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Convolution output function 
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#1
Jul2414, 09:07 PM

P: 12

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. 


#2
Jul2414, 09:37 PM

P: 12

Just realised this should be in the homework/CW area (by the rules listed in the pinned thread). Reposting.



#3
Jul2414, 09:59 PM

P: 866

$$ \begin{matrix} g & & & & 2 & 0 & 1 \\ f(n) & 1 & 2 & 3 & 4 \\ g*f & & & & 8 \end{matrix} \qquad \ldots \qquad \begin{matrix} & 2 & 0 & 1 \\ 1 & 2 & 3 & 4 \\ & 4 & 0 & & = 4 \end{matrix} \qquad \ldots \qquad \begin{matrix} & 2 & 0 & 1 \\ & & & 1 & 2 & 3 & 4 \\ & & & 1 \end{matrix} $$ 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. 


#4
Jul2414, 10:05 PM

P: 12

Convolution output function
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. 


#5
Jul2414, 10:40 PM

P: 866

Well, following your notation it seems to me that the summation should read
$$ y_k = \sum_{i} g_i \, f_{ki+1} . $$ That gives $$ \begin{align} y_1 &= g_1 \, f_1 + 0 = (2)(4) = 8 \\ & \vdots \\ y_{m+n1} &= y_6 = 0 + g_3 \, f_{63+1} = (1)(1) = 1, \end{align} $$ which looks correct. 


#6
Jul2414, 10:51 PM

P: 12

Yup seems to be. Guess I'll assume it an error within the text and move on. Cheers



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