---
title: "Nyquist criterion from given pole-zero and Nyquist plot"
subject: "Control Systems"
chapter: 6
key: "semester_3/control_systems:nyquist-criteria-from-plot"
source: https://www.bctengineer.com/q/semester_3/control_systems/nyquist-criteria-from-plot
site: BCT Engineering
---

# Nyquist criterion from given pole-zero and Nyquist plot

Control Systems · Chapter 6: Frequency Response Techniques

- Asked in: 2071 Chaitra, 4 marks
- Video: [Nyquist Stability Criterion, Part 1](https://www.youtube.com/watch?v=sof3meN96MA) — Brian Douglas
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_3/control_systems/nyquist-criteria-from-plot
- Whole chapter: https://www.bctengineer.com/md/semester_3/control_systems/chapter-6

## Question

State the Nyquist stability criteria for negative feedback control system. Using this concept determine whether the following system represented by figure below is stable. [Figure: open-loop pole-zero plot with poles marked at 3 (drawn on the negative real axis, i.e. s = −3) and at s = +2, and a zero at s = −1. Nyquist diagram: a closed contour crossing the real axis at about −1.25 and 0, extending to about ±1.6 on the imaginary axis, traversed counter-clockwise (arrow at top pointing left), enclosing the −1 point once]

## Short answer

Nyquist criterion: Z = P + N. Here P = 1 (open-loop pole at +2). The plot encircles −1 once counter-clockwise (N = −1), giving Z = 0. The closed-loop system is stable.

## Answer

#### Nyquist stability criterion
For a negative feedback system with open-loop transfer function $G(s)H(s)$:

$$
Z = P + N
$$

- $P$ = number of open-loop poles in the right half s-plane;
- $N$ = net number of **clockwise** encirclements of the point $(-1, j0)$ by the Nyquist plot ($N$ is negative for counter-clockwise encirclements);
- $Z$ = number of closed-loop poles in the right half-plane.

The closed loop is stable if and only if $Z = 0$, i.e. the Nyquist plot must encircle $(-1, j0)$ **counter-clockwise as many times as there are open-loop RHP poles**. If P = 0, it must not encircle −1 at all.

#### Applying it to the given system
- From the pole-zero plot: open-loop poles at $s = -3$ and $s = +2$, and a zero at $s = -1$. So $G(s)H(s) = \dfrac{K(s+1)}{(s+3)(s-2)}$, with **P = 1** (the pole at +2).
- From the Nyquist diagram: the plot crosses the real axis at about −1.25 and 0, so the point −1 lies **inside** the closed curve. The curve is traversed **counter-clockwise**, so it encircles −1 once anticlockwise: **N = −1**.

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

**The closed-loop system is stable.**

(Consistency check: at ω = 0, $G(0) = -K/6 = -1.25$ gives $K = 7.5$. Then $1 + G = 0$ gives $s^2 + 8.5s + 1.5 = 0$, whose roots are both in the LHP.)

## Check yourself

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

1. In Z = P + N, what does N represent?
   A. Number of open-loop zeros in the RHP
   B. Net clockwise encirclements of (-1, j0) by the Nyquist plot
   C. Number of closed-loop poles on the jw axis
   D. Number of counter-clockwise encirclements of the origin
   Answer: B. N is the net clockwise encirclement count of (-1, j0); counter-clockwise ones make N negative. (Key point: Meaning of Z = P + N)

2. When is the closed loop stable?
   A. P = 0 and N = 1
   B. Z = 0, i.e. the plot encircles -1 counter-clockwise P times
   C. N = P
   D. Z = P
   Answer: B. Stability needs Z = 0, so N = -P: counter-clockwise encirclements equal to the number of open-loop RHP poles. (Key point: Stability requirement)

3. The poles are at s = -3 and s = +2, and a zero at -1. What is P?
   A. P = 2
   B. P = 0
   C. P = 1, from the pole at +2
   D. P = 1, from the zero
   Answer: C. Only the pole at +2 is in the RHP. (Key point: Reading P from the pole-zero plot)

4. The diagram encircles -1 once counter-clockwise. What is Z?
   A. Z = 1 + (-1) = 0, stable
   B. Z = 1 + 1 = 2, unstable
   C. Z = 0 - 1 = -1
   D. Z = 1, unstable
   Answer: A. N = -1 and P = 1 so Z = 0 and the closed-loop system is stable. (Key point: Final result)

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