---
title: "ROC of (1/2)^k u(k) + (−1/3)^k u(k)"
subject: "Digital Control System"
chapter: 2
key: "semester_6/digital_control_system:roc-sum-of-exponentials"
source: https://www.bctengineer.com/q/semester_6/digital_control_system/roc-sum-of-exponentials
site: BCT Engineering
---

# ROC of (1/2)^k u(k) + (−1/3)^k u(k)

Digital Control System · Chapter 2: The Z-Transform

- Asked in: 2071 Magh, 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-sum-of-exponentials
- Whole chapter: https://www.bctengineer.com/md/semester_6/digital_control_system/chapter-2

## Question

Find ROC for x(k)=(1/2)ᵏ u(k)+(-1/3)ᵏ u(k), where u(k)=2 for t≥ 0 and u(k)=0 for t<0.

## Short answer

Both terms are causal exponentials with individual ROCs |z|>1/2 and |z|>1/3. Their intersection gives the overall ROC |z|>1/2. The transform is X(z)=z/(z-1/2)+z/(z+1/3).

## Answer

Both terms are **right-sided (causal)** exponentials. Each converges outside a circle, so the ROC of the sum is the overlap: $|z| > \tfrac12$.

$u(k)$ is taken as the unit step. The printed "$u(k) = 2$" would only multiply $X(z)$ by 2 and does not change the ROC.

#### Step 1: First term

$$
\mathcal{Z}\left[\left(\tfrac12\right)^k u(k)\right] = \sum_{k=0}^{\infty}\left(\tfrac12 z^{-1}\right)^k = \frac{1}{1 - \frac12 z^{-1}} = \frac{z}{z - \frac12}
$$

This converges when $\left|\tfrac12 z^{-1}\right| < 1$, i.e. $|z| > \tfrac12$.

#### Step 2: Second term

$$
\mathcal{Z}\left[\left(-\tfrac13\right)^k u(k)\right] = \sum_{k=0}^{\infty}\left(-\tfrac13 z^{-1}\right)^k = \frac{z}{z + \frac13}
$$

This converges when $|z| > \tfrac13$.

#### Step 3: Combine

$$
\begin{aligned}
X(z) &= \frac{z}{z - \frac12} + \frac{z}{z + \frac13} = \frac{z\left(z + \frac13\right) + z\left(z - \frac12\right)}{\left(z - \frac12\right)\left(z + \frac13\right)} \\
&= \frac{2z\left(z - \frac{1}{12}\right)}{\left(z - \frac12\right)\left(z + \frac13\right)}
\end{aligned}
$$

- Poles: $z = \tfrac12$, $z = -\tfrac13$
- Zeros: $z = 0$, $z = \tfrac{1}{12}$

#### Step 4: ROC of the sum

The ROC is the intersection of the two ROCs:

$$
\{|z| > \tfrac12\} \cap \{|z| > \tfrac13\} = \{|z| > \tfrac12\}
$$

```text
            Im z
             |   ROC: outside
        .----+----.   circle r=1/2
       /     |     \
  ----x------o--x---+---- Re z
     -1/3    0  1/2  1
       \     |     /
        '----+----'
```

#### Remarks

- For a causal sequence the ROC lies outside the circle through the **outermost pole** ($|z| = \tfrac12$ here).
- The ROC contains the unit circle, so the sequence is absolutely summable (stable).

**Answer:** $X(z) = \dfrac{z}{z-\frac12} + \dfrac{z}{z+\frac13}$, ROC $|z| > \tfrac12$.

## Check yourself

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

1. For (1/2)^k u(k), the ROC is:
   A. |z| > 1/2
   B. |z| < 1/2
   C. |z| > 2
   D. |z| = 1/2
   Answer: A. The series converges when |(1/2)z^{-1}| < 1, i.e. |z| > 1/2. (Key point: ROC of a causal exponential)

2. How is the ROC of the sum of two terms found?
   A. Intersection of the individual ROCs
   B. Union of the individual ROCs
   C. The ROC of the term with the smaller pole
   D. The ROC is the whole z-plane
   Answer: A. Both series must converge, so the overlap {|z|>1/2} and {|z|>1/3} gives |z|>1/2. (Key point: ROC of a sum)

3. For a causal sequence, the ROC lies outside the circle through:
   A. The outermost pole
   B. The innermost pole
   C. The zeros
   D. The unit circle always
   Answer: A. Here the outermost pole is at 1/2, so |z| > 1/2. (Key point: Outermost pole rule)