--- title: "Absolute stability analysis in the z-plane" subject: "Digital Control System" chapter: 3 key: "semester_6/digital_control_system:absolute-stability-z-plane" source: https://www.bctengineer.com/q/semester_6/digital_control_system/absolute-stability-z-plane site: BCT Engineering --- # Absolute stability analysis in the z-plane Digital Control System · Chapter 3: Analysis of Discrete Time Control System - Asked in: 2072 Magh, 8 marks; 2068 Magh, 4 marks (asked 2 times) - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_6/digital_control_system/absolute-stability-z-plane - Whole chapter: https://www.bctengineer.com/md/semester_6/digital_control_system/chapter-3 ## Question Discuss with example an absolute stability analysis method of a closed loop system in the Z-plane. ## Short answer Absolute stability requires all closed-loop poles inside |z|=1. Methods: Jury test (conditions: |a_n|0, (-1)^n P(-1)>0, etc.) or bilinear transformation plus Routh. Example: P(z)=z^3−1.3z^2−0.08z+0.24 fails P(1)>0, so the system is unstable. ## Answer **Absolute stability** asks only whether the system is stable or unstable. A closed-loop discrete-time system with pulse transfer function $C(z)/R(z) = G(z)/[1+GH(z)]$ is stable if all roots of the characteristic equation $$ P(z) = 1 + GH(z) = 0 $$ lie **inside the unit circle** $|z| < 1$. A simple pole on $|z| = 1$ makes the system critically stable. Any pole outside the circle, or a repeated pole on it, makes the system unstable. Factoring a high-order $P(z)$ is tedious, so two direct tests are used: 1. **Jury stability test**, applied directly in the z-plane. 2. **Bilinear transformation + Routh–Hurwitz**: substitute $z = (w+1)/(w-1)$. This maps the inside of the unit circle onto the left half of the w-plane, so the ordinary Routh test can be used. #### Jury stability test Write $P(z) = a_0z^n + a_1z^{n-1} + \dots + a_n$ with $a_0 > 0$. The system is stable if and only if all of these hold: 1. $|a_n| < a_0$ 2. $P(1) > 0$ 3. $(-1)^nP(-1) > 0$ 4. In the Jury table, $|b_{n-1}| > |b_0|$, $|c_{n-2}| > |c_0|$, … down to the row with three elements. The rows of the Jury table are built as: $$ b_k = \begin{vmatrix} a_n & a_{n-1-k} \\ a_0 & a_{k+1} \end{vmatrix},\qquad c_k = \begin{vmatrix} b_{n-1} & b_{n-2-k} \\ b_0 & b_{k+1} \end{vmatrix} $$ #### Example 1 Test $P(z) = z^3 - 1.3z^2 - 0.08z + 0.24 = 0$. Here $a_0 = 1$, $a_1 = -1.3$, $a_2 = -0.08$, $a_3 = 0.24$. | Condition | Value | Result | |---|---|---| | $\lvert a_3\rvert < a_0$ | $0.24 < 1$ | satisfied | | $P(1) > 0$ | $1 - 1.3 - 0.08 + 0.24 = -0.14$ | **violated** | | $(-1)^3P(-1) > 0$ | $-(-1 - 1.3 + 0.08 + 0.24) = 1.98$ | satisfied | | $\lvert b_2\rvert > \lvert b_0\rvert$ | $b_2 = -0.9424$, $b_0 = -0.232$ | satisfied | The condition $P(1) > 0$ fails, so the system is **unstable**. Factoring confirms this: the roots are $z = 1.2, 0.5, -0.4$, and $z = 1.2$ lies outside the unit circle. #### Example 2: stable range of gain Take $G(z) = \dfrac{K(0.3679z + 0.2642)}{(z-1)(z-0.3679)}$ with unity feedback. The characteristic equation is: $$ z^2 + (0.3679K - 1.3679)z + (0.3679 + 0.2642K) = 0 $$ - $|a_2| < 1$: $0.3679 + 0.2642K < 1$, so $K < 2.392$ - $P(1) = 0.6321K > 0$, so $K > 0$ - $P(-1) = 2.7358 - 0.1037K > 0$, so $K < 26.38$ **Answer:** The system is stable for $0 < K < 2.392$. At $K = 2.392$ it is critically stable and oscillates. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. When is a closed-loop discrete system absolutely stable? A. All roots of 1 + GH(z) = 0 lie inside the unit circle B. All roots lie on the unit circle C. All roots lie in the left half of the z-plane D. All zeros of GH(z) lie inside the unit circle Answer: A. Stability depends on the roots of the characteristic equation being inside |z| = 1. Zeros do not decide it. (Key point: Absolute stability criterion) 2. What does z = (w+1)/(w-1) allow? A. It maps the unit circle interior to the left half of the w-plane, so the Routh test can be used B. It maps the unit circle interior to the right half of the w-plane C. It removes the need for a characteristic equation D. It converts a discrete system into a continuous one exactly Answer: A. Under the bilinear transformation the inside of the unit circle becomes the left half-plane, so Routh-Hurwitz applies. (Key point: Bilinear transform) 3. Why is P(z) = z^3 - 1.3z^2 - 0.08z + 0.24 unstable? A. |a3| < a0 is violated B. P(1) = -0.14 is not greater than 0 C. (-1)^3 P(-1) is negative D. b2 is smaller than b0 Answer: B. P(1) = 1 - 1.3 - 0.08 + 0.24 = -0.14 < 0. This confirms the root at z = 1.2 outside the unit circle. (Key point: Example 1: failed condition) 4. For the second-order example, what is the stable gain range? A. 0 < K < 26.38 B. 0 < K < 2.392, critically stable at 2.392 C. K > 2.392 D. -5.178 < K < 26.38 Answer: B. The tightest condition is |a2| < 1, giving K < 2.392, together with K > 0 from P(1) > 0. 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