---
title: "Nyquist stability of 10/[s(s²+2s+4)] and GM"
subject: "Control Systems"
chapter: 6
key: "semester_3/control_systems:nyquist-10-s-s2-2s-4"
source: https://www.bctengineer.com/q/semester_3/control_systems/nyquist-10-s-s2-2s-4
site: BCT Engineering
---

# Nyquist stability of 10/[s(s²+2s+4)] and GM

Control Systems · Chapter 6: Frequency Response Techniques

- Asked in: 2066 Bhadra (old course), 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-10-s-s2-2s-4
- Whole chapter: https://www.bctengineer.com/md/semester_3/control_systems/chapter-6

## Question

Use Nyquist stability criteria to evaluate the stability of the system with open loop transfer function G(s) = 10/[s(s²+2s+4)]. Identify phase cross-over frequency and gain margin from the Nyquist plot.

## Short answer

Nyquist plot crosses the negative real axis at −1.25 at ωpc=2 rad/s; GM = 1/1.25 = 0.8 (−1.94 dB). With P=0 open-loop RHP poles, the plot encircles −1 twice clockwise (N=2), giving Z=2 closed-loop RHP poles. The closed-loop system is unstable.

## Answer

#### Frequency response

$$
G(j\omega)=\frac{10}{j\omega\left[(4-\omega^2)+j2\omega\right]} = \frac{10}{-2\omega^2 + j\omega(4-\omega^2)}
$$

$$
|G| = \frac{10}{\omega\sqrt{(4-\omega^2)^2+4\omega^2}},\qquad \angle G = -90^\circ-\tan^{-1}\frac{2\omega}{4-\omega^2}
$$

| $\omega$ | 0+ | 0.5 | 1 | 1.5 | 2 | 3 | 5 | $\infty$ |
|---|---|---|---|---|---|---|---|---|
| $|G|$ | $\infty$ | 5.15 | 2.77 | 1.92 | 1.25 | 0.43 | 0.09 | 0 |
| $\angle G$ | $-90^\circ$ | $-105^\circ$ | $-124^\circ$ | $-150^\circ$ | $-180^\circ$ | $-220^\circ$ | $-245^\circ$ | $-270^\circ$ |

- **Low-frequency asymptote:** $\text{Re}\,G = \dfrac{-20\omega^2}{4\omega^4+\omega^2(4-\omega^2)^2} \to -\dfrac{20}{16} = -1.25$ as $\omega\to0$.

#### Phase crossover frequency

The imaginary part of the denominator is zero when $4-\omega^2=0$:

$$
\omega_{pc}=2\ \text{rad/s},\qquad G(j2)=\frac{10}{-2(4)}=-1.25
$$

So the plot cuts the negative real axis at $-1.25$.

#### Nyquist plot

```text
              Im
     w<0 mirror  |
       ____      |
      /    \     |
 ----o--x---+----+----- Re
   -1.25 -1 \__/ |
      w=2        |
   w->0+ : to -inf j
   (asymptote Re=-1.25)
  small semicircle at s=0 maps
  to infinite arc, clockwise
```

- $\omega: 0^+\to\infty$: starts at $-1.25-j\infty$, passes through $-1.25$ at $\omega=2$, ends at the origin at $-270^\circ$.
- $\omega: -\infty\to0^-$: mirror image.
- Indentation around the pole at $s=0$ maps to an infinite semicircle clockwise (through the positive real axis).
- Infinite semicircle of the $s$-plane maps to the origin.

#### Nyquist criterion

$Z = P + N$

- $P=0$ (open-loop poles: $0$, $-1\pm j1.732$; none in the RHP).
- The crossing at $-1.25$ lies to the left of $-1$, so the point $-1+j0$ is **encircled twice clockwise**: $N=2$.
- $Z = 0+2 = 2$ closed-loop poles in the RHP.

**The closed-loop system is unstable.** (Check by Routh: $s^3+2s^2+4s+10$: $2\times4=8<10$, two sign changes.)

#### Gain margin

$$
GM = \frac{1}{|G(j\omega_{pc})|}=\frac{1}{1.25}=0.8,\qquad GM_{dB}=20\log0.8=-1.94\ \text{dB}
$$

The negative GM confirms instability; the gain must be reduced below $10\times0.8=8$ for stability.

**Answer: $\omega_{pc}=2$ rad/s, GM = 0.8 (−1.94 dB); N = 2, Z = 2, so the closed-loop system is unstable.**

## Check yourself

The answer's key points as multiple-choice questions, with the correct option and why.

1. For G(s) = 10/[s(s^2+2s+4)], where does the plot cut the negative real axis?
   A. w = 4 rad/s
   B. w = 1 rad/s
   C. w = 2 rad/s
   D. w = 0.5 rad/s
   Answer: C. The imaginary part of the denominator is w(4 - w^2), which is zero at w = 2 rad/s. (Key point: Phase crossover frequency)

2. What is G(j2)?
   A. -1.25
   B. -0.8
   C. +1.25
   D. -2.5
   Answer: A. G(j2) = 10 / (-2 x 4) = -1.25. (Key point: Crossing value)

3. With P = 0 and the plot crossing at -1.25, what are N and Z?
   A. N = 0, Z = 0
   B. N = 2, Z = 2
   C. N = 1, Z = 1
   D. N = -2, Z = -2
   Answer: B. -1 lies inside the loops (crossing is to its left), so there are 2 clockwise encirclements and Z = P + N = 2. (Key point: Nyquist criterion result)

4. What is the gain margin and what does it mean?
   A. 1.25 (1.94 dB), stable
   B. 0.8 (-1.94 dB), unstable; gain must be below 8
   C. 0.8 (1.94 dB), stable
   D. 1.25, unstable; gain must be above 12.5
   Answer: B. GM = 1/1.25 = 0.8 = -1.94 dB; a negative GM means unstable, and stability needs gain below 10 x 0.8 = 8. (Key point: Gain margin)

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