Highlights

• Definitive Instability: Both the Jury stability criterion and the modified Routh criterion (via bilinear transformation) conclusively determine that the system characterized by \( M(z) = z^4 - 3.0z^3 + 4.1z^2 - 3.0z + 4.0 \) is unstable.
• Jury Criterion Failure: The system fails a fundamental necessary condition of the Jury test: \(|a_0| < a_n\) is not satisfied (\(|4.0| \not< 1.0\)), immediately indicating instability.
• Modified Routh Confirmation: Applying the bilinear transformation \(z = \frac{s+1}{s-1}\) yields an s-domain polynomial whose Routh array reveals sign changes in the first column, confirming instability by indicating roots outside the stable region (left-half plane for 's', which corresponds to inside the unit circle for 'z').

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Understanding Stability in Discrete-Time Systems

The Unit Circle Condition

For a linear time-invariant (LTI) discrete-time system, stability is a crucial property. It essentially means that a bounded input will always produce a bounded output (BIBO stability). Mathematically, this translates to a condition on the roots of the system's characteristic polynomial, which is given in the z-domain. For the system to be stable, all the roots (poles) of its characteristic polynomial \( M(z) \) must lie strictly inside the unit circle in the complex z-plane. That is, the magnitude of every root \( z_i \) must satisfy \( |z_i| < 1 \).

Visual representation of the stability region (inside the unit circle) for discrete-time systems in the z-plane.

We are given the characteristic polynomial:

\[ M(z) = z^4 - 3.0z^3 + 4.1z^2 - 3.0z + 4.0 \]

To determine if the system represented by \( M(z) \) is stable, we will use two algebraic methods: the Jury stability criterion and the modified Routh criterion.

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Method 1: Jury Stability Criterion

Understanding the Jury Criterion

The Jury stability criterion is a direct method for checking the stability of discrete-time systems without explicitly calculating the roots of the characteristic polynomial. It's the discrete-time equivalent of the Routh-Hurwitz criterion used for continuous-time systems. It involves examining the coefficients of the polynomial \( M(z) = a_n z^n + a_{n-1} z^{n-1} + \cdots + a_1 z + a_0 \) through a series of necessary and sufficient conditions.

Applying the Jury Criterion to \( M(z) \)

Coefficients Identification

For the given polynomial \( M(z) = z^4 - 3.0z^3 + 4.1z^2 - 3.0z + 4.0 \), the degree is \( n=4 \) and the coefficients are:

• \( a_4 = 1.0 \) (leading coefficient)
• \( a_3 = -3.0 \)
• \( a_2 = 4.1 \)
• \( a_1 = -3.0 \)
• \( a_0 = 4.0 \) (constant term)

Condition 1: \( M(1) > 0 \)

We evaluate the polynomial at \( z = 1 \):

\[ M(1) = (1)^4 - 3.0(1)^3 + 4.1(1)^2 - 3.0(1) + 4.0 \] \[ M(1) = 1 - 3.0 + 4.1 - 3.0 + 4.0 = 3.1 \]

Since \( M(1) = 3.1 > 0 \), this condition is satisfied.

Condition 2: \( (-1)^n M(-1) > 0 \)

Since \( n = 4 \) is even, the condition simplifies to \( M(-1) > 0 \). We evaluate the polynomial at \( z = -1 \):

\[ M(-1) = (-1)^4 - 3.0(-1)^3 + 4.1(-1)^2 - 3.0(-1) + 4.0 \] \[ M(-1) = 1 - 3.0(-1) + 4.1(1) - 3.0(-1) + 4.0 \] \[ M(-1) = 1 + 3.0 + 4.1 + 3.0 + 4.0 = 15.1 \]

Since \( M(-1) = 15.1 > 0 \), this condition is also satisfied.

