the demonstration that states that the FT of the ACF function is the square of the DTFT of the signal

Question

I am following the book The Intuitive Guide to Fourier Analysis & Spectral Estimation with MATLAB. I am trying to selflearn the fourier analysis in matlab. I got lost in one passage in the demonstration that states that the FT of the ACF function is the square of the DTFT of the signal. I have attached it here As you can see in the passage that I named 1. a delta_tau is missed in my opinion. Can you confirm that? More important in the point 2. in my opinion there should be a tau and not t. So I don't know how it is demonstrated this formula, because if in the passage 2. there is t the demonstration can be ended. I hope you can help me

Answer 1

You are right, the derivation is full of typos. The first equation below Eq. $(8.39)$ should read

$$\int_{-\infty}^{\infty}x(t+\tau)e^{\color{red}{-}j\omega\tau}d\tau=X(\omega)e^{j\omega \color{red}{t}}\tag{1}$$

Substituting into $(8.39)$ gives

$$\begin{align}\mathcal{F}\big\{R(\tau)\big\}&=\int_{-\infty}^{\infty}x(t)X(\omega)e^{j\omega t}dt\\&=X(\omega)X(-\omega)=|X(\omega)|^2\end{align}\tag{2}$$

where the last equality is only true for real-valued $x(t)$. However, the overall result is also true for complex-valued $x(t)$ because in that case the ACF is defined differently:

$$\mathcal{F}\big\{R(\tau)\big\}=\int_{-\infty}^{\infty}x^*(t)x(t+\tau)dt\tag{3}$$