Parseval's theorem for CT aperiodic signals

Statement

For a continuous-time aperiodic (energy) signal $x(t)$ with Fourier transform $X(\omega)$, the total energy found in the time domain equals the total energy found in the frequency domain:

$$E = \int_{-\infty}^{\infty}|x(t)|^2dt = \frac{1}{2\pi}\int_{-\infty}^{\infty}|X(\omega)|^2d\omega$$

In terms of hertz ($\omega = 2\pi f$): $\int|x(t)|^2dt = \int|X(f)|^2df$.

Proof

Write $|x(t)|^2 = x(t)x^*(t)$:

$$E = \int_{-\infty}^{\infty} x(t)x^*(t)\,dt$$

Replace $x^*(t)$ using the conjugate of the inverse Fourier transform:

$$x(t) = \frac{1}{2\pi}\int_{-\infty}^{\infty}X(\omega)e^{j\omega t}d\omega \;\Rightarrow\; x^*(t) = \frac{1}{2\pi}\int_{-\infty}^{\infty}X^*(\omega)e^{-j\omega t}d\omega$$

Then

$$E = \int_{-\infty}^{\infty} x(t)\left[\frac{1}{2\pi}\int_{-\infty}^{\infty}X^*(\omega)e^{-j\omega t}d\omega\right]dt$$

Interchange the order of integration:

$$E = \frac{1}{2\pi}\int_{-\infty}^{\infty}X^*(\omega)\left[\int_{-\infty}^{\infty}x(t)e^{-j\omega t}dt\right]d\omega$$

The bracket is $X(\omega)$ by definition, so

$$E = \frac{1}{2\pi}\int_{-\infty}^{\infty}X^*(\omega)X(\omega)\,d\omega = \frac{1}{2\pi}\int_{-\infty}^{\infty}|X(\omega)|^2d\omega$$

Meaning

• $|X(\omega)|^2$ is the energy spectral density. The energy in a band $\omega_1$ to $\omega_2$ is $\frac{1}{2\pi}\int_{\omega_1}^{\omega_2}|X(\omega)|^2d\omega$.
• Energy can be computed in whichever domain is easier.
• The Fourier transform preserves energy (apart from the $1/2\pi$ factor), so the phase of $X(\omega)$ has no effect on energy.

Example

For $x(t) = e^{-at}u(t)$ with $a > 0$, $X(\omega) = \frac{1}{a+j\omega}$.

• Time domain: $E = \int_0^\infty e^{-2at}dt = \frac{1}{2a}$.
• Frequency domain: $E = \frac{1}{2\pi}\int_{-\infty}^{\infty}\frac{d\omega}{a^2+\omega^2} = \frac{1}{2\pi}\cdot\frac{\pi}{a} = \frac{1}{2a}$.

The two results agree.