A question about Conv2DTranspose kernel size (W3 Quiz #8)

In Week 3 quiz #8, what key change is required if this is the last layer of FCN-16. The answer is we’ll have to change kernel size to (16,16). How is that calculated?


From what I remember, One has to design the architecture in such a way that we get the desired shape in the output. Transpose Convolution output depends on the filter size and stride size.

For FCN16, we will need to upsample from 2x2 to 32x32, right? If we only change filter size to (16,16), but keep strides (8,8), then since there will be 8 steps overlapped when we move the filter, don’t we only get 24x24 output? I don’t quite understand about this. Maybe I’ve misunderstood how Transpose Convolution works.