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Convolve {1/4, −1, 2} and {2, 2/4, −2}

Short answer

Taking origins at n=0, the convolution y[n]=x[n]∗h[n] results in y[n]={0.5, -1.875, 3, 3, -4} (i.e., 1/2, -15/8, 3, 3, -4) with y[0]=0.5.

Find the output of an LTI system, for input x[n] = {1/4, −1, 2} and h[n] = {2, 2/4, −2}.

No origin is marked, so take the first sample at n=0n = 0: x[n]={0.25↑,−1,2}x[n] = \{\underset{\uparrow}{0.25}, -1, 2\} and h[n]={2↑,0.5,−2}h[n] = \{\underset{\uparrow}{2}, 0.5, -2\} (note 2/4=0.52/4 = 0.5).

The output of an LTI system is y[n]=x[n]∗h[n]=∑kx[k] h[n−k]y[n] = x[n]*h[n] = \sum_k x[k]\,h[n-k].

Output starts at n=0n = 0, length 3+3−1=53 + 3 - 1 = 5, so 0≤n≤40 \le n \le 4.

Tabular method

Rown=0n=0n=1n=1n=2n=2n=3n=3n=4n=4
x[0] h[n]x[0]\,h[n]0.50.125-0.5
x[1] h[n−1]x[1]\,h[n-1]-2-0.52
x[2] h[n−2]x[2]\,h[n-2]41-4
y[n]y[n]0.5-1.87533-4

Direct working

  • y[0]=0.25(2)=0.5y[0] = 0.25(2) = 0.5
  • y[1]=0.25(0.5)+(−1)(2)=0.125−2=−1.875y[1] = 0.25(0.5) + (-1)(2) = 0.125 - 2 = -1.875
  • y[2]=0.25(−2)+(−1)(0.5)+2(2)=−0.5−0.5+4=3y[2] = 0.25(-2) + (-1)(0.5) + 2(2) = -0.5 - 0.5 + 4 = 3
  • y[3]=(−1)(−2)+2(0.5)=2+1=3y[3] = (-1)(-2) + 2(0.5) = 2 + 1 = 3
  • y[4]=2(−2)=−4y[4] = 2(-2) = -4

Sum check: (∑x)(∑h)=1.25×0.5=0.625(\sum x)(\sum h) = 1.25 \times 0.5 = 0.625, and 0.5−1.875+3+3−4=0.6250.5 - 1.875 + 3 + 3 - 4 = 0.625.

Answer:

y[n]={12↑, −158, 3, 3, −4}={0.5↑, −1.875, 3, 3, −4}y[n] = \left\{\underset{\uparrow}{\tfrac{1}{2}},\ -\tfrac{15}{8},\ 3,\ 3,\ -4\right\} = \{\underset{\uparrow}{0.5},\ -1.875,\ 3,\ 3,\ -4\}

✓ Answer checked by Claude Opus 5.5 (high effort)

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