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Output for 5e^(jπn/2) input

Short answer

H(e^{jω}) = 1/(1 - (1/3)e^{-jω}). At ω=π/2, H = 1/(1 + j/3) = 0.9 - 0.3j, magnitude 0.9487, phase -18.43°. Output: y[n] = 5·0.9487 e^{j(πn/2 - 0.3217)} = 4.743 e^{j(πn/2 - 0.3217)}.

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

Eigenfunction property: for x[n] = e^(jω₀n) the output is y[n] = H(e^(jω₀)) e^(jω₀n).

Frequency response: H(e^(jω)) = Σ(n=0 to ∞) (⅓)ⁿ e^(−jωn) = 1/(1 − ⅓e^(−jω))

At ω₀ = π/2: e^(−jπ/2) = −j

H(e^(jπ/2)) = 1/(1 + j/3) = (1 − j/3)/(1 + 1/9) = 0.9 − 0.3j

|H| = √(0.81 + 0.09) = √0.9 = 0.9487, ∠H = −tan⁻¹(1/3) = −18.43° = −0.3217 rad.

Output (input amplitude 5):

y[n] = 5(0.9 − 0.3j) e^(jπn/2) = (4.5 − 1.5j) e^(jπn/2)

Answer: y[n] = 4.743 e^(j(πn/2 − 0.3217)), −∞ < n < ∞.

✓ Answer checked by Claude Opus 5.5 (high effort)

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