Condition 3: \( |a_0| < a_n \)

This condition compares the magnitude of the constant term \( a_0 \) with the leading coefficient \( a_n \). Note that for this condition, \( a_n \) must be positive, which it is (\(a_4 = 1.0\)).

\[ |a_0| = |4.0| = 4.0 \] \[ a_n = a_4 = 1.0 \]

The condition requires \( |a_0| < a_n \), which translates to \( 4.0 < 1.0 \). This inequality is clearly false.

Conclusion from Jury Criterion

The Jury stability criterion requires all conditions to be met for stability. Since Condition 3 (\( |a_0| < a_n \)) failed, we can immediately conclude that the system is unstable based on the Jury criterion. There is no need to construct the full Jury table or check further conditions.

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Method 2: Modified Routh Criterion via Bilinear Transformation

Understanding the Approach

The standard Routh-Hurwitz criterion applies to continuous-time systems, checking for roots in the left-half of the s-plane. To adapt it for discrete-time systems (checking for roots inside the unit circle of the z-plane), we use a bilinear transformation. This transformation maps the interior of the unit circle in the z-plane to the left-half of the s-plane.

The Bilinear Transformation

A common bilinear transformation is:

\[ z = \frac{s+1}{s-1} \]

By substituting this into the characteristic polynomial \( M(z) \), we obtain a new polynomial \( P(s) \) in the s-domain. If the original system \( M(z) \) is stable (all roots inside the unit circle), the transformed system \( P(s) \) will also be stable according to the Routh-Hurwitz criterion (all roots in the left-half s-plane).

Applying the Transformation to \( M(z) \)

Substituting z

Substitute \( z = \frac{s+1}{s-1} \) into \( M(z) = z^4 - 3.0z^3 + 4.1z^2 - 3.0z + 4.0 = 0 \):

\[ \left(\frac{s+1}{s-1}\right)^4 - 3.0\left(\frac{s+1}{s-1}\right)^3 + 4.1\left(\frac{s+1}{s-1}\right)^2 - 3.0\left(\frac{s+1}{s-1}\right) + 4.0 = 0 \]

To clear the denominators, multiply the entire equation by \( (s-1)^4 \):

\[ (s+1)^4 - 3.0(s+1)^3(s-1) + 4.1(s+1)^2(s-1)^2 - 3.0(s+1)(s-1)^3 + 4.0(s-1)^4 = 0 \]

Deriving the s-domain Polynomial P(s)

Expanding and collecting terms (a somewhat tedious algebraic process):

• \( (s+1)^4 = s^4 + 4s^3 + 6s^2 + 4s + 1 \)
• \( (s+1)^3(s-1) = (s^3 + 3s^2 + 3s + 1)(s-1) = s^4 + 2s^3 - 2s - 1 \)
• \( (s+1)^2(s-1)^2 = ((s+1)(s-1))^2 = (s^2 - 1)^2 = s^4 - 2s^2 + 1 \)
• \( (s+1)(s-1)^3 = (s+1)(s^3 - 3s^2 + 3s - 1) = s^4 - 2s^3 + 2s - 1 \)
• \( (s-1)^4 = s^4 - 4s^3 + 6s^2 - 4s + 1 \)

Substituting these back into the equation:

\[ (s^4 + 4s^3 + 6s^2 + 4s + 1) - 3.0(s^4 + 2s^3 - 2s - 1) + 4.1(s^4 - 2s^2 + 1) - 3.0(s^4 - 2s^3 + 2s - 1) + 4.0(s^4 - 4s^3 + 6s^2 - 4s + 1) = 0 \]

Combine coefficients for each power of \( s \):

