--- title: "Polar plot of 10/[s(s+1)²] with GM and PM" subject: "Control Systems" chapter: 6 key: "semester_3/control_systems:polar-10-s-s1sq" source: https://www.bctengineer.com/q/semester_3/control_systems/polar-10-s-s1sq site: BCT Engineering --- # Polar plot of 10/[s(s+1)²] with GM and PM Control Systems · Chapter 6: Frequency Response Techniques - Asked in: 2080 Bhadra, 8 marks; 2068 Baisakh (old course), 8 marks; 2078 Kartik, 8 marks (asked 3 times) - Video: [3. Draw Polar Plot, Calculate Phase Cross-over Frequency, Gain Margin & Comment on Stability.](https://www.youtube.com/watch?v=RvqDLSEsqZI) — ELECTRICAL HUB - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_3/control_systems/polar-10-s-s1sq - Whole chapter: https://www.bctengineer.com/md/semester_3/control_systems/chapter-6 ## Question Sketch an approximate polar plot for a unity feedback system with a feedforward transfer function: G(s) = 10/[s(s+1)²]. And obtain (i) GM (ii) PM (iii) gcf (iv) pcf and (v) Stability. ## Short answer Phase crossover occurs at 1 rad/s with gain margin 0.2 (-13.98 dB). Gain crossover at 2 rad/s gives phase margin -36.87°. The closed-loop system is unstable with two RHP poles. ## Answer $$ G(s) = \frac{10}{s(s+1)^2}, \qquad G(j\omega) = \frac{10}{\omega(1+\omega^2)}\angle\left(-90^\circ - 2\tan^{-1}\omega\right) $$ #### Key points of the polar plot | ω | \|G\| | ∠G | Re | Im | |---|---|---|---|---| | 0⁺ | ∞ | −90° | −20 (asymptote) | −∞ | | 0.2 | 48.1 | −112.6° | −18.5 | −44.4 | | 0.5 | 16.0 | −143.1° | −12.8 | −9.6 | | 1 | 5.0 | −180° | −5.0 | 0 | | 2 | 1.0 | −216.9° | −0.8 | 0.6 | | 5 | 0.077 | −247.4° | −0.03 | 0.07 | | ∞ | 0 | −270° | 0 | 0 | - **Low-frequency asymptote:** $\text{Re}\,G \to -K\sum T = -10(1+1) = -20$. The plot starts at $\infty\angle -90^\circ$ along the line Re = −20. - **Real-axis crossing:** $\angle G = -180^\circ \Rightarrow 2\tan^{-1}\omega = 90^\circ \Rightarrow \omega = 1$, where $|G| = 10/(1 \times 2) = 5$. The plot cuts the negative real axis at **−5**. - **High frequency:** the plot reaches the origin along $-270^\circ$ (from the second quadrant, tangent to the +j axis). ```text . | .. | ... | .. | ... | ... | ... | .... **** | -------------------------------*******--@**------- **** .... | *** | *** | *** | ** | *** | ** | * | -20 -5 0 * w > 0 . w < 0 (mirror) @ point (-1, j0) crosses real axis at -5 (w = 1); asymptote Re = -20 ``` #### (iv) Phase crossover frequency (pcf) $\omega_{pc} = 1$ rad/s. #### (i) Gain margin $$ GM = \frac{1}{|G(j\omega_{pc})|} = \frac{1}{5} = 0.2 = 20\log 0.2 = -13.98\ \text{dB} $$ #### (iii) Gain crossover frequency (gcf) $$ |G| = 1 \Rightarrow \omega(1 + \omega^2) = 10 \Rightarrow \omega^3 + \omega - 10 = 0 \Rightarrow \omega_{gc} = 2\ \text{rad/s} $$ (Check: $8 + 2 = 10$.) #### (ii) Phase margin $$ \begin{aligned} \angle G(j2) &= -90^\circ - 2\tan^{-1}2 = -90^\circ - 126.87^\circ = -216.87^\circ \\ PM &= 180^\circ + \angle G(j\omega_{gc}) = -36.87^\circ \end{aligned} $$ #### (v) Stability GM < 0 dB and PM < 0. The polar plot crosses the real axis at −5, to the left of −1, so the $(-1, j0)$ point is enclosed. With P = 0, the complete Nyquist plot encircles −1 twice clockwise, so Z = 2. The **closed-loop system is unstable**, with two RHP poles. Check with Routh: $s^3 + 2s^2 + s + 10$ gives $2 \times 1 < 10$, so it is unstable. For stability $K$ must be below 2, i.e. the gain must be reduced by a factor of 5. **Answer:** GM = **−13.98 dB**, PM = **−36.87°**, gcf = **2 rad/s**, pcf = **1 rad/s**. The system is **unstable**. