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Find x3[n] from X3=X1X2 (5-point)

Short answer

X3(k)=X1(k)X2(k) means x3[n] is the 5-point circular convolution of x1 and x2. Computing it gives x3[n] = {6, 1, 1, 0, 2}.

If X1(k) and X2(k) are DFT of sequences x1[n] = {1, 2, 0, 1, −2} and x2[n] = {1, 0, 1, 1, 2} respectively, find the sequence x3[n], if DFT of x3[n] is given by X3(k) = X1(k) X2(k).

Since X3(k) = X1(k)·X2(k), x3[n] is the 5-point circular convolution of x1[n] and x2[n] (multiplication of DFTs ⇔ circular convolution). Both sequences already have 5 points, so no padding is needed:

  • x1[n] = {1, 2, 0, 1, −2}
  • x2[n] = {1, 0, 1, 1, 2}

x3[n] = Σ (m = 0 to 4) x1[m] · x2[(n − m) mod 5]

Written as a matrix product (the first column of the circulant is x1, and each later column is the previous one shifted down circularly):

|    1   -2    1    0    2 |   | 1 |
|    2    1   -2    1    0 |   | 0 |
|    0    2    1   -2    1 |   | 1 |
|    1    0    2    1   -2 |   | 1 |
|   -2    1    0    2    1 |   | 2 |

Row by row:

  • x3[0] = (1)(1) + (−2)(0) + (1)(1) + (0)(1) + (2)(2) = 6
  • x3[1] = (2)(1) + (1)(0) + (−2)(1) + (1)(1) + (0)(2) = 1
  • x3[2] = (0)(1) + (2)(0) + (1)(1) + (−2)(1) + (1)(2) = 1
  • x3[3] = (1)(1) + (0)(0) + (2)(1) + (1)(1) + (−2)(2) = 0
  • x3[4] = (−2)(1) + (1)(0) + (0)(1) + (2)(1) + (1)(2) = 2

Check using DFTs: the inverse DFT of X1(k)·X2(k), k = 0 … 4, is {6, 1, 1, 0, 2} (Σx3 = 10 = X1(0)·X2(0) = 2 × 5).

Answer: x3[n] = {6, 1, 1, 0, 2}, n = 0 … 4.

✓ Answer checked by Claude Opus 5.5 (high effort)

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