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Find x3[n] from X3=X1X2, 4-point

Short answer

Pad x1 to four points and perform 4-point circular convolution with x2. The result is x3[n] = {-7, 6, 5, 1}.

Find x3[n] if DFT of x3[n] is given by X3(k) = X1(k) X2(k) where X1(k) and X2(k) are 4-point DFT of x1[n] = {1, 2, −2} and x2[n] = {1, 2, 3, −1} respectively.

Since X3(k) = X1(k)·X2(k), x3[n] is the 4-point circular convolution of x1[n] and x2[n] (multiplication of DFTs ⇔ circular convolution). x1[n] has only 3 points, so it is padded with one zero to N = 4:

  • x1[n] = {1, 2, -2, 0}
  • x2[n] = {1, 2, 3, -1}

x3[n] = Σ (m = 0 to 3) x1[m] · x2[(n − m) mod 4]

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

|    1    0   -2    2 |   | 1 |
|    2    1    0   -2 |   | 2 |
|   -2    2    1    0 |   | 3 |
|    0   -2    2    1 |   | -1 |

Row by row:

  • x3[0] = (1)(1) + (0)(2) + (-2)(3) + (2)(-1) = -7
  • x3[1] = (2)(1) + (1)(2) + (0)(3) + (-2)(-1) = 6
  • x3[2] = (-2)(1) + (2)(2) + (1)(3) + (0)(-1) = 5
  • x3[3] = (0)(1) + (-2)(2) + (2)(3) + (1)(-1) = 1

Check using DFTs: the inverse DFT of X1(k)·X2(k) for k = 0 … 3 is {-7, 6, 5, 1}. Also, Σx3[n] = 5 = X1(0)·X2(0) = 1 × 5 = 5.

Answer: x3[n] = {-7, 6, 5, 1}, n = 0 … 3.

✓ Answer checked by Claude Opus 5.5 (high effort)

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