From a book, I learned that image interpolation (or upsampling) can be written as a convolution like this: $$g(i,j) = \sum_{(x,y)}f(x,y)h(i-rx,j-ry)$$ while $r$ is the upsampling rate.

But I have a time understanding this formula, since AFAIK, the basic idea of image interpolation is to bring near-by pixels into consideration, so I thought a convolution like below might suffice: $$g(i,j) = \sum_{(x,y)}f(x,y)h(i-x,j-y)$$ What is the point of $r$ in the formula?


1 Answer 1


Image interpolation is not (exactly) the same as upsampling. Suppose that you have a grid $\Lambda$ which is a $640 \times 480$ grid of pixels. Each pixel has a location $(x,y) \in \Lambda$. Each pixel also has an intensity $I(x,y)$. This could be a gray-scale value or an RGB triplet. Now suppose that you want to know the intensity of a location that is not centered on one of the pixels. For example, suppose that you wanted the intensity at location $(17.35, 188.42)$. In this case, you would use the neighboring integer pixel intensities in an appropriate interpolation scheme (nearest neighbor, bilinear, bicubic, Lanczos etc.) to get the sub-pixel intensity. This is image interpolation.

Image upsampling is when you want to go from a grid $\Lambda$ that is $640 \times 480$ to a larger grid $\Gamma$ that is for example, $1280 \times 960$ pixels. In such a case, a simple algorithm is create the new grid filled with zeros. Then, if there is a pixel value in the original image at $(x,y)$, then take its intensity I(x,y) and fill it in the upsampled image at $I'(2x,2y)$. The $2$ is the $r$ (rate) referred in your formula. Then, we use an appropriate kernel (the $h$ in your formula) to filter the image and complete the upsampling. NOTE: This is a simple explanation. Please see Gozalez, Woods and Eddins, Digital Image Processing for a better treatment.


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