# Cross Correlation, how can any signals except the trivial cases be uncorrelated?

I have been working on a problem where I am trying to subtract two Power Spectral Densities (PSD's) from one another in the following way. $$|F(k) - G(k)|^2 = |F(k)|^2 - |G(k)|^2$$

Which is only valid if the cross correlation terms are 0.

It's my understanding that for signals $$f$$ and $$g$$, the cross correlation is given by $$(f \star g)(\tau) = \int_{-\infty}^{\infty} \overline{f(t)}g(t+\tau) dt$$ And now that the definition for these to be uncorrelated, $$(f \star g) = 0$$ has to be true for all $$\tau$$. But what i don't understand is that unless either $$f$$ or $$g$$ are just zero for all $$t$$ there will be a value of $$\tau$$ where $$\overline{f(t)}g(t+\tau) \neq 0$$ It then follows that if there are any non-zero points then there must exist other ones which cancel so that the integral ends up being 0.

It seems to me that the chances of this happening for any useful finite signal is next to nothing and so it's highly unlikely that anybody could ever use the spectral subtraction.

Can anyone explain what I am missing?

• You're right that getting exactly zero correlation for all $\tau$ is unusual for finite-energy deterministic signals. However, you can still say that if the correlation is small, then the error in the PSD calculation will also be small, and live with that small error.
– MBaz
May 30 '19 at 13:39
• Oh, please, please, please don't use $*$ to denote crosscorrelation; that operator is already overloaded with the commonly accepted meanings of convolution and (as a superscript) complex conjugation at least one of which you will be using since you are using $|\cdot|^2$ in your Fourier transforms. How about something like $\star$ for crosscorrelation? That's $\star$ by the way... May 30 '19 at 14:49

If the crosscorrelation function $$x\star y$$ of $$x$$ and $$y$$ is zero everywhere, then the Fourier transform of $$x\star y$$, which is $$X(f)Y^*(f)$$ (or $$X^*(f)Y(f)$$ for left-handed folks), must also have value $$0$$ for all $$f$$. But, finite-duration signals have Fourier transforms with support $$(-\infty, \infty)$$ and so the product $$X(f)Y^*(f)$$ cannot be zero everywhere.