Circular convolution of two sequences

$x[n]$ has 4 samples and $y[n]$ has 3, so use $N = 4$ and pad $y[n]$ with one zero:

$$x[n] = \{1, 0, 0, 1\}, \qquad y[n] = \{2, 0, 2, 0\}$$

Formula:

$$z[n] = x[n] \circledast y[n] = \sum_{m=0}^{3}x[m]\,y[(n-m)_4], \quad n = 0,1,2,3$$

Matrix (circulant) method: columns of the matrix are circular shifts of $y[n]$:

$$\begin{bmatrix} z[0] \\ z[1] \\ z[2] \\ z[3] \end{bmatrix} = \begin{bmatrix} 2 & 0 & 2 & 0 \\ 0 & 2 & 0 & 2 \\ 2 & 0 & 2 & 0 \\ 0 & 2 & 0 & 2 \end{bmatrix} \begin{bmatrix} 1 \\ 0 \\ 0 \\ 1 \end{bmatrix}$$

Each sample:

$$\begin{aligned} z[0] &= x[0]y[0] + x[1]y[3] + x[2]y[2] + x[3]y[1] = 2 + 0 + 0 + 0 = 2 \\ z[1] &= x[0]y[1] + x[1]y[0] + x[2]y[3] + x[3]y[2] = 0 + 0 + 0 + 2 = 2 \\ z[2] &= x[0]y[2] + x[1]y[1] + x[2]y[0] + x[3]y[3] = 2 + 0 + 0 + 0 = 2 \\ z[3] &= x[0]y[3] + x[1]y[2] + x[2]y[1] + x[3]y[0] = 0 + 0 + 0 + 2 = 2 \end{aligned}$$

Answer: $z[n] = x[n] \circledast y[n] = \{2, 2, 2, 2\}$

Check with linear convolution: $x * y = \{2, 0, 2, 2, 0, 2\}$ (length 6). Wrapping samples $n = 4, 5$ onto $n = 0, 1$ (time aliasing for $N = 4$): $\{2+0,\; 0+2,\; 2,\; 2\} = \{2,2,2,2\}$. Also, the sum of $z$ (8) equals sum of $x$ (2) times sum of $y$ (4).