--- title: "Nyquist stability of 50/[(s+1)(s+2)] and phase margin" subject: "Control Systems" chapter: 6 key: "semester_3/control_systems:nyquist-50-s1-s2-pm" source: https://www.bctengineer.com/q/semester_3/control_systems/nyquist-50-s1-s2-pm site: BCT Engineering --- # Nyquist stability of 50/[(s+1)(s+2)] and phase margin Control Systems · Chapter 6: Frequency Response Techniques - Asked in: 2079 Bhadra, 8+2 marks; 2067 Asar (old course), 8 marks; 2080 Bhadra, 6+2 marks (asked 3 times) - Video: [Nyquist Stability Criteria Part 1](https://www.youtube.com/watch?v=qJXoZ2AE_bI) — TutorialsPoint - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_3/control_systems/nyquist-50-s1-s2-pm - Whole chapter: https://www.bctengineer.com/md/semester_3/control_systems/chapter-6 ## Question Using Nyquist criteria, determine the stability of the closed loop system whose open loop transfer function is given by G(s)H(s) = 50/[(s+1)(s+2)]; also find the phase margin. ## Short answer The open-loop has no RHP poles (P=0) and the Nyquist plot never encircles -1 (N=0), so the closed-loop system is stable. Phase margin is 24.4° at gain crossover 6.89 rad/s, and gain margin is infinite. ## Answer $$ G(s)H(s) = \frac{50}{(s+1)(s+2)}, \qquad G(j\omega) = \frac{50}{\sqrt{(1+\omega^2)(4+\omega^2)}}\angle\left(-\tan^{-1}\omega - \tan^{-1}\tfrac{\omega}{2}\right) $$ #### Open-loop poles $-1$ and $-2$, so **P = 0**. There are no poles on the $j\omega$-axis. #### Polar (ω: 0 → ∞) data | ω | \|G\| | ∠G | Re | Im | |---|---|---|---|---| | 0 | 25 | 0° | 25 | 0 | | 1 | 15.81 | −71.6° | 5.0 | −15.0 | | 2 | 7.91 | −108.4° | −2.5 | −7.5 | | 5 | 1.82 | −146.9° | −1.53 | −0.99 | | 6.89 | 1.00 | −155.6° | −0.91 | −0.41 | | 10 | 0.49 | −163.0° | −0.47 | −0.14 | | ∞ | 0 | −180° | 0 | 0 | - The plot starts at 25 on the positive real axis, goes through the 4th and 3rd quadrants, and reaches the origin along $-180^\circ$. - The phase reaches $-180^\circ$ only at ω = ∞, so the plot **never cuts the negative real axis** at a finite point. - The $\omega < 0$ part is its mirror image. The infinite semicircle of the s-plane maps to the origin. ```text | ................ | ...... ....... .... .... ...| .... .. | .. .. | ... .. | .. .. | . -----*@**-------------------------------------*--- ** | * ** | ** ** | *** ** | ** ***| **** **** **** | ****** ******* | **************** 0 25 * w > 0 . w < 0 (mirror) @ point (-1, j0) w = 0 at 25; ends at origin along -180 deg ``` #### Nyquist stability **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 closed curve meets the real axis only at the origin and at $+25$, so the point $-1$ lies outside it and $N = 0$. $$ Z = P + N = 0 + 0 = 0 $$ The **closed-loop system is stable** for this gain (and in fact for any $K > 0$, because the plot never crosses the negative real axis, so GM = ∞). #### Phase margin Gain crossover: $|G| = 1$: $$ \begin{aligned} (1+\omega^2)(4+\omega^2) &= 2500 \\ \omega^4 + 5\omega^2 - 2496 &= 0 \\ \omega^2 &= \frac{-5 + \sqrt{25 + 9984}}{2} = 47.52 \\ \omega_{gc} &= 6.894\ \text{rad/s} \end{aligned} $$ $$ \begin{aligned} \angle G(j\omega_{gc}) &= -\tan^{-1}6.894 - \tan^{-1}3.447 = -81.75^\circ - 73.82^\circ = -155.57^\circ \\ PM &= 180^\circ - 155.57^\circ = 24.43^\circ \end{aligned} $$ **Answer:** N = 0, P = 0, so Z = 0 and the closed loop is **stable**. **PM ≈ 24.4°** at $\omega_{gc}$ = 6.89 rad/s (GM = ∞). ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. Why does the polar plot of 50/[(s+1)(s+2)] never cut the negative real axis at a finite frequency? A. The magnitude is always above 1 B. The phase only reaches -180 degrees as omega tends to infinity C. The poles are in the RHP D. The plot is a straight line Answer: B. Two simple poles give a total phase that approaches but never reaches -180 degrees, so the plot reaches the negative real axis only at the origin; hence GM is infinite. (Key point: Real-axis crossing) 2. With P = 0 and -1 outside the plot, what is the closed-loop conclusion? A. N = 0, Z = 0, stable B. N = 2, Z = 2, unstable C. N = 1, Z = 1, unstable D. N = 0, Z = 2, unstable Answer: A. No encirclement of (-1, j0) gives N = 0, and Z = P + N = 0, so the system is stable. (Key point: Stability by Nyquist) 3. How is the phase margin obtained, and what is it? A. Find the omega where angle = -180; PM = 0 B. Find omega where |G| = 1 (6.89 rad/s), then PM = 180 + angle = about 24.4 degrees C. Find omega where |G| = 1, then PM = angle = -155.6 degrees D. Use omega = 2 rad/s; PM = 71.6 degrees Answer: B. |G| = 1 gives omega_gc = 6.89 rad/s where the angle is -155.57 degrees, so PM = 180 - 155.57 = 24.43 degrees. 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