• \( s^4 \): \( 1 - 3.0 + 4.1 - 3.0 + 4.0 = 3.1 \)
• \( s^3 \): \( 4 - 3.0(2) + 4.1(0) - 3.0(-2) + 4.0(-4) = 4 - 6 + 0 + 6 - 16 = -12 \)
• \( s^2 \): \( 6 - 3.0(0) + 4.1(-2) - 3.0(0) + 4.0(6) = 6 + 0 - 8.2 + 0 + 24 = 21.8 \) *Correction: Answer A had 13.6, let's recheck.* * Check (s+1)^2(s-1)^2 = (s^2+2s+1)(s^2-2s+1) = s^2(s^2-2s+1)+2s(s^2-2s+1)+1(s^2-2s+1) = s^4-2s^3+s^2 + 2s^3-4s^2+2s + s^2-2s+1 = s^4 - 2s^2 + 1. This is correct. * Check term collection: \( s^2 \): \( 6 - 3.0(0) + 4.1(-2) - 3.0(0) + 4.0(6) = 6 - 8.2 + 24 = 21.8 \). Yes, 21.8 is correct. Answer A's derivation had an error here. Let's use 21.8.
• \( s \): \( 4 - 3.0(-2) + 4.1(0) - 3.0(2) + 4.0(-4) = 4 + 6 + 0 - 6 - 16 = -12 \)
• Constant: \( 1 - 3.0(-1) + 4.1(1) - 3.0(-1) + 4.0(1) = 1 + 3 + 4.1 + 3 + 4 = 15.1 \)

The resulting polynomial in the s-domain is:

\[ P(s) = 3.1s^4 - 12s^3 + 21.8s^2 - 12s + 15.1 = 0 \]

Applying the Routh-Hurwitz Criterion to P(s)

Constructing the Routh Array

We use the coefficients of \( P(s) \) to build the Routh array:

• \( a_4 = 3.1 \)
• \( a_3 = -12 \)
• \( a_2 = 21.8 \)
• \( a_1 = -12 \)
• \( a_0 = 15.1 \)

The Routh array is constructed as follows:

Row s⁴: 3.1 | 21.8 | 15.1

Row s³: -12 | -12 | 0

Row s²: b₁ | b₂ | b₃

Row s¹: c₁ | c₂ | c₃

Row s⁰: d₁ | d₂ | d₃

Calculate the elements:

\[ b_1 = \frac{(-12)(21.8) - (3.1)(-12)}{-12} = \frac{-261.6 + 37.2}{-12} = \frac{-224.4}{-12} = 18.7 \] \[ b_2 = \frac{(-12)(15.1) - (3.1)(0)}{-12} = \frac{-181.2}{-12} = 15.1 \] \[ b_3 = 0 \]

Row s²: 18.7 | 15.1 | 0

\[ c_1 = \frac{(18.7)(-12) - (-12)(15.1)}{18.7} = \frac{-224.4 + 181.2}{18.7} = \frac{-43.2}{18.7} \approx -2.31 \] \[ c_2 = 0 \] \[ c_3 = 0 \]

Row s¹: -2.31 | 0 | 0

\[ d_1 = \frac{(-2.31)(15.1) - (18.7)(0)}{-2.31} = \frac{-34.881}{-2.31} = 15.1 \]

Row s⁰: 15.1 | 0 | 0

Analyzing the First Column

The first column of the completed Routh array is: [3.1, -12, 18.7, -2.31, 15.1]

The Routh-Hurwitz stability criterion states that a system is stable if and only if all elements in the first column of the Routh array have the same sign. If there are sign changes, the system is unstable, and the number of sign changes equals the number of roots of \( P(s) \) in the right-half of the s-plane.

In our first column, the signs change from:

• 3.1 (+) to -12 (-) : Sign change 1
• -12 (-) to 18.7 (+) : Sign change 2
• 18.7 (+) to -2.31 (-) : Sign change 3
• -2.31 (-) to 15.1 (+) : Sign change 4

There are four sign changes in the first column.

Conclusion from Modified Routh Criterion

Since there are sign changes (specifically, four) in the first column of the Routh array for \( P(s) \), the polynomial \( P(s) \) has four roots with positive real parts (in the right-half s-plane). This implies that the original polynomial \( M(z) \) has four roots outside the unit circle. Therefore, the system is unstable according to the modified Routh criterion.

