Energy density spectrum of CT signals

The energy density spectrum (energy spectral density, ESD) of an aperiodic signal shows how the signal's total energy is spread over frequency. It is defined as $|X(\omega)|^2$.

Derivation

The energy of a CT aperiodic signal is

$$E = \int_{-\infty}^{\infty}|x(t)|^2dt = \int_{-\infty}^{\infty}x(t)x^*(t)\,dt$$

Express $x^*(t)$ through the inverse Fourier transform:

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

Substitute and change the order of integration:

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

This is Parseval's relation. Write it as

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

In hertz: $E = \int_{-\infty}^{\infty}|X(f)|^2df$, with ESD $|X(f)|^2$ in J/Hz.

Explanation

• $\frac{1}{2\pi}|X(\omega)|^2 d\omega$ is the energy carried by the frequency components in a small band $d\omega$ around $\omega$. The energy between $\omega_1$ and $\omega_2$ is therefore $\frac{1}{\pi}\int_{\omega_1}^{\omega_2}|X(\omega)|^2d\omega$ for a real signal, counting both the positive and the negative band.
• The ESD is real, non-negative, and holds no phase information. Signals with different phase but the same $|X(\omega)|$ have the same ESD.
• For real $x(t)$, $|X(-\omega)| = |X(\omega)|$, so the ESD is even.
• The ESD is the Fourier transform of the autocorrelation $R_{xx}(\tau) = \int x(t)x^*(t-\tau)dt$ (Wiener–Khinchin relation for energy signals).
• LTI systems: if $Y(\omega) = H(\omega)X(\omega)$, then $S_{yy}(\omega) = |H(\omega)|^2 S_{xx}(\omega)$.
• It applies only to energy signals (finite $E$). Power signals use the power spectral density instead.

Example

For $x(t) = e^{-at}u(t)$: $S_{xx}(\omega) = \frac{1}{a^2+\omega^2}$, and

$$E = \frac{1}{2\pi}\cdot\frac{\pi}{a} = \frac{1}{2a}$$

Half of this energy lies in the band $|\omega| < a$:

$$\frac{1}{2\pi}\int_{-a}^{a}\frac{d\omega}{a^2+\omega^2} = \frac{1}{2\pi}\cdot\frac{2}{a}\tan^{-1}(1) = \frac{1}{2\pi}\cdot\frac{2}{a}\cdot\frac{\pi}{4} = \frac{1}{4a}$$