--- title: "Range of K for stability (Jury test)" subject: "Digital Control System" chapter: 3 key: "semester_6/digital_control_system:range-of-k-jury" source: https://www.bctengineer.com/q/semester_6/digital_control_system/range-of-k-jury site: BCT Engineering --- # Range of K for stability (Jury test) Digital Control System · Chapter 3: Analysis of Discrete Time Control System - Asked in: 2082 Chaitra, 8 marks; 2081 Chaitra, 6 marks; 2076 Bhadra, 8 marks (asked 3 times) - Video: [Jury Stability in Z-PLANE](https://www.youtube.com/watch?v=Fc8gLJJg4hA) — Yash Jain - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_6/digital_control_system/range-of-k-jury - Whole chapter: https://www.bctengineer.com/md/semester_6/digital_control_system/chapter-3 ## Question Consider the discrete-time unity-feedback control system whose open-loop pulse transfer function is given by G(z)=K(0.3679z+0.2642)/((z-0.3679)(z-1)). Determine the range of gain K for stability by using the Jury stability test. ## Short answer Closed-loop characteristic equation is z^2+(0.3679K-1.3679)z+(0.3679+0.2642K)=0. Jury conditions for n=2 give |a2|<1 => K<2.3925, P(1)>0 => K>0, P(-1)>0 => K<26.38. So stable for 0 < K < 2.3925. ## Answer #### Step 1: Characteristic equation For unity feedback, $1 + G(z) = 0$: $$ (z - 0.3679)(z - 1) + K(0.3679z + 0.2642) = 0 $$ $$ P(z) = z^2 + (0.3679K - 1.3679)z + (0.3679 + 0.2642K) = 0 $$ So $n = 2$, $a_0 = 1$, $a_1 = 0.3679K - 1.3679$, $a_2 = 0.3679 + 0.2642K$. #### Step 2: Jury conditions for a second-order system For $n = 2$ there are only three conditions: $|a_2| < a_0$, $P(1) > 0$, and $P(-1) > 0$. No table rows are needed. **Condition 1: $|a_2| < a_0$** $$ |0.3679 + 0.2642K| < 1 $$ $$ \Rightarrow\; -1 < 0.3679 + 0.2642K < 1 \;\Rightarrow\; -5.178 < K < \frac{1 - 0.3679}{0.2642} = 2.3925 $$ **Condition 2: $P(1) > 0$** $$ P(1) = 1 + 0.3679K - 1.3679 + 0.3679 + 0.2642K = 0.6321K > 0 \;\Rightarrow\; K > 0 $$ **Condition 3: $(-1)^2P(-1) > 0$** $$ P(-1) = 1 - 0.3679K + 1.3679 + 0.3679 + 0.2642K = 2.7358 - 0.1037K > 0 \;\Rightarrow\; K < 26.38 $$ #### Step 3: Combine | Condition | Requirement | |---|---| | $\lvert a_2\rvert < a_0$ | $-5.178 < K < 2.3925$ | | $P(1) > 0$ | $K > 0$ | | $P(-1) > 0$ | $K < 26.38$ | The intersection of all three is $$ 0 < K < 2.3925 $$ #### Step 4: Check at the critical gain At $K = 2.3925$: $P(z) = z^2 - 0.4877z + 1 = 0$, which gives $z = 0.244 \pm j0.970$ with $|z| = 1.000$. The poles lie on the unit circle, so the system is marginally stable and oscillates. This $G(z)$ is the ZOH equivalent of $\dfrac{K}{s(s+1)}$ with $T = 1$ s. The continuous system is stable for every $K > 0$, but sampling limits the stable gain to $K < 2.39$. **Answer:** The closed-loop system is stable for $0 < K < 2.3925$. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. For unity feedback, which equation gives the characteristic polynomial P(z)? A. G(z) = 0 B. 1 + G(z) = 0 C. 1 - G(z) = 0 D. G(z)/(1+G(z)) = 0 Answer: B. The closed-loop poles are the roots of 1 + G(z) = 0, which gives z^2 + (0.3679K - 1.3679)z + (0.3679 + 0.2642K) = 0. (Key point: Characteristic equation) 2. Which conditions are needed for n = 2 in the Jury test? A. Only |a2| < a0 and P(1) > 0 B. |a2| < a0, P(1) > 0 and P(-1) > 0, with no table rows C. A full table with b and c rows D. Only P(1) > 0 and P(-1) > 0 Answer: B. For a second-order polynomial the three coefficient conditions are enough, because the Jury table has no rows to test. (Key point: Conditions for second order) 3. What does P(1) > 0 give here? A. K < 2.3925 B. K > 0, since P(1) = 0.6321K C. K < 26.38 D. K > 5.178 Answer: B. P(1) = 1 + (0.3679K - 1.3679) + (0.3679 + 0.2642K) = 0.6321K, so K must be positive. (Key point: P(1) condition gives lower bound) 4. What is the stable range of K, and why is the upper limit 2.3925? A. 0 < K < 26.38, from the P(-1) condition B. 0 < K < 2.3925, the intersection, where |a2| < 1 is the tightest condition C. -5.178 < K < 2.3925, from the first condition alone D. K > 0 with no upper limit, like the continuous system Answer: B. The three requirements are intersected. |a2| < 1 gives K < 2.3925, which is stricter than K < 26.38. At 2.3925 the poles sit on the unit circle. 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