---
title: "DTFT of aⁿu[n]"
subject: "Signals and Systems"
chapter: 3
key: "semester_4/signals_and_systems:dtft-geometric-sequence"
source: https://www.bctengineer.com/q/semester_4/signals_and_systems/dtft-geometric-sequence
site: BCT Engineering
---

# DTFT of aⁿu[n]

Signals and Systems · Chapter 3: Fourier Transform

- Asked in: 2082 Kartik, 6 marks; 2079 Jestha, 6 marks (asked 2 times)
- 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-geometric-sequence
- Whole chapter: https://www.bctengineer.com/md/semester_4/signals_and_systems/chapter-3

## Question

Find the Fourier transform of signal x[n] = aⁿu[n], 0 < a < 1

## Short answer

The DTFT is the sum of a geometric series: X(e^{jw}) = sum_{n=0}^infinity (a e^{-jw})^n = 1/(1 - a e^{-jw}), for |a|<1.

## Answer

The signal $x[n] = a^n u[n]$ with $0 < a < 1$ is a right-sided decaying exponential: $x[n] = 1, a, a^2, \dots$ for $n = 0, 1, 2, \dots$

#### DTFT
$$
\begin{aligned}
X(e^{j\omega}) &= \sum_{n=-\infty}^{\infty}a^n u[n]e^{-j\omega n} = \sum_{n=0}^{\infty}\left(ae^{-j\omega}\right)^n
\end{aligned}
$$
This is a geometric series with ratio $r = ae^{-j\omega}$. Since $|r| = a < 1$, it converges to $\frac{1}{1-r}$:
$$
\boxed{X(e^{j\omega}) = \frac{1}{1 - ae^{-j\omega}}}
$$

#### Magnitude and phase
Write $1 - ae^{-j\omega} = (1 - a\cos\omega) + ja\sin\omega$:
$$
|X(e^{j\omega})| = \frac{1}{\sqrt{1 - 2a\cos\omega + a^2}}, \qquad \angle X(e^{j\omega}) = -\tan^{-1}\!\left(\frac{a\sin\omega}{1 - a\cos\omega}\right)
$$

| $\omega$ | 0 | $\pi/2$ | $\pi$ |
|---|---|---|---|
| $\lvert X\rvert$ | $\frac{1}{1-a}$ (max) | $\frac{1}{\sqrt{1+a^2}}$ | $\frac{1}{1+a}$ (min) |
| $\angle X$ | 0 | $-\tan^{-1}a$ | 0 |

For example, with $a = 0.5$: $|X| = 2$ at $\omega = 0$, $0.894$ at $\pi/2$, and $0.667$ at $\pi$.

```text
 |X(e^jw)|  (a = 0.5)
 2.0 |              *
     |            *   *
     |          *       *
0.89 |        *           *
0.67 |  * * *               * * *
     +--+-----------+-----------+--> w
       -pi          0           pi
 (repeats every 2pi)
```

#### Observations
- The magnitude is even and the phase is odd in $\omega$, because $x[n]$ is real.
- The spectrum is periodic with period $2\pi$.
- With $0 < a < 1$ the signal is **low-pass**: most of its energy is near $\omega = 0$. For $-1 < a < 0$ the peak moves to $\omega = \pi$ (high-pass).
- If $a \ge 1$ the sum diverges and the DTFT does not exist.

## Check yourself

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

1. X(e^{jω}) equals:
   A. 1/(1 - a e^{-jω})
   B. 1/(1 + a e^{-jω})
   C. a/(1 - e^{-jω})
   D. 1/(1 - a e^{jω})
   Answer: A. Sum of a geometric series 1/(1-r) with r = a e^{-jω}. (Key point: Result)

2. For 0 < a < 1, where is |X| maximum and what is its value?
   A. ω = 0, value 1/(1-a)
   B. ω = π, value 1/(1+a)
   C. ω = 0, value 1/(1+a)
   D. ω = π/2, value 1/(1-a)
   Answer: A. At ω = 0, |X| = 1/(1-a) (max); at ω = π it is 1/(1+a) (min). (Key point: Magnitude extremes)

3. For a = 0.5, |X| at ω = 0 is:
   A. 0.667
   B. 2
   C. 0.894
   D. 1
   Answer: B. 1/(1-0.5) = 2. (Key point: Numerical check)

4. For 0 < a < 1 the signal is:
   A. Low-pass, with energy near ω = 0
   B. High-pass
   C. Band-pass
   D. All-pass
   Answer: A. For -1 < a < 0 the peak moves to ω = π instead. (Key point: Filter type)

## Close variants set in other exams

- [Fourier transform of e^(−2|t−1|)](https://www.bctengineer.com/question.md?key=semester_4%2Fsignals_and_systems%3Aft-shifted-two-sided-exponential) — Signals and Systems, chapter 3, asked 4×
- [Fourier transform of e^(−at)cos ω₀t](https://www.bctengineer.com/question.md?key=semester_4%2Fsignals_and_systems%3Aft-damped-cosine) — Signals and Systems, chapter 3, asked 4×
- [DTFT of cos ω₀n](https://www.bctengineer.com/question.md?key=semester_4%2Fsignals_and_systems%3Adtft-cosine) — Signals and Systems, chapter 3, asked 4×

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