---
title: "Inverse Z-transform: double pole at z=1"
subject: "Digital Control System"
chapter: 2
key: "semester_6/digital_control_system:inverse-integral-z2-double-pole"
source: https://www.bctengineer.com/q/semester_6/digital_control_system/inverse-integral-z2-double-pole
site: BCT Engineering
---

# Inverse Z-transform: double pole at z=1

Digital Control System · Chapter 2: The Z-Transform

- Asked in: 2082 Chaitra (new course), 4 marks; 2075 Bhadra, 8 marks (asked 2 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-z2-double-pole
- Whole chapter: https://www.bctengineer.com/md/semester_6/digital_control_system/chapter-2

## Question

Find the inverse z-transform of X(z)=z²/((z-1)²(z-e⁻ᵃᵀ)) by using inversion integral method.

## Short answer

Inverse z-transform of X(z)=z²/((z-1)²(z-e^(-aT))) via inversion integral. Residues at double pole z=1 and simple pole z=e^(-aT) give x(k) = [k - (k+1)e^(-aT) + e^(-a(k+1)T)]/(1-e^(-aT))² for k≥0.

## Answer

Let $p = e^{-aT}$ (with $a > 0$, so $p \ne 1$). By the inversion integral method,

$$
x(k) = \sum \text{residues of } X(z)z^{k-1} = \sum \text{Res}\left[\frac{z^{k+1}}{(z-1)^2(z-p)}\right]
$$

For $k \ge 0$ there is no pole at the origin. Poles: a **double pole at $z=1$** and a **simple pole at $z=p$**.

#### Residue at $z = p$ (simple pole)

$$
K_1 = \left.\frac{z^{k+1}}{(z-1)^2}\right|_{z=p} = \frac{p^{k+1}}{(1-p)^2} = \frac{e^{-a(k+1)T}}{(1-e^{-aT})^2}
$$

#### Residue at $z = 1$ (double pole)

$$
\begin{aligned}
K_2 &= \lim_{z\to1}\frac{d}{dz}\left[\frac{z^{k+1}}{z-p}\right] \\
&= \lim_{z\to1}\frac{(k+1)z^k(z-p) - z^{k+1}}{(z-p)^2} \\
&= \frac{(k+1)(1-p) - 1}{(1-p)^2} = \frac{k+1}{1-p} - \frac{1}{(1-p)^2}
\end{aligned}
$$

#### Result

$$
\begin{aligned}
x(k) &= K_1 + K_2 = \frac{k+1}{1-e^{-aT}} - \frac{1 - e^{-a(k+1)T}}{(1-e^{-aT})^2} \\
&= \frac{k - (k+1)e^{-aT} + e^{-a(k+1)T}}{(1-e^{-aT})^2}
\end{aligned}
$$

#### Check

- $k=0$: numerator $= 0 - p + p = 0$, so $x(0) = 0$. Initial value theorem: $\lim_{z\to\infty}X(z) = 0$. ✔
- $k=1$: numerator $= 1 - 2p + p^2 = (1-p)^2$, so $x(1) = 1$. Long division gives $X(z) = z^{-1} + (2+p)z^{-2} + \cdots$ ✔
- $k=2$: $x(2) = \frac{2-3p+p^3}{(1-p)^2} = p + 2$ ✔

**Answer:** $x(k) = \dfrac{k - (k+1)e^{-aT} + e^{-a(k+1)T}}{(1-e^{-aT})^2}$, $k = 0, 1, 2, \dots$

## Check yourself

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

1. The function whose residues are summed is:
   A. z^2/((z-1)^2 (z-p))
   B. z^(k-1)/((z-1)(z-p))
   C. z^k/(z-p)
   D. z^(k+1)/((z-1)^2 (z-p))
   Answer: D. X(z) z^(k-1). (Key point: Function)

2. Poles are:
   A. Double at z=1, simple at z=p
   B. Simple at 1, double at p
   C. Both simple
   D. At the origin
   Answer: A. No pole at origin for k >= 0. (Key point: Poles)

3. K1 =
   A. p^k/(1-p)
   B. p^(k+1)/(1-p)^2
   C. -p^(k+1)
   D. p^(k+1)/(1-p)
   Answer: B. Simple pole. (Key point: Residue at p)

4. x(1) =
   A. 0
   B. 2+p
   C. 1
   D. p
   Answer: C. Numerator (1-p)^2 divided by (1-p)^2. (Key point: Check)

## Close variants set in other exams

- [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×
- [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 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×

## Related in Digital Control System

- [Inverse Z-transform: double pole at z=2](https://www.bctengineer.com/question.md?key=semester_6%2Fdigital_control_system%3Ainverse-z2-double-pole-at-2) — Digital Control System, chapter 2
- [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 of (1+z⁻¹−z⁻²)/(1−z⁻¹)](https://www.bctengineer.com/question.md?key=semester_6%2Fdigital_control_system%3Ainverse-z-negative-powers) — Digital Control System, chapter 2
- [Inverse Z-transform of z⁻³/((1−z⁻¹)(1−0.2z⁻¹))](https://www.bctengineer.com/question.md?key=semester_6%2Fdigital_control_system%3Ainverse-z3-over-product) — Digital Control System, chapter 2
- [Inverse transform of −3.894z/(z²+0.6065)](https://www.bctengineer.com/question.md?key=semester_6%2Fdigital_control_system%3Ainverse-minus-3894) — Digital Control System, chapter 2

## Related in other subjects

- [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-transform with repeated pole](https://www.bctengineer.com/question.md?key=semester_7%2Fdigital_signal_analysis_and_processing%3Ainverse-z-double-pole) — Digital Signal Analysis and Processing, chapter 2, asked 5×
- [Inverse z-transform, ROC |z| < 0.4](https://www.bctengineer.com/question.md?key=semester_7%2Fdigital_signal_analysis_and_processing%3Ainverse-z-roc-less-0-4) — Digital Signal Analysis and Processing, chapter 2, asked 5×