Circular convolution of {1,2} and u[n]−u[n−4]

y[n] = u[n] − u[n−4] = {1, 1, 1, 1} (n = 0 … 3) and x[n] = {1, 2}. The longer sequence has 4 points, so take the circular convolution with N = 4; pad x[n] to {1, 2, 0, 0}.

x₃[n] = Σ (m = 0 to 3) x[m] y[(n − m) mod 4]

Since y[n] = 1 for all four indices, y[(n − m) mod 4] = 1 for every n and m. Hence

x₃[n] = x[0] + x[1] + x[2] + x[3] = 1 + 2 + 0 + 0 = 3, for every n.

Matrix method (circulant of x, columns shifted):

 | 1 0 0 2 |   | 1 |   | 3 |
 | 2 1 0 0 | · | 1 | = | 3 |
 | 0 2 1 0 |   | 1 |   | 3 |
 | 0 0 2 1 |   | 1 |   | 3 |

Check by DFT: Y(k) = 4δ(k) (the sum of y), X(0) = 3; so X(k)Y(k) = 12 δ(k), whose inverse DFT is 12/4 = 3 for all n.

Answer: x₃[n] = {3, 3, 3, 3} (4-point circular convolution).

(The linear convolution is {1, 3, 3, 3, 2}; the wrap-around of the last sample, 2, onto n = 0 gives 1 + 2 = 3.)