The question is if the cross-correlation of two signals $x$ and $y$ can be non-zero even if $x$ and $y$ do not share frequencies.

My approach is the following: The cross-correlation $\rho_{xy}\left(\tau\right):=x\left(\tau\right)\star y\left(\tau\right)$ can be expressed in terms of a convolution:

$$ \begin{array}{rcl} \rho_{xy}\left(\tau\right) & = & x\left(\tau\right)\star y\left(\tau\right)\\ & = & \displaystyle\sum_{t=-\infty}^{^{\infty}}x\left(t\right)\bar{y}\left(t+\tau\right))\\ & = & \left(\bar{x}\right)\left(-\tau\right)*y\left(\tau\right) \end{array}$$

i.e. the cross-correlation of $ x\left(\tau\right) $ and $ y\left(\tau\right) $ equals the convolution of $ \bar{x}\left[-\tau\right] $ and $ y\left(\tau\right).$

The convolution theorem $\mathcal{F}\left(x*y\right)=\mathcal{F}\left(x\right)\mathcal{F}\left(y\right) $ allows us to express the cross-correlation in terms of it's frequency content and replace the convolution by multiplication: $$ \mathcal{F}\left(\rho_{xy}\right)\left[\omega\right]=\mathcal{F}\left(\left(\bar{x}\right)\left(-\tau\right)*y\right) \left[\omega\right]=\bar{\mathcal{F}\left(x\right)}\left[\omega\right]\mathcal{F}\left(y\right)\left[\omega\right] $$ Due to Parseval's Theorem, signal energy in frequency and time domain corresponds to: $$ \sum_{\omega=-\infty}^{\infty}\left|\mathcal{F}\left(\rho_{xy}\right)\left[\omega\right]\right|^{2}=\sum_{\tau=-\infty}^{\infty}\left|\rho_{xy}\left(\tau\right)\right|^{2}.$$

Hence the energy of the cross-correlation of the signals corresponds in time and frequency dimension. Thus, unless two different signals share frequencies, they cannot interfere. Two orthogonal signals, i.e. $\left\langle x,y\right\rangle =0 $, can have shared frequencies without having a non-zero cross-correlation.

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    $\begingroup$ You are correct, but note that the cross correlation can be negative in general; so, what you found is that the correlation is necessarily zero, not just non-positive. $\endgroup$
    – MBaz
    Commented Aug 30, 2017 at 14:22
  • 3
    $\begingroup$ The notion of "shared frequencies" suggests that you are thinking only of periodic signals that are represented by Fourier series rather than Fourier transforms. On the other hand, cross-correlation as you define it is applicable only to signals with finite energy, and for that what you say is correct: if at most one $G(\omega)$ and $H(\omega)$ can be nonzero for each $\omega$, then $g(t)$ and $h(t)$ are orthogonal signals whose cross-correlation is zero. The reverse implication is invalid; cross-correlation being zero does not imply at most one of $G(\omega)$ and $H(\omega)$ is nonzero $\endgroup$ Commented Aug 30, 2017 at 14:26
  • $\begingroup$ How does one close the thread and accept the answers? :-) $\endgroup$
    – Marcel
    Commented Aug 30, 2017 at 16:53

1 Answer 1


The solution to the self-proposed problem was correct, as it stands.


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