Autocorrelation and ESD as Fourier transform pair

Auto-correlation function

The auto-correlation function measures the similarity between a signal and a time-shifted copy of itself. For an energy signal $x(t)$:

$$R_x(\tau) = \int_{-\infty}^{\infty} x(t)\,x^*(t-\tau)\,dt$$

where $\tau$ is the time shift (lag).

Properties (energy signal)

1. Value at origin equals energy: $R_x(0) = \int |x(t)|^2dt = E$.
2. Conjugate symmetry: $R_x(\tau) = R_x^*(-\tau)$; for a real signal it is an even function, $R_x(\tau) = R_x(-\tau)$.
3. Maximum at origin: $|R_x(\tau)| \le R_x(0)$ for all $\tau$.
4. Fourier pair with ESD: $R_x(\tau) \leftrightarrow \Psi_x(f) = |X(f)|^2$.
5. $R_x(\tau) \to 0$ as $|\tau| \to \infty$ for energy signals.

Proof: $R_x(\tau)$ and ESD form a Fourier transform pair

Take the Fourier transform of $R_x(\tau)$:

$$\begin{aligned} \mathcal{F}\{R_x(\tau)\} &= \int_{-\infty}^{\infty} R_x(\tau)e^{-j2\pi f\tau}\,d\tau \\ &= \int_{-\infty}^{\infty}\int_{-\infty}^{\infty} x(t)\,x^*(t-\tau)\,e^{-j2\pi f\tau}\,dt\,d\tau \end{aligned}$$

Change the variable in the inner integral: let $\lambda = t - \tau$, so $\tau = t - \lambda$ and $d\tau = -d\lambda$ (limits swap, cancelling the sign):

$$\begin{aligned} \mathcal{F}\{R_x(\tau)\} &= \int_{-\infty}^{\infty} x(t)\left[\int_{-\infty}^{\infty} x^*(\lambda)e^{-j2\pi f(t-\lambda)}\,d\lambda\right]dt \\ &= \int_{-\infty}^{\infty} x(t)e^{-j2\pi ft}\,dt \cdot \left[\int_{-\infty}^{\infty} x(\lambda)e^{-j2\pi f\lambda}\,d\lambda\right]^* \\ &= X(f)\,X^*(f) = |X(f)|^2 = \Psi_x(f) \end{aligned}$$

Hence

$$R_x(\tau) \;\xleftrightarrow{\ \mathcal{F}\ }\; \Psi_x(f) = |X(f)|^2$$

i.e. the auto-correlation function and the energy spectral density are a Fourier transform pair.

Check: putting $\tau = 0$ in the inverse transform, $R_x(0) = \int \Psi_x(f)\,df = \int |X(f)|^2df = E$, which is Rayleigh's energy theorem.

Example: for $x(t) = e^{-at}u(t)$, $R_x(\tau) = \frac{e^{-a|\tau|}}{2a}$ and $\Psi_x(f) = \frac{1}{a^2 + 4\pi^2f^2}$, which are indeed a Fourier pair.