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

# Inverse Z-transform: double pole at z=2

Digital Control System · Chapter 2: The Z-Transform

- Asked in: 2072 Asoj, 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-z2-double-pole-at-2
- Whole chapter: https://www.bctengineer.com/md/semester_6/digital_control_system/chapter-2

## Question

Obtain the inverse z-transform of X(z)=z²/((z-2)²(z-e⁻ᵃᵀ)).

## Short answer

Use the residue method for the double pole at z=2 and simple pole at z=e^{-aT}. The inverse is x(k)=[(2-e^{-aT})k 2^k - e^{-aT}2^k + e^{-a(k+1)T}]/(2-e^{-aT})^2 for k≥0.

## Answer

The inverse z-transform is found by the **residue (inversion integral) method**, since $X(z)$ has a double pole at $z = 2$ and a simple pole at $z = e^{-aT}$. Let $b = e^{-aT}$ for short.

$$
x(k) = \sum \text{residues of } X(z)\,z^{k-1} = \sum \text{residues of } \frac{z^{k+1}}{(z-2)^2 (z-b)}
$$

For $k \ge 0$ the factor $z^{k+1}$ adds no pole at $z = 0$, so only two poles count.

#### Residue at the simple pole $z = b$

$$
R_1 = \left[\frac{z^{k+1}}{(z-2)^2}\right]_{z=b} = \frac{b^{k+1}}{(b-2)^2} = \frac{e^{-a(k+1)T}}{(2-e^{-aT})^2}
$$

#### Residue at the double pole $z = 2$

$$
\begin{aligned}
R_2 &= \frac{d}{dz}\left[\frac{z^{k+1}}{z-b}\right]_{z=2} = \left[\frac{(k+1)z^{k}(z-b) - z^{k+1}}{(z-b)^2}\right]_{z=2} \\
&= \frac{2^{k}\left[(k+1)(2-b) - 2\right]}{(2-b)^2} = \frac{2^{k}\left[(2-b)k - b\right]}{(2-b)^2}
\end{aligned}
$$

#### Result

$$
x(k) = \frac{1}{(2-e^{-aT})^2}\Big[(2-e^{-aT})\,k\,2^{k} - e^{-aT}\,2^{k} + e^{-a(k+1)T}\Big], \quad k = 0,1,2,\dots
$$

#### Check

- $k = 0$: $\dfrac{-b + b}{(2-b)^2} = 0$. Since the numerator degree (2) is one less than the denominator degree (3), $X(z)$ starts with $z^{-1}$, so $x(0) = 0$ is correct.
- $k = 1$: $\dfrac{2(2-b) - 2b + b^2}{(2-b)^2} = \dfrac{(2-b)^2}{(2-b)^2} = 1$, which matches the first long-division term $z^{-1}$.
- Numerical check with $b = 0.5$: the formula gives $0, 1, 4.5, 14.25, 39.125$ for $k = 0$ to $4$, the same as long division of $X(z)$.

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

## Check yourself

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

1. Which method is used for X(z) with a double pole at z = 2 and simple pole at z = e^-aT?
   A. Initial value theorem
   B. Only the table with no expansion
   C. Final value theorem
   D. Residue (inversion integral) method
   Answer: D. x(k) is the sum of residues of X(z) z^(k-1). (Key point: Method)

2. Which function has its residues summed?
   A. X(z) z^(k-1) = z^(k+1)/((z-2)^2(z-b))
   B. X(z) z^(k+1)
   C. X(z)/z^(k)
   D. X(z) z^-k
   Answer: A. Inversion integral uses X(z) z^(k-1); here that equals z^(k+1)/((z-2)^2 (z-b)). (Key point: Function to take residues of)

3. How is the residue at the double pole z = 2 computed?
   A. Multiply by (z-2) and set z = 2
   B. Evaluate the function at z = 2
   C. Differentiate z^(k+1)/(z-b) once and set z = 2
   D. Differentiate twice
   Answer: C. For a pole of order 2 take d/dz of (z-2)^2 times the function. (Key point: Double pole residue)

4. Why is x(0) = 0 for this X(z)?
   A. The residue at b is zero
   B. k = 0 is excluded
   C. Numerator degree is one less than denominator degree
   D. Poles lie outside the unit circle
   Answer: C. Numerator degree 2, denominator degree 3, so X(z) starts with z^-1. (Key point: Check x(0))

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