--- title: "Nyquist plot of 2.2/[s(s+1)(s²+2s+2)]" subject: "Control Systems" chapter: 6 key: "semester_3/control_systems:nyquist-22-s-s1-s2-2s" source: https://www.bctengineer.com/q/semester_3/control_systems/nyquist-22-s-s1-s2-2s site: BCT Engineering --- # Nyquist plot of 2.2/[s(s+1)(s²+2s+2)] Control Systems · Chapter 6: Frequency Response Techniques - Asked in: 2081 Baisakh, 8 marks - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_3/control_systems/nyquist-22-s-s1-s2-2s - Whole chapter: https://www.bctengineer.com/md/semester_3/control_systems/chapter-6 ## Question Draw the Nyquist plot for the open loop transfer function given below and comment on closed loop stability. G(s) = 2.2/[s(s+1)(s²+2s+2)] ## Short answer The plot crosses at –0.99, just right of –1, so N=0 and Z=P+N=0. The closed-loop system is stable, but only marginally (GM ≈ 0.09 dB, PM ≈ 0.5°). ## Answer $$ G(s) = \frac{2.2}{s(s+1)(s^2 + 2s + 2)} $$ Poles: $0, -1, -1 \pm j1$, so **P = 0**. The pole at the origin is indented. #### Frequency response $$ \angle G = -90^\circ - \tan^{-1}\omega - \tan^{-1}\frac{2\omega}{2 - \omega^2}, \qquad |G| = \frac{2.2}{\omega\sqrt{1 + \omega^2}\sqrt{(2 - \omega^2)^2 + 4\omega^2}} $$ | ω | \|G\| | ∠G | Re | Im | |---|---|---|---|---| | 0⁺ | ∞ | −90° | −2.2 | −∞ | | 0.1 | 10.95 | −101.5° | −2.17 | −10.73 | | 0.3 | 3.51 | −124.1° | −1.97 | −2.90 | | 0.5 | 1.95 | −146.3° | −1.62 | −1.08 | | 0.816 | 0.990 | −180° | −0.990 | 0 | | 1 | 0.696 | −198.4° | −0.66 | +0.22 | | 2 | 0.110 | −270° | 0 | +0.11 | | ∞ | 0 | −360° | 0 | 0 | - **Low-frequency asymptote:** $G = \dfrac{1.1}{s(1+s)(1 + s + 0.5s^2)}$, so Re → $-1.1(1 + 1) = -2.2$. - **Phase crossover:** $s^4 + 3s^3 + 4s^2 + 2s$ at $s = j\omega$ has imaginary part $-3\omega^3 + 2\omega = 0$, so $\omega_{pc} = \sqrt{2/3} = 0.816$ rad/s. The real part is $\omega^4 - 4\omega^2 = 0.444 - 2.667 = -2.222$, so $G = 2.2/(-2.222) = -0.990$. - The plot cuts the negative real axis at **−0.99**, then passes through the second quadrant (crossing the +j axis at ω = 2, where ∠G = −270°) and reaches the origin along −360°. ```text .. | . | .. | .. | .. | ... | ... | ..... **********| --------------------------**@****--------**------- ***** ..........| *** | *** | ** | ** | ** | * | ** | -2.2 -0.99 0 * w > 0 . w < 0 (mirror) @ point (-1, j0) crosses at -0.99 (w = 0.816), just right of -1 ``` #### 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 crossing at −0.99 lies just to the **right** of −1, so the point $(-1, j0)$ is not encircled: N = 0 and $Z = 0 + 0 = 0$. The **closed-loop system is stable, but only barely**: $$ GM = \frac{1}{0.99} = 1.0101 = 0.087\ \text{dB}, \qquad PM \approx 0.5^\circ \;(\omega_{gc} = 0.811\ \text{rad/s}) $$ The closed-loop poles are at $-0.0026 \pm j0.814$ and $-1.497 \pm j1.039$. The dominant pair is almost on the $j\omega$-axis, so the step response will be a very lightly damped oscillation at about 0.81 rad/s. The gain limit (Routh) is $K < 2.222$, so a 1% gain increase would make the system unstable. **Answer:** N = 0, Z = 0, so the system is **stable** (marginally: GM ≈ 0.09 dB, PM ≈ 0.5°). ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. How is the phase crossover frequency found for G(s) = 2.2/[s(s+1)(s^2+2s+2)]? A. Set the real part of the denominator polynomial to zero B. Set the magnitude equal to 1 C. Set the imaginary part of s^4+3s^3+4s^2+2s at s = jw to zero, giving w = sqrt(2/3) = 0.816 D. Set w = wn of the quadratic = sqrt(2) Answer: C. Phase is -180 deg when the denominator is real, i.e. -3w^3 + 2w = 0, so w = sqrt(2/3) = 0.816 rad/s. (Key point: Phase crossover frequency) 2. The plot cuts the negative real axis at -0.99. What does this give for N and Z? A. -0.99 is to the right of -1, so N = 0 and Z = 0 (stable) B. -0.99 is left of -1, so N = 2 and Z = 2 C. -0.99 is to the right of -1, so N = 1 and Z = 1 D. The crossing is irrelevant; N is always 2 for a type 1 system Answer: A. -0.99 lies between -1 and 0, so (-1, j0) is not encircled; with P = 0, Z = 0 and the system is stable. (Key point: Crossing and encirclement) 3. How would you describe the closed-loop stability? A. Strongly stable with large margins B. Unstable C. Stable but only barely: GM about 0.09 dB and PM about 0.5 deg D. Stable with infinite gain margin Answer: C. GM = 1/0.99 = 1.0101 (0.087 dB) and PM is about 0.5 deg, so poles are almost on the jw axis and a 1% gain rise makes it unstable. (Key point: Quality of stability) 4. What does the plot approach as w tends to 0+? A. Real part tends to -2.2 with imaginary part tending to -infinity B. Real part tends to 0 with imaginary part tending to -infinity C. Magnitude tends to 0 at angle -90 deg D. Real part tends to +2.2 Answer: A. The type 1 factor gives infinite magnitude at -90 deg, and the vertical asymptote lies at Re = -1.1(1+1) = -2.2. 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