ESD vs PSD; PSD of an arbitrary signal

ESD vs PSD

PSD of an arbitrary signal z(t)

A power signal $z(t)$ (periodic or random) has infinite energy, so its Fourier transform may not exist. We therefore use a truncated version:

$$z_T(t)=\begin{cases} z(t), & -T/2\le t\le T/2\\ 0, & \text{otherwise}\end{cases}$$

$z_T(t)$ has finite energy, so it has a Fourier transform $Z_T(f)$.

Step 1 – average power. The average power of $z(t)$ is

$$P=\lim_{T\to\infty}\frac{1}{T}\int_{-T/2}^{T/2}\lvert z(t)\rvert^2dt=\lim_{T\to\infty}\frac{1}{T}\int_{-\infty}^{\infty}\lvert z_T(t)\rvert^2dt$$

Step 2 – apply Rayleigh's energy theorem to the energy signal $z_T(t)$:

$$\int_{-\infty}^{\infty}\lvert z_T(t)\rvert^2dt=\int_{-\infty}^{\infty}\lvert Z_T(f)\rvert^2df$$

Step 3 – substitute and interchange limit and integral:

$$\begin{aligned} P&=\lim_{T\to\infty}\frac{1}{T}\int_{-\infty}^{\infty}\lvert Z_T(f)\rvert^2df\\ &=\int_{-\infty}^{\infty}\left[\lim_{T\to\infty}\frac{\lvert Z_T(f)\rvert^2}{T}\right]df \end{aligned}$$

Step 4 – identify PSD. Since $P=\int S_z(f)df$, the quantity in brackets is the power per unit bandwidth:

$$S_z(f)=\lim_{T\to\infty}\frac{\lvert Z_T(f)\rvert^2}{T}\quad\text{W/Hz}$$

For a random signal the expectation is also taken: $S_z(f)=\lim_{T\to\infty}\frac{E[\lvert Z_T(f)\rvert^2]}{T}$.

Wiener–Khinchin relation: the same PSD is also the Fourier transform of the time-averaged autocorrelation function:

$$S_z(f)=\int_{-\infty}^{\infty}R_z(\tau)e^{-j2\pi f\tau}d\tau,\qquad R_z(\tau)=\lim_{T\to\infty}\frac{1}{T}\int_{-T/2}^{T/2}z(t)z^*(t-\tau)dt$$

Properties: $S_z(f)\ge0$, it is even for real $z(t)$, the total area under $S_z(f)$ equals $R_z(0)=P$, and through an LTI filter $S_y(f)=\lvert H(f)\rvert^2S_z(f)$.