--- title: "Nyquist stability of (s+5)/[(s−2)(s+2)] and GM" subject: "Control Systems" chapter: 6 key: "semester_3/control_systems:nyquist-s5-s-2-s2-gm" source: https://www.bctengineer.com/q/semester_3/control_systems/nyquist-s5-s-2-s2-gm site: BCT Engineering --- # Nyquist stability of (s+5)/[(s−2)(s+2)] and GM Control Systems · Chapter 6: Frequency Response Techniques - Asked in: 2075 Asoj, 8 marks - Video: [Stability Analysis Using Nyquist Plot: Concepts and Techniques](https://www.youtube.com/watch?v=tqmLcnBCzGw) — Engineering Funda - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_3/control_systems/nyquist-s5-s-2-s2-gm - Whole chapter: https://www.bctengineer.com/md/semester_3/control_systems/chapter-6 ## Question Using Nyquist criterion determine the stability of the feedback system whose open loop transfer function is given by G(s)H(s) = (s+5)/[(s−2)(s+2)]. Also find GM. ## Short answer Nyquist plot makes one counter-clockwise encirclement of -1, so N=-1. With P=1 (pole at +2), Z=0, thus closed-loop stable. Gain margin is 0.8 (-1.94 dB). ## Answer $$ G(s)H(s) = \frac{s+5}{(s-2)(s+2)} = \frac{s+5}{s^2 - 4} $$ Open-loop poles: $+2, -2$, so **P = 1** (one RHP pole). No poles lie on the $j\omega$-axis. #### Frequency response $$ G(j\omega) = \frac{5 + j\omega}{-(\omega^2 + 4)} = \frac{-5}{\omega^2 + 4} - j\frac{\omega}{\omega^2 + 4} $$ | ω | Re | Im | \|G\| | ∠G | |---|---|---|---|---| | 0 | −1.25 | 0 | 1.25 | −180° | | 1 | −1.00 | −0.20 | 1.02 | −168.7° | | 2 | −0.625 | −0.25 | 0.67 | −158.2° | | 5 | −0.172 | −0.172 | 0.24 | −135° | | ∞ | 0 | 0 | 0 | −90° | - For ω > 0 the plot is in the third quadrant, from −1.25 to the origin (arriving along −90°). - For ω < 0 it is the mirror image in the second quadrant. - The infinite semicircle maps to the origin. ```text | | | | | ....................... | .... .....| ... .. ---------*-----@-------------------------*-------- *** ** **** *****| *********************** | | | | | | -1.25 0 * w > 0 . w < 0 (mirror) @ point (-1, j0) w = 0 at -1.25; w -> +-inf at origin ``` #### 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 curve runs from the origin through the upper half to −1.25 and back through the lower half. This is a **counter-clockwise** loop, and it contains −1 because −1.25 < −1. So N = −1. $$ Z = P + N = 1 - 1 = 0 $$ The **closed-loop system is stable**. Check: $s^2 - 4 + s + 5 = s^2 + s + 1 = 0$ has roots $-0.5 \pm j0.866$. ✓ #### Gain margin Phase crossover is at ω = 0, where $|G| = 5/4 = 1.25$. $$ GM = \frac{1}{1.25} = 0.8 = 20\log0.8 = -1.94\ \text{dB} $$ For this open-loop unstable system, the negative GM means the gain can be **reduced** by at most a factor of 0.8 (gain $K > 0.8$ is needed) before the system becomes unstable. **Answer:** P = 1, N = −1, Z = 0, so the system is **stable**. **GM = 0.8 (−1.94 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+5)/[(s-2)(s+2)], what is P? A. P = 0 B. P = 2 C. P = 1, from the pole at +2 D. P = 1, from the zero at -5 Answer: C. Only the pole at s = +2 is in the RHP; the system is open-loop unstable. (Key point: Open-loop RHP poles) 2. The Nyquist plot encircles -1 once counter-clockwise. What is N? A. N = +1 B. N = 0 C. N = -1 D. N = +2 Answer: C. N counts clockwise encirclements, so a counter-clockwise one is N = -1. (Key point: Direction of encirclement) 3. With P = 1 and N = -1, what is Z and the conclusion? A. Z = 2, unstable B. Z = 0, stable C. Z = 1, unstable D. Z = -1, stable Answer: B. Z = P + N = 0; the closed-loop characteristic equation s^2 + s + 1 has LHP roots. (Key point: Stability result) 4. At w = 0, |G| = 1.25, so GM = 0.8 (-1.94 dB). What does it mean? A. Gain can be raised indefinitely B. Gain must be raised by 25% to become unstable C. Gain can be reduced by at most a factor 0.8 before instability D. The system is already unstable Answer: C. For an open-loop unstable system the plot must keep encircling -1, so lowering the gain below 0.8 moves the crossing right of -1 and loses stability. 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