I wonder where that equation for the sigma comes from, I have never seen it. It is hard to define a cutoff frequency for the Gaussian.
The Gaussian filter is quite compact in both the spatial domain and the frequency domain, and therefore is an extremely good low-pass filter. But it has no clear point at which it attenuates all higher frequencies sufficiently to no longer produce visible aliasing artifacts, without also attenuating lower frequencies so much that the downsampled image looks blurry.
Of course we can follow the tradition from the field of electronics, and define the cutoff frequency as the frequency above which the signal gets attenuated with at least 3dB. I think this definition might have lead to the equation in the OP, though I don’t feel like attempting to replicate that computation.
From personal experience, I find 0.5 times the subsampling factor to be a good compromise for regular images. For example, to downsample by a factor of 2, I’d apply a Gaussian filter with sigma 1.0 first. For OP’s example of going from 0.3 to 0.5 m per pixel, the downsampling factor is 0.5/0.3 = 1.667, half that is 0.833.
Note that a Gaussian kernel with a sigma below 0.8 cannot be sampled properly without excessive aliasing, applying a Gaussian filter with a smaller sigma should be done through multiplication in the frequency domain.
Finally, the kernel size. The Gaussian is infinite in size, but it becomes nearly zero very quickly, and we can truncate it without too much loss. The calculation 2 * ceil(2 * sigma) + 1 takes the central portion of the Gaussian of at least four sigma, two sigma to either side. The ceiling operation is the “at least”, it needs to be an integer size of course. The +1 accounts for the central pixel. This equation always produces an odd size kernel, so it can be symmetric around the origin.
However, two sigma is quite small for a Gaussian filter, it cuts off too much of the bell shape, affecting some of the good qualities of the filter. I always recommend using three sigma to either side: 2 * ceil(3 * sigma) + 1. For some applications the difference might not matter, but if your goal is to quantify, I would certainly try to avoid any sources of error.
I am working with satellite images with different spatial resolutions, understood as pixel/meter. For experiments I want to artificially down-sample these images, keeping the image size constant. For example I have a 512x512 image with spatial resolution 0.3m/pixel. I want to downsample it to 0.5m/pixel 512x512. I got advised to apply a Gaussian kernel to blur the image. But how do I calculate the standard deviation and kernel size of a Gaussian kernel to approximate the desired lower resolution? I can't find a rigorous method to do that calculation. Any help really much appreciated!
ChatGTP says that the formula is:
sigma = (desired_resolution / current_resolution) / (2 * sqrt(2 * log(2)))
and kernel_size = 2 * ceil(2 * sigma) + 1
But can't explain why. Can someone explain how standard deviation (sigma) and desired output resolution are connected? And how do I know which sigma to use? Oftentimes these existing resizing functions ask for a sigma, but in their documentation don't explain how to derive it.
Calculating Gaussian Kernel sigma and width to approximate a desired lower resolution pixel/m for satellite images