Parseval's theorem for DT aperiodic signals

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.