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

# DTFT of (1/2)^(n−1)u[n−1]

Signals and Systems · Chapter 3: Fourier Transform

- Asked in: 2078 Baisakh, 4 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-shifted-geometric
- 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−1) u[n−1]

## Short answer

x[n] = (1/2)^{n−1} u[n−1] is a delayed version of (1/2)^n u[n]. The DTFT of (1/2)^n u[n] is 1/(1−½ e^{−jω}); time-shift gives X(e^{jω}) = e^{−jω}/(1−½ e^{−jω}).

## Answer

The signal $x[n] = \left(\tfrac{1}{2}\right)^{n-1}u[n-1]$ is the sequence $g[n] = \left(\tfrac{1}{2}\right)^n u[n]$ delayed by one sample: $x[n] = g[n-1]$.

#### Method 1: time-shift property
$$
g[n] = \left(\tfrac{1}{2}\right)^n u[n] \leftrightarrow G(e^{j\omega}) = \sum_{n=0}^{\infty}\left(\tfrac{1}{2}e^{-j\omega}\right)^n = \frac{1}{1 - \frac{1}{2}e^{-j\omega}}
$$
The geometric series converges because $|\tfrac{1}{2}e^{-j\omega}| = \tfrac{1}{2} < 1$. Then, by $g[n - n_0] \leftrightarrow e^{-j\omega n_0}G(e^{j\omega})$ with $n_0 = 1$:
$$
\boxed{X(e^{j\omega}) = \frac{e^{-j\omega}}{1 - \frac{1}{2}e^{-j\omega}}}
$$

#### Method 2: direct sum (check)
$$
X(e^{j\omega}) = \sum_{n=1}^{\infty}\left(\tfrac{1}{2}\right)^{n-1}e^{-j\omega n} = e^{-j\omega}\sum_{m=0}^{\infty}\left(\tfrac{1}{2}e^{-j\omega}\right)^m = \frac{e^{-j\omega}}{1 - \frac{1}{2}e^{-j\omega}}
$$
Here $m = n - 1$.

**Magnitude:** $|X(e^{j\omega})| = \frac{1}{\sqrt{1.25 - \cos\omega}}$. This is 2 at $\omega = 0$ and $2/3$ at $\omega = \pi$, the same as for $g[n]$. The delay only adds a phase of $-\omega$.

## Check yourself

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

1. x[n] = (1/2)^(n-1) u[n-1] is related to g[n] = (1/2)^n u[n] by:
   A. x[n] = g[n-1]
   B. x[n] = g[n+1]
   C. x[n] = 2g[n]
   D. x[n] = g[-n]
   Answer: A. Replacing n by n-1 in g[n] gives (1/2)^(n-1)u[n-1]. (Key point: Recognising the shift)

2. The DTFT of (1/2)^n u[n] is:
   A. 1/(1 - (1/2)e^(-jω))
   B. 1/(1 + (1/2)e^(-jω))
   C. e^(-jω)/(1 - (1/2)e^(-jω))
   D. 1/(1 - (1/2)e^(jω))
   Answer: A. The geometric series Σ((1/2)e^(-jω))^n converges since |(1/2)e^(-jω)| = 1/2 < 1. (Key point: DTFT of g)

3. The DTFT of x[n] is:
   A. 1/(1 - (1/2)e^(-jω))
   B. e^(-jω)/(1 - (1/2)e^(-jω))
   C. e^(jω)/(1 - (1/2)e^(-jω))
   D. (1/2)e^(-jω)/(1 - e^(-jω))
   Answer: B. The one-sample delay multiplies G by e^(-jω). (Key point: Result)

4. Compared with g[n], the magnitude of X(e^(jω)) is:
   A. Halved
   B. Unchanged, only a phase of -ω is added
   C. Doubled
   D. Shifted in frequency
   Answer: B. |e^(-jω)| = 1, so only the phase changes. (Key point: Effect on magnitude)

## Close variants set in other exams

- [DTFT of (1/2)^(−n)u[−n−1]](https://www.bctengineer.com/question.md?key=semester_4%2Fsignals_and_systems%3Adtft-anticausal-sequence) — 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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