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

Short answer

Frequency response H(e^{jω}) = 1/(1 - 0.5 e^{-jω}). At ω=0: H=2; ω=π/2: H=0.8944∠-26.57°; ω=π: H=2/3. DC output 20, -5 sin becomes -4.472 sin(πn/2 - 26.57°), 20 cos becomes 13.33 cos(πn). Output: y[n]=20 - 4.472 sin(πn/2 - 26.57°) + 13.33 cos(πn).

Determine the output sequence of the system with impulse response h[n] = (1/2)^n u[n] when the input signal is x[n] = 10 − 5 sin(πn/2) + 20 cos(πn), −∞ < n < ∞.

h[n] = (½)ⁿ u[n] has frequency response

H(e^(jω)) = 1/(1 − ½e^(−jω))

The input is a sum of sinusoids, and for each sinusoid A cos(ω₀n + φ) the output is A|H(ω₀)| cos(ω₀n + φ + ∠H(ω₀)). By linearity, add the outputs.

x[n] = 10 − 5 sin(πn/2) + 20 cos(πn): three components, at ω = 0, π/2 and π.

Componentω₀H(e^(jω₀))|H|∠H
10 (DC)01/(1 − ½) = 220
−5 sin(πn/2)π/21/(1 + 0.5j) = 0.8 − 0.4j0.8944−26.57° (−0.4636 rad)
20 cos(πn)π1/(1 + ½) = 2/30.66670

(At ω = π/2: e^(−jπ/2) = −j, so H = 1/(1 − ½(−j)) = 1/(1 + 0.5j).)

Outputs:

  • DC: 10 × 2 = 20
  • Sine: −5 × 0.8944 × sin(πn/2 − 0.4636) = −4.472 sin(πn/2 − 0.4636)
  • cos(πn): 20 × (2/3) cos(πn) = 13.33 cos(πn) = 13.33(−1)ⁿ

Answer:

y[n] = 20 − 4.472 sin(πn/2 − 26.57°) + 13.33 cos(πn), −∞ < n < ∞

(Since the input is defined for all n, this steady-state result is the complete output.)

✓ Answer checked by Claude Opus 5.5 (high effort)

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