I'm a bit confused by the use of i and j here. In the previous example, it seemed like they represented the coordinates of the kernel. However, I was wondering why we're extending from -$\infty$ to $\infty$, and what would the kernel look like in this case?
motoole2
The domain of the kernel/filter in this slide does extend from $-\infty$ to $\infty$. Only when the filter has finite support, can we reduce this expression to a finite sum. For example, if $f(i,j) = 0$ when $abs(i) > K$ or $abs(j) > K$, we can rewrite this expression as follows:
$$\sum_{i,j=-K}^{K} f(i,j) I(x-i,y-j)$$
Note that there are times where the function $f(i,j)$ does not have finite support. For example, a Gaussian filter is technically one such function.
coolbean
I see! So a filter like Gaussian, which accepts a domain from -$\infty$ to $\infty$ and doesn't have a bounded range (since it approaches but never reaches zero at -$\infty$ and $\infty$), would be a function without finite support.
I'm still a bit confused about how these i,j values plug into the image, since I thought that images would have finite dimensions? Could you explain how this interaction works?What would happen if the domain/ranges of the filter and image were different?
motoole2
I can give it a shot.
In practice, when performing convolutions between images and kernels, we almost always treat them as having finite support---at least when it comes to this course. But mathematically, convolutions can handle images/kernels of arbitrary size.
The domain of the filter and images are often different. For example, in this lecture, we focused on defining 3x3 kernels, whereas the images themselves could have relatively high resolution. The convolution operation on this slide still applies in such cases, but we have to make some assumptions about the value of the function outside of this domain. For example, one such option is to assume the function has value 0 (as stated above).
I'm a bit confused by the use of i and j here. In the previous example, it seemed like they represented the coordinates of the kernel. However, I was wondering why we're extending from -$\infty$ to $\infty$, and what would the kernel look like in this case?
The domain of the kernel/filter in this slide does extend from $-\infty$ to $\infty$. Only when the filter has finite support, can we reduce this expression to a finite sum. For example, if $f(i,j) = 0$ when $abs(i) > K$ or $abs(j) > K$, we can rewrite this expression as follows:
$$\sum_{i,j=-K}^{K} f(i,j) I(x-i,y-j)$$
Note that there are times where the function $f(i,j)$ does not have finite support. For example, a Gaussian filter is technically one such function.
I see! So a filter like Gaussian, which accepts a domain from -$\infty$ to $\infty$ and doesn't have a bounded range (since it approaches but never reaches zero at -$\infty$ and $\infty$), would be a function without finite support.
I'm still a bit confused about how these i,j values plug into the image, since I thought that images would have finite dimensions? Could you explain how this interaction works?What would happen if the domain/ranges of the filter and image were different?
I can give it a shot.
In practice, when performing convolutions between images and kernels, we almost always treat them as having finite support---at least when it comes to this course. But mathematically, convolutions can handle images/kernels of arbitrary size.
The domain of the filter and images are often different. For example, in this lecture, we focused on defining 3x3 kernels, whereas the images themselves could have relatively high resolution. The convolution operation on this slide still applies in such cases, but we have to make some assumptions about the value of the function outside of this domain. For example, one such option is to assume the function has value 0 (as stated above).