---
title: "Inverse z of (z-4)/((z-1)(z-2)²)"
subject: "Applied Mathematics"
chapter: 2
key: "semester_4/applied_mathematics:inverse-z-z-4-over-z-1-z-2-sq"
source: https://www.bctengineer.com/q/semester_4/applied_mathematics/inverse-z-z-4-over-z-1-z-2-sq
site: BCT Engineering
---

# Inverse z of (z-4)/((z-1)(z-2)²)

Applied Mathematics · Chapter 2: The Z-Transform

- Asked in: 2073 Magh, 5 marks; 2070 Bhadra, 5 marks (asked 2 times)
- Video: [11. Inverse Z-Transform by Partial Fraction Method | Problem #1 | Complete Concept](https://www.youtube.com/watch?v=j5WEJbPgiGc) — MKS TUTORIALS by Manoj Sir
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_4/applied_mathematics/inverse-z-z-4-over-z-1-z-2-sq
- Whole chapter: https://www.bctengineer.com/md/semester_4/applied_mathematics/chapter-2

## Question

Find the inverse Z-transform of f(z) = (z - 4)/((z - 1)(z - 2)²) by partial fraction method.

## Short answer

Partial fractions on F(z)/z give residues A=1, B=−3, D=−1, C=2. Then F(z)=1−3z/(z−1)+2z/(z−2)−z/(z−2)². Invert: f(k)=δ(k)−3+2·2^k−k·2^(k−1), k≥0.

## Answer

F(z) = (z − 4)/((z − 1)(z − 2)²).

**Step 1.** Divide by z. Because F(z) has no factor z in the numerator, a pole at z = 0 appears:

F(z)/z = (z − 4)/(z (z − 1)(z − 2)²) = A/z + B/(z − 1) + C/(z − 2) + D/(z − 2)²

- A = (z − 4)/((z − 1)(z − 2)²) at z = 0 = (−4)/((−1)(4)) = 1
- B = (z − 4)/(z (z − 2)²) at z = 1 = −3/1 = −3
- D = (z − 4)/(z (z − 1)) at z = 2 = −2/2 = −1
- C: for large z, F/z ~ 1/z³ has no 1/z term, so A + B + C = 0, giving C = 2

**Step 2.** Multiply by z:

F(z) = 1 − 3z/(z − 1) + 2z/(z − 2) − z/(z − 2)²

**Step 3.** Invert:

- Z⁻¹[1] = δ(k) (unit impulse, 1 at k = 0)
- Z⁻¹[z/(z − 1)] = 1
- Z⁻¹[z/(z − 2)] = 2ᵏ
- Z⁻¹[z/(z − 2)²] = k 2ᵏ⁻¹

**Answer:**

f(k) = δ(k) − 3 + 2 · 2ᵏ − k 2ᵏ⁻¹,  k ≥ 0

```text
k    : 0  1  2  3   4
f(k) : 0  0  1  1  -3
```

Check: F(z) ~ z⁻² for large z, so f(0) = f(1) = 0 and f(2) = 1, matching the table.

## Check yourself

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

1. Why a pole at z=0 in F(z)/z?
   A. F has a double pole at 0
   B. F has no factor of z in the numerator
   C. z=0 is a root of the denominator
   D. Because of z-4
   Answer: B. Gives constant 1 in F, i.e. δ(k). (Key point: Extra pole)

2. What is D, coefficient of 1/(z-2)²?
   A. 1
   B. 2
   C. -1
   D. -3
   Answer: C. (z-4)/(z(z-1)) at 2 = -2/2. (Key point: Coefficient D)

3. What is Z⁻¹[1]?
   A. 1
   B. u(k)
   C. 2^k
   D. δ(k)
   Answer: D. Unit impulse. (Key point: Impulse)

4. What is f(k)?
   A. δ(k) − 3 + 2·2^k − k 2^{k−1}
   B. −3 + 2·2^k + k 2^{k−1}
   C. δ(k) + 3 + 2^k
   D. δ(k) − 3 − 2·2^k − k 2^k
   Answer: A. f(0)=f(1)=0, f(2)=1. (Key point: Result)

## Close variants set in other exams

- [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 of (2z³+z)/((z-2)²(z-1))](https://www.bctengineer.com/question.md?key=semester_4%2Fapplied_mathematics%3Ainverse-z-2z3-z) — Applied Mathematics, chapter 2, asked 5×

## Related in Applied Mathematics

- [Inverse z of (3z³+2z)/((z-3)²(z-2)), inversion integral](https://www.bctengineer.com/question.md?key=semester_4%2Fapplied_mathematics%3Ainverse-z-3z3-2z-inversion) — Applied Mathematics, chapter 2, asked 3×
- [Inverse z of (z-4)/((z-1)(z-2)²), inversion integral](https://www.bctengineer.com/question.md?key=semester_4%2Fapplied_mathematics%3Ainverse-z-z-4-inversion-integral) — Applied Mathematics, chapter 2, asked 3×
- [Inverse z of z/(z²-3z+2)](https://www.bctengineer.com/question.md?key=semester_4%2Fapplied_mathematics%3Ainverse-z-z-over-z2-3z-2) — Applied Mathematics, chapter 2, asked 2×
- [Inverse z of z(3z²-6z+4)/((z-1)²(z-2))](https://www.bctengineer.com/question.md?key=semester_4%2Fapplied_mathematics%3Ainverse-z-3z2-6z-4) — Applied Mathematics, chapter 2
- [Inverse z of z/(z²+7z+10)](https://www.bctengineer.com/question.md?key=semester_4%2Fapplied_mathematics%3Ainverse-z-z-over-z2-7z-10) — Applied Mathematics, chapter 2
- [Inverse z of z/((z+1)²(z-1))](https://www.bctengineer.com/question.md?key=semester_4%2Fapplied_mathematics%3Ainverse-z-z-over-z1sq-z-1) — Applied Mathematics, 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-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
- [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×