Why do we use an integral in the 1D convolution, but a sum in the 2D convolution? If I understand convolution correctly, we're applying the kernel (g) to an image (f) at point x. Since each image consists of some number of pixels, aren't images always discrete in dimension? Then why are we taking the integral from -inf to inf?
Thanks!
motoole2
There are 1D, 2D, and even ND convolution operations for both the discrete or continuous cases. So while I only showed the definition of 1D convolutions for the continuous case, you can also define a 2D version as follows:
$(f*g)(x,y) = \iint f(u,v) g(x-u,y-v) du dv$
In the general case, convolution operations can be applied to continuous functions that have a domain that spans -inf to inf. But you're right that, when working with images, this often simplifies to a finite domain and a discrete number of pixel values.
Why do we use an integral in the 1D convolution, but a sum in the 2D convolution? If I understand convolution correctly, we're applying the kernel (g) to an image (f) at point x. Since each image consists of some number of pixels, aren't images always discrete in dimension? Then why are we taking the integral from -inf to inf?
Thanks!
There are 1D, 2D, and even ND convolution operations for both the discrete or continuous cases. So while I only showed the definition of 1D convolutions for the continuous case, you can also define a 2D version as follows:
$(f*g)(x,y) = \iint f(u,v) g(x-u,y-v) du dv$
In the general case, convolution operations can be applied to continuous functions that have a domain that spans -inf to inf. But you're right that, when working with images, this often simplifies to a finite domain and a discrete number of pixel values.