Calculate a convolution

This exercise does not need Python.

Consider the image

x = \begin{pmatrix} 1 & 1 & 0 & 0 \\ 1 & 1 & 0 & 0 \end{pmatrix}

Calculate by hand the convolution of $x$ with $h_1=\begin{pmatrix}1 & -1\end{pmatrix}$ .
Calculate by hand the convolution of $x$ with $h_2=\begin{pmatrix}1 & -1\end{pmatrix}^T$ .
Calculate by hand the convolution of $x$ with $h_3=\begin{pmatrix}1 & 2 & 1\end{pmatrix}$ .

Correction¶

Objectives¶

Know how to calculate a convolution product.
Interpret a simple convolution.

The convolution is commutative: $f_1 = h_1 * x = x * h_1$ and is defined as:

f_1(y,x) = \sum_m \sum_n x(y-m,x-n) h_1(m,n)

To apply the convolution product, it is necessary to define an origin in the images $h_1$ and $x$ . Suppose that the origin is given by the red pixel:

x = \begin{pmatrix} \color{red}{1} & 1 & 0 & 0 \\ 1 & 1 & 0 & 0 \end{pmatrix} \quad\text{and}\quad h_1=\begin{pmatrix}\color{red}{1} & -1\end{pmatrix}.

Note that $h_1$ has only two non-zero pixels: $h_1(0,0)=1$ and $h_1(0,1)=-1$ . Then we have:

f_1 = \begin{pmatrix} ? & 0 & -1 & 0 \\ ? & 0 & -1 & 0 \end{pmatrix}

As well, considering that 1 is the pixel at the origin in $h_2$ , we obtain:

f_2 = \begin{pmatrix} ? & ? & ? & ? \\ 0 & 0 & 0 & 0 \end{pmatrix},

and, considering that 2 is the pixel at the origin in $h_3$ , we obtain:

f_2 = \begin{pmatrix} ? & 3 & 1 & ? \\ ? & 3 & 1 & ? \end{pmatrix}.