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Convolve {1, 2, 2, 3} and {1, 4, 3}

Short answer

Here x[-1]=1, x[0]=2, x[1]=2, x[2]=3 and h={1,4,3} starting at n=0. Convolution by the tabular method gives y={1,6,13,17,18,9} for n=-1 to 4, so y[0]=6.

Perform the convolution and draw output y[n]: x[n] = {1, 2, 2, 3} (↑ at the first 2, i.e. x[0] = 2) and h[n] = {1, 4, 3} (↑ at 1, i.e. h[0] = 1).

Given x[n]={1,2,2,3}x[n]=\{1,2,2,3\} with x[0]x[0] = the first 2, so x[−1]=1, x[0]=2, x[1]=2, x[2]=3x[-1]=1,\ x[0]=2,\ x[1]=2,\ x[2]=3. h[n]={1,4,3}h[n]=\{1,4,3\} with h[0]=1h[0]=1, so h[0]=1, h[1]=4, h[2]=3h[0]=1,\ h[1]=4,\ h[2]=3.

Output starts at n=−1+0=−1n=-1+0=-1, ends at n=2+2=4n=2+2=4; length 4+3−1=64+3-1=6. y[n]=∑kx[k]h[n−k]y[n]=\sum_k x[k]h[n-k].

Tabular method

kk, x[k]x[k]n=−1n=-101234
−1-1, 1143
00, 2286
11, 2286
22, 33129
y[n]y[n]161317189

Step values

y[−1]=x[−1]h[0]=1y[0]=x[−1]h[1]+x[0]h[0]=4+2=6y[1]=x[−1]h[2]+x[0]h[1]+x[1]h[0]=3+8+2=13y[2]=x[0]h[2]+x[1]h[1]+x[2]h[0]=6+8+3=17y[3]=x[1]h[2]+x[2]h[1]=6+12=18y[4]=x[2]h[2]=9\begin{aligned} y[-1]&=x[-1]h[0]=1\\ y[0]&=x[-1]h[1]+x[0]h[0]=4+2=6\\ y[1]&=x[-1]h[2]+x[0]h[1]+x[1]h[0]=3+8+2=13\\ y[2]&=x[0]h[2]+x[1]h[1]+x[2]h[0]=6+8+3=17\\ y[3]&=x[1]h[2]+x[2]h[1]=6+12=18\\ y[4]&=x[2]h[2]=9 \end{aligned}

Sum check: ∑y=64=(∑x)(∑h)=8×8\sum y=64=(\sum x)(\sum h)=8\times8.

Answer: y[n]={1, 6, 13, 17, 18, 9}y[n]=\{1,\ 6,\ 13,\ 17,\ 18,\ 9\} for n=−1,…,4n=-1,\dots,4, so y[0]=6y[0]=6.

Output plot

    y[n]
     18                 o
     17             o   |
     13         o   |   |
      9         |   |   |   o
      6     o   |   |   |   |
      1 o   |   |   |   |   |
      0 +---+---+---+---+---+--> n
        -1  0   1   2   3   4

✓ Answer checked by Claude Opus 5.5 (high effort)

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