LTI output for h=2u[n]−2u[n−4]

LTI system

A system is LTI if it is both

• Linear: T{a x₁ + b x₂} = a T{x₁} + b T{x₂} (additivity and homogeneity), and
• Time-invariant: if x[n] → y[n] then x[n − n₀] → y[n − n₀] for any shift n₀.

Such a system is completely described by its impulse response h[n], with y[n] = x[n] * h[n].

Computation

h[n] = 2u[n] − 2u[n − 4] = 2 for n = 0, 1, 2, 3 (else 0). x[n] = (⅓)ⁿ u[n].

y[n] = Σ(k=0 to 3) 2 · x[n − k] = 2 Σ(k=0 to 3) x[n − k] (only terms with n − k ≥ 0 count).

For 0 ≤ n ≤ 3: y[n] = 2 Σ(m=0 to n) (⅓)^m = 2 · [1 − (⅓)^(n+1)] / (1 − ⅓) = 3[1 − (⅓)^(n+1)]

• y[0] = 2
• y[1] = 8/3 ≈ 2.667
• y[2] = 26/9 ≈ 2.889
• y[3] = 80/27 ≈ 2.963

For n ≥ 3: the window covers x[n−3] … x[n]:

y[n] = 2 (⅓)^(n−3) [1 + ⅓ + ⅑ + 1/27] = 2 · (40/27) · 3^(3−n) = 80 · (⅓)ⁿ

Check n = 3: 80/27 ✓. So y[4] = 80/81 ≈ 0.988, y[5] = 80/243 ≈ 0.329, …

Answer:

y[n] = 0                         , n < 0
y[n] = 3[1 − (1/3)^(n+1)]        , 0 ≤ n ≤ 3
y[n] = 80 (1/3)^n                , n ≥ 3

i.e. y[n] = {2, 2.667, 2.889, 2.963, 0.988, 0.329, …} for n = 0, 1, 2, ….