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

# Inverse z of (z-4)/((z-1)(z-2)²), inversion integral

Applied Mathematics · Chapter 2: The Z-Transform

- Asked in: 2068 Magh, 5 marks
- 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-inversion-integral
- Whole chapter: https://www.bctengineer.com/md/semester_4/applied_mathematics/chapter-2

## Question

Find inverse z-transform of (z - 4)/((z - 1)(z - 2)²) by using inversion integral method.

## Short answer

Use inversion integral: x(k)=residues of (z-4)z^{k-1}/[(z-1)(z-2)^2]. Poles at z=1 and z=2 give x(k)=(4-k)2^{k-1}-3 for k≥1, with x(0)=0.

## Answer

**Inversion integral method.** x(k) = (1/2πj) ∮ X(z) z^{k−1} dz = sum of residues of X(z) z^{k−1} at all poles inside the contour (which encloses every pole).

Here X(z) = (z − 4)/[(z − 1)(z − 2)²], so

X(z) z^{k−1} = (z − 4) z^{k−1} / [(z − 1)(z − 2)²]

Poles: z = 1 (simple), z = 2 (double), and z = 0 for k = 0 only.

#### Residue at z = 1 (simple pole)

```text
R₁ = (z − 4) z^{k−1} / (z − 2)²  at z = 1
   = (−3)(1)/(1) = −3
```

#### Residue at z = 2 (double pole)

R₂ = d/dz [ (z − 4) z^{k−1} / (z − 1) ] at z = 2

```text
d/dz = z^{k−1}/(z−1) + (k−1)(z−4) z^{k−2}/(z−1) − (z−4) z^{k−1}/(z−1)²
at z = 2:  2^{k−1} − 2(k−1) 2^{k−2} + 2 · 2^{k−1}
         = 2^{k−1} [ 1 − (k − 1) + 2 ]
         = (4 − k) 2^{k−1}
```

#### Result

For k ≥ 1:

**x(k) = −3 + (4 − k) 2^{k−1}**

For k = 0 there is an extra pole at z = 0 with residue (−4)/[(−1)(4)] = 1, giving x(0) = −3 + 2 + 1 = 0 (as expected, because X(z) → 0 when z → ∞).

**Check:** x(0), x(1), x(2), … = 0, 0, 1, 1, −3, −19, −67, … Expanding X(z) in powers of z⁻¹ gives the same sequence.

**Answer:** x(k) = (4 − k) 2^{k−1} − 3 for k ≥ 1, and x(0) = 0.

## Check yourself

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

1. Inversion integral gives x(k) as
   A. sum of residues of X(z) z^(k-1) at poles inside the contour
   B. sum of residues of X(z) z^k
   C. X(z) z^(k-1) at z=0 only
   D. integral of X(z) from 0 to 1
   Answer: A. x(k) = (1/2 pi j) contour integral of X z^(k-1) dz. (Key point: Formula)

2. The poles of X(z)z^(k-1) are
   A. z = 1 and z = 2 simple
   B. z = 1 simple, z = 2 double, and z = 0 for k = 0
   C. z = -1 and z = -2
   D. z = 4 only
   Answer: B. Denominator (z-1)(z-2)^2; z^(k-1) adds pole at 0 when k = 0. (Key point: Poles)

3. Residue at z = 1 is
   A. 3
   B. 1
   C. -3
   D. -4
   Answer: C. (z-4) z^(k-1)/(z-2)^2 at 1 = -3. (Key point: Residue at 1)

4. The residue at the double pole z = 2 uses
   A. no derivative, plain evaluation
   B. second derivative
   C. the derivative of (z-2)^2 only
   D. the derivative d/dz[(z-4) z^(k-1)/(z-1)] at z = 2, giving (4-k)2^(k-1)
   Answer: D. Double pole needs one derivative. (Key point: Double pole)

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