PSDF of an arbitrary signal

Power spectral density function (PSDF)

The power spectral density $S_x(f)$ of a power signal describes how its average power is distributed over frequency. Its unit is W/Hz, and the area under it equals the average power:

$$P = \lim_{T\to\infty}\frac{1}{T}\int_{-T/2}^{T/2}|x(t)|^2dt = \int_{-\infty}^{\infty} S_x(f)\,df$$

Properties: $S_x(f) \ge 0$; even in $f$ for real signals; $S_x(f)$ is the Fourier transform of the autocorrelation $R_x(\tau)$ (Wiener–Khinchine theorem); for an LTI system $S_y(f) = |H(f)|^2S_x(f)$.

Derivation of PSDF for an arbitrary signal $x(t)$

A power signal (e.g. a non-periodic random-like waveform) has infinite energy, so its Fourier transform may not exist. We therefore use a truncated version:

$$x_T(t) = \begin{cases} x(t), & -T/2 \le t \le T/2 \\ 0, & \text{elsewhere} \end{cases}$$

$x_T(t)$ has finite energy, so it has a Fourier transform $X_T(f) = \int_{-T/2}^{T/2} x(t)e^{-j2\pi ft}dt$.

Step 1: energy of the truncated signal (Rayleigh's theorem):

$$E_T = \int_{-\infty}^{\infty}|x_T(t)|^2dt = \int_{-T/2}^{T/2}|x(t)|^2dt = \int_{-\infty}^{\infty}|X_T(f)|^2df$$

Step 2: average power over the interval $T$:

$$P_T = \frac{E_T}{T} = \frac{1}{T}\int_{-T/2}^{T/2}|x(t)|^2dt = \int_{-\infty}^{\infty}\frac{|X_T(f)|^2}{T}\,df$$

Step 3: let $T \to \infty$. The left side becomes the average power of $x(t)$:

$$P = \lim_{T\to\infty}\frac{1}{T}\int_{-T/2}^{T/2}|x(t)|^2dt = \int_{-\infty}^{\infty}\left[\lim_{T\to\infty}\frac{|X_T(f)|^2}{T}\right]df$$

Step 4: identify the PSD. Since $P = \int S_x(f)\,df$, the integrand is the power spectral density:

$$S_x(f) = \lim_{T\to\infty}\frac{|X_T(f)|^2}{T}$$

The quantity $|X_T(f)|^2/T$ is called the periodogram; its limit is the PSD.

Special case, periodic signal: with Fourier coefficients $c_n$ and fundamental $f_0$, the PSD is a set of impulses,

$$S_x(f) = \sum_{n=-\infty}^{\infty}|c_n|^2\,\delta(f - nf_0)$$

Example: for $x(t) = A\cos 2\pi f_0 t$, $c_{\pm1} = A/2$, so $S_x(f) = \frac{A^2}{4}[\delta(f-f_0) + \delta(f+f_0)]$ and $P = A^2/4 + A^2/4 = A^2/2$, as expected.

Relation to autocorrelation: it can also be shown that $S_x(f) = \int R_x(\tau)e^{-j2\pi f\tau}d\tau$ where $R_x(\tau) = \lim_{T\to\infty}\frac1T\int_{-T/2}^{T/2}x(t)x^*(t-\tau)dt$.