How to Find the Kernel of the Convolution for Linear Interpolation?

Question

I'm trying to solve the following exercise:

Image A was doubled by linear interpolation. The magnification was performed in two stages. In the first stage, add about zero pixel to the image between any two existing pixels. In the second stage, perform a convolution. Find the kernel of the convolution.

In the solution they said the kernel should be: $$ 0.25\begin{bmatrix}1 & 2 & 1\\ 2 & 4 & 2\\ 1 & 2 & 1 \end{bmatrix} $$ But I can't seem to figure why and how did they got to that specific kernel. Is it possible to explain in technical terms?

Answer 1

You may notice that the kernel is a separable kernel.
So we can analyze it in 1D:

$$ \boldsymbol{K} = \frac{1}{4} \begin{bmatrix} 1 & 2 & 1 \\ 2 & 4 & 2 \\ 1 & 2 & 1 \end{bmatrix} = \boldsymbol{k} \boldsymbol{k}^{T}, \; \boldsymbol{k} = \frac{1}{2} \begin{bmatrix} 1 \\ 2 \\ 1 \end{bmatrix} $$

Now, let's see what the kernel vK = 0.5 * [1, 2, 1] does in 1D.
Assume we have the signal vX = [1, 7, 3].
Then, let's follow the steps:

1. Upsample the Data with Zeros
   We add zero between each 2 samples: vU = [1, 0, 7, 0, 3].
2. Convolve the Upsampled Data with the Kernel
   Since the kernel is symmetric, we can just see whet it does.
   For the 0 between 1 and 7 it basically yields: 0.5 * 1 + 1 * 0 + 0.5 * 7 = 4. The is basically linear interpolation between 1 and 7.
   When it works on the 7 it multiplies it by 1 and each size is zero, hence have no meaning, so nothing happens. For the zero between 7 and 3 it will again do a linear interpolation (The average as the sampling is uniform) and yield 5 so the result: conv(vK, vX) = [1, 4, 7, 5, 3].

So indeed this is the kernel which applies a liner interpolation.