---
title: "Inverse z-transform with repeated pole"
subject: "Digital Signal Analysis and Processing"
chapter: 2
key: "semester_7/digital_signal_analysis_and_processing:inverse-z-double-pole"
source: https://www.bctengineer.com/q/semester_7/digital_signal_analysis_and_processing/inverse-z-double-pole
site: BCT Engineering
---

# Inverse z-transform with repeated pole

Digital Signal Analysis and Processing · Chapter 2: Z-transform

- Asked in: 2073 Shrawan, 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-double-pole
- 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) = z/((z − 0.4)(z + 1.5)²), ROC: |z| < 0.4.

## Short answer

Use partial fractions of X(z)/z = A/(z-0.4)+B/(z+1.5)+C/(z+1.5)^2; find A=100/361, C=-10/19, B=-100/361. With ROC |z|<0.4 (left-sided), invert using -a^n u[-n-1] and -n a^(n-1) u[-n-1] to get x[n]=[ (100/361)((-1.5)^n - (0.4)^n) + (10/19)n(-1.5)^(n-1) ] u[-n-1].

## Answer

X(z) = z/((z − 0.4)(z + 1.5)²), ROC |z| < 0.4. Poles: 0.4 and −1.5 (double). The ROC is inside all the poles, so x[n] is **left-sided (anticausal)**.

**Partial fractions of X(z)/z = 1/((z − 0.4)(z + 1.5)²):**

1/((z − 0.4)(z + 1.5)²) = A/(z − 0.4) + B/(z + 1.5) + C/(z + 1.5)²

- A = 1/(0.4 + 1.5)² = 1/3.61 = 100/361
- C = 1/(−1.5 − 0.4) = −1/1.9 = −10/19
- B = −A = −100/361 (from the large-z behaviour: the coefficients of the 1/z terms must add to zero)

X(z) = (100/361) z/(z − 0.4) − (100/361) z/(z + 1.5) − (10/19) z/(z + 1.5)²

**Inverse transforms, left-sided (ROC inside the poles):**
- z/(z − a) = 1/(1 − az⁻¹) ↔ −aⁿ u[−n−1]
- z/(z − a)² ↔ −n a^(n−1) u[−n−1]

With a = 0.4 and a = −1.5:

x[n] = [ −(100/361)(0.4)ⁿ + (100/361)(−1.5)ⁿ + (10/19) n (−1.5)^(n−1) ] u[−n−1]

**Check (series in z):** expanding X(z) about z = 0 gives X(z) = −1.1111z − 1.2963z² − 4.7222z³ − …, so x[−1] = −1.1111, x[−2] = −1.2963, x[−3] = −4.7222 and x[n] = 0 for n ≥ 0. The formula gives x[−1] = −1.1111, x[−2] = −1.2963, x[−3] = −4.7222. ✓

**Answer:** x[n] = [ (100/361)((−1.5)ⁿ − (0.4)ⁿ) + (10/19) n (−1.5)^(n−1) ] u[−n−1]

## Check yourself

The answer's key points as multiple-choice questions, with the correct option and why.

1. ROC |z| < 0.4 means x[n] is
   A. right-sided
   B. left-sided
   C. two-sided
   D. finite
   Answer: B. ROC is inside all poles. (Key point: ROC direction)

2. Partial fractions are done on
   A. X(z)/z
   B. X(z)z
   C. X(z)^2
   D. X(z)-1
   Answer: A. So each term has a factor z/(z-a). (Key point: Partial fraction)

3. z/(z-a)^2 for left-sided inverts to
   A. n a^(n-1) u[n]
   B. a^n u[-n-1]
   C. -a^n u[n]
   D. -n a^(n-1) u[-n-1]
   Answer: D. Left-sided counterpart of n a^(n-1) u[n]. (Key point: Double pole pair)

4. x[n] = 0 for n >= 0 because
   A. X(z) has no zeros
   B. poles are inside
   C. X(z) expands in positive powers of z
   D. ROC includes infinity
   Answer: C. Series about z = 0 has only positive powers, hence negative n. (Key point: Zero for n >= 0)

## 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, 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×
- [Inverse z-transform for three ROCs](https://www.bctengineer.com/question.md?key=semester_7%2Fdigital_signal_analysis_and_processing%3Ainverse-z-three-rocs) — 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

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