Convolve {1, 2, 0, −1} and {2, 0, 2}

No origin is marked, so take the first sample of each sequence at $n = 0$: $x[n] = \{\underset{\uparrow}{1}, 2, 0, -1\}$ for $0 \le n \le 3$ and $h[n] = \{\underset{\uparrow}{2}, 0, 2\}$ for $0 \le n \le 2$.

Convolution sum:

$$y[n] = \sum_{k} x[k]\,h[n-k]$$

Output range: starts at $0 + 0 = 0$; length $= 4 + 3 - 1 = 6$, so $0 \le n \le 5$.

Tabular method

Each row is one input sample times the shifted $h[n]$; add the columns:

Check by direct calculation

• $y[0] = x[0]h[0] = 1\times2 = 2$
• $y[1] = x[0]h[1] + x[1]h[0] = 0 + 4 = 4$
• $y[2] = x[0]h[2] + x[1]h[1] + x[2]h[0] = 2 + 0 + 0 = 2$
• $y[3] = x[1]h[2] + x[2]h[1] + x[3]h[0] = 4 + 0 - 2 = 2$
• $y[4] = x[2]h[2] + x[3]h[1] = 0 + 0 = 0$
• $y[5] = x[3]h[2] = -1\times2 = -2$

Sum check: $\sum y = (\sum x)(\sum h) = 2 \times 4 = 8$, and $2 + 4 + 2 + 2 + 0 - 2 = 8$.

Answer:

$$y[n] = \{\underset{\uparrow}{2},\ 4,\ 2,\ 2,\ 0,\ -2\}$$

i.e. $y[n] = 2\delta[n] + 4\delta[n-1] + 2\delta[n-2] + 2\delta[n-3] - 2\delta[n-5]$.

(If $x$ and $h$ had different origins, the values stay the same and only the starting index shifts: start of $y$ = start of $x$ + start of $h$.)