---
title: "ROC of x(k) = −a^k u(−k−1)"
subject: "Digital Control System"
chapter: 2
key: "semester_6/digital_control_system:roc-left-sided-sequence"
source: https://www.bctengineer.com/q/semester_6/digital_control_system/roc-left-sided-sequence
site: BCT Engineering
---

# ROC of x(k) = −a^k u(−k−1)

Digital Control System · Chapter 2: The Z-Transform

- Asked in: 2075 Bhadra, 4 marks
- Video: [ROC and its Properties](https://www.youtube.com/watch?v=LyOQVvPVIuE) — Neso Academy
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_6/digital_control_system/roc-left-sided-sequence
- Whole chapter: https://www.bctengineer.com/md/semester_6/digital_control_system/chapter-2

## Question

Find the region of convergence for x(k)=-aᵏu(-k-1), where u(k) is the unit step function.

## Short answer

The sequence is left-sided; its z-transform is X(z)=z/(z-a). The region of convergence is the interior of a circle: |z| < |a|.

## Answer

$x(k) = -a^k u(-k-1)$ is a **left-sided (anti-causal)** sequence. It is nonzero only for $k \le -1$, where it equals $-a^k$. Its z-transform is $\dfrac{z}{z-a}$ with **ROC $|z| < |a|$**.

#### Step 1: Apply the definition

$$
X(z) = \sum_{k=-\infty}^{\infty}x(k)z^{-k} = -\sum_{k=-\infty}^{-1}a^k z^{-k}
$$

Put $m = -k$, so $m$ runs from 1 to $\infty$:

$$
X(z) = -\sum_{m=1}^{\infty}a^{-m}z^{m} = -\sum_{m=1}^{\infty}\left(\frac{z}{a}\right)^m
$$

#### Step 2: Sum the geometric series

The series converges only when $\left|\dfrac{z}{a}\right| < 1$, i.e. $|z| < |a|$. Then

$$
X(z) = -\frac{z/a}{1 - z/a} = -\frac{z}{a - z} = \frac{z}{z-a} = \frac{1}{1 - az^{-1}}
$$

#### Step 3: Region of convergence

$$
\text{ROC: } |z| < |a|
$$

This is the inside of a circle of radius $|a|$, centred at the origin. The pole at $z = a$ lies on the outer boundary.

```text
           Im z
            |
        .---+---.
      /     |     \
     |  ROC: |z|<|a|
  ---+------0------x---- Re z
     |      |      a  (pole on
      \     |     /    boundary)
        '---+---'
```

#### Comparison with the right-sided sequence

| Sequence | $X(z)$ | ROC |
|---|---|---|
| $a^k u(k)$ (causal) | $\dfrac{z}{z-a}$ | $\lvert z\rvert > \lvert a\rvert$ |
| $-a^k u(-k-1)$ (anti-causal) | $\dfrac{z}{z-a}$ | $\lvert z\rvert < \lvert a\rvert$ |

Both sequences have the **same algebraic $X(z)$**. Only the ROC tells them apart, so an ROC must always be stated with a z-transform. For a left-sided sequence the ROC is always the inside of a circle bounded by the innermost pole.

**Answer:** $X(z) = \dfrac{z}{z-a}$, ROC $|z| < |a|$.

## Check yourself

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

1. What kind of sequence is x(k) = -a^k u(-k-1)?
   A. Left-sided (anti-causal), nonzero only for k <= -1
   B. Right-sided (causal), nonzero for k >= 0
   C. Two-sided for all k
   D. Finite length
   Answer: A. u(-k-1) is 1 only when k <= -1. (Key point: Type of sequence)

2. What is the ROC of this sequence?
   A. |z| < |a|
   B. |z| > |a|
   C. |z| = |a|
   D. 0 < |z| < infinity
   Answer: A. The series sum of (z/a)^m converges only when |z/a| < 1. (Key point: Region of convergence)

3. Why must an ROC always be stated with a z-transform?
   A. a^k u(k) and -a^k u(-k-1) have the same X(z) = z/(z-a) but different ROCs
   B. The ROC changes the pole locations
   C. The ROC changes the zeros of X(z)
   D. Without the ROC the transform is zero
   Answer: A. Only the ROC distinguishes the causal from the anti-causal sequence. (Key point: Role of ROC)

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