---
title: "Nyquist stability of 60/[(s+1)(s+2)(s+5)]"
subject: "Control Systems"
chapter: 6
key: "semester_3/control_systems:nyquist-60-s1-s2-s5"
source: https://www.bctengineer.com/q/semester_3/control_systems/nyquist-60-s1-s2-s5
site: BCT Engineering
---

# Nyquist stability of 60/[(s+1)(s+2)(s+5)]

Control Systems · Chapter 6: Frequency Response Techniques

- Asked in: 2081 Baisakh, 8 marks
- Video: [#215 Problem on Nyquist plot in control systems || EC Academy](https://www.youtube.com/watch?v=5PEVdDxZWjA) — EC Academy
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_3/control_systems/nyquist-60-s1-s2-s5
- 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) = 60/[(s+1)(s+2)(s+5)].

## Short answer

P=0. Nyquist plot crosses real axis at –0.476, no encirclement of –1 (N=0), so Z=0. Closed-loop stable with GM≈6.4 dB.

## Answer

#### Open-loop poles

$G(s)H(s)=\dfrac{60}{(s+1)(s+2)(s+5)}$ has poles at $-1,-2,-5$: none in the right half plane, so $P=0$. There is no pole on the $j\omega$ axis, so no indentation is needed.

#### Frequency response

Denominator: $(s+1)(s+2)(s+5)=s^3+8s^2+17s+10$. With $s=j\omega$:

$$
G(j\omega)H(j\omega)=\frac{60}{(10-8\omega^2)+j(17\omega-\omega^3)}
$$

$$
|GH|=\frac{60}{\sqrt{1+\omega^2}\sqrt{4+\omega^2}\sqrt{25+\omega^2}},\quad \angle GH=-\tan^{-1}\omega-\tan^{-1}\frac{\omega}{2}-\tan^{-1}\frac{\omega}{5}
$$

| $\omega$ | 0 | 0.5 | 1.118 | 2 | 2.84 | 4.123 | 10 | $\infty$ |
|---|---|---|---|---|---|---|---|---|
| $|GH|$ | 6 | 5.18 | 3.41 | 1.76 | 1.0 | 0.476 | 0.052 | 0 |
| $\angle GH$ | $0^\circ$ | $-46^\circ$ | $-90^\circ$ | $-130^\circ$ | $-155^\circ$ | $-180^\circ$ | $-226^\circ$ | $-270^\circ$ |

#### Axis crossings

- **Imaginary axis** (real part of denominator zero): $10-8\omega^2=0\Rightarrow\omega=1.118$ rad/s, $GH=-j3.41$.
- **Negative real axis** (imaginary part zero): $17\omega-\omega^3=0\Rightarrow\omega=\sqrt{17}=4.123$ rad/s.
$$
GH(j4.123)=\frac{60}{10-8(17)}=\frac{60}{-126}=-0.476
$$

#### Nyquist plot

```text
               Im
         w<0   |  ___
            ___|_/   \
 ----o--x--+---+------o--- Re
    -1 -0.476  |     6 (w=0)
           \___|_    /
               | \__/ w>0
               | -j3.41
```

- $\omega=0\to\infty$: starts at $+6$, goes through $-j3.41$, cuts the negative real axis at $-0.476$, reaches the origin at $-270^\circ$.
- $\omega=-\infty\to0$: mirror image.
- Infinite semicircle of the $s$-plane maps to the origin.

#### Nyquist criterion

$Z=P+N$. The plot cuts the real axis at $-0.476$, to the right of $-1$, so the point $-1+j0$ is **not encircled**: $N=0$.

$$
Z=0+0=0
$$

No closed-loop poles in the RHP, so the **closed-loop system is stable**.

Relative stability: $GM=1/0.476=2.1$ (6.44 dB), PM ≈ 25° at $\omega_{gc}=2.84$ rad/s. The system becomes unstable if the gain 60 is raised above $60\times2.1=126$ (Routh: $8\times17>10+K \Rightarrow K<126$).

**Answer: N = 0, P = 0, Z = 0 – the closed-loop system is stable (GM ≈ 6.4 dB).**

## Check yourself

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

1. For G(s)H(s) = 60/[(s+1)(s+2)(s+5)], what is P, and is an indentation needed?
   A. P = 3, indentation needed
   B. P = 1, no indentation
   C. P = 0, no indentation
   D. P = 0, indentation needed
   Answer: C. All poles are in the left half plane and none lie on the jw axis, so P = 0 with no indentation. (Key point: Open-loop RHP poles)

2. Where does the plot cross the negative real axis?
   A. -0.476 at w = sqrt(17) rad/s
   B. -0.476 at w = 1.118 rad/s
   C. -2.1 at w = sqrt(17) rad/s
   D. -3.41 at w = 4.123 rad/s
   Answer: A. The imaginary part 17w - w^3 is zero at w = sqrt(17) = 4.123, where GH = 60/(10 - 136) = -0.476. (Key point: Real axis crossing)

3. Since the crossing at -0.476 lies to the right of -1, what is Z?
   A. Z = 2
   B. Z = 1
   C. Z = -1
   D. Z = 0, system stable
   Answer: D. -1 is not encircled, so N = 0 and Z = P + N = 0; the closed loop is stable. (Key point: Applying the criterion)

4. What is the gain margin and the maximum gain before instability?
   A. GM = 2.1 (6.4 dB), gain up to 126
   B. GM = 0.476, gain up to 28.6
   C. GM = 2.1, gain up to 12.6
   D. GM = 0.476, gain up to 126
   Answer: A. GM = 1/0.476 = 2.1 (6.44 dB), so the gain can rise to 60 x 2.1 = 126. (Key point: Gain margin)

## Close variants set in other exams

- [Nyquist stability of (s+5)/[(s−2)(s+2)] and GM](https://www.bctengineer.com/question.md?key=semester_3%2Fcontrol_systems%3Anyquist-s5-s-2-s2-gm) — Control Systems, chapter 6, asked 2×

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