---
title: "Output for 5e^(jπn/2) input"
subject: "Digital Signal Analysis and Processing"
chapter: 3
key: "semester_7/digital_signal_analysis_and_processing:lti-output-exp-pi2"
source: https://www.bctengineer.com/q/semester_7/digital_signal_analysis_and_processing/lti-output-exp-pi2
site: BCT Engineering
---

# Output for 5e^(jπn/2) input

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

- Asked in: 2070 Chaitra, 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-pi2
- 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] = (1/3)^n u[n] and input signal x[n] = 5e^(jπn/2) for −∞ < n < ∞.

## 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)}.

## Answer

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 < ∞.

## Check yourself

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

1. What does an LTI system do to e^(j w0 n)?
   A. Changes its frequency to w0/3
   B. Scales it by H(e^(j w0)), leaving the frequency unchanged
   C. Adds a DC term
   D. Attenuates it by exactly h[0]
   Answer: B. Complex exponentials are eigenfunctions of LTI systems. (Key point: Eigenfunction property)

2. For h[n] = (1/3)^n u[n], H(e^(jw)) is:
   A. 1/(1 - 3e^(-jw))
   B. 1/(1 + (1/3)e^(-jw))
   C. 1/(1 - (1/3)e^(-jw))
   D. (1/3)/(1 - e^(-jw))
   Answer: C. Geometric series with ratio (1/3)e^(-jw). (Key point: Frequency response)

3. What is H(e^(j pi/2))?
   A. 0.9 + 0.3j
   B. 0.75 - 0.25j
   C. 0.75 + 0.433j
   D. 0.9 - 0.3j
   Answer: D. e^(-j pi/2) = -j, so H = 1/(1 + j/3) = (1 - j/3)/(10/9) = 0.9 - 0.3j. (Key point: Evaluate at pi/2)

4. What is y[n]?
   A. 4.743 e^(j(pi n/2 - 0.3217))
   B. 4.743 e^(j(pi n/2 + 0.3217))
   C. 5 e^(j(pi n/2 - 0.3217))
   D. 0.9487 e^(j pi n/2)
   Answer: A. 5 x 0.9487 = 4.743 with phase -0.3217 rad. (Key point: Output)

## Close variants set in other exams

- [Output for 5e^(jπn/3) input](https://www.bctengineer.com/question.md?key=semester_7%2Fdigital_signal_analysis_and_processing%3Alti-output-exp-pi3) — Digital Signal Analysis and Processing, chapter 3, asked 2×

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