Find x3[n] from X3=X1X2 (1,-2,5,1,2)

x2[n] has 4 points; for the 5-point DFTs it is padded with one zero. 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):

• x1[n] = {1, −2, 5, 1, 2}
• x2[n] = {1, 2, −3, −2, 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    2    1    5   -2 |   |  1 |
|   -2    1    2    1    5 |   |  2 |
|    5   -2    1    2    1 |   | -3 |
|    1    5   -2    1    2 |   | -2 |
|    2    1    5   -2    1 |   |  0 |

Row by row:

• x3[0] = (1)(1) + (2)(2) + (1)(−3) + (5)(−2) + (−2)(0) = −8
• x3[1] = (−2)(1) + (1)(2) + (2)(−3) + (1)(−2) + (5)(0) = −8
• x3[2] = (5)(1) + (−2)(2) + (1)(−3) + (2)(−2) + (1)(0) = −6
• x3[3] = (1)(1) + (5)(2) + (−2)(−3) + (1)(−2) + (2)(0) = 15
• x3[4] = (2)(1) + (1)(2) + (5)(−3) + (−2)(−2) + (1)(0) = −7

Check using DFTs: the inverse DFT of X1(k)·X2(k), k = 0 … 4, is {−8, −8, −6, 15, −7} (Σx3 = −14 = X1(0)·X2(0) = 7 × (−2)).

Answer: x3[n] = {−8, −8, −6, 15, −7}, n = 0 … 4.