---
title: "Jury test: z³−1.1z²−0.1z+0.2"
subject: "Digital Control System"
chapter: 3
key: "semester_6/digital_control_system:jury-z3-1-1z2"
source: https://www.bctengineer.com/q/semester_6/digital_control_system/jury-z3-1-1z2
site: BCT Engineering
---

# Jury test: z³−1.1z²−0.1z+0.2

Digital Control System · Chapter 3: Analysis of Discrete Time Control System

- Asked in: 2077 Chaitra, 8 marks; 2075 Baisakh, 4 marks; 2074 Bhadra, 4 marks (asked 3 times)
- Video: [Digital Control - Stability Methods - Jury's Test](https://www.youtube.com/watch?v=a_GXpxDMjB0) — Hossam Magdy Balaha
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_6/digital_control_system/jury-z3-1-1z2
- Whole chapter: https://www.bctengineer.com/md/semester_6/digital_control_system/chapter-3

## Question

Determine the stability of system having characteristic equation using Jury test: P(z)=z³-1.1z²-0.1z+0.2=0.

## Short answer

Jury test yields |a3|=0.2<1, P(-1)>0, but P(1)=0, indicating a root exactly at z=1. The system is critically (marginally) stable, not asymptotically stable.

## Answer

#### Jury stability test (Ogata form)

For $P(z) = a_0z^n + a_1z^{n-1} + \dots + a_n$ with $a_0 > 0$, all roots lie inside the unit circle if and only if:

1. $|a_n| < a_0$
2. $P(1) > 0$
3. $(-1)^nP(-1) > 0$
4. $|b_{n-1}| > |b_0|$, $|c_{n-2}| > |c_0|$, and so on, for the rows of the Jury table

The table entries are

$$
b_k = \begin{vmatrix} a_n & a_{n-1-k} \\ a_0 & a_{k+1}\end{vmatrix}, \qquad
c_k = \begin{vmatrix} b_{n-1} & b_{n-2-k} \\ b_0 & b_{k+1}\end{vmatrix}
$$

#### Coefficients

$P(z) = z^3 - 1.1z^2 - 0.1z + 0.2$, so $n = 3$, $a_0 = 1$, $a_1 = -1.1$, $a_2 = -0.1$, $a_3 = 0.2$.

#### Conditions 1 to 3

- $|a_3| = 0.2 < a_0 = 1$: satisfied
- $P(1) = 1 - 1.1 - 0.1 + 0.2 = 0$: **not greater than 0**. The equality means there is a root exactly at $z = 1$.
- $(-1)^3P(-1) = -(-1 - 1.1 + 0.1 + 0.2) = 1.8 > 0$: satisfied

#### Jury table (for completeness)

$$
\begin{aligned}
b_0 &= a_3a_1 - a_0a_2 = 0.2(-1.1) - 1(-0.1) = -0.12 \\
b_1 &= a_3a_2 - a_0a_1 = 0.2(-0.1) - 1(-1.1) = 1.08 \\
b_2 &= a_3a_3 - a_0a_0 = 0.04 - 1 = -0.96
\end{aligned}
$$

| Row | $z^0$ | $z^1$ | $z^2$ | $z^3$ |
|---|---|---|---|---|
| 1 | 0.2 | −0.1 | −1.1 | 1 |
| 2 | 1 | −1.1 | −0.1 | 0.2 |
| 3 | $b_2 = -0.96$ | $b_1 = 1.08$ | $b_0 = -0.12$ | |

- $|b_2| = 0.96 > |b_0| = 0.12$: satisfied

#### Conclusion

All conditions hold except $P(1) > 0$, which is exactly zero. So one root lies **on** the unit circle at $z = 1$ and the others lie inside. The system is **critically (marginally) stable**, not asymptotically stable.

Check by factorising:

$$
P(z) = (z - 1)(z^2 - 0.1z - 0.2) = (z - 1)(z - 0.5)(z + 0.4)
$$

The roots are $z = 1$, $0.5$ and $-0.4$, which confirms the result.

