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Convolution dilation specifies the spacing between neighboring kernel sample points in a convolutional layer. Dilation one samples adjacent positions; larger dilation separates those positions while keeping the filter's learned coefficient count fixed. Convolution stride instead controls the spacing between successive filter origins.^[ref]

For an own real one-dimensional example, takeand coefficients $(1,1,1)$, with stride one, no padding and no bias. Use the unflipped cross-correlation convention. Dilation one sums three adjacent values and gives $(7,14,28,56,112)$. Dilation two instead uses offsets $0,2,4$:The second filter evaluates a wider input span while still using three coefficients.

For $k$ sampled coefficients separated by $d$ positions, the first-to-last displacement is $d(k-1)$. Including the first position gives the effective convolution kernel sizeThus $k=3,d=2$ spans five positions. At each origin only three of those five positions are sampled directly.

The same linear operation can be represented by a dense width-five kernel $(1,0,1,0,1)$. The inserted zeros are fixed structure, not additional trainable coefficients. Dilation changes the footprint without making the original three-parameter filter a general five-parameter one.^[ref]

For input length $L$, symmetric zero padding $p$ and stride $s$, the usual per-axis size formula becomeswhen the padded input accommodates the footprint. The example gives $\lfloor (7-4-1)/1\rfloor +1=3$ outputs. The floor function counts origins whose final sample remains in range.^[ref]

Dilation can enlarge a unit's spatial neural network receptive field without evaluating every intervening input position. Whether later layers fill those gaps depends on the architecture. A wider span alone does not establish greater numerical influence at every position or better performance on a learning task.

The convolution stride specifies how far the filter's origin advances between neighboring output positions of a convolutional layer. It controls which spatial positions are evaluated. A larger stride usually produces a smaller output grid without changing the number of coefficients in each filter.^[ref]

For a real one-dimensional example, take the inputand a size-three all-ones filter, no bias and no padding. Using the unflipped cross-correlation convention, stride one evaluates windows beginning at indices $0,1,2,3,4$ and givesStride two evaluates windows beginning at indices $0,2,4$, givingThe second result keeps every second origin from this first grid; it still combines three input values for each retained origin.

For an undilated length-$L$ input and length-$k$ filter, no padding and stride $s$, a zero-indexed output $t$ uses input indices $st$ through $st+k-1$. Requiring its last index not to exceed $L-1$ gives $st\le L-k$. Counting the admissible nonnegative integers $t$ therefore giveswhen $L\ge k$ and $s$ is a positive integer. The floor function is essential when the last candidate origin would extend beyond the input.

With symmetric zero padding of $p$ positions per side and convolution dilation $d$, the corresponding standard shape expression isHere dilation changes the spacing of samples within a filter footprint; stride changes the spacing between filter origins. The two parameters have different roles.^[ref]

For two-dimensional data, specify a stride for each spatial axis; unequal strides need not reduce height and width by the same amount. The weights of the filter remain shared across the visited positions. Stride also changes how neighboring outputs' neural network receptive fields are spaced in the original input, so it affects more than just a displayed array size.

The effective convolution kernel size is the bounding extent of a dilated convolution kernel along a spatial axis. For $k$ sampled coefficients with convolution dilation $d$, their offsets are $0,d,2d,\dots ,(k-1)d$. The first-to-last displacement is $d(k-1)$, so including the first position givesThis extent determines the kernel footprint used in spatial shape calculations; it is not generally the number of coefficients.^[ref]

For a two-dimensional example, choose a sampled kernel shape $(3,2)$ with dilation $(2,3)$. Its row offsets are $\{0,2,4\}$ and column offsets are $\{0,3\}$. The bounding shape is thereforeThere are six sampled spatial positions inside a twenty-position bounding box. Fourteen intermediate positions are absent from the sampling pattern.

With three input neural network channels and four output channels, ordinary dense channel connectivity usesweights and optionally four biases, giving $76$ parameters. A general dense five-by-four kernel with the same channel counts would have $240$ weights. Representing the dilated operation that way would require $168$ of them to be fixed zeros, rather than independently learned parameters.^[ref]

For an unpadded $9$-by-$10$ input and convolution stride one, the output grid isMore generally, for one axis with symmetric zero padding $p$ and stride $s$,when the padded input accommodates the footprint. The floor function counts admissible filter origins.^[ref]

The effective size is a geometric quantity for one layer. The complete neural network receptive field through several layers additionally depends on their kernels, strides and connections. Likewise, changing stride moves the spacing of origins; changing dilation changes the footprint at each origin. Keeping bounding extent, sampled-position count and learned-parameter count separate avoids confusing three different sizes.

A neural network receptive field describes which input positions can influence a specified output unit through the network's computational connections. A local convolutional layer uses only a small input region for each output, but successive layers can combine information from a larger region. This dependence is relative to a chosen input layer; an immediate previous-layer neighborhood and the corresponding original-input region have different sizes.^[ref]

For an own one-dimensional example, use two kernel-size-three convolutional layers, with convolution stride one, no padding and no convolution dilation. Intermediate units $0$, $1$ and $2$ depend on input sets $\{0,1,2\}$, $\{1,2,3\}$ and $\{2,3,4\}$ respectively. The first output of the second layer combines these three units, so its possible original-input dependency isTwo size-three filters therefore give a size-five receptive region, not size nine.

A recurrence tracks the geometry of an undilated serial network. Let $r_{\ell }$ be the span of the receptive region in original-input positions and $j_{\ell }$ the original-input index separation between adjacent layer-$\ell$ outputs. With kernel size $k_{\ell }$, stride $s_{\ell }$ and initial values $r_0=j_0=1$,The last contributing previous-layer unit lies $(k_{\ell }-1)j_{\ell -1}$ input positions beyond the first, which proves the span increment. This derivation assumes the stated local serial connectivity, no padding and no dilation.

If the first size-three layer instead has stride two, then $r_1=3$ and $j_1=2$. A second size-three, stride-one layer has $r_2=7$: its three input supports are $\{0,1,2\}$, $\{2,3,4\}$ and $\{4,5,6\}$. Their union reaches from zero through six. For other sampling choices, a span can include gaps, so span and number of contributing positions need not agree.

This is a structural dependency calculation. Zero weights or the current behavior of an activation function can suppress an influence that the architecture permits; the calculation does not measure how strongly each position affects a particular output. A unit-stride pointwise convolution mixes neural network channels at the same location and does not enlarge this spatial region. The distinction helps explain why many channels alone do not guarantee access to distant input context.