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We can express 2D convolution between $f(m,n)$ and $h(m,n)$ as following

\begin{align} g(m,n) &= \displaystyle \sum_{k_1=-\infty}^{\infty} \sum_{k_2=-\infty}^{\infty} f_{\!_{k_1,k_2}}h(m-k_1,n-k_2)\tag{1}\\ &=f(m,n)*h(m,n)\tag{2} \end{align} where \begin{equation} h(m,n) = \begin{cases} h_{m,n} & {0\leq m \leq L_r, 0\leq n \leq L_c}\\ 0 & \text{otherwise} \end{cases} \end{equation}

and

\begin{equation} f(m,n) = f_{m,n} \forall {-\infty\leq m \leq \infty, -\infty\leq n \leq \infty} \end{equation}

From (1), it seems as if the value of g at location (m,n) is equal to the double sum. But (2) is more like the input-output relationship that can be represent as. here $f(m,n)$ and $g(m,n)$ represent the entire 2D signals. Can someone please figure out where I am making a mistake enter image description here

NAASI
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  • Are the functions $f$, $g$ and $h$ with integer domains? – peterwhy Jul 11 '15 at 00:49
  • And I don't understand why you think there is a mistake between lines (1) and (2). – peterwhy Jul 11 '15 at 00:55
  • (1) n (2) are correct but I am confused about what they mean. From (2), I conclude that the 2D convolution of entire 2D data f(m,n) with filter h(m,n) gives the complete 2D matrix g(m,n), whereas from (1) it seems that the value of function g at single location (m,n) can be computed using the double sum. so (1) is taking about a single point and (2) is about the whole data. Thats where I am missing the point – NAASI Jul 11 '15 at 14:20
  • I looked for 1D case. at https://en.wikipedia.org/wiki/Convolution short hand notation is written as $(fh)[n]$ but at http://cnx.org/contents/febae08d-0391-4973-a2e1-fcac232bcb1c@4/Convolution---Discrete-time it is written as $y(n)=f(n)h(n)$ . what is the difference between these two notations. – NAASI Jul 11 '15 at 14:33
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    It is like saying differentiation $\frac d{dx}$ is an operation on a function but not a particular value of the function: $$f'(x) = \frac{d}{dx}\ f(x)\tag3$$ and the three $x$'s do not even have the same meaning, unlike $$f'(x_0) = \left.\frac d{dx} \ f(x)\right|{x = x_0}\tag4$$ But it is possible to write the formula of $f'$ at a single location $x$, without using $\frac d{dx}$: $$f'(x) = \lim{h\to 0}\frac{f(x+h)-f(x)}{h}\tag5$$ And indeed, $(5)$ is a definition of the $\frac d{dx}$ operation in $(3)$, like $(1)$ is a definition of the (2D discrete-time) $*$ operation in $(2)$. – peterwhy Jul 11 '15 at 18:50
  • thanks … you made an effort here. – NAASI Jul 11 '15 at 19:58

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