Output for e^(jπn) input, h=(1/2)^n u[n]

For a complex exponential input, an LTI system output is the same exponential scaled by the frequency response (eigenfunction property):

x[n] = e^(jω₀n) → y[n] = H(e^(jω₀)) e^(jω₀n)

Frequency response:

H(e^(jω)) = Σ h[n] e^(−jωn) = Σ(n=0 to ∞) (½ e^(−jω))ⁿ = 1/(1 − ½e^(−jω)) (converges since |½| < 1)

At ω₀ = π: e^(−jπ) = −1

H(e^(jπ)) = 1/(1 − ½(−1)) = 1/(1.5) = 2/3

Output:

y[n] = A · H(e^(jπ)) · e^(jπn) = (2A/3) e^(jπn) = (2A/3)(−1)ⁿ

Check: Direct convolution y[n] = Σ(k=0 to ∞) (½)ᵏ A e^(jπ(n−k)) = A e^(jπn) Σ (½ e^(−jπ))ᵏ = A e^(jπn) Σ (−½)ᵏ = A e^(jπn)/(1 + ½) = (2A/3) e^(jπn). ✓