Hey, I've begun going through a book called "An introduction to geophysical exploration" by Phillip Kearey and Michael Brooks and I've come across a problem I can't for the life of me see how they got their answer. Essentially, given an input function g_{i} (i = 1,2.... m), and a convolution operator f_{j} (j = 1,2 ......n) the convolution output is given by: y_{k} = [itex]\Sigma[/itex]g_{i}f_{k - i} (k = 1,2 ..... m + n - 1) with (Sigma sum starting at i = 1 and going up to m). Their example is with an input of g(2,0,1) and operator of f(4,3,2,1) the output is y(8,6,8,5,2,1). Not only can I not see how this is obtained, but based on the function if I'm trying to find y_{1} I end up with negative index of f_{j} as I perform f_{k - i} as i increases from 1 to 3 with the largest index being f_{0} (from f_{1 - 1} where k = 1 and i = 1). How can this be correct? Any help would be appreciated.
Just realised this should be in the homework/CW area (by the rules listed in the pinned thread). Reposting.
The convolution g*f basically applies a "sliding filter" to g, where the filter is just a reversed f (or f by a reversed g; it's symmetric). In this case f(-n) = (1,2,3,4), so the terms in g*f are just $$ \begin{matrix} g & & & & 2 & 0 & 1 \\ f(-n) & 1 & 2 & 3 & 4 \\ g*f & & & & 8 \end{matrix} \qquad \ldots \qquad \begin{matrix} & 2 & 0 & 1 \\ 1 & 2 & 3 & 4 \\ & 4 & 0 & & = 4 \end{matrix} \qquad \ldots \qquad \begin{matrix} & 2 & 0 & 1 \\ & & & 1 & 2 & 3 & 4 \\ & & & 1 \end{matrix} $$ At the endpoints you see that f(-n) is allowed to "fall off the ends," so you get as many outputs as both inputs combined, less 1. This is just because f and g have to overlap by at least one term to make an output. There are some nice visual demonstrations at the Wikipedia page for Convolution.
Yeah they have a demonstration of this on the next page of the textbook, which is easy enough to follow (although it didn't mention anything about the reversed filter which is helpful, I had wondered if it was a mistake originally). It's just frustrating I can't get the same result using the Sigma Summation given. Using the sigma summation I get y(0,8,6,8,5,2) as if I regard any impossible index's i.e. f(n) where n < 1 as 0, y1 can only ever be 0 as opposed to 8. So my solution is shifted to the right by one place from the actual solution.
Well, following your notation it seems to me that the summation should read $$ y_k = \sum_{i} g_i \, f_{k-i+1} . $$ That gives $$ \begin{align} y_1 &= g_1 \, f_1 + 0 = (2)(4) = 8 \\ & \vdots \\ y_{m+n-1} &= y_6 = 0 + g_3 \, f_{6-3+1} = (1)(1) = 1, \end{align} $$ which looks correct.