Skip to main content

Convolve |n| and 2n sequences

Short answer

Convolving x[n] = |n| (-2 ≤ n ≤ 2) and h[n] = 2n (2 ≤ n ≤ 5) gives c[n] = {8, 16, 22, 32, 24, 20, 26, 20} for n = 0 to 7.

Find and plot the convolution between x[n] = |n| for −2 ≤ n ≤ 2, 0 otherwise and y[n] = 2n for 2 ≤ n ≤ 5, 0 otherwise.

Write both sequences with their sample positions. (The second signal is called y[n]y[n] in the question; to avoid confusion it is renamed h[n]h[n] and the result is c[n]=x[n]∗h[n]c[n]=x[n]*h[n].)

x[n]=∣n∣:x[−2]=2, x[−1]=1, x[0]=0, x[1]=1, x[2]=2x[n]=|n|:\quad x[-2]=2,\ x[-1]=1,\ x[0]=0,\ x[1]=1,\ x[2]=2 h[n]=2n:h[2]=4, h[3]=6, h[4]=8, h[5]=10h[n]=2n:\quad h[2]=4,\ h[3]=6,\ h[4]=8,\ h[5]=10

The result starts at n=−2+2=0n=-2+2=0, ends at n=2+5=7n=2+5=7, and has 5+4−1=85+4-1=8 samples.

Tabular method

Rows are x[k] h[n−k]x[k]\,h[n-k] for each kk:

kk, x[k]x[k]n=0n=01234567
−2-2, 28121620
−1-1, 146810
00, 00000
11, 146810
22, 28121620
c[n]c[n]816223224202620

Example check: c[3]=x[−2]h[5]+x[−1]h[4]+x[0]h[3]+x[1]h[2]=20+8+0+4=32c[3]=x[-2]h[5]+x[-1]h[4]+x[0]h[3]+x[1]h[2]=20+8+0+4=32. Sum check: ∑c=168=(∑x)(∑h)=6×28\sum c = 168 = (\sum x)(\sum h)=6\times 28.

Answer: c[n]={8,16,22,32,24,20,26,20}c[n]=\{8,16,22,32,24,20,26,20\} for n=0,1,…,7n=0,1,\dots,7, zero elsewhere.

Plot

    c[n]
     32             o
     26             |           o
     24             |   o       |
     22         o   |   |       |
     20         |   |   |   o   |   o
     16     o   |   |   |   |   |   |
      8 o   |   |   |   |   |   |   |
      0 +---+---+---+---+---+---+---+--> n
        0   1   2   3   4   5   6   7

✓ Answer checked by Claude Opus 5.5 (high effort)

All Signals and Systems chapter 6 questions →