 when the two arctans sum to 90 degrees, i.e. 2w x w = 1. (Key point: Phase crossover) 3. The plot crosses the negative real axis at -0.667. What is the gain margin? A. 0.667 (-3.52 dB) B. 1.5 (3.52 dB) C. 3 (9.5 dB) D. 0.5 (-6 dB) Answer: B. GM = 1/|G(j w_pc)| = 1/0.667 = 1.5, which is 20 log 1.5 = 3.52 dB. (Key point: Gain margin) 4. What is the stability conclusion? A. Unstable, since the pole at the origin counts in P B. Unstable, N = 2 C. Stable, since -0.667 is to the right of -1, so N = 0 and Z = P + N = 0 D. Marginally stable Answer: C. The crossing at -0.667 is between -1 and 0, so -1 is not encircled. With P = 0, Z = 0 and the closed loop is stable (Routh check 2s^3 + 3s^2 + s + 1 agrees). 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