---
title: "DTFT of (1/2)^(−n)u[−n−1]"
subject: "Signals and Systems"
chapter: 3
key: "semester_4/signals_and_systems:dtft-anticausal-sequence"
source: https://www.bctengineer.com/q/semester_4/signals_and_systems/dtft-anticausal-sequence
site: BCT Engineering
---

# DTFT of (1/2)^(−n)u[−n−1]

Signals and Systems · Chapter 3: Fourier Transform

- Asked in: 2069 Bhadra, 5 marks
- Video: [Determine DTFT of given sequences](https://www.youtube.com/watch?v=DnhOxaJSV3E) — Engg-Course-Made-Easy
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_4/signals_and_systems/dtft-anticausal-sequence
- Whole chapter: https://www.bctengineer.com/md/semester_4/signals_and_systems/chapter-3

## Question

Compute discrete time Fourier transform of the discrete time signal x[n] = (1/2)^(−n) u[−n−1]

## Short answer

x[n] = 2^n u[–n–1] decays for n→ –∞. Summing the geometric series gives X(e^{jω}) = ½ e^{jω} / (1 – ½ e^{jω}) = e^{jω}/(2 – e^{jω}).

## Answer

**Simplify the signal:** $\left(\frac12\right)^{-n} = 2^n$, and $u[-n-1] = 1$ for $n \le -1$. So

$$
x[n] = 2^n \text{ for } n \le -1, \quad 0 \text{ for } n \ge 0
$$

i.e. $x[n] = \{\dots, \frac18, \frac14, \frac12\}$ ending at $n = -1$. It decays as $n \to -\infty$, so it is absolutely summable and the DTFT exists.

**DTFT:**

$$
\begin{aligned}
X(e^{j\omega}) &= \sum_{n=-\infty}^{-1}2^n e^{-j\omega n} \quad (\text{put } m = -n) \\
&= \sum_{m=1}^{\infty}2^{-m}e^{j\omega m} = \sum_{m=1}^{\infty}\left(\tfrac12 e^{j\omega}\right)^m \\
&= \frac{\frac12 e^{j\omega}}{1 - \frac12 e^{j\omega}}, \qquad \left|\tfrac12e^{j\omega}\right| < 1
\end{aligned}
$$

**Answer:**

$$
X(e^{j\omega}) = \frac{0.5\,e^{j\omega}}{1 - 0.5\,e^{j\omega}} = \frac{e^{j\omega}}{2 - e^{j\omega}}
$$

**Magnitude:** $|X(e^{j\omega})| = \frac{1}{\sqrt{5 - 4\cos\omega}}$, giving $1$ at $\omega = 0$ (equal to $\sum x[n] = \frac12 + \frac14 + \dots = 1$) and $\frac13$ at $\omega = \pi$.

**Phase:** $\angle X(e^{j\omega}) = \omega + \tan^{-1}\frac{\sin\omega}{2 - \cos\omega}$.

(Check with z-transform: $-a^n u[-n-1] \leftrightarrow \frac{1}{1-az^{-1}}$, $|z|<|a|$. With $a = 2$: $X(z) = -\frac{1}{1-2z^{-1}} = \frac{z}{2 - z}$, ROC $|z|<2$, which contains the unit circle. Putting $z = e^{j\omega}$ gives the same result.)

## Check yourself

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

1. What does (1/2)^(-n) u[-n-1] equal?
   A. 2^n for n ≥ 0
   B. (1/2)^n for n ≤ -1
   C. 2^n for n ≤ -1, zero otherwise
   D. 2^n for all n
   Answer: C. (1/2)^(-n) = 2^n, and u[-n-1] is 1 for n ≤ -1. (Key point: Simplify)

2. Why does the DTFT exist?
   A. It is causal
   B. It is bounded for n ≥ 0
   C. 2^n grows
   D. 2^n decays as n → -∞, so it is absolutely summable (sum = 1)
   Answer: D. For n ≤ -1, 2^n = 2^{-|n|} decays. (Key point: Existence)

3. What does putting m = -n give?
   A. Σ_{m=1}^{∞} (0.5 e^{jω})^m
   B. Σ_{m=0}^{∞} (0.5 e^{-jω})^m
   C. Σ_{m=1}^{∞} (2 e^{jω})^m
   D. Σ_{m=1}^{∞} (0.5 e^{-jω})^m
   Answer: A. 2^{-m} e^{jωm} = (0.5e^{jω})^m. (Key point: Substitution)

4. What is X(e^{jω})?
   A. 1/(2 - e^{jω})
   B. e^{jω}/(2 - e^{jω})
   C. e^{-jω}/(2 - e^{-jω})
   D. e^{jω}/(1 - 2e^{jω})
   Answer: B. Sum = 0.5e^{jω}/(1 - 0.5e^{jω}) = e^{jω}/(2 - e^{jω}); |X(0)| = 1. (Key point: Result)

## Close variants set in other exams

- [DTFT of (1/2)^(n−1)u[n−1]](https://www.bctengineer.com/question.md?key=semester_4%2Fsignals_and_systems%3Adtft-shifted-geometric) — Signals and Systems, chapter 3, asked 3×
- [DTFT of u[n−2] − u[n−6]](https://www.bctengineer.com/question.md?key=semester_4%2Fsignals_and_systems%3Adtft-step-difference) — Signals and Systems, chapter 3, asked 3×

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