--- title: "Nyquist stability of (4s+1)/[s²(s+1)(2s+1)]" subject: "Control Systems" chapter: 6 key: "semester_3/control_systems:nyquist-4s1-s2-s1-2s1" source: https://www.bctengineer.com/q/semester_3/control_systems/nyquist-4s1-s2-s1-2s1 site: BCT Engineering --- # Nyquist stability of (4s+1)/[s²(s+1)(2s+1)] Control Systems · Chapter 6: Frequency Response Techniques - Asked in: 2074 Asoj, 8 marks; 2079 Bhadra, 8 marks (asked 2 times) - 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-4s1-s2-s1-2s1 - Whole chapter: https://www.bctengineer.com/md/semester_3/control_systems/chapter-6 ## Question The open loop transfer function of a control system is G(s)H(s) = (4s+1)/[s²(s+1)(2s+1)]. Using Nyquist criterion, determine the open loop and closed loop stability of this system. ## Short answer Open-loop has no RHP poles (P=0), but double integrator. Nyquist plot crosses real at -10.67 and with the infinite semicircle encircles -1 twice clockwise (N=2). Thus Z=2, closed-loop unstable. ## Answer $$ G(s)H(s) = \frac{4s+1}{s^2(s+1)(2s+1)} $$ #### Open-loop stability Open-loop poles: $0, 0, -1, -0.5$. None lie in the RHP, so **P = 0**. However, the **double pole at the origin** makes the open-loop system itself not stable: its impulse response contains a term $t$ that grows without bound. In Nyquist terms we take P = 0 and indent the contour around the origin. #### Mapping the Nyquist contour $$ \angle G(j\omega) = -180^\circ + \tan^{-1}4\omega - \tan^{-1}\omega - \tan^{-1}2\omega $$ | ω | \|G\| | ∠G | Re | Im | |---|---|---|---|---| | 0.1 | 105.1 | −175.2° | −104.7 | −8.76 | | 0.2 | 29.1 | −174.5° | −29.0 | −2.82 | | 0.354 | 10.66 | −180° | −10.66 | 0 | | 0.5 | 5.66 | −188.1° | −5.60 | +0.80 | | 1 | 1.30 | −212.5° | −1.10 | +0.70 | | 2 | 0.22 | −236.5° | −0.12 | +0.18 | | ∞ | 0 | −270° | 0 | 0 | 1. **ω = 0⁺ → ∞:** the plot starts at infinity just below the negative real axis (angle slightly above −180°). It cuts the negative real axis where $$ \tan^{-1}4\omega = \tan^{-1}\omega + \tan^{-1}2\omega \Rightarrow 4\omega = \frac{3\omega}{1 - 2\omega^2} \Rightarrow \omega_{pc} = \frac{1}{\sqrt8} = 0.354\ \text{rad/s} $$ $$ |G(j0.354)| = \frac{\sqrt{1+2}}{0.125\sqrt{1.125}\sqrt{1.5}} = 10.67 $$ So the crossing is at **−10.67**. The plot then enters the second quadrant and reaches the origin along −270°. 2. **ω = −∞ → 0⁻:** the mirror image. 3. **Infinite semicircle:** maps to the origin. 4. **Small semicircle around the double pole:** $G \approx 1/s^2 = (1/\varepsilon^2)e^{-j2\theta}$. As θ goes from −90° to +90°, $G$ traces a full **360° clockwise circle of infinite radius**, from +180° through 0° to −180°. ```text | | | | | *************** | ... ******** **** | ....... ******* **| --------*******---------------------------@--**--- ******* ....... ..| *** ........ .... | ............... | | | | | | -10.67 0 * w > 0 . w < 0 (mirror) @ point (-1, j0) crosses at -10.67 (w = 0.354) ``` (An infinite clockwise circle joins the ω = 0⁻ and ω = 0⁺ ends.) #### Closed-loop 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 point −1 lies between the crossing at −10.67 and the origin. Following the plot together with the infinite 360° clockwise arc, the point (−1, j0) is encircled **twice clockwise**: N = 2. $$ Z = P + N = 0 + 2 = 2 $$ The **closed-loop system is unstable**, with 2 poles in the RHP. **Check (Routh):** $2s^4 + 3s^3 + s^2 + 4s + 1 = 0$ | Row | Col 1 | Col 2 | Col 3 | |---|---|---|---| | $s^4$ | 2 | 1 | 1 | | $s^3$ | 3 | 4 | | | $s^2$ | −1.667 | 1 | | | $s^1$ | 5.8 | | | | $s^0$ | 1 | | | There are two sign changes, so there are 2 RHP poles (they are at $0.26 \pm j1.02$). ✓ **Answer:** Open loop: no RHP poles (P = 0), but the repeated pole at the origin makes it **unstable in the BIBO sense**. Closed loop: N = 2, Z = 2, so it is **unstable**. