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Output for 5e^(jπn/3) input

Short answer

H(e^{jω}) = 1/(1 - 0.5 e^{-jω}). At ω=π/3, H = 1/(0.75 + j0.433) = 1 - 0.577j, magnitude 1.155, phase -30°. Output: y[n] = 5·1.155 e^{j(πn/3 - π/6)} = 5.774 e^{j(πn/3 - π/6)}.

Find the output of LTI system having impulse response h[n] = (1/2)^n u[n] and input signal x[n] = 5e^(jπn/3) for −∞ < n < ∞.

For an LTI system, a complex exponential input produces the same exponential scaled by the frequency response: x[n] = e^(jω₀n) → y[n] = H(e^(jω₀)) e^(jω₀n).

Frequency response: H(e^(jω)) = Σ(n=0 to ∞) (½)ⁿ e^(−jωn) = 1/(1 − ½e^(−jω))

At ω₀ = π/3: e^(−jπ/3) = cos 60° − j sin 60° = 0.5 − 0.866j

H(e^(jπ/3)) = 1/(1 − 0.25 + 0.433j) = 1/(0.75 + 0.433j)

= (0.75 − 0.433j)/(0.75² + 0.433²) = (0.75 − 0.433j)/0.75 = 1 − 0.577j

|H| = √(1 + 0.3333) = 1.1547, ∠H = −tan⁻¹(0.577) = −30° = −π/6.

Output:

y[n] = 5 · 1.1547 · e^(j(πn/3 − π/6)) = 5.774 e^(j(πn/3 − π/6))

Answer: y[n] = (5 − 2.887j) e^(jπn/3) = 5.774 e^(j(πn/3 − π/6)), −∞ < n < ∞. The input is scaled by 1.155 and delayed in phase by 30°.

✓ Answer checked by Claude Opus 5.5 (high effort)

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