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

# Nyquist plot of (s+2)/[(s+1)(s−1)]

Control Systems · Chapter 6: Frequency Response Techniques

- Asked in: 2082 Baisakh, 6+2+2 marks; 2078 Kartik, 8 marks; 2070 Chaitra, 8 marks (asked 3 times)
- Video: [Final Exam Tutorial - Nyquist Plot Example](https://www.youtube.com/watch?v=lZbUp-ywSq8) — Ceresapien
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_3/control_systems/nyquist-s2-over-s2-1
- Whole chapter: https://www.bctengineer.com/md/semester_3/control_systems/chapter-6

## Question

Construct complete Nyquist plot for a unity feedback control system whose open loop transfer function is G(s) = (s+2)/(s²−1). Check stability of the system as per Nyquist Criterion. What is its gain margin?

## Short answer

Open-loop has one RHP pole (P=1). Nyquist plot encircles -1 once counterclockwise (N=-1), so Z=0 and the closed-loop system is stable. Gain margin is 0.5 (-6.02 dB) at phase crossover frequency 0 rad/s.

## Answer

$$
G(s) = \frac{s+2}{s^2-1} = \frac{s+2}{(s-1)(s+1)}
$$

The open-loop poles are $+1$ and $-1$, so **P = 1** (one RHP pole). No poles lie on the $j\omega$-axis, so the Nyquist contour needs no indentation.

#### Frequency response
$$
G(j\omega) = \frac{2 + j\omega}{-(1+\omega^2)} = \frac{-2}{1+\omega^2} - j\frac{\omega}{1+\omega^2}
$$

| ω | Re | Im | \|G\| | ∠G |
|---|---|---|---|---|
| 0 | −2 | 0 | 2 | −180° |
| 0.5 | −1.60 | −0.40 | 1.65 | −166° |
| 1 | −1.00 | −0.50 | 1.12 | −153.4° |
| 2 | −0.40 | −0.40 | 0.57 | −135° |
| 5 | −0.08 | −0.19 | 0.21 | −111.8° |
| ∞ | 0 | 0 | 0 | −90° |

- For $\omega > 0$ the imaginary part is always negative, so the plot lies in the third quadrant. It runs from $-2$ (at ω = 0) to the origin (at ω = ∞), arriving along $-90^\circ$.
- The part for $\omega < 0$ is the mirror image about the real axis.
- The large semicircle ($s = Re^{j\theta}$, $R \to \infty$) maps to the origin, because $G \to 1/s \to 0$.

#### Complete Nyquist plot
```text
                                         |
                                         |
                                         |
                                         |
                 ................        |
            ......              ......   |
         ....                        ....|
        ..                              ..
--------*---------------@----------------*--------
        **                              **
         ****                        ****|
            ******              ******   |
                 ****************        |
                                         |
                                         |
                                         |
                                         |
       -2              -1                0
* w > 0   . w < 0 (mirror)   @ point (-1, j0)
w = 0 at -2; w -> +-inf at origin
```

#### Stability by Nyquist criterion
**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 $-\infty$ to $+\infty$, the plot runs from the origin through the upper half to $-2$, then through the lower half back to the origin. This is a closed curve traced **counter-clockwise**.
- The point $-1$ lies inside it, since the plot crosses the real axis at $-2$, to the left of $-1$.
- So $N = -1$ (one counter-clockwise encirclement).

$$
Z = P + N = 1 + (-1) = 0
$$

The **closed-loop system is stable**. Check: $1 + G = 0 \Rightarrow s^2 + s + 1 = 0$, whose roots $-0.5 \pm j0.866$ are in the LHP. ✓

#### Gain margin
The plot cuts the negative real axis at ω = 0 (phase crossover frequency $\omega_{pc} = 0$), where $|G| = 2$.

$$
GM = \frac{1}{|G(j\omega_{pc})|} = \frac{1}{2} = 0.5 \;\Rightarrow\; GM = 20\log_{10}0.5 = -6.02\ \text{dB}
$$

Because the open loop is unstable, a negative GM does **not** mean instability here. It means the loop gain can be **reduced** by a factor of 2 before the system becomes unstable. (With gain $K$: $s^2 + Ks + 2K - 1 = 0$ is stable only for $K > 0.5$.)

**Answer:** P = 1, N = −1 (one CCW encirclement), Z = 0, so the closed loop is **stable**. **GM = 0.5 (−6.02 dB)**.

## Check yourself

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

1. For G(s) = (s+2)/(s^2-1), what is P, the number of open-loop RHP poles?
   A. 0
   B. 2
   C. 1, from the pole at +1
   D. 1, from the zero at -2
   Answer: C. The poles are +1 and -1, so exactly one is in the right half plane; zeros do not count in P. (Key point: Open-loop poles P)

2. The plot crosses the negative real axis at -2 and is traced counter-clockwise. What is N for the point -1?
   A. N = -1 (one counter-clockwise encirclement)
   B. N = +1
   C. N = 0
   D. N = -2
   Answer: A. Since -1 lies inside the closed curve (the crossing at -2 is to its left) and the curve is counter-clockwise, N = -1. (Key point: Encirclement count)

3. Applying Z = P + N, what is the conclusion?
   A. Z = 2, unstable
   B. Z = 0, stable
   C. Z = 1, unstable
   D. Z = 1, marginally stable
   Answer: B. Z = 1 + (-1) = 0, so the closed loop has no RHP poles; this matches s^2 + s + 1 = 0 having roots -0.5 +- j0.866. (Key point: Nyquist criterion)

4. What is the gain margin and how is it interpreted?
   A. GM = 2 (6 dB); the gain can be doubled before instability
   B. GM = 0.5 (-6.02 dB); the gain can be reduced by half before instability
   C. GM = 0.5; negative GM means the system is unstable
   D. GM = infinity
   Answer: B. At the phase crossover (omega = 0), |G| = 2 so GM = 1/2 = -6.02 dB. Because the open loop is unstable, a negative GM is fine; the loop needs K > 0.5. (Key point: Gain margin)

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