---
title: "Inverse Z-transform by partial fractions"
subject: "Digital Control System"
chapter: 2
key: "semester_6/digital_control_system:inverse-pfe-2z3-z"
source: https://www.bctengineer.com/q/semester_6/digital_control_system/inverse-pfe-2z3-z
site: BCT Engineering
---

# Inverse Z-transform by partial fractions

Digital Control System · Chapter 2: The Z-Transform

- Asked in: 2082 Chaitra, 5 marks; 2082 Kartik, 5 marks; 2078 Chaitra, 6 marks; 2071 Bhadra, 4 marks (asked 4 times)
- Video: [Z-Transform Inversion by Partial Fraction Expansion (PFE)](https://www.youtube.com/watch?v=RprzYUDKrrA) — Adam Panagos
- 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-pfe-2z3-z
- Whole chapter: https://www.bctengineer.com/md/semester_6/digital_control_system/chapter-2

## Question

Obtain the inverse z-transform of X(z)=(2z³+z)/((z-2)²(z-1)) by using partial fraction expansion method.

## Short answer

Expand X(z)/z = (2z²+1)/((z-2)²(z-1)) into partial fractions: 3/(z-1) + 9/(z-2)² - 1/(z-2). Then X(z) = 3z/(z-1) + 9z/(z-2)² - z/(z-2). Inverse Z-transform gives x(k) = 3 + 9k·2^(k-1) – 2^k.

## Answer

Since $X(z)$ has a zero at $z=0$, expand $X(z)/z$ in partial fractions.

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

#### Coefficients

$$
\begin{aligned}
A &= \left.\frac{2z^2+1}{(z-2)^2}\right|_{z=1} = \frac{2+1}{1} = 3 \\
B &= \left.\frac{2z^2+1}{z-1}\right|_{z=2} = \frac{8+1}{1} = 9 \\
C &= \left.\frac{d}{dz}\left[\frac{2z^2+1}{z-1}\right]\right|_{z=2} = \left.\frac{4z(z-1) - (2z^2+1)}{(z-1)^2}\right|_{z=2} \\
&= \frac{8 - 9}{1} = -1
\end{aligned}
$$

So

$$
X(z) = \frac{3z}{z-1} + \frac{9z}{(z-2)^2} - \frac{z}{z-2}
$$

#### Inverse transforms used

- $\mathcal{Z}^{-1}\left[\frac{z}{z-1}\right] = 1$
- $\mathcal{Z}^{-1}\left[\frac{z}{z-a}\right] = a^k$
- $\mathcal{Z}^{-1}\left[\frac{z}{(z-a)^2}\right] = k\,a^{k-1}$

#### Result

$$
x(k) = 3 + 9k\,2^{k-1} - 2^k, \quad k = 0, 1, 2, \dots
$$

#### Check (first few values)

| k | 0 | 1 | 2 | 3 | 4 |
|---|---|---|---|---|---|
| $x(k)$ | 2 | 10 | 35 | 103 | 275 |

$x(0) = \lim_{z\to\infty}X(z) = 2$ (initial value theorem), which agrees. Long division of $X(z) = \frac{2z^3+z}{z^3-5z^2+8z-4}$ also gives $2 + 10z^{-1} + 35z^{-2} + 103z^{-3} + \cdots$

**Answer:** $x(k) = 3 + 9k(2^{k-1}) - 2^k$, which can also be written $x(k) = 3 + (4.5k - 1)\,2^k$.

## Check yourself

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

1. Why expand X(z)/z rather than X(z)?
   A. So each term becomes z/(z-a) form with known inverse
   B. To remove poles
   C. To make the degree lower than zero
   D. Because X(z) has no zeros
   Answer: A. Multiply back by z after expanding. (Key point: Method)

2. A for the pole at z=1 is:
   A. 9
   B. 3
   C. -1
   D. 1
   Answer: B. (2+1)/(1-2)^2 = 3. (Key point: Coefficient A)

3. z/(z-a)^2 inverts to:
   A. a^k
   B. k a^k
   C. k a^(k-1)
   D. a^(k+1)
   Answer: C. Used for the 9z/(z-2)^2 term. (Key point: Repeated pole)

4. The result is:
   A. x(k) = 3 + 9k 2^k + 2^k
   B. x(k) = 3 - 9k 2^(k-1) + 2^k
   C. x(k) = 1 + 2^k
   D. x(k) = 3 + 9k 2^(k-1) - 2^k
   Answer: D. Check: x(0) = 2 by initial value theorem. (Key point: Result)

## Close variants set in other exams

- [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×

## Related in Digital Control System

- [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 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×
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- [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=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 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

## Related in other subjects

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- [Inverse z of (2z³+z)/((z-2)²(z-1))](https://www.bctengineer.com/question.md?key=semester_4%2Fapplied_mathematics%3Ainverse-z-2z3-z) — Applied Mathematics, chapter 2, asked 5×
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