---
title: "DTFT of (n+1)aⁿu[n]"
subject: "Signals and Systems"
chapter: 3
key: "semester_4/signals_and_systems:dtft-n-plus-one-geometric"
source: https://www.bctengineer.com/q/semester_4/signals_and_systems/dtft-n-plus-one-geometric
site: BCT Engineering
---

# DTFT of (n+1)aⁿu[n]

Signals and Systems · Chapter 3: Fourier Transform

- Asked in: 2072 Asoj, 6 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-n-plus-one-geometric
- Whole chapter: https://www.bctengineer.com/md/semester_4/signals_and_systems/chapter-3

## Question

Obtain the DTFT of the signal x[n] = (n+1)aⁿu[n], 0 < a < 1

## Short answer

The signal can be expressed as a^n u[n] convolved with itself. DTFT of a^n u[n] is 1/(1−a e^{-jω}), so by convolution property X(e^{jω}) = 1/(1−a e^{-jω})^2.

## Answer

**Known pair:** $a^n u[n] \leftrightarrow \frac{1}{1-ae^{-j\omega}}$, for $|a|<1$.

**Property used (differentiation in frequency):** $n\,x[n] \leftrightarrow j\frac{dX(e^{j\omega})}{d\omega}$.

**Step 1: split the signal.**

$$
x[n] = (n+1)a^n u[n] = n\,a^n u[n] + a^n u[n]
$$

**Step 2: DTFT of $n a^n u[n]$.**

$$
\begin{aligned}
j\frac{d}{d\omega}\left[\frac{1}{1-ae^{-j\omega}}\right] &= j\cdot\frac{-\left(ja\,e^{-j\omega}\right)}{(1-ae^{-j\omega})^2} \\
&= \frac{a\,e^{-j\omega}}{(1-ae^{-j\omega})^2}
\end{aligned}
$$

(derivative of $1 - ae^{-j\omega}$ is $ja\,e^{-j\omega}$).

**Step 3: add the two parts.**

$$
\begin{aligned}
X(e^{j\omega}) &= \frac{ae^{-j\omega}}{(1-ae^{-j\omega})^2} + \frac{1}{1-ae^{-j\omega}} \\
&= \frac{ae^{-j\omega} + 1 - ae^{-j\omega}}{(1-ae^{-j\omega})^2} \\
&= \frac{1}{(1-ae^{-j\omega})^2}
\end{aligned}
$$

**Answer:**

$$
(n+1)a^n u[n] \;\longleftrightarrow\; \frac{1}{\left(1 - a e^{-j\omega}\right)^2}, \quad 0<a<1
$$

**Check by convolution:** $(n+1)a^n u[n] = a^n u[n] * a^n u[n]$, since $\sum_{k=0}^{n}a^k a^{n-k} = (n+1)a^n$. Convolution in time gives the product $\frac{1}{1-ae^{-j\omega}}\cdot\frac{1}{1-ae^{-j\omega}}$, the same result.

Magnitude: $|X(e^{j\omega})| = \frac{1}{1 - 2a\cos\omega + a^2}$, maximum $\frac{1}{(1-a)^2}$ at $\omega = 0$. Phase: $-2\tan^{-1}\frac{a\sin\omega}{1-a\cos\omega}$.

## Check yourself

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

1. How is (n+1)a^n u[n] split?
   A. n a^n u[n] - a^n u[n]
   B. a^(n+1) u[n]
   C. a^n u[n+1]
   D. n a^n u[n] + a^n u[n]
   Answer: D. (n+1)a^n = n a^n + a^n. (Key point: Splitting)

2. What is the DTFT of n x[n]?
   A. j dX(e^{jω})/dω
   B. -j dX/dω
   C. dX/dω
   D. jω X(e^{jω})
   Answer: A. Differentiating Σx[n]e^(-jωn) gives -jn, so multiplying by j gives n x[n]. (Key point: Frequency differentiation)

3. What is the DTFT of (n+1)a^n u[n]?
   A. 1/(1 - a e^(-jω))
   B. 1/(1 - a e^(-jω))^2
   C. a e^(-jω)/(1 - a e^(-jω))^2
   D. 1/(1 - a e^(-j2ω))
   Answer: B. The numerator a e^(-jω) + 1 - a e^(-jω) = 1. (Key point: Final result)

4. What identity cross-checks the answer?
   A. (n+1)a^n u[n] = a^n u[n] + a^n u[n]
   B. = a^n u[n] · a^n u[n]
   C. (n+1)a^n u[n] = a^n u[n] * a^n u[n]
   D. = a^(2n) u[n]
   Answer: C. The convolution sum Σa^k a^(n-k) = (n+1)a^n, so the DTFT is the square of 1/(1-ae^(-jω)). (Key point: Convolution check)

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