Convolve {1,2,1,−1} and {1,2,3,1}

Given (origin marked):

• $x_1[n] = \{1, \underset{\uparrow}{2}, 1, -1\}$, so $x_1[-1] = 1$, $x_1[0] = 2$, $x_1[1] = 1$, $x_1[2] = -1$ ($-1 \le n \le 2$).
• $x_2[n] = \{\underset{\uparrow}{1}, 2, 3, 1\}$, so $0 \le n \le 3$.

$$y[n] = x_1[n]*x_2[n] = \sum_k x_1[k]\,x_2[n-k]$$

Output range: start $= -1 + 0 = -1$; length $= 4 + 4 - 1 = 7$; so $-1 \le n \le 5$.

Tabular method

Each row is one sample of $x_1$ times $x_2$ shifted to that sample's position:

Direct check of some values

• $y[-1] = x_1[-1]x_2[0] = 1$
• $y[0] = x_1[-1]x_2[1] + x_1[0]x_2[0] = 2 + 2 = 4$
• $y[1] = 1(3) + 2(2) + 1(1) = 8$
• $y[2] = 1(1) + 2(3) + 1(2) + (-1)(1) = 8$
• $y[3] = 2(1) + 1(3) + (-1)(2) = 3$
• $y[4] = 1(1) + (-1)(3) = -2$
• $y[5] = (-1)(1) = -1$

Sum check: $(\sum x_1)(\sum x_2) = 3 \times 7 = 21$, and $1 + 4 + 8 + 8 + 3 - 2 - 1 = 21$.

Answer:

$$y[n] = \{1,\ \underset{\uparrow}{4},\ 8,\ 8,\ 3,\ -2,\ -1\}$$

for $-1 \le n \le 5$ (the arrow marks $n = 0$, so $y[0] = 4$).