--- title: "Nyquist stability of 1/[s²(1+2s)(1+s)]" subject: "Control Systems" chapter: 6 key: "semester_3/control_systems:nyquist-1-s2-1-2s-1" source: https://www.bctengineer.com/q/semester_3/control_systems/nyquist-1-s2-1-2s-1 site: BCT Engineering --- # Nyquist stability of 1/[s²(1+2s)(1+s)] Control Systems · Chapter 6: Frequency Response Techniques - Asked in: 2078 Bhadra, 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-1-s2-1-2s-1 - Whole chapter: https://www.bctengineer.com/md/semester_3/control_systems/chapter-6 ## Question Using Nyquist criteria, determine the stability of the feedback system whose OLTF is given by G(s)H(s) = 1/[s²(1+2s)(1+s)]. ## Short answer P=0, type-2 system. Infinite clockwise arc at origin plus plot yield two clockwise encirclements of –1 (N=2). Z=2, closed-loop unstable for any positive gain. ## Answer #### Open-loop data $$ G(s)H(s)=\frac{1}{s^2(1+2s)(1+s)} $$ Poles: double pole at $s=0$ (type-2), and $-0.5$, $-1$. No poles in the RHP: $P=0$. The double pole at the origin is bypassed by a small semicircle in the Nyquist contour. #### Section I: $\omega=0^+\to\infty$ $$ |GH|=\frac{1}{\omega^2\sqrt{1+4\omega^2}\sqrt{1+\omega^2}},\qquad \angle GH=-180^\circ-\tan^{-1}2\omega-\tan^{-1}\omega $$ | $\omega$ | $0^+$ | 0.2 | 0.4 | 0.707 | 1 | 2 | $\infty$ | |---|---|---|---|---|---|---|---| | $|GH|$ | $\infty$ | 22.8 | 4.53 | 0.943 | 0.316 | 0.027 | 0 | | $\angle GH$ | $-180^\circ$ | $-213^\circ$ | $-240^\circ$ | $-270^\circ$ | $-288^\circ$ | $-319^\circ$ | $-360^\circ$ | - The phase is always below $-180^\circ$, so the plot lies in the **second quadrant** at low frequency. For small $\omega$: $GH\approx-\dfrac{1}{\omega^2}+j\dfrac{3}{\omega}$ (real part $\to-\infty$, imaginary part $\to+\infty$). - It cuts the **positive imaginary axis** where $\tan^{-1}2\omega+\tan^{-1}\omega=90^\circ\Rightarrow2\omega^2=1\Rightarrow\omega=0.707$, at $GH=+j0.943$. - It then enters the first quadrant and reaches the origin at $-360^\circ$. - It never cuts the negative real axis for finite $\omega>0$. #### Section II: infinite semicircle $s=Re^{j\theta}$, $R\to\infty$: $GH\to0$, maps to the origin. #### Section III: $\omega=-\infty\to0^-$ Mirror image of section I (third and fourth quadrants). #### Section IV: indentation at the origin $s=\epsilon e^{j\theta}$, $\theta:-90^\circ\to+90^\circ$: $$ GH\approx\frac{1}{\epsilon^2}e^{-j2\theta} $$ Angle goes from $+180^\circ$ to $-180^\circ$: a **circle of infinite radius, 360° clockwise** (two half circles), through the positive real axis. ```text Im w>0 | j0.943 (from -inf|+j inf) \_______|__ | \_ to origin ----x------+------ Re -1 | _/ _______|_/ / w<0 | plus a full infinite clockwise circle at w=0 ``` #### Nyquist criterion The $-1$ point lies between the $\omega>0$ branch (above it) and the $\omega<0$ branch (below it), and the 360° infinite clockwise arc closes the curve around it. So $-1+j0$ is **encircled twice clockwise**: $N=2$. $$ Z=P+N=0+2=2 $$ The closed-loop system has two poles in the RHP and is **unstable**. Check by Routh: $1+GH=0\Rightarrow 2s^4+3s^3+s^2+1=0$. The $s$ term is missing, so the system cannot be stable; the roots are $-1.04\pm j0.50$ and $0.29\pm j0.54$ (two in the RHP). A type-2 system with only extra lags needs a zero (lead action) to be stabilised. **Answer: P = 0, N = 2, Z = 2 – the closed-loop system is unstable for any positive gain.** ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. For the double pole at the origin, how does the small semicircle of the contour map? A. To a small circle near the origin B. To an infinite circle traversed 360 degrees clockwise C. To an infinite semicircle traversed anticlockwise D. To the point -1 Answer: B. GH is about (1/eps^2) e^(-j2 theta), and theta runs through 180 degrees, so the angle sweeps 360 degrees clockwise at infinite radius. (Key point: Indentation for a double pole) 2. Why does the plot lie in the second quadrant at low frequency? A. The phase is always above -180 degrees B. The phase starts at -180 degrees and goes below it, due to the extra lags C. The magnitude is zero D. The gain is negative Answer: B. Phase = -180 - tan^-1 2w - tan^-1 w is always below -180 degrees, putting the plot in the second quadrant. (Key point: Low-frequency quadrant) 3. Where does the plot cross the imaginary axis? A. At w = 1, GH = j0.316 B. At w = 0.707, GH = +j0.943 C. At w = 0.707, GH = -j0.943 D. It crosses the negative real axis at w = 1 Answer: B. Phase -270 degrees needs 2w^2 = 1, so w = 0.707, where GH = +j0.943; it never crosses the negative real axis. (Key point: Axis crossing) 4. What are N and Z, and what is the conclusion? A. N = 0, Z = 0, stable B. N = 1, Z = 1, unstable C. N = 2, Z = 2, unstable for any positive gain D. N = 2, Z = 0, stable Answer: C. -1 is encircled twice clockwise; with P = 0, Z = 2 so the system is unstable for any positive gain. 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