---
title: "Output for 5e^(jπn/3) input"
subject: "Digital Signal Analysis and Processing"
chapter: 3
key: "semester_7/digital_signal_analysis_and_processing:lti-output-exp-pi3"
source: https://www.bctengineer.com/q/semester_7/digital_signal_analysis_and_processing/lti-output-exp-pi3
site: BCT Engineering
---

# Output for 5e^(jπn/3) input

Digital Signal Analysis and Processing · Chapter 3: Analysis of LTI System in Frequency Domain

- Asked in: 2072 Kartik, 4 marks
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_7/digital_signal_analysis_and_processing/lti-output-exp-pi3
- Whole chapter: https://www.bctengineer.com/md/semester_7/digital_signal_analysis_and_processing/chapter-3

## Question

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 < ∞.

## 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)}.

## Answer

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°.

## Check yourself

The answer's key points as multiple-choice questions, with the correct option and why.

1. How is the output for 5e^(j pi n/3) obtained?
   A. Convolve numerically with h[n] over finite n
   B. Multiply the input by H(e^(j pi/3))
   C. Multiply the input by h[0]
   D. Divide the input by H(e^(j pi/3))
   Answer: B. The exponential passes through unchanged in form, scaled by H at its frequency. (Key point: Eigenfunction approach)

2. What is H(e^(j pi/3)) for h[n] = (1/2)^n u[n]?
   A. 1 + 0.577j
   B. 0.75 + 0.433j
   C. 1 - 0.577j
   D. 2/3
   Answer: C. 1/(0.75 + 0.433j) = (0.75 - 0.433j)/0.75 = 1 - 0.577j. (Key point: Evaluating H)

3. What are |H| and the phase of H at w = pi/3?
   A. 0.866 and +30 degrees
   B. 1.155 and +30 degrees
   C. 1.5 and -60 degrees
   D. 1.155 and -30 degrees
   Answer: D. |H| = sqrt(1 + 1/3) = 1.155 and angle = -arctan(0.577) = -30 degrees. (Key point: Magnitude and phase)

4. What is y[n]?
   A. 5.774 e^(j(pi n/3 - pi/6))
   B. 5 e^(j(pi n/3 - pi/6))
   C. 5.774 e^(j(pi n/3 + pi/6))
   D. 2.887 e^(j pi n/3)
   Answer: A. Amplitude 5 x 1.1547 = 5.774 and phase shift of -pi/6. (Key point: Final answer)

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