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What is the maximum count of non-zero elements, that can a linear convolution of discrete signals of "lengths" 5 and 7 have?

When I label the length of signal $x[n]$ as $M$, and the length of signal $h[n]$ as $N$, then the length of their convolution is the signal $y[n]=M+N-1$. However this is not that thing I was looking for. So how can I find the maximum count of non-zero elements?

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    $\begingroup$ hm, why do you think that's not the thing you're looking for? $\endgroup$ Nov 30, 2017 at 21:24

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It seems like you have already the correct answer, but try to visualize what's going on

First understand that signals of length $n_0$ are really infinite length, but have nonzero values at $n = 0$ and $n = n_0 - 1$. The values in between can be anything, but for the purposes of this problem take them to be nonzero as well.

Now perform the discrete convolution by literally shifting the length-5 signal and dot multiplying it with the length-7 signal. Your result will also be an infinite length signal with nonzero values only where the two signals overlap (when they dont overlap, you should find the convolution to be zero). In this case, there are 11 points where the signals actually overlap.

If some parts within the signal are zero, it is possible that you get fewer nonzero values in the result. However, in the max case where the full signal is nonzero you get this max, $11 = 7 + 5 - 1$ samples

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