So I was reading this textbook - Digital Signal Processing by Proakis and Manolakis. The authors define the cross-correlation operation for 2 energy signals $x(n)$ and $y(n)$ in section 2.6.1 as:

$$ r_{xy}(l)=\sum_{n=-\infty}^{\infty}x(n)y(n-l) $$

In section 2.6.3, they define the cross-correlation operation of two power signals as:

$$ r_{xy}(l)=\lim_{M\to\infty}\frac{1}{2M+1}\sum_{n=-M}^{M}x(n)y(n-l) $$

My question is, why do we need to have a separate definition for power signals? Are these two definitions the same? If they aren't same, what is the intuition behind defining them like that?


Energy signals (such as an exponentially decaying pulse) are bounded in total energy as time goes to infinity in which case you can use the first definition. Power signals (such as a sine wave) have finite energy over finite time intervals but infinite energy for all time (power is energy/time) so the first equation will not converge in that case.

  • $\begingroup$ Thank you @DanBoschen for answering. I understand everything that you have written, but I don't quite see how this leads to power signals having a separate definition. I mean, the equations don't have an energy term, it is just $x(n)\times y(n-l)$. Can you please elaborate? $\endgroup$ – Aditya DS May 6 '20 at 12:35
  • $\begingroup$ The integral of a product term is proportional to energy which is what you see in the first equation, while power is energy per unit time resulting in the averaging of the power term as done in the second equation. $\endgroup$ – Dan Boschen May 6 '20 at 12:53
  • $\begingroup$ But the point is power signals as we described them cannot possibly use the first equation, right? So we need the more complicated second equation to bound it. We don’t need that for energy signals so why not take the easier path? $\endgroup$ – Dan Boschen May 6 '20 at 12:59
  • $\begingroup$ Right, now I get it. So $ x(n)y(n)\propto x^{2}(n) $ or $ x(n)y(n)\propto y^{2}(n) $. And since we are summing it, it is similar to calculating the energy. And if that diverges (in case of power signal), we can't use the first equation. Did I understand properly? Thanks a lot!! $\endgroup$ – Aditya DS May 6 '20 at 13:04
  • $\begingroup$ Yes to the extent there is actual correlation which means there is some x in y. I think it is much clearer if you look at the similar auto-correlation formula where you would have the clear example of x^2 which is the energy of x if you accumulate (integrate) all samples. So this is the case here of a cross-correlation so same thing but with two different signals. $\endgroup$ – Dan Boschen May 6 '20 at 13:10

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