---
title: "Inverse Z-transform of z⁻³/((1−z⁻¹)(1−0.2z⁻¹))"
subject: "Digital Control System"
chapter: 2
key: "semester_6/digital_control_system:inverse-z3-over-product"
source: https://www.bctengineer.com/q/semester_6/digital_control_system/inverse-z3-over-product
site: BCT Engineering
---

# Inverse Z-transform of z⁻³/((1−z⁻¹)(1−0.2z⁻¹))

Digital Control System · Chapter 2: The Z-Transform

- Asked in: 2079 Chaitra, 4 marks
- Video: [11. Inverse Z-Transform by Partial Fraction Method | Problem #1 | Complete Concept](https://www.youtube.com/watch?v=j5WEJbPgiGc) — MKS TUTORIALS by Manoj Sir
- 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-z3-over-product
- Whole chapter: https://www.bctengineer.com/md/semester_6/digital_control_system/chapter-2

## Question

Obtain the inverse z-transform of X(z)=z⁻³/((1-z⁻¹)(1-0.2z⁻¹)).

## Short answer

Factor X(z)=z^{-3}F(z). Partial fractions give F(z)=1.25/(1-z^{-1})-0.25/(1-0.2z^{-1}), so f(k)=1.25-0.25(0.2)^k. Shifting by 3 yields x(k)=1.25[1-(0.2)^{k-2}] for k≥3, zero otherwise.

## Answer

Remove the pure delay $z^{-3}$ first. Invert the remaining part by partial fractions, then shift the result by 3 samples.

#### Step 1: Write $X(z) = z^{-3}F(z)$

$$
F(z) = \frac{1}{(1-z^{-1})(1-0.2z^{-1})} = \frac{z^2}{(z-1)(z-0.2)}
$$

#### Step 2: Partial fractions of $F(z)/z$

$$
\frac{F(z)}{z} = \frac{z}{(z-1)(z-0.2)} = \frac{A}{z-1} + \frac{B}{z-0.2}
$$

$$
A = \frac{1}{1-0.2} = 1.25, \qquad B = \frac{0.2}{0.2-1} = -0.25
$$

$$
F(z) = \frac{1.25z}{z-1} - \frac{0.25z}{z-0.2}
\;\Rightarrow\;
f(k) = 1.25 - 0.25(0.2)^k = 1.25\left[1 - (0.2)^{k+1}\right]
$$

#### Step 3: Apply the shift theorem

$z^{-3}F(z) \leftrightarrow f(k-3)\,1(k-3)$:

$$
x(k) = \begin{cases} 0, & k = 0, 1, 2 \\[4pt] 1.25\left[1 - (0.2)^{k-2}\right], & k \ge 3 \end{cases}
$$

Equivalently $x(k) = \left[1.25 - 0.25(0.2)^{k-3}\right]1(k-3)$.

#### Values

| $k$ | 0 | 1 | 2 | 3 | 4 | 5 | 6 | $\infty$ |
|---|---|---|---|---|---|---|---|---|
| $x(k)$ | 0 | 0 | 0 | 1 | 1.2 | 1.24 | 1.248 | 1.25 |

#### Check

- Long division: $X(z) = z^{-3} + 1.2z^{-4} + 1.24z^{-5} + 1.248z^{-6} + \dots$, which matches.
- Final value theorem: $\lim_{z\to1}(1-z^{-1})X(z) = \dfrac{1}{1-0.2} = 1.25$, which matches.

**Answer:** $x(k) = 1.25\left[1 - (0.2)^{k-2}\right]$ for $k \ge 3$, and $x(k) = 0$ for $k < 3$.

## Check yourself

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

1. What is the first step for z^-3/((1-z^-1)(1-0.2z^-1))?
   A. Multiply by z^3 only at end of long division
   B. Use the final value theorem
   C. Remove z^-3, invert the rest, then shift 3 samples
   D. Ignore the delay
   Answer: C. X = z^-3 F(z), so x(k) = f(k-3). (Key point: Pure delay)

2. What are A and B in F(z)/z = A/(z-1) + B/(z-0.2)?
   A. A = 1, B = -1
   B. A = 0.8, B = 0.2
   C. A = -1.25, B = 0.25
   D. A = 1.25, B = -0.25
   Answer: D. A = 1/(1-0.2) = 1.25, B = 0.2/(0.2-1) = -0.25. (Key point: Coefficients)

3. What is f(k)?
   A. 1.25 - 0.25(0.2)^k
   B. 1.25 + 0.25(0.2)^k
   C. 1 - 0.2^k
   D. 0.25 - 1.25(0.2)^k
   Answer: A. Inverse of 1.25z/(z-1) - 0.25z/(z-0.2). (Key point: f(k))

4. What does x(k) settle to?
   A. 0
   B. 1.25
   C. 1
   D. 0.2
   Answer: B. Final value = 1/(1-0.2) = 1.25. (Key point: Final value)

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