---
title: "Parseval's theorem for DT aperiodic signals"
subject: "Signals and Systems"
chapter: 3
key: "semester_4/signals_and_systems:dt-parseval-aperiodic"
source: https://www.bctengineer.com/q/semester_4/signals_and_systems/dt-parseval-aperiodic
site: BCT Engineering
---

# Parseval's theorem for DT aperiodic signals

Signals and Systems · Chapter 3: Fourier Transform

- Asked in: 2080 Chaitra, 4 marks; 2070 Magh, 5 marks; 2082 Bhadra (new course), 3 marks (asked 3 times)
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_4/signals_and_systems/dt-parseval-aperiodic
- Whole chapter: https://www.bctengineer.com/md/semester_4/signals_and_systems/chapter-3

## Question

State and prove Parseval's theorem for a DT aperiodic signal (Parseval's relation for discrete time Fourier transforms).

## Short answer

Parseval's theorem: sum |x[n]|^2 = (1/(2pi)) integral_{2pi} |X(e^{jw})|^2 dw. Proof: write energy as sum x[n]x*[n], replace x*[n] with its inverse DTFT, interchange sum and integral to obtain the integral of |X|^2.

## Answer

#### Statement
For a discrete-time aperiodic (finite-energy) signal $x[n]$ with DTFT $X(e^{j\omega})$:
$$
E = \sum_{n=-\infty}^{\infty}|x[n]|^2 = \frac{1}{2\pi}\int_{2\pi}|X(e^{j\omega})|^2d\omega
$$
The energy in the time domain equals the energy in the frequency domain. $|X(e^{j\omega})|^2$ is the **energy density spectrum**.

#### Proof
$$
E = \sum_{n=-\infty}^{\infty}x[n]x^*[n]
$$
Use the inverse DTFT for $x^*[n]$:
$$
x^*[n] = \left[\frac{1}{2\pi}\int_{2\pi}X(e^{j\omega})e^{j\omega n}d\omega\right]^* = \frac{1}{2\pi}\int_{2\pi}X^*(e^{j\omega})e^{-j\omega n}d\omega
$$
Substitute and interchange the sum and the integral:
$$
\begin{aligned}
E &= \sum_{n}x[n]\cdot\frac{1}{2\pi}\int_{2\pi}X^*(e^{j\omega})e^{-j\omega n}d\omega \\
&= \frac{1}{2\pi}\int_{2\pi}X^*(e^{j\omega})\left[\sum_{n}x[n]e^{-j\omega n}\right]d\omega \\
&= \frac{1}{2\pi}\int_{2\pi}X^*(e^{j\omega})X(e^{j\omega})d\omega = \frac{1}{2\pi}\int_{2\pi}|X(e^{j\omega})|^2d\omega
\end{aligned}
$$

#### Example
For $x[n] = a^n u[n]$ with $|a| < 1$:
- Time domain: $E = \sum_{n=0}^{\infty}a^{2n} = \frac{1}{1-a^2}$.
- Frequency domain: $|X(e^{j\omega})|^2 = \frac{1}{1 - 2a\cos\omega + a^2}$, and the standard integral $\frac{1}{2\pi}\int_{-\pi}^{\pi}\frac{d\omega}{1-2a\cos\omega+a^2} = \frac{1}{1-a^2}$ gives the same value.

The integral covers only one period of $2\pi$ because $X(e^{j\omega})$ is periodic.

## Check yourself

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

1. Parseval's relation for a DT aperiodic signal is:
   A. Σ|x[n]|^2 = (1/2π)∫_{2π}|X(e^{jω})|^2 dω
   B. Σ|x[n]|^2 = ∫_{-∞}^{∞}|X(e^{jω})|^2 dω
   C. Σ|x[n]| = (1/2π)∫|X|dω
   D. Σ|x[n]|^2 = 2π∫|X|^2 dω
   Answer: A. The integral covers one period of 2π with a 1/2π factor. (Key point: Statement)

2. In the proof, x*[n] is replaced by:
   A. (1/2π)∫X*(e^{jω})e^{-jωn}dω
   B. (1/2π)∫X(e^{jω})e^{jωn}dω
   C. Σ X*(e^{jω})
   D. x[-n]
   Answer: A. Conjugating the inverse DTFT gives X* and e^{-jωn}. (Key point: Proof step)

3. After interchanging sum and integral, Σ x[n]e^{-jωn} is:
   A. X*(e^{jω})
   B. X(e^{jω})
   C. |X(e^{jω})|^2
   D. x[0]
   Answer: B. It is the definition of the DTFT, leaving X* X = |X|^2. (Key point: After interchange)

4. For x[n] = a^n u[n], |a| < 1, the energy is:
   A. 1/(1-a)
   B. 1/(1-a^2)
   C. 1/(1+a^2)
   D. a/(1-a^2)
   Answer: B. E = Σ a^{2n} = 1/(1-a^2). (Key point: Example)

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