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Given a random signal $ Z \left( t \right) $ which is addition of two independent signals $ X \left( t \right) $ and $ Y \left( t \right) $ with constant parameters $ a $ and $ b $:

$$ Z (t) = aX(t) + bY (t) $$

If the auto correlation function of $ R_{XX} \left( \tau \right) $ and $ R_{YY} \left( \tau \right) $ is known (Assume both are Stationary Signals).
What would be the Auto correlation function of $ Z \left( t \right) $?

Is it given by (I skipped the steps):

$$ {R}_{ZZ} \left( \tau \right) = a^2 R_{XX} \left( \tau \right) + b^2 R_{YY}\left( \tau \right) $$

I neglected the $ {R}_{XY} \left( \tau \right) $ contribution because it is zero as $ X \left( t \right) \perp Y \left( t \right) $ (Independent Random Signals).

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You're correct as the Cross Correlation function vanishes.
This has the implicit assumption the process has zero mean (Actually, at least one of them).

Namely, in order to have $ {R}_{XY} \left( \tau \right) = 0 $ having $ X \left( t \right) \perp Y \left( t \right) $ isn't enough. It is required that at least of them has zero mean (Namely, $ \mathbb{E} \left[ X \left( t \right) \right] = 0 $, $ \mathbb{E} \left[ Y \left( t \right) \right] = 0 $ or both).

Pay attention that in practice for a single realization the empirical cross correlation won't vanish.

Remark
At the above we use correlation as defined in the context of signal processing.

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    $\begingroup$ just a minor point. If you're going to use $\tau$ to denote Z's correlation lag, then you should use it for x and y also. otherwise, the assumption is usually that $\tau = 0$ which can possibly confuse readers. $\endgroup$
    – mark leeds
    Commented May 31, 2018 at 21:32
  • $\begingroup$ I disagree, E(x y)=E(x)E(y) when x and y are independent. $\endgroup$
    – user28715
    Commented Jun 2, 2018 at 0:16
  • $\begingroup$ What here not to agree? He states the variables are independent. I implicitly assume each is centered as well. $\endgroup$
    – Royi
    Commented Jun 2, 2018 at 8:33
  • $\begingroup$ @Royi You implicitly assumed that it was centered. Assumptions should be explicit. Neither you or the OP state that it is "centered." If the processes are not "centered," do you contend that your answer is correct of not? So what's not to agree, It's wrong as is. $\endgroup$
    – user28715
    Commented Jun 2, 2018 at 21:50
  • $\begingroup$ @StanleyPawlukiewicz, I updated the question to make the implicit assumption explicit. Yet I'd say that no one would use the Auto Correlation Function for non centered signals (What's the point of using the ACF for analysis of the data in that case?). So think the above is all about the Autocovariance Function. $\endgroup$
    – Royi
    Commented Jun 2, 2018 at 22:50

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