**Answer:** Critically (marginally) stable, with a root at $z = 1$ (because $P(1) = 0$).

## Check yourself

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

1. What is P(1) for P(z) = z^3 - 1.1z^2 - 0.1z + 0.2?
   A. 0
   B. 0.2
   C. 1.8
   D. -0.2
   Answer: A. P(1) = 1 - 1.1 - 0.1 + 0.2 = 0. This is a boundary case, not a strict pass. (Key point: P(1) evaluation)

2. What does P(1) = 0 tell you about the roots?
   A. A root lies exactly at z = 1, on the unit circle
   B. A root lies at z = 0
   C. All roots are inside the unit circle
   D. All roots are outside the unit circle
   Answer: A. P(1) = 0 means z = 1 is a root, and that point is on the unit circle. (Key point: Meaning of P(1) = 0)

3. What is the stability verdict?
   A. Asymptotically stable
   B. Unstable, since P(1) > 0 fails
   C. Critically (marginally) stable
   D. Cannot be determined
   Answer: C. One simple root is on the unit circle and the rest are inside, so the system is critically stable, not asymptotically stable. (Key point: Conclusion)

4. What are the roots of the polynomial?
   A. 1, 0.5 and 0.4
   B. 1, 0.5 and -0.4
   C. -1, 0.5 and -0.4
   D. 1.1, 0.1 and 0.2
   Answer: B. P(z) = (z - 1)(z - 0.5)(z + 0.4). Only z = 1 is on the circle, the others have magnitude below 1. (Key point: Factorised check)

## Close variants set in other exams

- [Jury test: 2z⁴+7z³+10z²+4z+1](https://www.bctengineer.com/question.md?key=semester_6%2Fdigital_control_system%3Ajury-2z4-7z3) — Digital Control System, chapter 3, asked 8×

## Related in Digital Control System

- [Jury test: z³+2.1z²+1.44z+0.32](https://www.bctengineer.com/question.md?key=semester_6%2Fdigital_control_system%3Ajury-z3-2-1z2) — Digital Control System, chapter 3, asked 4×
- [Range of K for stability (Jury test)](https://www.bctengineer.com/question.md?key=semester_6%2Fdigital_control_system%3Arange-of-k-jury) — Digital Control System, chapter 3, asked 3×
- [Jury stability test explained](https://www.bctengineer.com/question.md?key=semester_6%2Fdigital_control_system%3Awhat-is-jury-test) — Digital Control System, chapter 3
- [Jury test: z⁴−1.2z³+0.07z²+0.3z−0.08](https://www.bctengineer.com/question.md?key=semester_6%2Fdigital_control_system%3Ajury-z4-1-2z3-0-3z) — Digital Control System, chapter 3, asked 3×
- [Jury test: z⁴−1.2z³+0.07z²+1.5z−0.06](https://www.bctengineer.com/question.md?key=semester_6%2Fdigital_control_system%3Ajury-z4-1-2z3-1-5z) — Digital Control System, chapter 3, asked 2×
- [How to study stability; test a cubic](https://www.bctengineer.com/question.md?key=semester_6%2Fdigital_control_system%3Astability-study-method-cubic) — Digital Control System, chapter 3

## Related in other subjects

- [Characteristic equation from state model [−5 −1; 3 −1]](https://www.bctengineer.com/question.md?key=semester_3%2Fcontrol_systems%3Ass-char-eq-5-1-3) — Control Systems, chapter 8, asked 2×
- [Characteristic strength from test results](https://www.bctengineer.com/question.md?key=semester_5%2Fconcrete_technology_and_masonry_structure%3Acharacteristic-strength-30-28) — Concrete Technology and Masonry Structure, chapter 5
- [Characteristic equation and verification [1,0,2;0,2,1;…]](https://www.bctengineer.com/question.md?key=semester_3%2Fmathematics_iii%3Acharacteristic-equation-and-verification-1-0) — Mathematics III, chapter 1
- [JK characteristic equation and toggle mode](https://www.bctengineer.com/question.md?key=semester_3%2Fdigital_logic%3Ajk-characteristic-toggle) — Digital Logic, chapter 6