---
title: "How to study stability; test a cubic"
subject: "Digital Control System"
chapter: 3
key: "semester_6/digital_control_system:stability-study-method-cubic"
source: https://www.bctengineer.com/q/semester_6/digital_control_system/stability-study-method-cubic
site: BCT Engineering
---

# How to study stability; test a cubic

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

- Asked in: 2079 Chaitra, 8 marks
- Video: [Jury's Stability Test with examples|| EET305 Signals and systems || Control System](https://www.youtube.com/watch?v=SKugo4Ml1xk) — Spark Lab
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_6/digital_control_system/stability-study-method-cubic
- Whole chapter: https://www.bctengineer.com/md/semester_6/digital_control_system/chapter-3

## Question

How do you perform the stability study of discrete time control system? Perform the stability test of the digital control system having characteristic equation given below: P(z)=z³-1.1z²-0.1z+0.2.

## Short answer

Stability of a discrete-time system is checked by verifying all characteristic roots lie inside |z|=1, using Jury test, bilinear transformation, or factoring. For P(z)=z^3−1.1z^2−0.1z+0.2, Jury test gives P(1)=0, so a pole is at z=1; the system is critically stable. Roots: 1, 0.5, −0.4.

## Answer

#### How the stability of a discrete-time control system is studied
A linear discrete-time system is studied through the roots (poles) of its closed-loop characteristic equation $P(z) = 1 + GH(z) = 0$.

| Location of closed-loop poles | Nature of system |
|---|---|
| All inside the unit circle $\lvert z\rvert < 1$ | asymptotically stable |
| Simple pole(s) on $\lvert z\rvert = 1$, rest inside | critically (marginally) stable |
| Any pole outside, or a repeated pole on $\lvert z\rvert = 1$ | unstable |

Zeros do not affect absolute stability. The usual methods are:
1. **Direct root finding:** factor $P(z)$. This is practical only for low order.
2. **Jury stability test:** a tabular test on the coefficients of $P(z)$, done directly in the z-plane.
3. **Bilinear transformation + Routh test:** put $z = \frac{w+1}{w-1}$ and apply the Routh array in the w-plane.
4. **Root locus and Nyquist/Bode methods in the z-plane**, mainly for relative stability and gain design.

#### Jury test for $P(z) = z^3 - 1.1z^2 - 0.1z + 0.2$
Here $a_0 = 1$, $a_1 = -1.1$, $a_2 = -0.1$, $a_3 = 0.2$, and $n = 3$.

| Condition | Calculation | Result |
|---|---|---|
| $\lvert a_3\rvert < a_0$ | $0.2 < 1$ | satisfied |
| $P(1) > 0$ | $1 - 1.1 - 0.1 + 0.2 = 0$ | **not satisfied** (equal to 0) |
| $(-1)^3P(-1) > 0$ | $P(-1) = -1 - 1.1 + 0.1 + 0.2 = -1.8$, so $1.8 > 0$ | satisfied |

Jury table, row 3:

$$
\begin{aligned}
b_2 &= a_3^2 - a_0^2 = 0.04 - 1 = -0.96 \\
b_1 &= a_3a_2 - a_0a_1 = -0.02 + 1.1 = 1.08 \\
b_0 &= a_3a_1 - a_0a_2 = -0.22 + 0.1 = -0.12
\end{aligned}
$$

| Row | | | | |
|---|---|---|---|---|
| 1 | 0.2 | −0.1 | −1.1 | 1 |
| 2 | 1 | −1.1 | −0.1 | 0.2 |
| 3 | −0.96 | 1.08 | −0.12 | |

$|b_2| = 0.96 > |b_0| = 0.12$, satisfied.

#### Conclusion
Every condition holds except $P(1) > 0$, because $P(1) = 0$. This means there is a closed-loop pole exactly at $z = 1$. So the system is **not asymptotically stable**. It is **critically (marginally) stable**.

**Check by factoring:** $P(z) = (z-1)(z^2 - 0.1z - 0.2) = (z-1)(z-0.5)(z+0.4)$. The roots are $1, 0.5, -0.4$. There is one simple root on the unit circle and the other two are inside. The output will contain a non-decaying constant component.

## Check yourself

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

1. What happens if a simple closed-loop pole lies on |z| = 1 and the rest are inside?
   A. Asymptotically stable
   B. Critically (marginally) stable
   C. Unstable
   D. Not defined
   Answer: B. A simple pole on the unit circle with all others inside gives critical stability. Any pole outside, or a repeated one on the circle, gives instability. (Key point: Pole locations and stability)

2. Which is a direct method for stability in the z-plane without solving for roots?
   A. Jury stability test
   B. Partial fractions
   C. Final value theorem
   D. Inverse z-transform
   Answer: A. The Jury test uses only the coefficients of P(z). Bilinear transformation plus Routh is the other tabular option. (Key point: Methods of stability study)

3. For P(z) = z^3 - 1.1z^2 - 0.1z + 0.2, which condition is not satisfied?
   A. |a3| < a0
   B. P(1) > 0, since P(1) = 0
   C. (-1)^3 P(-1) > 0
   D. |b2| > |b0|
   Answer: B. P(1) = 1 - 1.1 - 0.1 + 0.2 = 0, so a pole sits at z = 1. All other conditions are satisfied. (Key point: Condition that fails)

4. What is the verdict, and what does the output contain?
   A. Stable; the output decays to zero
   B. Critically stable; the output has a non-decaying constant component
   C. Unstable; the output grows without limit
   D. Unstable; the output oscillates with growing amplitude
   Answer: B. Roots 1, 0.5 and -0.4 mean one simple root on the unit circle, so the system is not asymptotically stable and a constant component persists. (Key point: Conclusion)

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