---
title: "Nyquist contour mapping of 300/[(s+3)(s+1)(s+2)]"
subject: "Control Systems"
chapter: 6
key: "semester_3/control_systems:nyquist-contour-300"
source: https://www.bctengineer.com/q/semester_3/control_systems/nyquist-contour-300
site: BCT Engineering
---

# Nyquist contour mapping of 300/[(s+3)(s+1)(s+2)]

Control Systems · Chapter 6: Frequency Response Techniques

- Asked in: 2080 Baisakh, 8 marks
- Video: [Lec 49 GATE Problem on Nyquist Plot](https://www.youtube.com/watch?v=GxZzVdyV-FE) — Gatematic by Sohail Sir
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_3/control_systems/nyquist-contour-300
- Whole chapter: https://www.bctengineer.com/md/semester_3/control_systems/chapter-6

## Question

What is Nyquist Contour? Map the Nyquist contour of open loop transfer function G(s) = 300/[(s+3)(s+1)(s+2)] into G(s) plane and apply Nyquist criterion to check the stability of closed loop system.

## Short answer

For G(s)=300/[(s+1)(s+2)(s+3)], P=0; plot crosses real axis at –5 (ω=3.317 rad/s), encircling –1 twice clockwise (N=2). Hence Z=2, system unstable.

## Answer

#### Nyquist contour

The **Nyquist contour** is a closed path in the $s$-plane that encloses the entire right half plane. It runs up the whole $j\omega$ axis from $-j\infty$ to $+j\infty$ and returns along a semicircle of infinite radius, traversed **clockwise**. If $G(s)H(s)$ has poles on the $j\omega$ axis, the contour goes around them by small semicircles of radius $\epsilon\to0$ so they are not on the path.

```text
        jw
        |\  +j inf
        | \
        |  \  R -> inf
   -----+---)----- sigma
        |  /  (clockwise)
        | /
        |/  -j inf
```

Mapping this contour through $G(s)H(s)$ gives the Nyquist plot; by the principle of the argument, $N=Z-P$ (clockwise encirclements of $-1$), so $Z=P+N$.

#### Mapping $G(s)=\dfrac{300}{(s+1)(s+2)(s+3)}$

Poles at $-1,-2,-3$: $P=0$, none on the $j\omega$ axis.

$$
G(j\omega)=\frac{300}{(6-6\omega^2)+j(11\omega-\omega^3)}
$$

**Section I ($\omega=0\to+\infty$):**

| $\omega$ | 0 | 0.5 | 1 | 2 | 3.317 | 5 | $\infty$ |
|---|---|---|---|---|---|---|---|
| $|G|$ | 50 | 42.8 | 30 | 13.2 | 5.0 | 1.87 | 0 |
| $\angle G$ | $0^\circ$ | $-50^\circ$ | $-90^\circ$ | $-142^\circ$ | $-180^\circ$ | $-206^\circ$ | $-270^\circ$ |

- Imaginary-axis crossing: $6-6\omega^2=0\Rightarrow\omega=1$, $G=\dfrac{300}{j10}=-j30$.
- Real-axis crossing: $11\omega-\omega^3=0\Rightarrow\omega=\sqrt{11}=3.317$, $G=\dfrac{300}{6-66}=-5$.

**Section II (infinite semicircle, $s=Re^{j\theta}$):** $G\approx\dfrac{300}{R^3}e^{-j3\theta}\to0$; maps to the origin.

**Section III ($\omega=-\infty\to0$):** mirror image of section I about the real axis.

```text
              Im
              |   ___ w<0
         _____|__/   \
 --o--o--+----+-------o--- Re
  -5  -1 \____|__    / 50
    w=3.317   |  \__/ w>0
              | -j30 (w=1)
```

#### Nyquist criterion

The plot crosses the negative real axis at $-5$, to the left of $-1$. Following the curve, the point $-1+j0$ is **encircled twice in the clockwise direction**: $N=2$.

$$
Z=P+N=0+2=2
$$

Two closed-loop poles lie in the right half plane, so the **closed-loop system is unstable**. (Check: $s^3+6s^2+11s+306=0$ has roots $-8.74$ and $1.37\pm j5.75$.) GM $=1/5=0.2$ ($-14$ dB); the gain must be below $300/5=60$ for stability.

**Answer: N = 2, P = 0, Z = 2 – the closed-loop system is unstable.**

## Check yourself

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

1. What is the Nyquist contour?
   A. The unit circle in the s-plane
   B. A closed clockwise path enclosing the entire right half s-plane
   C. A closed path enclosing the left half plane
   D. The jw axis from 0 to infinity only
   Answer: B. It runs up the whole jw axis and back along an infinite semicircle, clockwise, enclosing the full RHP. (Key point: Nyquist contour definition)

2. For G(s) = 300/[(s+1)(s+2)(s+3)], how does the infinite semicircle of the contour map?
   A. To a large circle
   B. To the point -1
   C. To the origin
   D. To the positive real axis
   Answer: C. G is about 300/R^3 for large R, which tends to 0, so the arc maps to the origin. (Key point: Infinite semicircle mapping)

3. Where does the plot cross the negative real axis?
   A. -5 at w = sqrt(11)
   B. -30 at w = 1
   C. -50 at w = 0
   D. -0.2 at w = 3.317
   Answer: A. 11w - w^3 = 0 gives w = sqrt(11) = 3.317, where G = 300/(6 - 66) = -5. (Key point: Real axis crossing)

4. The crossing at -5 lies left of -1. What does the criterion say?
   A. N = 0, Z = 0, stable
   B. N = 2, Z = 2, unstable
   C. N = 1, Z = 1, unstable
   D. N = -2, Z = 0, stable
   Answer: B. -1 is encircled twice clockwise, so Z = P + N = 0 + 2 = 2 RHP closed-loop poles: unstable. (Key point: Stability result)

## Close variants set in other exams

- [Nyquist contour mapping of 200/[(s+3)(s+1)(s+2)]](https://www.bctengineer.com/question.md?key=semester_3%2Fcontrol_systems%3Anyquist-contour-200) — Control Systems, chapter 6

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