---
title: "Inverse Z-transform of (1+z⁻¹−z⁻²)/(1−z⁻¹)"
subject: "Digital Control System"
chapter: 2
key: "semester_6/digital_control_system:inverse-z-negative-powers"
source: https://www.bctengineer.com/q/semester_6/digital_control_system/inverse-z-negative-powers
site: BCT Engineering
---

# Inverse Z-transform of (1+z⁻¹−z⁻²)/(1−z⁻¹)

Digital Control System · Chapter 2: The Z-Transform

- Asked in: 2081 Chaitra, 3 marks
- 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_6/digital_control_system/inverse-z-negative-powers
- Whole chapter: https://www.bctengineer.com/md/semester_6/digital_control_system/chapter-2

## Question

Obtain the inverse z-transform of X(z)=(1+z⁻¹-z⁻²)/(1-z⁻¹).

## Short answer

Divide numerator by denominator in powers of z^{-1} to get X(z)=z^{-1}+1/(1-z^{-1}). Inverse yields x(k)=δ(k-1)+1(k), so x(0)=1, x(1)=2, and x(k)=1 for k≥2.

## Answer

Split $X(z)$ into a polynomial part plus a proper part, then read each term from the z-transform table.

#### Step 1: Divide the numerator by the denominator (in powers of $z^{-1}$)

Write $1 + z^{-1} - z^{-2} = (1 - z^{-1})(\alpha + \beta z^{-1}) + \gamma$:

$$
(1 - z^{-1})(\alpha + \beta z^{-1}) + \gamma = (\alpha + \gamma) + (\beta - \alpha)z^{-1} - \beta z^{-2}
$$

Comparing coefficients:

- $z^{-2}$: $-\beta = -1 \Rightarrow \beta = 1$
- $z^{-1}$: $\beta - \alpha = 1 \Rightarrow \alpha = 0$
- constant: $\alpha + \gamma = 1 \Rightarrow \gamma = 1$

$$
X(z) = z^{-1} + \frac{1}{1 - z^{-1}}
$$

#### Step 2: Inverse transform

- $z^{-1} \;\leftrightarrow\; \delta(k-1)$ (a unit pulse at $k = 1$)
- $\dfrac{1}{1-z^{-1}} \;\leftrightarrow\; 1(k)$ (a unit step)

$$
x(k) = 1(k) + \delta(k-1)
$$

So:

| $k$ | 0 | 1 | 2 | 3 | 4 | ... |
|---|---|---|---|---|---|---|
| $x(k)$ | 1 | 2 | 1 | 1 | 1 | ... |

#### Check by direct division

$(1 + z^{-1} - z^{-2}) \div (1 - z^{-1}) = 1 + 2z^{-1} + z^{-2} + z^{-3} + \dots$, which gives the same sequence.

**Answer:** $x(0) = 1$, $x(1) = 2$, $x(k) = 1$ for $k \ge 2$, i.e. $x(k) = 1(k) + \delta(k-1)$.

## Check yourself

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

1. X(z) = (1 + z^-1 - z^-2)/(1 - z^-1) splits into?
   A. 1 + z^-1/(1 - z^-1)
   B. z^-2 + 1/(1 - z^-1)
   C. 1/(1 - z^-1) - z^-1
   D. z^-1 + 1/(1 - z^-1)
   Answer: D. Matching coefficients gives alpha = 0, beta = 1, gamma = 1. (Key point: Division)

2. What time sequence does z^-1 represent?
   A. Unit step
   B. Unit pulse delta(k-1)
   C. Unit ramp
   D. Pulse at k = 0
   Answer: B. z^-1 is a unit pulse delayed by one sample. (Key point: Pulse term)

3. What is x(k)?
   A. delta(k) + 1(k-1)
   B. 1(k) + delta(k-1): 1, 2, 1, 1, ...
   C. 1(k) - delta(k-1): 1, 0, 1, ...
   D. 1, 1, 2, 1, 1, ...
   Answer: B. Step plus a pulse at k = 1 gives 1, 2, 1, 1, ... (Key point: Result)

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