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

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

Digital Control System · Chapter 2: The Z-Transform

- Asked in: 2076 Bhadra, 4 marks
- 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-integral-z-minus-cube
- Whole chapter: https://www.bctengineer.com/md/semester_6/digital_control_system/chapter-2

## Question

By using inversion integral method, obtain inverse Z-transform of X(z)=z⁻¹(1-z⁻²)/(1+z⁻¹)².

## Short answer

Using inversion integral, x(0)=0, x(1)=1, and x(k)=-2(-1)^k for k>=2.

## Answer

In the **inversion integral method**, $x(k)$ is the sum of the residues of $X(z)z^{k-1}$ at all its poles inside the contour (which encloses every pole):

$$
x(k) = \frac{1}{2\pi j}\oint_C X(z)z^{k-1}\,dz = \sum \text{residues of } X(z)z^{k-1}
$$

#### Step 1: Simplify $X(z)$

$1 - z^{-2} = (1 - z^{-1})(1 + z^{-1})$, so

$$
X(z) = \frac{z^{-1}(1 - z^{-1})(1 + z^{-1})}{(1 + z^{-1})^2} = \frac{z^{-1}(1 - z^{-1})}{1 + z^{-1}} = \frac{z - 1}{z(z+1)}
$$

$$
X(z)z^{k-1} = \frac{(z-1)z^{k-2}}{z+1}
$$

The pole at $z = 0$ exists only for $k = 0$ and $k = 1$, so those cases are handled separately.

#### Case $k = 0$: $\dfrac{z-1}{z^2(z+1)}$

- Double pole at $z = 0$: $\dfrac{d}{dz}\left[\dfrac{z-1}{z+1}\right]_{z=0} = \left[\dfrac{2}{(z+1)^2}\right]_{z=0} = 2$
- Pole at $z = -1$: $\left[\dfrac{z-1}{z^2}\right]_{z=-1} = -2$

$$
x(0) = 2 - 2 = 0
$$

#### Case $k = 1$: $\dfrac{z-1}{z(z+1)}$

- Pole at $z = 0$: $\left[\dfrac{z-1}{z+1}\right]_{z=0} = -1$
- Pole at $z = -1$: $\left[\dfrac{z-1}{z}\right]_{z=-1} = 2$

$$
x(1) = -1 + 2 = 1
$$

#### Case $k \ge 2$: only the pole at $z = -1$

$$
x(k) = \left[(z-1)z^{k-2}\right]_{z=-1} = (-2)(-1)^{k-2} = -2(-1)^k
$$

#### Result

$$
x(k) = \begin{cases} 0, & k = 0 \\ 1, & k = 1 \\ -2(-1)^k, & k \ge 2 \end{cases}
$$

| $k$ | 0 | 1 | 2 | 3 | 4 | 5 |
|---|---|---|---|---|---|---|
| $x(k)$ | 0 | 1 | −2 | 2 | −2 | 2 |

#### Check by long division

$\dfrac{z^{-1}(1-z^{-1})}{1+z^{-1}} = z^{-1}(1 - 2z^{-1} + 2z^{-2} - 2z^{-3} + \dots) = z^{-1} - 2z^{-2} + 2z^{-3} - \dots$, which matches.

The pole at $z = -1$ lies on the unit circle, so the sequence keeps oscillating between $\pm2$.

**Answer:** $x(0) = 0$, $x(1) = 1$, $x(k) = -2(-1)^k$ for $k \ge 2$.

## Check yourself

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

1. In the inversion integral method, x(k) equals:
   A. The sum of residues of X(z)z^{k-1} at the poles inside the contour
   B. The sum of residues of X(z)z^{k} at poles outside the contour
   C. The derivative of X(z) at z = 1
   D. The product of the residues of X(z)
   Answer: A. The contour encloses all poles and x(k) is the sum of residues of X(z)z^{k-1}. (Key point: Inversion integral principle)

2. What does X(z) = z^{-1}(1 - z^{-2})/(1 + z^{-1})^2 simplify to?
   A. (z-1)/(z(z+1))
   B. (z+1)/(z(z-1))
   C. (z-1)/(z+1)^2
   D. 1/(z(z+1))
   Answer: A. Cancelling (1 + z^{-1}) from 1 - z^{-2} = (1 - z^{-1})(1 + z^{-1}) leaves z^{-1}(1 - z^{-1})/(1 + z^{-1}). (Key point: Simplifying X(z))

3. Why are k = 0 and k = 1 handled separately?
   A. X(z)z^{k-1} has an extra pole at z = 0 only for those k
   B. The pole at z = -1 disappears for those k
   C. The residue formula fails for k >= 2
   D. The sequence is zero for those k
   Answer: A. X(z)z^{k-1} = (z-1)z^{k-2}/(z+1), which has a pole at the origin only when k < 2. (Key point: Why k = 0 and 1 are special)

4. What is x(k) for k >= 2?
   A. -2(-1)^k
   B. 2(-1)^k
   C. -2
   D. (-1)^k
   Answer: A. The residue at z = -1 is (-2)(-1)^{k-2} = -2(-1)^k, giving -2, 2, -2, ... (Key point: Final result)

## Close variants set in other exams

- [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×
- [Inverse Z-transform of z(z+2)/(z−1)²](https://www.bctengineer.com/question.md?key=semester_6%2Fdigital_control_system%3Ainverse-integral-z-z-plus-2) — Digital Control System, chapter 2, asked 10×
- [Inverse Z-transform: double pole at z=1](https://www.bctengineer.com/question.md?key=semester_6%2Fdigital_control_system%3Ainverse-integral-z2-double-pole) — Digital Control System, chapter 2, asked 10×
- [Inverse Z-transform by inversion integral (cubic)](https://www.bctengineer.com/question.md?key=semester_6%2Fdigital_control_system%3Ainverse-integral-z-1-cubic) — Digital Control System, chapter 2, asked 10×

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