Are there any signals with brickwall autocorrelation?

Question

Are there any signals whose autocorrelation $R(\tau)$ has the following form? Assuming $\tau_c > 0$ and $R_0 > 0$ a constant,

$$R(\tau) = \begin{cases}R_0, \text{ for $|\tau| < \tau_c$} \\ 0, \text{ otherwise}\end{cases}$$

Answer 1

It is not possible for a (WSS) random process $\{X(t)\}$to have an autocorrelation function of the form $$R_X(\tau) = E[X(t)X(t+\tau)] = \begin{cases}\sigma^2, & |\tau|\leq T,\\0, & |\tau| > T. \end{cases}$$

Assuming a zero-mean process for simplicity, note that $R_X(0) = E[X^2(t)] = \sigma^2$ where $\sigma^2$ is the common variance of the random variables. Suppose that the highlighted statement above is true and that $T = 2$ and thus $R_X(1) = E[X(t)X(t+1)] = \operatorname{cov}(X(t), X(t+1))$ also has the same value $\sigma^2$. Then, $X(t)$ and $X(t+1)$ are perfectly correlated random variables and thus are equal in the mean-square sense -- their difference has variance $0$: \begin{align}\operatorname{var}(X(t) - X(t+1)) &= \operatorname{var}(X(t))+\operatorname{var}(X(t+1)) - 2\operatorname{cov}(X(t), X(t+1))\\ &= \sigma^2+\sigma^2-2\sigma^2\\ &=0.\end{align} and thus $X(t) = X(t+1)$ in the mean-square sense. But then, since $t$ is arbitrary, we have that \begin{align}X(t+1) &= X((t+1) + 1) = X(t+2)\\ X(t+2) &= X((t+2) + 1) = X(t+3). \end{align} So, $X(t) = X(t+1) = X(t+2) = X(t+3)$ so that $R_X(3) = E[X(t)X(t+3)]=\sigma^2$ also, contrary to the hypothesis that $R_X(\tau) = 0$ when $|\tau| > 2$.

In summary, $R_X(\tau)$ cannot have a fixed nonzero value in a finite-length interval including the origin and value $0$ outside this finite-length interval.