PSD expression and relation with autocorrelation

Power spectral density (PSD)

The power spectral density $S_x(f)$ of a power signal $x(t)$ is a function that shows how the signal's average power is distributed over frequency (W/Hz). The total average power is

$$P = \int_{-\infty}^{\infty} S_x(f)\,df$$

Expression for PSD

Truncate $x(t)$ to the interval $-T/2 \le t \le T/2$ to get $x_T(t)$, which has finite energy and a Fourier transform $X_T(f)$. By Rayleigh's theorem,

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

Divide by $T$ and let $T \to \infty$:

$$P = \lim_{T\to\infty}\frac1T\int_{-T/2}^{T/2}|x(t)|^2dt = \int_{-\infty}^{\infty}\lim_{T\to\infty}\frac{|X_T(f)|^2}{T}\,df$$

Comparing with $P = \int S_x(f)df$:

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

Relationship with autocorrelation (Wiener–Khinchine theorem)

The time-averaged autocorrelation of a power signal is

$$R_x(\tau) = \lim_{T\to\infty}\frac1T\int_{-T/2}^{T/2}x(t)\,x^*(t-\tau)\,dt$$

For the truncated signal, $\frac1T\int x_T(t)x_T^*(t-\tau)dt$ is a time-autocorrelation of an energy signal divided by $T$. For an energy signal we know (shown below) $\mathcal{F}\{\int x_T(t)x_T^*(t-\tau)dt\} = |X_T(f)|^2$:

$$\begin{aligned} \int\!\!\int x_T(t)x_T^*(t-\tau)e^{-j2\pi f\tau}dt\,d\tau &= \int x_T(t)e^{-j2\pi ft}\left[\int x_T^*(\lambda)e^{j2\pi f\lambda}d\lambda\right]dt \\ &= X_T(f)\,X_T^*(f) = |X_T(f)|^2 \end{aligned}$$

(using $\lambda = t - \tau$). Dividing by $T$ and taking the limit:

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

Therefore the PSD and the autocorrelation function form a Fourier transform pair:

$$S_x(f) = \int_{-\infty}^{\infty}R_x(\tau)e^{-j2\pi f\tau}d\tau, \qquad R_x(\tau) = \int_{-\infty}^{\infty}S_x(f)e^{j2\pi f\tau}df$$

At $\tau = 0$: $R_x(0) = \int S_x(f)df = P$, the average power.

Example: $x(t) = A\cos 2\pi f_0t$ gives $R_x(\tau) = \frac{A^2}{2}\cos 2\pi f_0\tau$ and hence $S_x(f) = \frac{A^2}{4}[\delta(f-f_0) + \delta(f+f_0)]$, with $P = R_x(0) = A^2/2$.