--- title: "Nyquist plot of (s+10)/[(s−3)(s+3)] and gain margin" subject: "Control Systems" chapter: 6 key: "semester_3/control_systems:nyquist-s10-s-3-s3" source: https://www.bctengineer.com/q/semester_3/control_systems/nyquist-s10-s-3-s3 site: BCT Engineering --- # Nyquist plot of (s+10)/[(s−3)(s+3)] and gain margin Control Systems · Chapter 6: Frequency Response Techniques - Asked in: 2076 Chaitra, 5+1+2 marks; 2074 Chaitra, 8 marks (asked 2 times) - Video: [Nyquist Plot (Problems) - Frequency Response Analysis - Control System](https://www.youtube.com/watch?v=dXRQSm2vb7M) — Ekeeda - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_3/control_systems/nyquist-s10-s-3-s3 - Whole chapter: https://www.bctengineer.com/md/semester_3/control_systems/chapter-6 ## Question Sketch the Nyquist contour and plot of unity feedback system having open loop transfer function G(s)H(s) = (s+10)/[(s−3)(s+3)]. (i) Comment on stability. (ii) Determine gain margin. ## Short answer Open-loop has an unstable pole at s=3 (P=1). Nyquist plot encircles -1 once counterclockwise (N=-1), so Z=0, closed-loop stable. Gain margin is 0.9 (-0.92 dB), measured at ω=0 phase crossover. ## Answer $$ G(s)H(s) = \frac{s+10}{(s-3)(s+3)} = \frac{s+10}{s^2 - 9} $$ Open-loop poles: $+3$ and $-3$, so **P = 1**. #### Nyquist contour The contour encloses the whole right half of the s-plane: - the $j\omega$-axis from $-j\infty$ to $+j\infty$; - a semicircle of infinite radius on the right, traversed clockwise. No poles lie on the $j\omega$-axis, so no indentation is needed. The RHP pole at $s = +3$ lies inside the contour. ```text jw ^ +j inf |\ | \ R -> inf | \ ------+---x----> σ | / +3 (inside) | / |/ -j inf ``` #### Mapping $$ G(j\omega) = \frac{10 + j\omega}{-(\omega^2 + 9)} = \frac{-10}{\omega^2 + 9} - j\frac{\omega}{\omega^2 + 9} $$ | ω | Re | Im | ∠G | |---|---|---|---| | 0 | −1.111 | 0 | −180° | | 1 | −1.000 | −0.100 | −174.3° | | 3 | −0.556 | −0.167 | −163.3° | | 5 | −0.294 | −0.147 | −153.4° | | 10 | −0.092 | −0.092 | −135° | | ∞ | 0 | 0 | −90° | - For ω > 0 the plot lies in the third quadrant, from $-1.111$ to the origin. - The ω < 0 part is the mirror image. - The infinite arc maps to the origin. ```text | | | | | | ....................... | ..... ....| -----------*--@-------------------------**-------- ***** ****| *********************** | | | | | | | -1.11 0 * w > 0 . w < 0 (mirror) @ point (-1, j0) w = 0 at -1.111; w -> +-inf at origin ``` #### (i) 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 $Z = 0$. As ω goes from −∞ to +∞, the closed curve runs from the origin through the upper half to $-1.111$, then through the lower half back to the origin. This is **counter-clockwise**, and $-1$ lies inside because $-1.111 < -1$. So N = −1. $$ Z = P + N = 1 - 1 = 0 $$ The **closed-loop system is stable**. Check: $s^2 - 9 + s + 10 = s^2 + s + 1$, with roots in the LHP. ✓ #### (ii) Gain margin Phase crossover is at ω = 0, where $|G| = 10/9 = 1.111$. $$ GM = \frac{1}{1.111} = 0.9 \;\Rightarrow\; 20\log0.9 = -0.92\ \text{dB} $$ Since the open loop is unstable, this means the gain may be **reduced** only by a factor of 0.9 before instability. With a gain $K$, stability needs $K > 0.9$. **Answer:** P = 1, N = −1, Z = 0, so the system is **stable**. **GM = 0.9 (−0.92 dB)**. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. For G(s)H(s) = (s+10)/[(s-3)(s+3)], what is P and why does the contour need no indentation? A. P = 2; the contour is indented around both poles B. P = 0; both poles are inside the contour C. P = 1 (pole at +3); no pole lies on the jw-axis D. P = 1 (zero at -10); poles are at the origin Answer: C. Only the pole at +3 is in the RHP, and no pole sits on the jw-axis, so the contour is not indented. (Key point: Open-loop RHP poles) 2. The plot starts at -1.111 (w = 0) and ends at the origin, tracing a closed curve as w goes from -infinity to +infinity. What is N? A. N = +1, clockwise B. N = 0 C. N = -1: the curve is counter-clockwise and encloses -1 since -1.111 < -1 D. N = -2 Answer: C. The curve crosses the real axis at -1.111, left of -1, and is traced counter-clockwise, giving one counter-clockwise encirclement. (Key point: Nyquist plot direction) 3. What is Z, and what does it mean? A. Z = 0, stable B. Z = 2, unstable C. Z = 1, unstable D. Z = 0, unstable Answer: A. Z = P + N = 1 - 1 = 0, so the closed loop is stable; the closed-loop polynomial s^2 + s + 1 has LHP roots. 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