--- title: "Inverse z-transform for three ROCs" subject: "Digital Signal Analysis and Processing" chapter: 2 key: "semester_7/digital_signal_analysis_and_processing:inverse-z-three-rocs" source: https://www.bctengineer.com/q/semester_7/digital_signal_analysis_and_processing/inverse-z-three-rocs site: BCT Engineering --- # Inverse z-transform for three ROCs Digital Signal Analysis and Processing · Chapter 2: Z-transform - Asked in: 2071 Chaitra, 5 marks - Video: [Z-Transform Inversion by Partial Fraction Expansion (PFE)](https://www.youtube.com/watch?v=RprzYUDKrrA) — Adam Panagos - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_7/digital_signal_analysis_and_processing/inverse-z-three-rocs - Whole chapter: https://www.bctengineer.com/md/semester_7/digital_signal_analysis_and_processing/chapter-2 ## Question Find the inverse z-transform of X(z) = 1/((z − 0.5)(z + 2)) if i) ROC: 0.5 < |z| < 2 ii) ROC: |z| < 0.5 iii) ROC: |z| > 2. ## Short answer Using partial fractions: X(z) = 0.4/(z-0.5) - 0.4/(z+2). For ROC 0.5<|z|<2: x[n] = 0.4(0.5)^{n-1}u[n-1] + 0.4(-2)^{n-1}u[-n]. For |z|<0.5: x[n] = 0.4[(-2)^{n-1} - (0.5)^{n-1}]u[-n]. For |z|>2: x[n] = 0.4[(0.5)^{n-1} - (-2)^{n-1}]u[n-1]. ## Answer X(z) = 1/((z − 0.5)(z + 2)). Poles at z = 0.5 and z = −2. **Partial fractions:** 1/((z − 0.5)(z + 2)) = 0.4/(z − 0.5) − 0.4/(z + 2) (A at z = 0.5: 1/2.5 = 0.4; B at z = −2: 1/(−2.5) = −0.4.) Write each term in z⁻¹ form: 1/(z − a) = z⁻¹/(1 − az⁻¹): X(z) = 0.4 z⁻¹/(1 − 0.5z⁻¹) − 0.4 z⁻¹/(1 + 2z⁻¹) Standard pairs (with the one-sample delay): - Causal (ROC outside pole a): z⁻¹/(1 − az⁻¹) ↔ a^(n−1) u[n−1] - Anticausal (ROC inside pole a): z⁻¹/(1 − az⁻¹) ↔ −a^(n−1) u[−n] #### i) ROC 0.5 < |z| < 2 Pole 0.5 is inside the ring (causal part), pole −2 is outside (anticausal part). x[n] = 0.4 (0.5)^(n−1) u[n−1] + 0.4 (−2)^(n−1) u[−n] #### ii) ROC |z| < 0.5 Both poles are outside the ROC: both terms left-sided. x[n] = −0.4 (0.5)^(n−1) u[−n] + 0.4 (−2)^(n−1) u[−n] = 0.4 [(−2)^(n−1) − (0.5)^(n−1)] u[−n] #### iii) ROC |z| > 2 Both poles are inside the ROC: both terms right-sided. x[n] = 0.4 [(0.5)^(n−1) − (−2)^(n−1)] u[n−1] **Checks:** (iii) X(z) = 1/(z² + 1.5z − 1) = z⁻² − 1.5z⁻³ + …, so x[2] = 1 and x[3] = −1.5; the formula gives 0.4(0.5 + 2) = 1 and 0.4(0.25 − 4) = −1.5. ✓ (ii) X(0) = 1/((−0.5)(2)) = −1, and the formula gives x[0] = 0.4[(−2)⁻¹ − (0.5)⁻¹] = −1. ✓ Only case (i) includes the unit circle, so it is the stable (but non-causal) system. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. 1/((z-0.5)(z+2)) = A. 0.4/(z+0.5) - 0.4/(z-2) B. 0.4/(z-0.5) - 0.4/(z+2) C. 0.5/(z-0.5) - 0.5/(z+2) D. 1/(z-0.5) + 1/(z+2) Answer: B. A = 1/2.5 = 0.4, B = -0.4. (Key point: Partial fractions) 2. For 0.5<|z|<2, pole 0.5 gives A. a causal term; pole -2 gives anticausal B. anticausal; -2 causal C. both causal D. both anticausal Answer: A. ROC outside 0.5, inside -2. (Key point: Ring ROC) 3. Both terms are A. left-sided B. two-sided C. finite D. right-sided Answer: D. Both poles inside the ROC. (Key point: ROC |z|>2) 4. The stable system corresponds to A. |z|<0.5 B. |z|>2 C. 0.5<|z|<2 D. none Answer: C. Only the ring includes the unit circle. (Key point: Stable case) ## Close variants set in other exams - [Inverse z-transform, ROC |z| < 0.4](https://www.bctengineer.com/question.md?key=semester_7%2Fdigital_signal_analysis_and_processing%3Ainverse-z-roc-less-0-4) — Digital Signal Analysis and Processing, chapter 2, asked 5× - [Inverse z-transform with repeated pole](https://www.bctengineer.com/question.md?key=semester_7%2Fdigital_signal_analysis_and_processing%3Ainverse-z-double-pole) — Digital Signal Analysis and Processing, chapter 2, asked 5× - [Inverse z-transform, ROC |z| < 1](https://www.bctengineer.com/question.md?key=semester_7%2Fdigital_signal_analysis_and_processing%3Ainverse-z-z-minus-1-z-minus-2) — Digital Signal Analysis and Processing, chapter 2, asked 5× ## Related in Digital Signal Analysis and Processing - [Properties of ROC of z-transform](https://www.bctengineer.com/question.md?key=semester_7%2Fdigital_signal_analysis_and_processing%3Aroc-properties) — Digital Signal Analysis and Processing, chapter 2, asked 6× - [Z-transform and ROC of two-sided exponential](https://www.bctengineer.com/question.md?key=semester_7%2Fdigital_signal_analysis_and_processing%3Az-transform-two-sided-exp) — Digital Signal Analysis and Processing, chapter 2 - [Inverse z-transform of factored X(z)](https://www.bctengineer.com/question.md?key=semester_7%2Fdigital_signal_analysis_and_processing%3Ainverse-z-factored-form) — Digital Signal Analysis and Processing, chapter 2 - [Inverse z-transform by partial fractions](https://www.bctengineer.com/question.md?key=semester_7%2Fdigital_signal_analysis_and_processing%3Ainverse-z-partial-fraction) — Digital Signal Analysis and Processing, chapter 2 ## Related in other subjects - [Inverse Z-transform of function](https://www.bctengineer.com/question.md?key=semester_3%2Fmathematics_iii%3Ainverse-z-transform-of-function) — Mathematics III, chapter 3 - [Define z-transform and ROC](https://www.bctengineer.com/question.md?key=semester_4%2Fapplied_mathematics%3Az-transform-definition-roc) — Applied Mathematics, chapter 2 - [Inverse Z-transform by inversion integral](https://www.bctengineer.com/question.md?key=semester_6%2Fdigital_control_system%3Ainverse-integral-step-exponential) — Digital Control System, chapter 2, asked 10× - [Inverse Z-transform of z(z+2)/(z−1)²](https://www.bctengineer.com/question.md?key=semester_6%2Fdigital_control_system%3Ainverse-integral-z-z-plus-2) — Digital Control System, chapter 2, asked 10×