LTI output, h[n]=(1/2)^n window

The output is y[n] = x[n] * h[n] = Σ x[k] h[n − k].

Impulse response: u[n+2] − u[n−2] is 1 for −2 ≤ n ≤ 1, so h[n] = (½)ⁿ for n = −2, −1, 0, 1:

h[−2] = 4, h[−1] = 2, h[0] = 1, h[1] = ½.

Input: x[n] = {2, 1, 0.5, −1}. No origin is marked, so I take the first sample at n = 0: x[0] = 2, x[1] = 1, x[2] = 0.5, x[3] = −1.

Output starts at n = −2 + 0 = −2, ends at n = 1 + 3 = 4 (length 4 + 4 − 1 = 7).

Using y[n] = Σ x[k] h[n − k]:

y[−2] = x0·h[−2]                          = 2·4              = 8
y[−1] = x0·h[−1] + x1·h[−2]               = 4 + 4            = 8
y[ 0] = x0·h[0] + x1·h[−1] + x2·h[−2]     = 2 + 2 + 2        = 6
y[ 1] = x0·h[1] + x1·h[0] + x2·h[−1] + x3·h[−2]
                                          = 1 + 1 + 1 − 4    = −1
y[ 2] = x1·h[1] + x2·h[0] + x3·h[−1]      = 0.5 + 0.5 − 2    = −1
y[ 3] = x2·h[1] + x3·h[0]                 = 0.25 − 1         = −0.75
y[ 4] = x3·h[1]                           = −0.5

Answer: y[n] = {8, 8, 6, −1, −1, −0.75, −0.5} for n = −2 … 4.

Check

1. Sum rule: Σy = (Σx)(Σh). Σx = 2.5, Σh = 7.5, product = 18.75. Σy = 8 + 8 + 6 − 1 − 1 − 0.75 − 0.5 = 18.75. ✓
2. Polynomial (z-transform) method: X(z) = 2 + z⁻¹ + 0.5z⁻² − z⁻³, H(z) = 4z² + 2z + 1 + 0.5z⁻¹. Multiplying gives 8z² + 8z + 6 − z⁻¹ − z⁻² − 0.75z⁻³ − 0.5z⁻⁴, the same coefficients. ✓