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How to study stability; test a cubic

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.

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.

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)=0P(z) = 1 + GH(z) = 0.

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

Zeros do not affect absolute stability. The usual methods are:

  1. Direct root finding: factor P(z)P(z). This is practical only for low order.
  2. Jury stability test: a tabular test on the coefficients of P(z)P(z), done directly in the z-plane.
  3. Bilinear transformation + Routh test: put z=w+1w−1z = \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)=z3−1.1z2−0.1z+0.2P(z) = z^3 - 1.1z^2 - 0.1z + 0.2

Here a0=1a_0 = 1, a1=−1.1a_1 = -1.1, a2=−0.1a_2 = -0.1, a3=0.2a_3 = 0.2, and n=3n = 3.

ConditionCalculationResult
∣a3∣<a0\lvert a_3\rvert < a_00.2<10.2 < 1satisfied
P(1)>0P(1) > 01−1.1−0.1+0.2=01 - 1.1 - 0.1 + 0.2 = 0not satisfied (equal to 0)
(−1)3P(−1)>0(-1)^3P(-1) > 0P(−1)=−1−1.1+0.1+0.2=−1.8P(-1) = -1 - 1.1 + 0.1 + 0.2 = -1.8, so 1.8>01.8 > 0satisfied

Jury table, row 3:

b2=a32−a02=0.04−1=−0.96b1=a3a2−a0a1=−0.02+1.1=1.08b0=a3a1−a0a2=−0.22+0.1=−0.12\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
10.2−0.1−1.11
21−1.1−0.10.2
3−0.961.08−0.12

∣b2∣=0.96>∣b0∣=0.12|b_2| = 0.96 > |b_0| = 0.12, satisfied.

Conclusion

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

Check by factoring: P(z)=(z−1)(z2−0.1z−0.2)=(z−1)(z−0.5)(z+0.4)P(z) = (z-1)(z^2 - 0.1z - 0.2) = (z-1)(z-0.5)(z+0.4). The roots are 1,0.5,−0.41, 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.

✓ Answer checked by Claude Opus 5.5 (high effort)

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