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. How is the phase crossover frequency of G(s) = 10/[s(s+1)^2] found and what is it? A. Set |G| = 1; omega = 2 rad/s B. Set the angle to -180 degrees, i.e. 2 tan^-1(omega) = 90 degrees; omega = 1 rad/s C. Set the angle to -90 degrees; omega = 0 D. Set the angle to -180 degrees; omega = 2 rad/s Answer: B. Angle = -90 - 2 tan^-1(omega) = -180 requires tan^-1(omega) = 45 degrees, so omega = 1 rad/s. (Key point: Phase crossover frequency) 2. What is the gain margin? A. 5 (13.98 dB) B. 0.2 (-13.98 dB), because |G(j1)| = 5 C. 10 (20 dB) D. 0.5 (-6 dB) Answer: B. At omega = 1, |G| = 10/(1 x 2) = 5, so GM = 1/5 = 0.2, which is -13.98 dB. (Key point: Gain margin) 3. What are the gain crossover frequency and phase margin? A. omega_gc = 2 rad/s, PM = -36.87 degrees B. omega_gc = 1 rad/s, PM = 0 degrees C. omega_gc = 2 rad/s, PM = +36.87 degrees D. omega_gc = 5 rad/s, PM = -36.87 degrees Answer: A. |G| = 1 gives omega^3 + omega - 10 = 0, so omega = 2. The angle there is -216.87 degrees, so PM = 180 - 216.87 = -36.87 degrees. (Key point: Gain crossover and phase margin) 4. Why is the closed-loop system unstable? A. The plot crosses the real axis at -5, left of -1, enclosing -1; GM and PM are both negative B. The plot has an asymptote at -20 C. P = 2 open-loop poles are in the RHP D. The plot reaches the origin along -270 degrees Answer: A. The plot cuts the negative real axis at -5, so (-1, j0) is encircled; with P = 0 this gives Z = 2, and the negative margins agree. (Key point: Stability conclusion) ## Related in Control Systems - [Transfer function with G3 feedforward, H1 and unity loop](https://www.bctengineer.com/question.md?key=semester_3%2Fcontrol_systems%3Abdr-g1g2g3-h1-unity-2070o) — Control Systems, chapter 3 - [Nyquist stability of 10/[s(s²+2s+4)] and GM](https://www.bctengineer.com/question.md?key=semester_3%2Fcontrol_systems%3Anyquist-10-s-s2-2s-4) — Control Systems, chapter 6 - [GM and PM from Bode plot of 1/[s(0.1s+1)(0.2s+1)]](https://www.bctengineer.com/question.md?key=semester_3%2Fcontrol_systems%3Abode-gm-pm-1-s-01s) — Control Systems, chapter 6 - [Lead compensator for k/[s²(1+0.25s)], Ka = 10, PM 35°](https://www.bctengineer.com/question.md?key=semester_3%2Fcontrol_systems%3Alead-k-s2-025s-ka10-pm35) — Control Systems, chapter 7, asked 5× - [Nyquist plot of (s+10)/[(s−3)(s+3)] and gain margin](https://www.bctengineer.com/question.md?key=semester_3%2Fcontrol_systems%3Anyquist-s10-s-3-s3) — Control Systems, chapter 6, asked 2× - [Stability of 56/[s(s⁴+7s³+6s²+42s+8)] feedback system](https://www.bctengineer.com/question.md?key=semester_3%2Fcontrol_systems%3Astability-56-s5) — Control Systems, chapter 4 ## Related in other subjects - [Bode design: PM 55°, GM 10 dB, Kv = 5](https://www.bctengineer.com/question.md?key=semester_6%2Fdigital_control_system%3Abode-design-phase-margin) — Digital Control System, chapter 4 - [DT transfer function; drawing a Bode plot](https://www.bctengineer.com/question.md?key=semester_4%2Fsignals_and_systems%3Adt-transfer-function-bode) — Signals and Systems, chapter 6 - [Excitation system with stabilizing transformer: transfer function](https://www.bctengineer.com/question.md?key=semester_7%2Fpower_plant_equipment%3Aexcitation-stabilizing-transformer-model) — Power Plant Equipment, chapter 3, asked 13× - [Function of excitation system and its transfer function](https://www.bctengineer.com/question.md?key=semester_7%2Fpower_plant_equipment%3Aexcitation-function-tf) — Power Plant Equipment, chapter 3, asked 2×