--- title: "Polar and Nyquist plot of K/[s(s+1)(0.1s+1)]" subject: "Control Systems" chapter: 6 key: "semester_3/control_systems:polar-nyquist-k-s-s1-01s1" source: https://www.bctengineer.com/q/semester_3/control_systems/polar-nyquist-k-s-s1-01s1 site: BCT Engineering --- # Polar and Nyquist plot of K/[s(s+1)(0.1s+1)] Control Systems · Chapter 6: Frequency Response Techniques - Asked in: 2071 Shrawan, 3+3+2 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/polar-nyquist-k-s-s1-01s1 - Whole chapter: https://www.bctengineer.com/md/semester_3/control_systems/chapter-6 ## Question Construct the polar plot of unity feedback system with G(s) = K/[s(s+1)(0.1s+1)]. Then, upgrade the plot to make it Nyquist plot. Hence find range of k for stable operation. ## Short answer Polar plot crosses negative real axis at –K/11 (ω = √10 ≈ 3.16 rad/s). With P = 0, Nyquist plot must not encircle –1, requiring –K/11 > –1. The closed-loop system is stable for 0 < K < 11. ## Answer The **polar plot** is the locus of $G(j\omega)$ in the complex plane as $\omega$ goes from 0 to $\infty$. The **Nyquist plot** adds the mirror image for $-\infty<\omega<0$ and the map of the small indentation around the pole at the origin, so that the whole Nyquist contour is mapped. #### Polar plot $$ G(j\omega)=\frac{K}{j\omega(1+j\omega)(1+j0.1\omega)} $$ $$ |G(j\omega)|=\frac{K}{\omega\sqrt{1+\omega^2}\sqrt{1+0.01\omega^2}},\qquad \angle G(j\omega)=-90^\circ-\tan^{-1}\omega-\tan^{-1}0.1\omega $$ | $\omega$ | Magnitude | Phase | |---|---|---| | $0^+$ | $\infty$ | $-90^\circ$ | | $\sqrt{10}=3.162$ | $K/11$ | $-180^\circ$ | | $\infty$ | 0 | $-270^\circ$ | - **Low-frequency asymptote:** as $\omega\to0$, $\text{Re}\,G \to -K(T_1+T_2) = -K(1+0.1) = -1.1K$. The plot starts at infinity parallel to the negative imaginary axis, along the line $\text{Re}=-1.1K$. - **Negative real axis crossing:** set phase $=-180^\circ$: $\tan^{-1}\omega+\tan^{-1}0.1\omega=90^\circ \Rightarrow 0.1\omega^2=1 \Rightarrow \omega_{pc}=\sqrt{10}=3.162$ rad/s. $$ \begin{aligned} |G(j\omega_{pc})| &= \frac{K}{3.162\times\sqrt{11}\times\sqrt{1.1}} = \frac{K}{11} \end{aligned} $$ So the plot crosses the negative real axis at $-K/11$ and ends at the origin, tangent to the positive imaginary axis ($-270^\circ$). ```text Im | -K/11 | ----o-------+--------> Re / \_____/| | w=3.16 | | | | w ->0+ (to -inf j, | asymptote Re=-1.1K) ``` #### Nyquist plot 1. Draw the polar plot for $\omega = 0^+ \to \infty$ (third quadrant, crosses at $-K/11$, ends at origin). 2. Draw its mirror image about the real axis for $\omega=-\infty \to 0^-$. 3. The infinite semicircle of the Nyquist contour ($s=Re^{j\theta}$, $R\to\infty$) maps to the origin. 4. The small semicircle around the pole at $s=0$ ($s=\epsilon e^{j\theta}$, $\theta: -90^\circ\to+90^\circ$) maps to $\frac{K}{\epsilon}e^{-j\theta}$, i.e. an **infinite-radius semicircle traversed clockwise** from $\omega=0^-$ to $\omega=0^+$ through the positive real axis. The resulting closed curve is the Nyquist plot. It cuts the negative real axis only at $-K/11$. #### Range of K for stability Nyquist criterion: $Z = P + N$, where $P$ = open-loop RHP poles, $N$ = clockwise encirclements of $-1+j0$. - $P = 0$ (poles at $0,-1,-10$; the origin pole is excluded by the indentation). - For stability, $Z=0$, so $N=0$: the point $-1$ must not be enclosed, i.e. it must lie to the left of the crossing point. $$ \frac{K}{11} < 1 \;\Rightarrow\; K < 11 $$ If $K>11$, the curve encircles $-1$ twice clockwise ($N=2$, $Z=2$): unstable. At $K=11$ the system oscillates at $\omega=3.162$ rad/s. Check by Routh: $0.1s^3+1.1s^2+s+K=0$ gives $1.1\times1 > 0.1K \Rightarrow K<11$. **Answer: The closed-loop system is stable for $0 < K < 11$.** ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. What must be added to the polar plot (w from 0 to infinity) to obtain the Nyquist plot? A. Only the mirror image for negative frequencies B. The mirror image for negative frequencies and the map of the indentation around the pole at the origin C. Only the unit circle D. A shift of the plot by -1 Answer: B. The Nyquist plot maps the whole contour: the mirror image for w < 0 plus the infinite arc from the small indentation at s = 0. (Key point: Polar vs Nyquist plot) 2. For G(s) = K/[s(s+1)(0.1s+1)], where does the phase equal -180 degrees? A. w = 1 rad/s B. w = 10 rad/s C. w = sqrt(10) = 3.16 rad/s D. w = 0.316 rad/s Answer: C. tan^-1 w + tan^-1 0.1w = 90 degrees requires 0.1 w^2 = 1, so w = sqrt(10). (Key point: Phase crossover frequency) 3. Where does the plot cross the negative real axis? A. -K/11 B. -K/10 C. -1.1K D. -11K Answer: A. At w = 3.162 the magnitude is K/(3.162 x sqrt(11) x sqrt(1.1)) = K/11. (Key point: Crossing point on the real axis) 4. What condition on K gives a stable closed loop, and what does K greater than 11 cause? A. K < 11; the curve encircles -1 twice clockwise B. K > 11; the curve encircles -1 once anticlockwise C. K < 1.1; the curve passes through -1 D. K < 11; the curve encircles -1 once clockwise Answer: A. Stability needs the crossing -K/11 to lie to the right of -1, so K < 11; for K > 11, N = 2 and Z = 2. 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