Find x3[n] from X3=X1X2, x1 5-point, x2 4-point

x2[n] has only 4 points; since the DFTs are 5-point, it is padded with one zero. Because 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):

• x1[n] = {1, −2, 2, 1, 4}
• x2[n] = {2, 1, −3, −1, 0}

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    4    1    2   -2 |   |  2 |
|   -2    1    4    1    2 |   |  1 |
|    2   -2    1    4    1 |   | -3 |
|    1    2   -2    1    4 |   | -1 |
|    4    1    2   -2    1 |   |  0 |

Row by row:

• x3[0] = (1)(2) + (4)(1) + (1)(−3) + (2)(−1) + (−2)(0) = 1
• x3[1] = (−2)(2) + (1)(1) + (4)(−3) + (1)(−1) + (2)(0) = −16
• x3[2] = (2)(2) + (−2)(1) + (1)(−3) + (4)(−1) + (1)(0) = −5
• x3[3] = (1)(2) + (2)(1) + (−2)(−3) + (1)(−1) + (4)(0) = 9
• x3[4] = (4)(2) + (1)(1) + (2)(−3) + (−2)(−1) + (1)(0) = 5

Check using DFTs: the inverse DFT of X1(k)·X2(k), k = 0 … 4, is {1, −16, −5, 9, 5} (Σx3 = −6 = X1(0)·X2(0) = 6 × (−1)).

Answer: x3[n] = {1, −16, −5, 9, 5}, n = 0 … 4.