Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Convolution output function

  1. Jul 24, 2014 #1
    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 gi (i = 1,2.... m), and a convolution operator fj (j = 1,2 ......n) the convolution output is given by:

    yk = [itex]\Sigma[/itex]gifk - 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 y1 I end up with negative index of fj as I perform fk - i as i increases from 1 to 3 with the largest index being f0 (from f1 - 1 where k = 1 and i = 1). How can this be correct?

    Any help would be appreciated.
     
  2. jcsd
  3. Jul 24, 2014 #2
    Just realised this should be in the homework/CW area (by the rules listed in the pinned thread). Reposting.
     
  4. Jul 24, 2014 #3
    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.
     
  5. Jul 24, 2014 #4
    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.
     
  6. Jul 24, 2014 #5
    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.
     
  7. Jul 24, 2014 #6
    Yup seems to be. Guess I'll assume it an error within the text and move on. Cheers
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted
Similar Discussions: Convolution output function
Loading...