--- title: "Jury test: z⁴−1.2z³+0.07z²+1.5z−0.06" subject: "Digital Control System" chapter: 3 key: "semester_6/digital_control_system:jury-z4-1-2z3-1-5z" source: https://www.bctengineer.com/q/semester_6/digital_control_system/jury-z4-1-2z3-1-5z site: BCT Engineering --- # Jury test: z⁴−1.2z³+0.07z²+1.5z−0.06 Digital Control System · Chapter 3: Analysis of Discrete Time Control System - Asked in: 2072 Asoj, 8 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/jury-z4-1-2z3-1-5z - Whole chapter: https://www.bctengineer.com/md/semester_6/digital_control_system/chapter-3 ## Question Examine the stability of given system with characteristic equation by Jury stability test: p(z)=z⁴-1.2z³+0.07z²+1.5z-0.06=0. ## Short answer For P(z)=z⁴-1.2z³+0.07z²+1.5z-0.06, the first three Jury conditions pass (|a₄|0, P(-1)>0). However |b₃|=0.9964 < |b₀|=1.428 and |c₂|=1.0464 < |c₀|=1.659, so the system is unstable. ## Answer We apply the Jury test to $P(z) = a_0z^4 + a_1z^3 + a_2z^2 + a_3z + a_4$, where $a_0 = 1$, $a_1 = -1.2$, $a_2 = 0.07$, $a_3 = 1.5$, $a_4 = -0.06$, and $n = 4$. #### Coefficient conditions 1. $|a_4| < a_0$: $0.06 < 1$, satisfied. 2. $P(1) = 1 - 1.2 + 0.07 + 1.5 - 0.06 = 1.31 > 0$, satisfied. 3. $(-1)^4P(-1) = 1 + 1.2 + 0.07 - 1.5 - 0.06 = 0.71 > 0$, satisfied. #### Jury table $$ b_k = \begin{vmatrix} a_4 & a_{3-k} \\ a_0 & a_{k+1} \end{vmatrix} = a_4a_{k+1} - a_0a_{3-k},\qquad c_k = \begin{vmatrix} b_3 & b_{2-k} \\ b_0 & b_{k+1} \end{vmatrix} $$ $$ \begin{aligned} b_3 &= a_4^2 - a_0^2 = 0.0036 - 1 = -0.9964 \\ b_2 &= a_4a_3 - a_0a_1 = -0.09 + 1.2 = 1.11 \\ b_1 &= a_4a_2 - a_0a_2 = -0.0042 - 0.07 = -0.0742 \\ b_0 &= a_4a_1 - a_0a_3 = 0.072 - 1.5 = -1.428 \end{aligned} $$ $$ \begin{aligned} c_2 &= b_3^2 - b_0^2 = 0.9928 - 2.0392 = -1.0464 \\ c_1 &= b_3b_2 - b_0b_1 = -1.1060 - 0.1060 = -1.2120 \\ c_0 &= b_3b_1 - b_0b_2 = 0.0739 + 1.5851 = 1.6590 \end{aligned} $$ | Row | $z^0$ | $z^1$ | $z^2$ | $z^3$ | $z^4$ | |---|---|---|---|---|---| | 1 | −0.06 | 1.5 | 0.07 | −1.2 | 1 | | 2 | 1 | −1.2 | 0.07 | 1.5 | −0.06 | | 3 | −0.9964 | 1.11 | −0.0742 | −1.428 | | | 4 | −1.428 | −0.0742 | 1.11 | −0.9964 | | | 5 | −1.0464 | −1.2120 | 1.6590 | | | #### Remaining conditions 4. $|b_3| > |b_0|$: $0.9964 > 1.428$ is **false**. 5. $|c_2| > |c_0|$: $1.0464 > 1.6590$ is **false**. #### Conclusion The first three conditions hold, but the table conditions fail. Hence **the system is unstable**. **Check by root finding:** the roots are $z = 1.0084 \pm j0.8572$ (with $|z| = 1.3235$), $z = -0.8568$ and $z = 0.0400$. Two complex poles lie outside the unit circle. This also shows why checking only $P(1)$ and $P(-1)$ is not enough. ## 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 + 1.5z - 0.06, which statement is correct? A. P(1) = 1.31 and P(-1) = 0.71, so conditions 1 to 3 hold B. P(1) = -1.31, so condition 2 fails C. P(-1) = -0.71, so condition 3 fails D. |a4| > a0, so condition 1 fails Answer: A. P(1) = 1.31 > 0, P(-1) = 0.71 > 0 and |a4| = 0.06 < 1, so all three pass. (Key point: Conditions 1 to 3) 2. What is the b-row test result? A. |b3| = 0.9964 > |b0| = 1.428, satisfied B. |b3| = 0.9964 is not greater than |b0| = 1.428, so it fails C. |b3| = 1.428 > |b0| = 0.9964, satisfied D. It is not applicable Answer: B. b3 = a4^2 - a0^2 = -0.9964 and b0 = a4 a1 - a0 a3 = -1.428. Since 0.9964 < 1.428, the condition fails. (Key point: b-row condition) 3. What is the conclusion? A. Stable, because P(1) and P(-1) are positive B. Unstable, because the table conditions fail C. Critically stable D. Stable, because |a4| < a0 Answer: B. Failing any Jury condition means instability. Here both the b and c row conditions fail. (Key point: Verdict) 4. What does the root check z = 1.0084 +/- j0.8572 (|z| = 1.3235) show? A. Checking P(1) and P(-1) alone is enough B. Passing conditions 1 to 3 does not guarantee stability, because complex poles can still be outside the unit circle C. Complex poles are always inside the circle D. The polynomial has a root at z = 1 Answer: B. Two complex poles are outside the unit circle even though conditions 1 to 3 hold, so the table conditions must be checked too. 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