# Generating a random signal for an autocorrelation that is not square integrable

Suppose I have a function that is not square integrable such as the zeroth Bessel function of the first kind. How do I generate a random sequence, with mean zero of course, such that its autocorrelation satisfies this function?

If $$r_{xx}(t)$$ is the autocorrelation of $$x(t)$$ it's Fourier Transforms (FT) are given by

$$R_{xx}(\omega) = |X(\omega)|^2$$

That creates a handy way to create a random signal from a given autocorrelation

1. Calculate the FT of the autocorrelation (which is real)
2. Take the square root
3. Apply a random phase (uniformly distributed on $$[-\pi,\pi]$$) maintaining Hermitian symmetry, i.e. $$\varphi(-\omega) = -\varphi(\omega)$$
4. Take the inverse FT
• This method doesn't work because the Fourier transform of the $r_{xx}(t)$ is not always positive if $r_{xx}(t)$ is the zero-th order Bessel function. You will get negative, even complex, components when the ACF is Fourier-transformed. Feb 6, 2023 at 2:03
• @JulianOng : Nope. The FT of the autocorrelation is the power spectral density. It's always positive. (and it's real if $x(t)$ is real as well). Feb 6, 2023 at 12:46
• The problem is, what if the FT of some given and hypothetical autocorrelation (which is real and an even function) is not always non-negative? Nov 2, 2023 at 16:13