---
title: "Z-transform of a finite two-sided sequence"
subject: "Digital Control System"
chapter: 2
key: "semester_6/digital_control_system:z-transform-finite-sequence"
source: https://www.bctengineer.com/q/semester_6/digital_control_system/z-transform-finite-sequence
site: BCT Engineering
---

# Z-transform of a finite two-sided sequence

Digital Control System · Chapter 2: The Z-Transform

- Asked in: 2072 Magh, 4 marks
- Video: [Z - Transform and ROC of a Two Sided Sequence](https://www.youtube.com/watch?v=qCuRdhSRKxs) — Nitin Bhopale
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_6/digital_control_system/z-transform-finite-sequence
- Whole chapter: https://www.bctengineer.com/md/semester_6/digital_control_system/chapter-2

## Question

If x(k)=1/2ᵏ, for -4≤ k≤ 4, then find Z[x(k)].

## Short answer

X(z) = 16z^4 + 8z^3 + 4z^2 + 2z + 1 + 0.5z^{−1} + 0.25z^{−2} + 0.125z^{−3} + 0.0625z^{−4} = (512z^9 − 1)/(16z^4(2z−1)), ROC 0 < |z| < ∞.

## Answer

$x(k) = (1/2)^k = 2^{-k}$ is defined only for $-4 \le k \le 4$, so it is a **finite-length, two-sided** sequence. Use the two-sided definition $X(z) = \sum_k x(k)z^{-k}$.

#### Step 1: List the samples

| $k$ | −4 | −3 | −2 | −1 | 0 | 1 | 2 | 3 | 4 |
|---|---|---|---|---|---|---|---|---|---|
| $x(k)$ | 16 | 8 | 4 | 2 | 1 | 0.5 | 0.25 | 0.125 | 0.0625 |

#### Step 2: Write the sum

$$
\begin{aligned}
X(z) &= \sum_{k=-4}^{4}2^{-k}z^{-k} \\
&= 16z^4 + 8z^3 + 4z^2 + 2z + 1 + 0.5z^{-1} + 0.25z^{-2} + 0.125z^{-3} + 0.0625z^{-4}
\end{aligned}
$$

#### Step 3: Closed form (finite geometric series)

The ratio between successive terms is $r = (2z)^{-1}$, and the first term is $(2z)^4$:

$$
X(z) = \sum_{k=-4}^{4}(2z)^{-k} = \frac{(2z)^4\left[1 - (2z)^{-9}\right]}{1 - (2z)^{-1}} = \frac{(2z)^4 - (2z)^{-5}}{1 - \frac{1}{2z}}
$$

Multiplying by $\dfrac{2z}{2z}$:

$$
X(z) = \frac{(2z)^9 - 1}{(2z)^4\,(2z - 1)} = \frac{512z^9 - 1}{16z^4(2z-1)}
$$

The apparent pole at $z = \tfrac12$ is cancelled by a zero of the numerator, so $X(z)$ is really a finite polynomial in $z$ and $z^{-1}$.

#### Step 4: Region of convergence

A finite sum always converges except where a term becomes infinite:

- the positive powers $z^4, \dots, z$ blow up at $z = \infty$
- the negative powers $z^{-1}, \dots, z^{-4}$ blow up at $z = 0$

$$
\text{ROC: } 0 < |z| < \infty
$$

Check: at $z = 1$ the sum is $16+8+4+2+1+0.5+0.25+0.125+0.0625 = 31.9375$, and the closed form gives $\dfrac{511}{16} = 31.9375$.

**Answer:** $X(z) = 16z^4 + 8z^3 + 4z^2 + 2z + 1 + 0.5z^{-1} + 0.25z^{-2} + 0.125z^{-3} + 0.0625z^{-4} = \dfrac{512z^9 - 1}{16z^4(2z-1)}$, ROC $0 < |z| < \infty$.

## Check yourself

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

1. x(k) = 2^{-k} for -4 <= k <= 4 is:
   A. A finite-length two-sided sequence
   B. A causal sequence
   C. An anti-causal sequence
   D. An infinite right-sided sequence
   Answer: A. It is nonzero on both sides of k = 0 but only for 9 samples. (Key point: Type of sequence)

2. What is the highest positive power of z in X(z)?
   A. 16z^4
   B. 16z^{-4}
   C. z^4/16
   D. 0.0625z^4
   Answer: A. x(-4)=16 multiplies z^{4}, since the term is x(k)z^{-k}. (Key point: Expansion)

3. The ROC is:
   A. 0 < |z| < infinity
   B. |z| > 1/2
   C. |z| < 2
   D. The whole plane including 0 and infinity
   Answer: A. Positive powers blow up at infinity and negative powers at zero, so both are excluded. (Key point: Region of convergence)

4. In (512z^9 - 1)/(16z^4(2z-1)), the apparent pole at z = 1/2 is:
   A. Cancelled by a zero of the numerator
   B. A genuine pole
   C. Outside the ROC
   D. A double pole
   Answer: A. The numerator is zero at z = 1/2, so X(z) is really a finite Laurent polynomial. (Key point: Pole cancellation)

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