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Convolution is a linear process, which means that superposition holds. Thus, you can break up any convolution kernel $k$ into multiple parts ($k_1, k_2, ... k_N$) such that $k = \sum k_i$. … The sum would be equal to the convolution product of the original kernel. Doing that would be inefficient, though, both in terms of computations and memory. …

No, if the two sequences are random then convolving them is not useful.

Convolving that is circular convolution. In practice linear convolution and circular convolution are nearly the same, the difference happening at the beginning and the end of linear convolution. … With a few very rare exceptions we don't "choose" circular convolution. We almost always want linear convolution. …

Yes, in general, your #2 is correct. That being said, both of the filters stink (with your triangle filter being a little better). No, f3 does not have the same frequency response as f4. To get an …

I suspect it should be something like the following: $$ y[n] = h[n] * \delta[n + N] \Rightarrow H(z)\cdot z^N $$

Since convolution via DFT is circular convolution, it's not so much of a quadrant swap as it is a wrap around. I believe that if you put the non-zero taps in the middle that it will fix the problem. …

The figures are simply showing how the circular aspect of the convolution works in 2 dimensions. …