---
title: "Nyquist range of K for K/[s(s+1)(s+2)]"
subject: "Control Systems"
chapter: 6
key: "semester_3/control_systems:nyquist-k-s-s1-s2-range"
source: https://www.bctengineer.com/q/semester_3/control_systems/nyquist-k-s-s1-s2-range
site: BCT Engineering
---

# Nyquist range of K for K/[s(s+1)(s+2)]

Control Systems · Chapter 6: Frequency Response Techniques

- Asked in: 2066 Jestha (old course), 10 marks
- Video: [Nyquist plot | Range of  k for Stability | Example | CS | Control Systems | Lec-108](https://www.youtube.com/watch?v=6SUW_A_9rTE) — Education 4u
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_3/control_systems/nyquist-k-s-s1-s2-range
- Whole chapter: https://www.bctengineer.com/md/semester_3/control_systems/chapter-6

## Question

Use Nyquist stability criterion to find the range of K for which the unity feedback system represented by open loop transfer function G(s) = K/[s(s+1)(s+2)] is stable.

## Short answer

Real-axis crossing at –K/6 for ωpc=√2 rad/s. For stability –K/6 > –1, so 0 < K < 6. At K=6 marginal oscillation at 1.414 rad/s.

## Answer

#### Frequency response

$$
G(j\omega)=\frac{K}{j\omega(1+j\omega)(2+j\omega)}=\frac{K}{-3\omega^2+j\omega(2-\omega^2)}
$$

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

| $\omega$ | $0^+$ | 0.5 | 1 | $\sqrt2$ | 2 | $\infty$ |
|---|---|---|---|---|---|---|
| $|G|/K$ | $\infty$ | 0.868 | 0.316 | 0.167 | 0.079 | 0 |
| $\angle G$ | $-90^\circ$ | $-130.6^\circ$ | $-161.6^\circ$ | $-180^\circ$ | $-198.4^\circ$ | $-270^\circ$ |

#### Key points of the plot

- **Low-frequency asymptote:** $\text{Re}\,G=\dfrac{-3K\omega^2}{9\omega^4+\omega^2(2-\omega^2)^2}\to-\dfrac{3K}{4}$ as $\omega\to0$.
- **Negative real-axis crossing:** imaginary part of denominator zero: $2-\omega^2=0 \Rightarrow \omega_{pc}=\sqrt2=1.414$ rad/s.
$$
G(j\sqrt2)=\frac{K}{-3(2)}=-\frac{K}{6}
$$
- As $\omega\to\infty$, $G\to0$ at $-270^\circ$.

#### Nyquist plot

```text
            Im
   w<0 mirror|
     ___     |
    /   \    |
 --o-----+---+------- Re
 -K/6 \_/    |
   w=1.414   |
  w->0+: to -inf j along Re=-0.75K
  pole at s=0 -> infinite arc,
  clockwise from w=0- to w=0+
```

1. $\omega=0^+\to\infty$: polar plot above.
2. $\omega=-\infty\to0^-$: mirror image about the real axis.
3. Small indentation at $s=0$ maps to an infinite clockwise semicircle.
4. Infinite $s$-semicircle maps to the origin.

#### Applying the criterion

$Z=P+N$ with $P=0$ (poles at $0,-1,-2$).

- If $K/6<1$: $-1$ lies outside the plot, $N=0$, $Z=0$: **stable**.
- If $K/6>1$: $-1$ is encircled twice clockwise, $N=2$, $Z=2$: unstable.
- If $K=6$: the plot passes through $-1$: marginally stable, oscillation at $\omega=\sqrt2=1.414$ rad/s.

$$
\frac{K}{6}<1 \Rightarrow K<6
$$

Check by Routh: $s^3+3s^2+2s+K=0$ requires $3\times2>K$, i.e. $K<6$.

**Answer: the closed-loop system is stable for $0<K<6$; at $K=6$ it oscillates at 1.414 rad/s.**

## Check yourself

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

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

2. Where does the plot cross the negative real axis?
   A. -K/6
   B. -K/3
   C. -3K/4
   D. -K/2
   Answer: A. G(j sqrt 2) = K/(-3 x 2) = -K/6. (Key point: Crossing point)

3. How is the pole at s = 0 handled in the Nyquist contour?
   A. It is ignored
   B. A small semicircular indentation maps to an infinite clockwise semicircle in the G-plane
   C. It maps to a unit circle
   D. It adds one RHP pole to P
   Answer: B. The indentation avoids the pole and maps to an infinite-radius clockwise arc; P stays 0. (Key point: Effect of the origin pole)

4. What range of K gives a stable closed loop?
   A. 0 < K < 3
   B. 0 < K < 12
   C. K > 6
   D. 0 < K < 6
   Answer: D. Stability needs K/6 < 1, so 0 < K < 6, which agrees with Routh. (Key point: Range of K)

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