Note: Both methods consistently indicate instability, although the exact Routh array calculations differed slightly from Answer A due to a re-evaluation of the P(s) coefficients. The Jury criterion provided the quickest route to the conclusion in this specific case.

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Comparative Analysis of Stability Methods

Both Jury and the modified Routh (via bilinear transform) criteria are algebraic tests used to assess the stability of discrete-time systems without finding the roots. They differ in their approach and sometimes complexity.

• Jury Criterion: Works directly with the z-domain polynomial coefficients. It involves checking specific inequalities and potentially constructing a table. It can sometimes detect instability very early if necessary conditions fail.
• Modified Routh Criterion: Requires transforming the z-domain polynomial to the s-domain using bilinear transformation, which can be algebraically intensive. Then, the standard Routh-Hurwitz procedure is applied to the resulting s-domain polynomial.

The following chart provides a qualitative comparison of these methods based on common characteristics:

Radar Chart: Stability Test Characteristics

Interpretation: This chart subjectively rates the two methods (scale 1-10, higher is better/more). Jury is more direct for discrete systems but can be computationally intensive manually. Modified Routh involves complex setup (transformation) but leverages the well-known Routh-Hurwitz procedure, which is often well-supported in software and provides information about the number of unstable roots.

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Visualizing the Stability Concepts

Understanding the relationships between the polynomial, its roots, the stability criteria, and the final stability verdict is key. This mindmap illustrates these connections for discrete-time systems.

Mindmap: Key Concepts in Discrete System Stability

Interpretation: This mindmap outlines the process. The stability condition requires roots inside the unit circle. Both Jury and Modified Routh tests are applied. Jury fails early on a necessary condition. Modified Routh involves transformation and then Routh array analysis, which shows sign changes. Both methods lead to the same conclusion: instability.

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Summary Table of Results

This table summarizes the analysis performed on the given characteristic polynomial.

Aspect	Details
Characteristic Polynomial	\( M(z) = z^4 - 3.0z^3 + 4.1z^2 - 3.0z + 4.0 \)
System Type	Discrete-Time LTI System
Stability Requirement	All roots \(|z| < 1\)
Method 1: Jury Criterion	Applied directly to \( M(z) \) coefficients
Jury Conditions Checked	\( M(1) = 3.1 > 0 \) (OK) \( M(-1) = 15.1 > 0 \) (OK) \( |a_0| < a_n \implies 4.0 < 1.0 \) (FAILED)
Jury Conclusion	Unstable
Method 2: Modified Routh Criterion	Applied Routh-Hurwitz to \( P(s) \) after bilinear transform \( z = \frac{s+1}{s-1} \)
Transformed Polynomial	\( P(s) = 3.1s^4 - 12s^3 + 21.8s^2 - 12s + 15.1 \)
Routh Array First Column	[3.1, -12, 18.7, -2.31, 15.1]
Routh Analysis	4 sign changes detected
Modified Routh Conclusion	Unstable (4 roots outside the unit circle implied)
Overall Conclusion	The system is definitively UNSTABLE.

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Related Video Explanation

While the Jury test is specific to discrete systems, the modified Routh method relies on the standard Routh-Hurwitz criterion after transformation. Understanding the Routh-Hurwitz criterion itself is fundamental. The following video provides an introduction and a simple example of the Routh stability criterion, which forms the basis for the second part of our analysis.

Video explaining the basics of the Routh Stability Criterion.

This video explains the core concept of constructing the Routh array and interpreting the first column for stability in continuous-time systems (s-domain). This is precisely the procedure we applied to the transformed polynomial \( P(s) \) in the modified Routh criterion analysis.

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Frequently Asked Questions (FAQ)

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Recommended Further Exploration

• How do you construct the full Jury table for a 4th order polynomial?
• What are the practical implications of having unstable poles in a control system?
• Compare the Jury criterion with the Schur-Cohn test for discrete systems stability.
• How can an unstable discrete-time system be stabilized using feedback control?

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References