---
title: "z-transform of a^{|k|}"
subject: "Applied Mathematics"
chapter: 2
key: "semester_4/applied_mathematics:z-transform-a-abs-k"
source: https://www.bctengineer.com/q/semester_4/applied_mathematics/z-transform-a-abs-k
site: BCT Engineering
---

# z-transform of a^{|k|}

Applied Mathematics · Chapter 2: The Z-Transform

- Asked in: 2068 Magh, 2.5 marks
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_4/applied_mathematics/z-transform-a-abs-k
- Whole chapter: https://www.bctengineer.com/md/semester_4/applied_mathematics/chapter-2

## Question

Find the z-transform of a^{|k|}.

## Short answer

Split into k≥0 and k<0: the bilateral transform sums to z(1-a^2)/[(z-a)(1-az)] with ROC |a|<|z|<1/|a|. For one-sided k≥0 it is z/(z-a).

## Answer

**Definition.** a^{|k|} is a two-sided sequence (k from −∞ to ∞), so the bilateral transform is used:

X(z) = Σ_{k=−∞}^{∞} a^{|k|} z⁻ᵏ , taking |a| < 1.

Split into k ≥ 0 and k < 0:

```text
X(z) = Σ_{k=0}^{∞} (a/z)ᵏ + Σ_{k=1}^{∞} (a z)ᵏ
```

- First series: converges for |z| > |a| to z/(z − a).
- Second series (let m = −k): converges for |z| < 1/|a| to az/(1 − az).

```text
X(z) = z/(z − a) + az/(1 − az)
     = [ z(1 − az) + az(z − a) ] / [ (z − a)(1 − az) ]
     = z (1 − a²) / [ (z − a)(1 − az) ]
```

**Answer:** X(z) = z(1 − a²) / [(z − a)(1 − az)], region of convergence |a| < |z| < 1/|a|.

If only k ≥ 0 is considered, the transform is simply z/(z − a) with |z| > |a|. (Numerical check at a = 0.5, z = 1.2: both forms give 3.2143.)

## Check yourself

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

1. Why is the bilateral transform needed?
   A. a^|k| is nonzero for negative k as well
   B. a is complex
   C. a^|k| grows with k
   D. The sequence is periodic
   Answer: A. It is two-sided, so the sum runs from -infinity to infinity. (Key point: Transform type)

2. The sum is split into
   A. k even and k odd
   B. k >= 0: sum (a/z)^k and k < 0: sum (az)^m for m >= 1
   C. k >= 0 and k > 1
   D. real and imaginary parts
   Answer: B. For k<0, a^|k| z^-k = (az)^m with m = -k. (Key point: Splitting)

3. The negative-k series converges when
   A. |z| > 1/|a|
   B. |z| > |a|
   C. |z| < 1/|a|
   D. |z| < |a|
   Answer: C. A geometric series in (az) needs |az| < 1. (Key point: Convergence)

4. The transform and ROC are
   A. z/(z - a), |z| > |a|
   B. z(1 + a^2)/[(z - a)(1 - az)], |z| < 1/|a|
   C. z(1 - a^2)/[(z + a)(1 + az)], |z| > |a|
   D. z(1 - a^2)/[(z - a)(1 - az)], |a| < |z| < 1/|a|
   Answer: D. Adding the two parts gives this, valid in the overlap annulus. (Key point: Result and ROC)