---
title: "Inverse z-transform, ROC |z| < 0.4"
subject: "Digital Signal Analysis and Processing"
chapter: 2
key: "semester_7/digital_signal_analysis_and_processing:inverse-z-roc-less-0-4"
source: https://www.bctengineer.com/q/semester_7/digital_signal_analysis_and_processing/inverse-z-roc-less-0-4
site: BCT Engineering
---

# Inverse z-transform, ROC |z| < 0.4

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

- Asked in: 2074 Chaitra, 5 marks; 2073 Chaitra, 5 marks (asked 2 times)
- Video: [Inverse Z-Transform | Inverse Z-Transform Using Partial Fraction | Signals and System | Mathspedia |](https://www.youtube.com/watch?v=YI7Tun5J7Qw) — Mathspedia
- 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-roc-less-0-4
- 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) = (2z³ + 2z² + 3z + 5)/(z² − 0.1z − 0.2), ROC: |z| < 0.4.

## Short answer

Long-divide to get 2z+2.2 and a proper fraction, then partial fractions A/(z-0.5)+B/(z+0.4). With ROC |z|<0.4 (inside all poles), invert left-sided: -a^(n-1)u[-n] for 1/(z-a). Result: x[n]=2δ[n+1]+2.2δ[n]-(145/18)(0.5)^(n-1)u[-n]+(998/225)(-0.4)^(n-1)u[-n].

## Answer

**Step 1. Factor the denominator.** z² − 0.1z − 0.2 = (z − 0.5)(z + 0.4). Poles at z = 0.5 and z = −0.4.

**Step 2. Improper fraction, so divide first** (numerator degree 3 > denominator degree 2):

(2z³ + 2z² + 3z + 5) ÷ (z² − 0.1z − 0.2) = 2z + 2.2, remainder 3.62z + 5.44.

X(z) = 2z + 2.2 + (3.62z + 5.44)/((z − 0.5)(z + 0.4))

**Step 3. Partial fractions of the remainder term:**

(3.62z + 5.44)/((z − 0.5)(z + 0.4)) = A/(z − 0.5) + B/(z + 0.4)

- A = (3.62·0.5 + 5.44)/(0.5 + 0.4) = 7.25/0.9 = 145/18 ≈ 8.056
- B = (3.62·(−0.4) + 5.44)/(−0.4 − 0.5) = 3.992/(−0.9) = −998/225 ≈ −4.436

**Step 4. Invert with ROC |z| < 0.4.** Both poles are outside this ROC, so every term is left-sided (anticausal). Use

−aⁿ u[−n−1] ↔ 1/(1 − az⁻¹), |z| < |a|, and the delay rule 1/(z − a) = z⁻¹/(1 − az⁻¹) ↔ −a^(n−1) u[−n].

- 2z ↔ 2δ[n + 1]; 2.2 ↔ 2.2δ[n]
- A/(z − 0.5) ↔ −A (0.5)^(n−1) u[−n]
- B/(z + 0.4) ↔ −B (−0.4)^(n−1) u[−n] = +(998/225)(−0.4)^(n−1) u[−n]

**Answer:**
```text
x[n] = 2δ[n+1] + 2.2δ[n]
       − (145/18)(0.5)^(n−1) u[−n]
       + (998/225)(−0.4)^(n−1) u[−n]
```
**Check (expand X(z) in powers of z):** X(z) = −25 − 2.5z − 133.75z² + 44.375z³ + …, so x[0] = −25, x[−1] = −2.5, x[−2] = −133.75, x[−3] = 44.375. The formula gives x[0] = 2.2 − 27.2 = −25, x[−1] = 2 − 4.5 = −2.5, x[−2] = −133.75, x[−3] = 44.375. ✓ x[n] = 0 for n ≥ 1.

## Check yourself

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

1. Since numerator degree exceeds denominator degree, we first
   A. factor only
   B. take log
   C. divide
   D. ignore
   Answer: C. Polynomial division gives 2z + 2.2 plus a proper remainder. (Key point: Improper fraction)

2. The denominator factors as
   A. (z+0.5)(z-0.4)
   B. (z-0.5)(z+0.4)
   C. (z-0.5)(z-0.4)
   D. (z+0.5)(z+0.4)
   Answer: B. Check: product is z^2 - 0.1z - 0.2. (Key point: Poles)

3. With ROC |z| < 0.4, both poles are outside, so x[n] is
   A. left-sided (anticausal)
   B. right-sided
   C. two-sided
   D. finite
   Answer: A. Interior ROC implies anticausal. (Key point: ROC direction)

4. 2z + 2.2 inverts to
   A. 2 delta[n-1] + 2.2 delta[n]
   B. 2 u[n] + 2.2
   C. 2z delta[n]
   D. 2 delta[n+1] + 2.2 delta[n]
   Answer: D. z is an advance. (Key point: Polynomial part)

## Close variants set in other exams

- [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×
- [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×

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