--- title: "Jury stability test explained" subject: "Digital Control System" chapter: 3 key: "semester_6/digital_control_system:what-is-jury-test" source: https://www.bctengineer.com/q/semester_6/digital_control_system/what-is-jury-test site: BCT Engineering --- # Jury stability test explained Digital Control System · Chapter 3: Analysis of Discrete Time Control System - Asked in: 2071 Magh, 6 marks - 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/what-is-jury-test - Whole chapter: https://www.bctengineer.com/md/semester_6/digital_control_system/chapter-3 ## Question What is Jury stability test in digital control system? Explain. ## Short answer The Jury stability test is an algebraic test that determines if all roots of a discrete-time characteristic polynomial lie inside the unit circle without solving for roots. It uses determinants of the coefficients to build a table and checks conditions; if all hold, the system is stable. ## Answer The **Jury stability test** is an algebraic test for the absolute stability of a discrete-time (sampled-data) system. It works directly in the z-plane. Using only the coefficients of the characteristic polynomial $P(z)$, it tells whether all roots lie inside the unit circle $|z| = 1$, without solving for the roots. It does for the z-plane what the Routh–Hurwitz test does for the s-plane. #### Procedure Write the characteristic equation as $$ P(z) = a_0z^n + a_1z^{n-1} + \dots + a_{n-1}z + a_n = 0, \qquad a_0 > 0 $$ Form the Jury table: | Row | $z^0$ | $z^1$ | $\dots$ | $z^{n-1}$ | $z^n$ | |---|---|---|---|---|---| | 1 | $a_n$ | $a_{n-1}$ | $\dots$ | $a_1$ | $a_0$ | | 2 | $a_0$ | $a_1$ | $\dots$ | $a_{n-1}$ | $a_n$ | | 3 | $b_{n-1}$ | $b_{n-2}$ | $\dots$ | $b_0$ | | | 4 | $b_0$ | $b_1$ | $\dots$ | $b_{n-1}$ | | | 5 | $c_{n-2}$ | $\dots$ | $c_0$ | | | | ⋮ | | | | | | | $2n-3$ | $q_2$ | $q_1$ | $q_0$ | | | Each even row is the row above it written in reverse. The elements are $2\times2$ determinants: $$ b_k = \begin{vmatrix} a_n & a_{n-1-k} \\ a_0 & a_{k+1} \end{vmatrix},\quad c_k = \begin{vmatrix} b_{n-1} & b_{n-2-k} \\ b_0 & b_{k+1} \end{vmatrix},\ \dots $$ #### Stability conditions The system is stable if and only if **all** of these hold: 1. $|a_n| < a_0$ 2. $P(z)\big|_{z=1} > 0$ 3. $(-1)^nP(z)\big|_{z=-1} > 0$ 4. $|b_{n-1}| > |b_0|$, $|c_{n-2}| > |c_0|$, …, $|q_2| > |q_0|$ There are $n+1$ conditions in total. Check conditions 1–3 first. If any of them fails, the system is unstable and no table is needed. For $n = 2$ no table is needed at all. #### Example Take $G(z) = \dfrac{K(0.3679z+0.2642)}{(z-1)(z-0.3679)}$ with unity feedback. The characteristic equation is $$ P(z) = z^2 + (0.3679K - 1.3679)z + (0.3679 + 0.2642K) = 0 $$ 1. $|a_2| < 1$: $0.3679 + 0.2642K < 1$, so $K < 2.392$ 2. $P(1) = 0.6321K > 0$, so $K > 0$ 3. $P(-1) = 2.7358 - 0.1037K > 0$, so $K < 26.38$ The system is stable for $0 < K < 2.392$. #### Features - It works directly with z-domain coefficients, so no bilinear transformation is needed. - It also gives the range of a gain $K$ for stability, as in the example. - If a condition becomes an equality (for example $P(1) = 0$), a root lies on the unit circle and the system is critically stable. - It shows only absolute stability, not relative stability (damping, margins). ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. What does the Jury test do? A. Finds the exact roots of P(z) B. Shows from the coefficients whether all roots lie inside the unit circle, without solving for them C. Gives damping ratio and margins D. Converts P(z) to the s-domain Answer: B. It is the z-plane counterpart of Routh-Hurwitz and tests absolute stability from coefficients only. (Key point: Purpose of Jury test) 2. Which set of conditions must all hold for stability? A. |a_n| < a0, P(1) > 0, (-1)^n P(-1) > 0, and |b(n-1)| > |b0|, |c(n-2)| > |c0|, ... B. |a_n| > a0, P(1) < 0 and P(-1) < 0 C. Only P(1) > 0 and P(-1) > 0 D. All a_i positive Answer: A. There are n+1 conditions in total: three on the coefficients and the rest from the table rows. (Key point: Stability conditions) 3. How should the conditions be checked? A. Always build the whole table first B. Check conditions 1 to 3 first; if one fails the system is unstable and no table is needed C. Check the table only D. Check only |a_n| < a0 Answer: B. The first three need no table, and any failure already proves instability. For n = 2 the table is never needed. (Key point: Efficient use) 4. What does an equality such as P(1) = 0 mean, and what does the test not give? A. A root is on the unit circle (critically stable); it gives only absolute, not relative, stability B. A root is outside the circle; it gives relative stability C. All roots are inside; it gives damping D. Nothing; it gives phase margin Answer: A. An equality puts a root on the unit circle. The test does not give damping or margins. 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