PSD (Power spectral density) explanation

I'm trying to understand how the PSD is calculated. I've looked in a few of my Communication Engineering textbooks but to no avail. I've also looked online. Wikipedia seems to have the best explanation; however, I get lost at the part where they decide to make the CDF (Cumulative Distrubution Function) and then for some reason decide to related that to the autocorrelation function.

I guess what I don't understand is, how does autocorrelation having anything to do with calculating the PSD? I would've thought that the PSD simple be the Fourier Transform of $P(t)$ (where $P(t)$ is the power of the signal with respect to time).

Answer

You are right, PSD has to do with calculating the Fourier Transform of the power of the signal and guess what.....it does. But first let's look at the mathematical relationship between the PSD and the autocorrelation function.

1. 
   

   Notations:

   
   
   
   
   • Fourier Transform: $$ \mathcal{F}[ x(t)] = X(\omega) = \int_{-\infty}^{\infty} x(t)e^{-j\omega t}dt $$
   
   
   • (Time) Auto-Correlation Function: $$ R(\tau) = x(\tau) * x(-\tau) = \int_{-\infty}^{\infty} x(t)x(t + \tau)dt $$
   
   
   
   


2. 
   

   Let's prove that the Fourier Transform of the Auto-Correlation function does indeed equal to the Power Spectral density of our stochastic signal signal $x(t)$.

   
   
   

$$ \mathcal{F}[ R(\tau)]= \int_{-\infty}^{\infty} R(\tau)e^{-j\omega \tau}d\tau $$ $$ = \int_{-\infty}^{\infty} \int_{-\infty}^{\infty} x(t)x(t + \tau) e^{-j\omega \tau} dtd\tau $$ $$ = \int_{-\infty}^{\infty}x(t) \underbrace{\int_{-\infty}^{\infty} x(t + \tau) e^{-j\omega \tau} d\tau}_{\mathcal{F}[x(t + \tau) ] = X(\omega)e^{j\omega t}} dt $$ $$ = X(\omega) {\int_{-\infty}^{\infty}x(t)e^{j\omega t} dt} $$

$$ = X(\omega)X^*(\omega)= |X(\omega)|^2 $$

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What does it all mean? Note: This explanation is a bit "hacky". But here it goes

The Fourier transform tells us the spectral components of a signal. In our case, the signal is Stochastic; So, trying to calculate the spectral components of the signal will be pointless because, for every realisation of the random process, you will have different expressions for $ \mathcal{F}[x(t)]$.

What if you take the Expected Value of the Fourier transform then? This wouldn't work. Let's take a zero mean signal for example.

$$ \mathbb{E}\{ \mathcal{F}[x(t)] \} = \mathcal{F}[\mathbb{E}\{ x(t) \}] = 0$$

Instead, what if you take the Fourier transform of the square of the signal. $$ \mathbb{E}\{ \mathcal{F}[x^2(t)] \} = \mathcal{F}[\underbrace{\mathbb{E}\{ x^2(t) \}}_{\text{Av. Power of the Signal}}] $$

The autocorrelation function is essentially the $P(t) $ which you were alluding to.

References:

[1] Communications 1, P-L. Dragotti, Imperial College London

[2] White Noise and Estimation, F. Tobar [Unpublished Report]