---
title: "Jury test: 2z⁴+7z³+10z²+4z+1"
subject: "Digital Control System"
chapter: 3
key: "semester_6/digital_control_system:jury-2z4-7z3"
source: https://www.bctengineer.com/q/semester_6/digital_control_system/jury-2z4-7z3
site: BCT Engineering
---

# Jury test: 2z⁴+7z³+10z²+4z+1

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

- Asked in: 2082 Kartik, 8 marks; 2080 Chaitra, 6 marks; 2078 Chaitra, 6 marks; 2073 Bhadra, 8 marks; 2070 Magh, 8 marks (asked 5 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-2z4-7z3
- 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)=2z⁴+7z³+10z²+4z+1=0.

## Short answer

After checking |a4|<a0, P(1)>0, P(-1)>0, the Jury table gives |b3|>|b0| satisfied but |c2|=8 < |c0|=20 fails. Therefore the system is unstable.

## 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

$P(z) = 2z^4 + 7z^3 + 10z^2 + 4z + 1$, so $n = 4$, $a_0 = 2$, $a_1 = 7$, $a_2 = 10$, $a_3 = 4$, $a_4 = 1$.

#### Conditions 1 to 3

- $|a_4| = 1 < a_0 = 2$: satisfied
- $P(1) = 2 + 7 + 10 + 4 + 1 = 24 > 0$: satisfied
- $(-1)^4P(-1) = 2 - 7 + 10 - 4 + 1 = 2 > 0$: satisfied

#### Jury table

$$
\begin{aligned}
b_0 &= a_4a_1 - a_0a_3 = 1(7) - 2(4) = -1 \\
b_1 &= a_4a_2 - a_0a_2 = 1(10) - 2(10) = -10 \\
b_2 &= a_4a_3 - a_0a_1 = 1(4) - 2(7) = -10 \\
b_3 &= a_4a_4 - a_0a_0 = 1 - 4 = -3
\end{aligned}
$$

$$
\begin{aligned}
c_0 &= b_3b_1 - b_0b_2 = (-3)(-10) - (-1)(-10) = 20 \\
c_1 &= b_3b_2 - b_0b_1 = (-3)(-10) - (-1)(-10) = 20 \\
c_2 &= b_3b_3 - b_0b_0 = 9 - 1 = 8
\end{aligned}
$$

| Row | $z^0$ | $z^1$ | $z^2$ | $z^3$ | $z^4$ |
|---|---|---|---|---|---|
| 1 | 1 | 4 | 10 | 7 | 2 |
| 2 | 2 | 7 | 10 | 4 | 1 |
| 3 | $b_3 = -3$ | $b_2 = -10$ | $b_1 = -10$ | $b_0 = -1$ | |
| 4 | $b_0 = -1$ | $b_1 = -10$ | $b_2 = -10$ | $b_3 = -3$ | |
| 5 | $c_2 = 8$ | $c_1 = 20$ | $c_0 = 20$ | | |

#### Condition 4

- $|b_3| = 3 > |b_0| = 1$: satisfied
- $|c_2| = 8 > |c_0| = 20$: **not satisfied**

#### Conclusion

One Jury condition fails, so the system is **unstable**.

Check: the roots of $P(z)$ are $z = -1.526 \pm j1.078$ (magnitude 1.868, outside the unit circle) and $z = -0.224 \pm j0.305$ (magnitude 0.379, inside). Two roots lie outside the unit circle, which confirms the result.

**Answer:** Unstable, because $|c_2| = 8 < |c_0| = 20$.

## Check yourself

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

1. What does the Jury test establish about P(z)?
   A. Whether all roots of P(z) lie inside the unit circle |z| = 1
   B. Whether all roots lie in the left half of the z-plane
   C. Whether all roots are real
   D. Whether all roots lie outside the unit circle
   Answer: A. The Jury test checks, from the coefficients alone, whether every root of P(z) is inside the unit circle, which is the stability requirement of a discrete system. (Key point: Stability condition: roots inside unit circle)

2. For P(z) = 2z^4 + 7z^3 + 10z^2 + 4z + 1, what are P(1) and (-1)^4 P(-1)?
   A. P(1) = 24 and P(-1) = 24
   B. P(1) = 2 and P(-1) = 24
   C. P(1) = 24 and P(-1) = 2
   D. P(1) = -24 and P(-1) = 2
   Answer: C. P(1) = 2+7+10+4+1 = 24 and P(-1) = 2-7+10-4+1 = 2. Both are positive, so conditions 2 and 3 are satisfied. (Key point: Conditions 1 to 3)

3. Which Jury condition fails for this polynomial?
   A. |a4| < a0
   B. |b3| > |b0|
   C. P(1) > 0
   D. |c2| > |c0|, since 8 is less than 20
   Answer: D. The first three conditions and |b3| = 3 > |b0| = 1 hold, but |c2| = 8 is not greater than |c0| = 20, so the last table condition fails. (Key point: Table condition that fails)

4. What is the conclusion, and how is it confirmed?
   A. Stable, because conditions 1 to 3 are satisfied
   B. Unstable, because two roots lie outside the unit circle (magnitude 1.868)
   C. Critically stable, because a root lies on the unit circle
   D. Unstable, because P(1) is too large
   Answer: B. A single failed condition makes the system unstable. The roots -1.526 +/- j1.078 have magnitude 1.868 > 1, confirming it. (Key point: Conclusion and check)

## Close variants set in other exams

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