LTI output, h[n]=(1/3)^n window

Output: y[n] = Σ x[k] h[n − k].

Impulse response: u[n+1] − u[n−2] = 1 for −1 ≤ n ≤ 1, so h[n] = (⅓)ⁿ for n = −1, 0, 1:

h[−1] = 3, h[0] = 1, h[1] = ⅓.

Input: x[n] = {2, 1, 0.5, 3}, taking the first sample at n = 0.

The output runs from n = −1 to n = 1 + 3 = 4 (6 samples).

y[−1] = x0·h[−1]                         = 2·3                = 6
y[ 0] = x0·h[0] + x1·h[−1]               = 2 + 3              = 5
y[ 1] = x0·h[1] + x1·h[0] + x2·h[−1]     = 2/3 + 1 + 1.5      = 19/6 = 3.1667
y[ 2] = x1·h[1] + x2·h[0] + x3·h[−1]     = 1/3 + 0.5 + 9      = 59/6 = 9.8333
y[ 3] = x2·h[1] + x3·h[0]                = 1/6 + 3            = 19/6 = 3.1667
y[ 4] = x3·h[1]                          = 3·(1/3)            = 1

Answer: y[n] = {6, 5, 3.167, 9.833, 3.167, 1} for n = −1 … 4.

Check: Σh = 3 + 1 + ⅓ = 13/3, Σx = 6.5, product = 28.1667. Σy = 6 + 5 + 3.1667 + 9.8333 + 3.1667 + 1 = 28.1667. ✓