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This answer may be straight forward but I cannot figure out. One can understand the difference between cross-correlation and convolution from the link below:

What is the difference between convolution and cross-correlation?

The question is can there be two complex signal which are uncorrelated but their spectrum overlaps? In other words spectrum is not disjoint or product of power spectrum of each signal is not zero in each frequencies?

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If you mean correlation at the origin (no signal shifts), then yes so long as all the overlapping frequencies the two signals are 90 degrees out of phase.

That is, for all positive $\omega$ that are overlapping, $h_1(\omega) = k_\omega h_2(\omega) e^{\pm i \tfrac{\pi}{2}}$, where $h_1$ and $h_2$ are the spectrums and $k_\omega$ is just a real scalar. The negative frequencies will be the conjugate of these.

If you mean correlation with signal shifts (cross-correlation - similar to convolution) then no, because if both signals have the same non-zero frequency then at some shift the sinusoids will not be 90 degrees out of phase, and therefore correlated.

More simply

Imagine a signal consisting of a sine wave and another consisting of a cosine wave of the same frequency. They will have overlapping spectrums (same frequency) but have zero correlation.

Add to the first signal a sinusoid of a certain frequency, and add to the second signal the same sinusoid but 90 degrees phase shifted.

Example

Barbara image with DC component removed, and Barbara image with DC removed and half spectrum multiplied by $i$ and the other half by $-i$ (90 degrees phase shift and conjugate for the negative frequencies)

enter image description here enter image description here

Their spectrums overlap completely but they have zero correlation (at shift 0).

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    $\begingroup$ If you have two deterministic signals $x(t)$ and $y(t)$ with Fourier transforms $X(\omega)$ and $Y(\omega)$, the Fourier transform of their cross-correlation is given by $X(\omega)Y^*(\omega)$. If, as you suggested, $Y(\omega)=kX(\omega)e^{i\pi /2}$, then the Fourier transform of the cross-correlation becomes $k^*|X(\omega)|^2e^{-i\pi /2}$. I don't see how this becomes zero if $x(t)\neq 0$. $\endgroup$ – Matt L. Jul 2 '15 at 12:51
  • $\begingroup$ @MattL. Thanks, I just assumed the OP meant correlation of the signal vectors, not cross-correlation. Have updated answer. $\endgroup$ – geometrikal Jul 2 '15 at 13:31
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    $\begingroup$ But the cross-correlation at shift $0$ is just the integral over $X(\omega)Y^*(\omega)$, which is also generally non-zero under the condition you gave in your answer, because it's basically the integral over $|X(\omega)|^2$. $\endgroup$ – Matt L. Jul 3 '15 at 7:09
  • $\begingroup$ @MattL. I agree with you that this answer is incorrect. I hope that geometrikal either edits it into what needs to be said: $X(\omega)$ and $Y(\omega)$ need to be orthogonal as signals, that is, $$\int_{-\infty}^\infty X(\omega)[Y(\omega)]^*\,\mathrm d\omega = 0$$ or deletes it entirely (since the original idea of phase orthogonality is not salvageable). $\endgroup$ – Dilip Sarwate Jul 3 '15 at 12:37
  • $\begingroup$ @MattL. You are forgetting the negative frequencies. It will be something like $k|X(\omega)|^2 e^{-i\pi/2} + k|X(-\omega)|^2 e^{i\pi/2} = 0$ $\endgroup$ – geometrikal Jul 4 '15 at 0:53
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Let $x(t)$ and $y(t)$ denote two complex-valued signals. Their crosscorrelation function is given by $$R_{x,y}(t) = \int_{-\infty}^\infty x(\tau)[y(\tau-t)]^* \,\mathrm d\tau \tag{1}$$ where $\,^*$ denotes complex conjugate. The Fourier transform of the crosscorrelation function $R_{x,y}(t)$ is $S_{x,y}(f)$, which is sometimes called the cross-spectral density, happens to satisfy the identity $$S_{x,y}(f) = X(f)[Y(f)]^*, ~ -\infty < f < \infty, \tag{2}$$ where $X(f)$ and $Y(f)$ are the Fourier transforms of $x(t)$ and $y(t)$.

Signals $x(t)$ and $y(t)$ are said to be uncorrelated if $R_{x,y}(t)$ equals $0$ for all $t, -\infty < t < \infty$. I leave it for you to decide whether the property $R_{x,y}(t)=0$ for all $t$ is compatible with the claim that there is an interval $G$ on the frequency axis such that $X(f) \neq 0$ and $Y(f) \neq 0$ for all $f \in G$.

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  • $\begingroup$ I was discussing deterministic signals. For the definition of crosscorrelation, see Wikipedia. If $R_{x,y}(0) = 0$ but $R_{x,y}(t) \neq 0$ in general, then th signals are said to be orthogonal at zero shift. $\endgroup$ – Dilip Sarwate Jul 3 '15 at 2:34
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At the risk of being redundant, I add yet another answer because I think that not all things have been addressed as clearly as necessary.

As mentioned in Dilip Sarwate's answer, the cross-correlation function of two signals $x(t)$ and $y(t)$ cannot vanish (for all values of its argument) if their spectra overlap, because this would mean that their cross-spectral density given by $X(\omega)Y^*(\omega)$ must equal zero. And the latter can't be true if $X(\omega)$ and $Y(\omega)$ overlap.

The other question is if the cross-correlation can be zero for zero shifts if the spectra overlap. This would mean that

$$\int_{-\infty}^{\infty}x(t)y^*(t)dt=0\tag{1}$$

Or, equivalently,

$$\int_{-\infty}^{\infty}X(\omega)Y^*(\omega)d\omega=0\tag{2}$$

This condition is called orthogonality. It can be satisfied for signals that overlap in time and that have overlapping spectra, if they - or their spectra - form a Hilbert transform pair. E.g., if $x(t)$ and $y(t)$ are a Hilbert transform pair then their Fourier transforms are related by

$$Y(\omega)=-j\cdot\text{sign}(\omega)X(\omega)\tag{3}$$

(This is what geometrikal addressed in his answer.)

The integral in (2) then becomes

$$j\int_{-\infty}^{\infty}|X(\omega)|^2\text{sign}(\omega)d\omega\tag{4}$$

which, if it exists (if necessary as a Cauchy principal value), is indeed zero, even though both signals overlap in time as well as in frequency.

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  • $\begingroup$ Also for higher dimensions the Riesz transform can be used to generate orthogonal signals, assuming zero DC. $\endgroup$ – geometrikal Jul 4 '15 at 12:47

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