--- title: "Jury test: z⁴−1.2z³+0.07z²+0.3z−0.08" subject: "Digital Control System" chapter: 3 key: "semester_6/digital_control_system:jury-z4-1-2z3-0-3z" source: https://www.bctengineer.com/q/semester_6/digital_control_system/jury-z4-1-2z3-0-3z site: BCT Engineering --- # Jury test: z⁴−1.2z³+0.07z²+0.3z−0.08 Digital Control System · Chapter 3: Analysis of Discrete Time Control System - Asked in: 2075 Bhadra, 6 marks; 2071 Bhadra, 8 marks (asked 2 times) - 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/jury-z4-1-2z3-0-3z - Whole chapter: https://www.bctengineer.com/md/semester_6/digital_control_system/chapter-3 ## Question Examine the stability of the following characteristic equation using Jury test: P(z)=z⁴-1.2z³+0.07z²+0.3z-0.08=0. ## Short answer All Jury conditions hold: |a4|=0.08<1, P(1)=0.09>0, P(-1)=1.89>0, |b3|=0.9936>0.2040, |c2|=0.9456>0.3150. Hence the system is stable. ## Answer #### Jury stability test (Ogata form) For $P(z) = a_0z^n + a_1z^{n-1} + \dots + a_n$ with $a_0 > 0$, all roots lie inside the unit circle if and only if: 1. $|a_n| < a_0$ 2. $P(1) > 0$ 3. $(-1)^nP(-1) > 0$ 4. $|b_{n-1}| > |b_0|$, $|c_{n-2}| > |c_0|$, and so on, for the rows of the Jury table The table entries are $$ b_k = \begin{vmatrix} a_n & a_{n-1-k} \\ a_0 & a_{k+1}\end{vmatrix}, \qquad c_k = \begin{vmatrix} b_{n-1} & b_{n-2-k} \\ b_0 & b_{k+1}\end{vmatrix} $$ #### Coefficients $P(z) = z^4 - 1.2z^3 + 0.07z^2 + 0.3z - 0.08$, so $n = 4$, $a_0 = 1$, $a_1 = -1.2$, $a_2 = 0.07$, $a_3 = 0.3$, $a_4 = -0.08$. #### Conditions 1 to 3 - $|a_4| = 0.08 < a_0 = 1$: satisfied - $P(1) = 1 - 1.2 + 0.07 + 0.3 - 0.08 = 0.09 > 0$: satisfied - $(-1)^4P(-1) = 1 + 1.2 + 0.07 - 0.3 - 0.08 = 1.89 > 0$: satisfied #### Jury table $$ \begin{aligned} b_0 &= a_4a_1 - a_0a_3 = (-0.08)(-1.2) - 0.3 = -0.2040 \\ b_1 &= a_4a_2 - a_0a_2 = (-0.08)(0.07) - 0.07 = -0.0756 \\ b_2 &= a_4a_3 - a_0a_1 = (-0.08)(0.3) + 1.2 = 1.1760 \\ b_3 &= a_4^2 - a_0^2 = 0.0064 - 1 = -0.9936 \end{aligned} $$ $$ \begin{aligned} c_0 &= b_3b_1 - b_0b_2 = (-0.9936)(-0.0756) - (-0.2040)(1.1760) = 0.3150 \\ c_1 &= b_3b_2 - b_0b_1 = (-0.9936)(1.1760) - (-0.2040)(-0.0756) = -1.1839 \\ c_2 &= b_3^2 - b_0^2 = 0.9872 - 0.0416 = 0.9456 \end{aligned} $$ | Row | $z^0$ | $z^1$ | $z^2$ | $z^3$ | $z^4$ | |---|---|---|---|---|---| | 1 | −0.08 | 0.3 | 0.07 | −1.2 | 1 | | 2 | 1 | −1.2 | 0.07 | 0.3 | −0.08 | | 3 | −0.9936 | 1.1760 | −0.0756 | −0.2040 | | | 4 | −0.2040 | −0.0756 | 1.1760 | −0.9936 | | | 5 | 0.9456 | −1.1839 | 0.3150 | | | #### Condition 4 - $|b_3| = 0.9936 > |b_0| = 0.2040$: satisfied - $|c_2| = 0.9456 > |c_0| = 0.3150$: satisfied #### Conclusion All the Jury conditions are satisfied, so all the roots lie inside the unit circle and the system is **stable**. Check by factorising: $P(z) = (z - 0.8)(z - 0.5)(z + 0.5)(z - 0.4)$. The roots are $0.8$, $0.5$, $-0.5$ and $0.4$, all with $|z| < 1$. **Answer:** Stable. All conditions hold, with $|b_3| = 0.9936 > 0.2040$ and $|c_2| = 0.9456 > 0.3150$. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. For P(z) = z^4 - 1.2z^3 + 0.07z^2 + 0.3z - 0.08, what are P(1) and P(-1)? A. P(1) = 0.09 and P(-1) = 1.89 B. P(1) = 1.89 and P(-1) = 0.09 C. P(1) = -0.09 and P(-1) = 1.89 D. P(1) = 0.09 and P(-1) = -1.89 Answer: A. P(1) = 1 - 1.2 + 0.07 + 0.3 - 0.08 = 0.09 and P(-1) = 1 + 1.2 + 0.07 - 0.3 - 0.08 = 1.89. Both are positive. (Key point: Conditions 1 to 3) 2. What is checked at the b row, and what is the result? A. |b3| = 0.9936 > |b0| = 0.2040, satisfied B. |b3| = 0.2040 > |b0| = 0.9936, satisfied C. |b3| = 0.9936 < |b0| = 0.2040, violated D. |b1| > |b2|, satisfied Answer: A. The condition is |b(n-1)| > |b0|. Here b3 = -0.9936 and b0 = -0.2040, so it holds. (Key point: First table condition) 3. What is the c-row condition result? A. |c2| = 0.9456 < |c0| = 0.3150, violated B. |c2| = 0.9456 > |c0| = 0.3150, satisfied C. |c2| = 0.3150 > |c0| = 0.9456, satisfied D. It is not needed for n = 4 Answer: B. For n = 4 the table goes to the c row. c2 = 0.9456 exceeds c0 = 0.3150, so it holds. (Key point: Second table condition) 4. What is the conclusion? A. Unstable, because P(1) is small B. Stable, with roots 0.8, 0.5, -0.5 and 0.4 all inside the unit circle C. Critically stable, with a root at z = 1 D. Unstable, with a root at 0.8 Answer: B. All Jury conditions hold, so all roots are inside the unit circle. Factorising gives (z-0.8)(z-0.5)(z+0.5)(z-0.4). 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