Boundary Conditions
There are various conditions one can choose from to deal with boundaries when taking a fft.
FourierFilterFlux.originalSize — FunctionoriginalSize(sz, boundary::ConvBoundary)Given the size sz from the fft, return the size of the input.
FourierFilterFlux.effectiveSize — FunctioneffectiveSize(sz, boundary::ConvBoundary)Given the input size, figure out the resulting size
Periodic
The simplest to implement, this just assumes that the two edges are adjacent boundary=Periodic().
Pad{N}
The typical choice in convolutional neural networks, this assumes that the signal is zero beyond the values given. Currently, it is up to the user to figure out what the appropriate padding is; it is at worst the size of the space domain support of your filter. Suppose you have a 1D filter with 5 adjacent non-zero entries in the space domain, e.g. w = [0..., 0, 5,1,-9,3,2, 0..., 0]; then the most carry-over that is possible is [2, 0..., 0, 5,1,-9,3], so if we pad with at least 5 zeros, then no entries will carry over. To do this, set boundary = Pad((5,)). You can choose larger paddings, of course, but you pay the price of a larger effective signal size. This can be chosen automatically by setting the padding amount to -1, e.g. boundary = Pad((-1,)).
FourierFilterFlux.Pad — TypePad(x::Vararg{<:Integer, N}) where N = Pad{N}(x)N gives the number of dimensions of convolution, while x gives the specific amount to pad in each dimension (done on both sides). If the values in x are negative, then the support of the filters will be determined automataically
Symmetric
This is the assumption underlying the DCT type II[1] 
It assumes that the first and last value are repeated. This has two primary advantages: it makes the boundary continuous, and the derivative is less discontinuous than it would be with simply repeating the entries (as in the DCT type I above). The trade-off presently is that the implementation requires twice the space of the original image, which can be cost prohibitive.
To use this boundary condition, just set boundary=Sym().
- 1This image is from wikipedia.
