The cross-correlation of continuous $f,g$ is:
$$(f \star g)(\tau)=\int_{-\infty}^{\infty}f^*(t)g(t+\tau)dt$$
Why is it an integral?
Why doesn't
The cross-correlation of continuous $f,g$ is:
$$(f \star g)(\tau)=f^*(t)g(t+\tau)$$ suffice?
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2Well, how does one evaluate your definition? If I have $f$, $g$, and $\tau$, and I want to compute $(f\star g)(\tau)$, where do I get the $t$ to put in the right-hand side? – Dec 04 '15 at 21:56
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Surely one can evaluate functions piecewise using some partitioning? Or "shift" functions. – mavavilj Dec 04 '15 at 21:58
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I don't understand your response. My question is, what is $t$ in the right-hand side of your definition? – Dec 04 '15 at 21:59
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A value shift. E.g. $f(x)=x$, then $f(x+0.5) = (x+0.5)$, whatever $x$ is? For periodic functions this is the "phase shift". – mavavilj Dec 04 '15 at 22:02
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What @Rahul was trying to tell you, I think, is that in the LHS of your alternative definition you have a function that depends only of $\tau$ and in the RHS you have a $t$ that is kind of bothering us :) That is a simple check of why you must integrate over $t$: to get rid of it. Suppose that $f(t) = Asin(\omega t + \theta)$ and $g(t)$ is an arbitrary function. Look that $f(t)$ is a function of $t$, $\theta$ is just a known paramater. Then $(f \star g)(\tau)$ is still a function only of $\tau$. – Carlos H. Mendoza-Cardenas Dec 07 '15 at 23:48