# Are there any signals with brickwall autocorrelation?

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

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

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.