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. G(s)H(s) = (4s+1)/[s^2(s+1)(2s+1)] has poles 0, 0, -1, -0.5. What is P used in the Nyquist criterion? A. P = 2 (the double pole at the origin) B. P = 0, since no pole is in the RHP C. P = 4 D. P = 1 Answer: B. No poles are strictly in the RHP, so P = 0; the origin poles are handled by indenting the contour. (Key point: Open-loop P) 2. How does the small semicircle around the double pole at the origin map? A. Into a 180 degree arc counter-clockwise B. Into a point at the origin C. Into a 360 degree infinite-radius circle traced clockwise, since G is about 1/s^2 D. Into a 90 degree arc Answer: C. With G about (1/eps^2)e^(-j2 theta) and theta from -90 to +90 degrees, the angle sweeps 360 degrees clockwise. (Key point: Small semicircle at a double pole) 3. Where does the plot cut the negative real axis? A. At -1.0 at w = 1 B. At -10.67 at w = 0.354 rad/s C. At -0.5 at w = 0.5 D. It never cuts it Answer: B. Setting the phase to -180 gives 4w = 3w/(1 - 2w^2), so w = 1/sqrt8 = 0.354, where |G| = 10.67. (Key point: Real-axis crossing) 4. Since -1 lies between the crossing at -10.67 and the origin, what is the result? A. N = 0, Z = 0, stable B. N = 2 clockwise, Z = 2, unstable C. N = 1, Z = 1, unstable D. N = -2, Z = -2 Answer: B. The point is encircled twice clockwise, so Z = 0 + 2 = 2 RHP closed-loop poles; Routh on 2s^4 + 3s^3 + s^2 + 4s + 1 shows two sign changes. (Key point: Closed-loop stability) ## Close variants set in other exams - [Nyquist stability of 2/[s(s+1)(2s+1)]](https://www.bctengineer.com/question.md?key=semester_3%2Fcontrol_systems%3Anyquist-2-s-s1-2s1) — Control Systems, chapter 6, asked 3× ## Related in Control Systems - [Nyquist stability of 1/[s(1+2s)(1+s)] and gain margin](https://www.bctengineer.com/question.md?key=semester_3%2Fcontrol_systems%3Anyquist-1-s-1-2s-1) — Control Systems, chapter 6, asked 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 - [R-H stability of s⁵+2s⁴+2s³+4s²+s+2](https://www.bctengineer.com/question.md?key=semester_3%2Fcontrol_systems%3Arh-s5-2-2-4-1) — Control Systems, chapter 4 - [Traffic light system of Kathmandu: open or closed loop](https://www.bctengineer.com/question.md?key=semester_3%2Fcontrol_systems%3Atraffic-light-justify) — Control Systems, chapter 1 - [Control system block diagram and closed loop advantages](https://www.bctengineer.com/question.md?key=semester_3%2Fcontrol_systems%3Awhat-is-cs-bd-cl-advantages) — Control Systems, chapter 1 - [Closed loop control system block diagram explained](https://www.bctengineer.com/question.md?key=semester_3%2Fcontrol_systems%3Awhat-is-cs-bd-closed-loop) — Control Systems, chapter 1, asked 4× ## Related in other subjects - [Open loop versus closed loop control](https://www.bctengineer.com/question.md?key=semester_6%2Fembedded_system%3Aopen-vs-closed-loop) — Embedded System, chapter 7, asked 6× - [Steps to design closed loop control system](https://www.bctengineer.com/question.md?key=semester_6%2Fembedded_system%3Aclosed-loop-design-steps) — Embedded System, chapter 7 - [Closed loop better than open loop (cruise)](https://www.bctengineer.com/question.md?key=semester_6%2Fembedded_system%3Aclosed-loop-better-cruise-integral) — Embedded System, chapter 7 - [Open loop and closed loop V/f inverter-fed drive](https://www.bctengineer.com/question.md?key=semester_7%2Futilization_of_electrical_energy%3Aclosed-loop-vf-inverter-drive) — Utilization of Electrical Energy, chapter 3