3
$\begingroup$

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.

$\endgroup$
  • 1
    $\begingroup$ 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 Aug 30 '17 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 $\endgroup$ – Dilip Sarwate Aug 30 '17 at 14:26
  • $\begingroup$ How does one close the thread and accept the answers? :-) $\endgroup$ – Marcel Aug 30 '17 at 16:53
0
$\begingroup$

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

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.