ESD of an energy signal (ideal BPF example)

The energy spectral density (ESD) $\Psi_g(f)$ of an energy signal tells how its energy is distributed with frequency; $\Psi_g(f)\,df$ is the energy in a small band $df$.

Derivation using an ideal BPF

Let $g(t)$ be an energy signal with Fourier transform $G(f)$ and energy $E=\int\lvert g(t)\rvert^2dt$. To find how much energy lies near a frequency $f_0$, pass $g(t)$ through an ideal narrow band-pass filter centred at $f_0$ with bandwidth $\Delta f$:

$$H(f)=\begin{cases}1, & f_0-\frac{\Delta f}{2}\le\lvert f\rvert\le f_0+\frac{\Delta f}{2}\\0, & \text{otherwise}\end{cases}$$

              |H(f)|
     +--+     |     +--+
     |  |     |     |  |  <- width df each
 ----+--+-----+-----+--+----> f
       -f0    0       f0

Step 1 – output spectrum: $Y(f)=H(f)G(f)$.

Step 2 – output energy by Rayleigh's theorem:

$$E_y=\int_{-\infty}^{\infty}\lvert Y(f)\rvert^2df=\int_{-\infty}^{\infty}\lvert H(f)\rvert^2\lvert G(f)\rvert^2df$$

Step 3 – narrow filter: $\lvert H(f)\rvert^2=1$ only in the two narrow bands. If $\Delta f$ is small, $\lvert G(f)\rvert^2$ is nearly constant there, and for a real signal $\lvert G(-f_0)\rvert=\lvert G(f_0)\rvert$:

$$E_y\approx\lvert G(f_0)\rvert^2\Delta f+\lvert G(-f_0)\rvert^2\Delta f=2\lvert G(f_0)\rvert^2\Delta f$$

Step 4 – interpret: the energy in the band of total width $2\Delta f$ (positive plus negative frequencies) is $2\lvert G(f_0)\rvert^2\Delta f$. So the energy per unit bandwidth at $f_0$ is

$$\Psi_g(f)=\lvert G(f)\rvert^2\quad(\text{J/Hz})$$

Summing over all bands gives the total energy:

$$E=\int_{-\infty}^{\infty}\Psi_g(f)\,df=\int_{-\infty}^{\infty}\lvert G(f)\rvert^2df$$

Example: for $g(t)=A\,\text{rect}(t/T)$, $G(f)=AT\,\text{sinc}(fT)$ and $\Psi_g(f)=A^2T^2\text{sinc}^2(fT)$; its area is $A^2T$, equal to the time-domain energy.

Basic properties of ESD

1. Area gives energy: $\int\Psi_g(f)df=E$.
2. Real and non-negative: $\Psi_g(f)=\lvert G(f)\rvert^2\ge0$.
3. Even function for real $g(t)$: $\Psi_g(-f)=\Psi_g(f)$.
4. No phase information: time-shifted signals have the same ESD.
5. Wiener–Khinchin pair: $\Psi_g(f)$ is the Fourier transform of the energy autocorrelation $R_g(\tau)=\int g(t)g^*(t-\tau)dt$.
6. Through an LTI system: $\Psi_y(f)=\lvert H(f)\rvert^2\Psi_x(f)$.