When lowpass filtering with a Gaussian kernel, how can $\sigma$ be related to cycles/pixel?
The frequency response of a Gaussian is also a Gaussian, see here. The cutoff frequency $\sigma_f$ is defined as the standard deviation of the frequency response. This is given by $$ \sigma_f = \frac{1}{2\pi\sigma} $$ Given that $N$ is the wavelength of the frequency components you wish to remove, you can calculate the corresponding frequency and work out how much you want to reduce it by, work out the corresponding cut-off frequency and filter standard deviation.

A more back of the envelope approach is to say that Gaussians of standard deviation, $\sigma$, remove or strongly reduce components with lateral size $&lt;\sigma$.