---
title: "Inverse z-transform by partial fractions"
subject: "Digital Signal Analysis and Processing"
chapter: 2
key: "semester_7/digital_signal_analysis_and_processing:inverse-z-partial-fraction"
source: https://www.bctengineer.com/q/semester_7/digital_signal_analysis_and_processing/inverse-z-partial-fraction
site: BCT Engineering
---

# Inverse z-transform by partial fractions

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

- Asked in: 2071 Shrawan, 6 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-partial-fraction
- 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 + 2z⁻¹ + z⁻²)/(1 + 1.5z⁻¹ + 0.5z⁻²), |z| > 1, using partial fraction method.

## Short answer

Factor and cancel (1+z^{-1}) to obtain (1+z^{-1})/(1+0.5z^{-1}) = 2 - 1/(1+0.5z^{-1}). Since ROC |z|>1 contains |z|>0.5, invert causally: x[n]=2δ[n] - (-0.5)^n u[n].

## Answer

X(z) = (1 + 2z⁻¹ + z⁻²)/(1 + 1.5z⁻¹ + 0.5z⁻²), ROC |z| > 1 (right-sided, causal).

**Step 1. Factor.**
- Numerator: 1 + 2z⁻¹ + z⁻² = (1 + z⁻¹)²
- Denominator: 1 + 1.5z⁻¹ + 0.5z⁻² = (1 + z⁻¹)(1 + 0.5z⁻¹)

The factor (1 + z⁻¹) cancels, so X(z) = (1 + z⁻¹)/(1 + 0.5z⁻¹).

**Step 2. Partial fractions.** Numerator and denominator have the same degree, so there is a constant term. Write in z:

X(z) = (z + 1)/(z + 0.5) = 1 + 0.5/(z + 0.5)

Equivalently, in the form A + B/(1 + 0.5z⁻¹): X(z) = 2 − 1/(1 + 0.5z⁻¹). (Check: 2 + z⁻¹ − 1 = 1 + z⁻¹ over (1 + 0.5z⁻¹). ✓)

**Step 3. Invert (causal).** Using aⁿu[n] ↔ 1/(1 − az⁻¹) with a = −0.5:

x[n] = 2δ[n] − (−0.5)ⁿ u[n]

**Answer:** x[n] = 2δ[n] − (−0.5)ⁿ u[n]

**Check:** x[0] = 2 − 1 = 1; x[1] = 0.5; x[2] = −0.25; x[3] = 0.125. Long division of (1 + z⁻¹) by (1 + 0.5z⁻¹) gives 1 + 0.5z⁻¹ − 0.25z⁻² + 0.125z⁻³ …. ✓

The pole at z = −1 given by the denominator is cancelled by the zero, so the effective ROC is |z| > 0.5, which contains the stated ROC |z| > 1.

## Check yourself

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

1. The numerator (1+z^-1)^2 and denominator (1+z^-1)(1+0.5z^-1) share
   A. (1+0.5z^-1)
   B. (1+z^-1), which cancels
   C. nothing
   D. z^-1
   Answer: B. Leaves (1+z^-1)/(1+0.5z^-1). (Key point: Cancellation)

2. Equal degrees give
   A. a constant term in partial fractions
   B. no constant
   C. a delayed term
   D. a double pole
   Answer: A. X(z) = 2 - 1/(1+0.5z^-1). (Key point: Constant term)

3. x[n] =
   A. delta[n] + (-0.5)^n u[n]
   B. 2 delta[n] - (0.5)^n u[n]
   C. (-0.5)^n u[n]
   D. 2 delta[n] - (-0.5)^n u[n]
   Answer: D. Causal inverse with a = -0.5. (Key point: Answer)

4. x[0], x[1] are
   A. 2 and -1
   B. 1 and -0.5
   C. 1 and 0.5
   D. 0 and 1
   Answer: C. x[0] = 2-1 = 1; x[1] = 0.5. (Key point: Check)

## Related in Digital Signal Analysis and Processing

- [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 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×
- [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 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

## Related in other subjects

- [Inverse Z-transform by partial fractions](https://www.bctengineer.com/question.md?key=semester_6%2Fdigital_control_system%3Ainverse-pfe-2z3-z) — Digital Control System, chapter 2, asked 5×
- [Inverse Z-transform by partial fractions](https://www.bctengineer.com/question.md?key=semester_6%2Fdigital_control_system%3Ainverse-pfe-exponential-as-printed) — Digital Control System, chapter 2, asked 5×
- [Inverse z of (2z²-5z)/((z-2)(z-3))](https://www.bctengineer.com/question.md?key=semester_4%2Fapplied_mathematics%3Ainverse-z-2z2-5z) — Applied Mathematics, chapter 2, asked 5×
- [Inverse Z-transform of z⁻³/((1−z⁻¹)(1−0.2z⁻¹))](https://www.bctengineer.com/question.md?key=semester_6%2Fdigital_control_system%3Ainverse-z3-over-product) — Digital Control System, chapter 2