--- title: "Nyquist contour mapping of 200/[(s+3)(s+1)(s+2)]" subject: "Control Systems" chapter: 6 key: "semester_3/control_systems:nyquist-contour-200" source: https://www.bctengineer.com/q/semester_3/control_systems/nyquist-contour-200 site: BCT Engineering --- # Nyquist contour mapping of 200/[(s+3)(s+1)(s+2)] Control Systems · Chapter 6: Frequency Response Techniques - Asked in: 2080 Baisakh, 8 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-contour-200 - Whole chapter: https://www.bctengineer.com/md/semester_3/control_systems/chapter-6 ## Question What is Nyquist Contour? Map the Nyquist contour of open loop transfer function G(s) = 200/[(s+3)(s+1)(s+2)] into G(s) plane and apply Nyquist criterion to check the stability of closed loop system. ## Short answer The Nyquist contour encloses the RHP and maps G(s) to a plot crossing at –3.33, giving two clockwise encirclements of –1. With P=0, N=2, so Z=2 and the closed-loop system is unstable. ## Answer #### Nyquist contour The **Nyquist contour** is a closed path in the s-plane that encloses the entire right half-plane. It consists of: 1. the whole $j\omega$-axis from $-j\infty$ to $+j\infty$; 2. a semicircle of infinite radius ($s = Re^{j\theta}$, $R \to \infty$, θ from +90° to −90°), traversed **clockwise**; 3. small semicircular indentations (radius ε → 0) around any open-loop poles that lie on the $j\omega$-axis, so the path avoids them. Mapping this contour through $G(s)H(s)$ gives the Nyquist plot. By the argument principle, the encirclements of $(-1, j0)$ count the RHP zeros of $1 + G(s)H(s)$, i.e. the RHP closed-loop poles. #### Mapping $G(s) = \dfrac{200}{(s+1)(s+2)(s+3)}$ Open-loop poles: $-1, -2, -3$, so **P = 0**. No indentation is needed. $$ (j\omega+1)(j\omega+2)(j\omega+3) = (6 - 6\omega^2) + j(11\omega - \omega^3) $$ | Section | Mapping | |---|---| | ω = 0 | $G = 200/6 = 33.3\angle0^\circ$ | | ω = 1 | $20\angle -90^\circ$ (crosses the −j axis) | | ω = 2 | $8.77\angle -142.1^\circ$ | | ω = √11 = 3.317 | Im of denominator = 0, Re = $6 - 66 = -60$, so $G = -3.33$ | | ω = 5 | $1.25\angle -205.9^\circ$ (second quadrant) | | ω → +∞ | $0\angle -270^\circ$ | | ω < 0 | mirror image about the real axis | | infinite semicircle | maps to the origin | ```text |........................ ..... ...... ... | .... ... | ... .. | ... . | .. .. | .. .. | . ---*****@*-------------------------------------*-- ** | * ** | ** * | ** ** | *** *** | *** *** | **** ***** ****** |************************ -3.33 33.3 * w > 0 . w < 0 (mirror) @ point (-1, j0) w = 0 at 33.3; crosses at -3.33 (w = 3.317) ``` #### 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 **−3.33**, to the left of −1. The closed curve (ω from −∞ to +∞) goes round the point $(-1, j0)$ **twice clockwise**, so N = 2. $$ Z = P + N = 0 + 2 = 2 $$ The **closed-loop system is unstable**, with two closed-loop poles in the RHP. - Gain margin: $GM = 1/3.33 = 0.3 = -10.5$ dB (negative). - The gain would have to be reduced below $200/3.33 = 60$ for stability. - Check (Routh): $s^3 + 6s^2 + 11s + 206$ gives $6 \times 11 = 66 < 206$, so it is unstable. ✓ ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. What is the Nyquist contour? A. A closed s-plane path enclosing the entire right half-plane: the jw axis plus an infinite clockwise semicircle, with indentations around jw-axis poles B. A closed path enclosing only the left half-plane C. The locus of G(jw) for w from 0 to infinity D. A circle of unit radius around -1 Answer: A. The contour encloses the whole RHP so encirclements reveal RHP zeros of 1 + GH, i.e. unstable closed-loop poles. (Key point: Definition of Nyquist contour) 2. For G(s) = 200/[(s+1)(s+2)(s+3)], what are P and the indentation requirement? A. P = 3, indentation at each pole B. P = 1, indentation at -1 C. P = 0, no indentation needed D. P = 0, indentation at the origin Answer: C. All poles (-1, -2, -3) are in the LHP and none is on the jw axis. (Key point: P and indentation) 3. Where does the plot cross the negative real axis? A. At -3.33 when w = 3.317 rad/s B. At -33.3 when w = 0 C. At -1 when w = 1 D. At -60 when w = 3.317 Answer: A. The denominator is real when 11w - w^3 = 0, so w = sqrt(11); there the real part is 6 - 66 = -60 and G = 200/(-60) = -3.33. (Key point: Real-axis crossing) 4. The crossing at -3.33 is left of -1. What are N and Z? A. N = 0, Z = 0, stable B. N = 2 clockwise, Z = 2, unstable C. N = -2, Z = -2, stable D. N = 1, Z = 1, unstable Answer: B. The closed curve encircles (-1, j0) twice clockwise, so Z = 0 + 2 = 2 RHP closed-loop poles; GM = -10.5 dB agrees. 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