---
title: "Inverse Z-transform by inversion integral"
subject: "Digital Control System"
chapter: 2
key: "semester_6/digital_control_system:inverse-integral-step-exponential"
source: https://www.bctengineer.com/q/semester_6/digital_control_system/inverse-integral-step-exponential
site: BCT Engineering
---

# Inverse Z-transform by inversion integral

Digital Control System · Chapter 2: The Z-Transform

- Asked in: 2081 Chaitra, 5 marks; 2077 Chaitra, 4 marks; 2074 Bhadra, 4 marks (asked 3 times)
- 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-step-exponential
- Whole chapter: https://www.bctengineer.com/md/semester_6/digital_control_system/chapter-2

## Question

Obtain the inverse z-transform of X(z)=z(1-e⁻ᵃᵀ)/((z-1)(z-e⁻ᵃᵀ)) using inversion integral method.

## Short answer

For X(z) = z(1-e^(-aT))/((z-1)(z-e^(-aT))), compute residues of X(z)z^(k-1) at z=1 and z=e^(-aT). Residue at z=1 is 1; at z=e^(-aT) is –e^(-akT). Summing residues gives x(kT) = 1 – e^(-akT).

## Answer

The inversion integral gives

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

#### Form $X(z)z^{k-1}$

$$
X(z)z^{k-1} = \frac{(1-e^{-aT})\,z^k}{(z-1)(z-e^{-aT})}
$$

For $k = 0, 1, 2, \dots$ there is no pole at $z=0$; the poles are $z = 1$ and $z = e^{-aT}$ (both simple, assuming $a > 0$).

#### Residues

At $z = 1$:

$$
K_1 = \left.(z-1)\frac{(1-e^{-aT})z^k}{(z-1)(z-e^{-aT})}\right|_{z=1} = \frac{1-e^{-aT}}{1-e^{-aT}} = 1
$$

At $z = e^{-aT}$:

$$
K_2 = \left.\frac{(1-e^{-aT})z^k}{z-1}\right|_{z=e^{-aT}} = \frac{(1-e^{-aT})e^{-akT}}{e^{-aT}-1} = -e^{-akT}
$$

#### Result

$$
x(k) = K_1 + K_2 = 1 - e^{-akT}, \quad k = 0, 1, 2, \dots
$$

**Check:** $x(0) = 0$ and $\lim_{z\to\infty}X(z) = 0$ (initial value theorem) agree; $x(\infty) = 1$ and $\lim_{z\to1}(z-1)X(z) = 1$ (final value theorem) agree. This is the sampled form of $x(t) = 1 - e^{-at}$.

**Answer:** $x(kT) = 1 - e^{-akT}$.

## Check yourself

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

1. The inversion integral gives x(k) as:
   A. Sum of residues of X(z) z^(k-1) at its poles
   B. Sum of residues of X(z)
   C. Product of poles
   D. X(z) evaluated at z=1
   Answer: A. x(k) = (1/2 pi j) integral X(z) z^(k-1) dz. (Key point: Inversion integral)

2. The poles of X(z)z^(k-1) here are:
   A. z = 0 and z = 1
   B. z = 1 and z = e^-aT
   C. z = -1 only
   D. z = e^aT only
   Answer: B. No pole at 0 for k >= 0. (Key point: Poles)

3. The residue at z=1 is:
   A. -1
   B. e^-akT
   C. 1
   D. 1 - e^-aT
   Answer: C. The factors (1 - e^-aT) cancel. (Key point: Residue at z=1)

4. x(k) =
   A. e^(-akT)
   B. 1 + e^(-akT)
   C. akT
   D. 1 - e^(-akT)
   Answer: D. K1 + K2 = 1 - e^-akT, final value 1. (Key point: Result)

## Close variants set in other exams

- [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 of z⁻¹(1−z⁻²)/(1+z⁻¹)²](https://www.bctengineer.com/question.md?key=semester_6%2Fdigital_control_system%3Ainverse-integral-z-minus-cube) — 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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