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

# Inverse Z-transform by partial fractions

Digital Control System · Chapter 2: The Z-Transform

- Asked in: 2073 Magh, 6 marks
- 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-exponential-as-printed
- Whole chapter: https://www.bctengineer.com/md/semester_6/digital_control_system/chapter-2

## Question

Obtain the inverse z-transform of the following system using partial fraction expansion method: x(z)=z(1-eᵃᵀ)/((z-1)(z-eᵃᵀ)).

## Short answer

Partial fraction expansion of X(z)/z gives X(z)=z/(z-1)-z/(z-e^{aT}). Using standard pairs, the inverse is x(k)=1-e^{akT} for k≥0.

## Answer

The function is solved by **partial fractions of $X(z)/z$**, because every entry in the z-transform table has a factor $z$ in its numerator. The question is solved as printed, with $e^{aT}$. (The usual textbook form has $e^{-aT}$. That version is given at the end.)

Let $c = e^{aT}$:

$$
X(z) = \frac{(1-c)\,z}{(z-1)(z-c)}
$$

#### Step 1: Divide by $z$ and expand

$$
\frac{X(z)}{z} = \frac{1-c}{(z-1)(z-c)} = \frac{A}{z-1} + \frac{B}{z-c}
$$

$$
\begin{aligned}
A &= \left[\frac{1-c}{z-c}\right]_{z=1} = \frac{1-c}{1-c} = 1 \\
B &= \left[\frac{1-c}{z-1}\right]_{z=c} = \frac{1-c}{c-1} = -1
\end{aligned}
$$

#### Step 2: Multiply back by $z$

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

#### Step 3: Use standard pairs

| $X(z)$ | $x(k)$ |
|---|---|
| $\dfrac{z}{z-1}$ | $1$ |
| $\dfrac{z}{z-e^{aT}}$ | $e^{akT}$ |

$$
x(k) = 1 - e^{akT}, \qquad k = 0, 1, 2, \dots
$$

#### Check by long division

$X(z) = \dfrac{(1-c)z}{z^2 - (1+c)z + c} = (1-c)z^{-1} + (1-c)(1+c)z^{-2} + \dots$

- $x(0) = 1 - 1 = 0$ (matches)
- $x(1) = 1 - c$ (matches)
- $x(2) = 1 - c^2 = (1-c)(1+c)$ (matches)

#### Note on the usual form

For $a > 0$ the printed form grows without bound: the pole $z = e^{aT}$ lies outside the unit circle. The standard textbook function is $X(z) = \dfrac{(1-e^{-aT})z}{(z-1)(z-e^{-aT})}$, which is $\mathcal{Z}\left[\dfrac{a}{s(s+a)}\right]$. The same steps give $A = 1$, $B = -1$ and

$$
x(k) = 1 - e^{-akT}
$$

This is a sampled exponential rise toward 1.

**Answer (as printed):** $x(k) = 1 - e^{akT}$ for $k \ge 0$. With $e^{-aT}$: $x(k) = 1 - e^{-akT}$.

## Check yourself

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

1. Why is the partial fraction expansion done on X(z)/z?
   A. To remove the poles
   B. To make the poles real
   C. To reduce the order
   D. Table entries have z in the numerator
   Answer: D. Terms like z/(z-a) are in the table, so expand X(z)/z and multiply back by z. (Key point: Why X(z)/z)

2. With c = e^aT, what are A and B in X(z)/z = A/(z-1) + B/(z-c)?
   A. A = 1, B = -1
   B. A = -1, B = 1
   C. A = 1-c, B = c-1
   D. A = 1, B = 1
   Answer: A. A = (1-c)/(1-c) = 1, B = (1-c)/(c-1) = -1. (Key point: Coefficients)

3. What is x(k) for the printed function?
   A. e^(akT) - 1 for k > 0 only
   B. 1 - e^(akT)
   C. 1 + e^(akT)
   D. 1 - e^(-akT)
   Answer: B. z/(z-1) gives 1 and z/(z-e^aT) gives e^(akT). (Key point: Inverse)

4. Why does the printed form grow without bound for a > 0?
   A. Numerator has e^aT
   B. T is too large
   C. Pole z = e^aT is outside the unit circle
   D. Pole at z = 1
   Answer: C. A pole outside the unit circle gives a divergent term. (Key point: Stability note)

## 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-2z3-z) — Digital Control System, chapter 2, asked 5×

## Related in Digital Control System

- [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×
- [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
- [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

## Related in other subjects

- [Inverse z-transform by partial fractions](https://www.bctengineer.com/question.md?key=semester_7%2Fdigital_signal_analysis_and_processing%3Ainverse-z-partial-fraction) — Digital Signal Analysis and Processing, chapter 2