This was a question on one of my past year paper questions.

Question: Prove

$$f(m,n)*g(m-r,n-s) = f(m-r,n-s) *g(m,n)$$

where $f(m,n)$ and $g(m,n)$ are 2-D discrete functions, $r$ and $s$ are integers and $*$ is the 2-D convolution operator.

Not sure how to start on this and need some guidance on it.

  • 6
    $\begingroup$ Hint: Your mission, if you choose to accept it, is to try to do a one-dimensional case first. Write the sum to compute the first convolution at $k$, and play with the index of summation $i$ (doing things like introducing a new variable $\ell = k - i$ or $k-i-r$ or something similar -- be creative) to see if you can rearrange things to get the second convolution sum. Then become bolder. Write the $2$-D sum and make two similar changes on the two indices of summation. Good luck! This comment will self-destruct in $15$ hours. $\endgroup$ – Dilip Sarwate Nov 13 '11 at 12:33

In addition to the above comment:

Show that the 2D Z-transforms of both sides are equal.

Step-by-step: If $F(z_1, z_2)$ is the 2D Z-transform of $f(m,n)$, then what is the 2D Z-transform of

  1. $f(m-r, n)$?

  2. $f(m-r, n-s)$?

  3. $f(m, n) * g(m, n)$?

  4. $f(m-r, n-s) * g(m, n)$?

(Yes, I realize that the 2D Z-transform may not be taught before this exam was presented, but it is a valid solution, nonetheless.)

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1-D convolution x(t) * h(t) = Summation x(r)h(t-r)dr let t-r =s = - Inverse Summation x(t-s)h(s)ds = Summation x(t-s)h(s)ds = h(t)*x(t) 2-D convolution

f(m,n)∗g(m−r,n−s)=Summation Summationf(h,b)g(m-r-h,n-s-b) Let m-r-h = e n-s-b = c = Summation Summation f(m-r-e,n-s-c)g(h,b) = f(m−r,n−s)∗g(m,n)

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