---
title: "Jury test: z³+2.1z²+1.44z+0.32"
subject: "Digital Control System"
chapter: 3
key: "semester_6/digital_control_system:jury-z3-2-1z2"
source: https://www.bctengineer.com/q/semester_6/digital_control_system/jury-z3-2-1z2
site: BCT Engineering
---

# Jury test: z³+2.1z²+1.44z+0.32

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

- Asked in: 2082 Chaitra (new course), 4 marks
- 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-z3-2-1z2
- Whole chapter: https://www.bctengineer.com/md/semester_6/digital_control_system/chapter-3

## Question

Examine the stability of the given characteristic equation using Jury stability test: P(z)=z³+2.1z²+1.44z+0.32=0.

## Short answer

Applying the Jury test to P(z)=z^3+2.1z^2+1.44z+0.32: |a3|<a0 (0.32<1), P(1)=4.86>0, (-1)^3P(-1)=0.02>0, |b2|>|b0| (0.8976>0.768). All conditions satisfied; the system is stable. Roots are −0.5, −0.8, −0.8.

## Answer

We apply the Jury test to $P(z) = a_0z^3 + a_1z^2 + a_2z + a_3$, where $a_0 = 1$, $a_1 = 2.1$, $a_2 = 1.44$, $a_3 = 0.32$, and $n = 3$.

#### Conditions on the coefficients
1. $|a_3| < a_0$: $0.32 < 1$, satisfied.
2. $P(1) = 1 + 2.1 + 1.44 + 0.32 = 4.86 > 0$, satisfied.
3. $(-1)^3P(-1) > 0$: $P(-1) = -1 + 2.1 - 1.44 + 0.32 = -0.02$, so $(-1)^3P(-1) = 0.02 > 0$, satisfied.

#### Jury table
$$
b_k = \begin{vmatrix} a_3 & a_{2-k} \\ a_0 & a_{k+1} \end{vmatrix}
$$

| Row | $z^0$ | $z^1$ | $z^2$ | $z^3$ |
|---|---|---|---|---|
| 1 | $a_3 = 0.32$ | $a_2 = 1.44$ | $a_1 = 2.1$ | $a_0 = 1$ |
| 2 | $a_0 = 1$ | $a_1 = 2.1$ | $a_2 = 1.44$ | $a_3 = 0.32$ |
| 3 | $b_2 = -0.8976$ | $b_1 = -1.6392$ | $b_0 = -0.768$ | |

$$
\begin{aligned}
b_2 &= a_3^2 - a_0^2 = 0.1024 - 1 = -0.8976 \\
b_1 &= a_3a_2 - a_0a_1 = 0.32(1.44) - 2.1 = -1.6392 \\
b_0 &= a_3a_1 - a_0a_2 = 0.32(2.1) - 1.44 = -0.768
\end{aligned}
$$

4. $|b_2| > |b_0|$: $0.8976 > 0.768$, satisfied.

#### Conclusion
All the Jury conditions are satisfied, so **the system is stable**.

**Check by factoring:** $z = -0.5$ is a root, since $-0.125 + 0.525 - 0.72 + 0.32 = 0$. Then $P(z) = (z+0.5)(z^2 + 1.6z + 0.64) = (z+0.5)(z+0.8)^2$. The roots are $-0.5, -0.8, -0.8$, all inside $|z| = 1$. The margin is small: $(-1)^3P(-1)$ is only 0.02 because two roots sit at $-0.8$, close to $z = -1$.

## Check yourself

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

1. For P(z) = z^3 + 2.1z^2 + 1.44z + 0.32, what is (-1)^3 P(-1)?
   A. -0.02
   B. 0.02
   C. 4.86
   D. -4.86
   Answer: B. P(-1) = -1 + 2.1 - 1.44 + 0.32 = -0.02, so (-1)^3 P(-1) = 0.02 > 0, which satisfies the condition. (Key point: Condition 3 sign)

2. What is the result of the b-row condition?
   A. |b2| = 0.8976 > |b0| = 0.768, satisfied
   B. |b2| = 0.768 > |b0| = 0.8976, satisfied
   C. |b2| = 0.8976 < |b0| = 0.768, violated
   D. No table is needed
   Answer: A. b2 = a3^2 - a0^2 = -0.8976 and b0 = a3 a1 - a0 a2 = -0.768. The magnitude of b2 is larger, so it holds. (Key point: Table condition)

3. What is the conclusion, and why is the margin small?
   A. Unstable, since P(-1) < 0
   B. Stable, but two roots at -0.8 are close to z = -1
   C. Critically stable, with a root at -1
   D. Stable, because all roots are at 0
   Answer: B. P(z) = (z+0.5)(z+0.8)^2, so all roots are inside the unit circle. The double root at -0.8 is near -1, giving the tiny 0.02. (Key point: Conclusion and margin)

## Close variants set in other exams

- [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×
- [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×

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