CNN backpropagation with stride>1 I read that to compute the derivative of the error with respect to the input of a convolution layer is the same to make of a convolution between deltas of the next layer and the weight matrix rotated by $180°$, i.e. something like

$$\delta^l_{ij}=\delta^{l+1}_{ij} * rot180(W^{l+1})f'(x^l_{ij})$$

with $*$ convolution operator; This is valid with $stride=1$; but what happens when stride is greater than $1$? is still a convolution with a kernel rotation or I can't make this simplification?
Backpropagation with stride > 1 involves dilation of the gradient tensor with stride-1 zeroes. I created a blog post that describes this in greater detail.