---
title: "Output for e^(jπn) input, h=(1/2)^n u[n]"
subject: "Digital Signal Analysis and Processing"
chapter: 3
key: "semester_7/digital_signal_analysis_and_processing:lti-output-complex-exp-half"
source: https://www.bctengineer.com/q/semester_7/digital_signal_analysis_and_processing/lti-output-complex-exp-half
site: BCT Engineering
---

# Output for e^(jπn) input, h=(1/2)^n u[n]

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

- Asked in: 2071 Shrawan, 5 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-complex-exp-half
- Whole chapter: https://www.bctengineer.com/md/semester_7/digital_signal_analysis_and_processing/chapter-3

## Question

Consider an LTI system with impulse response h[n] = (1/2)^n u[n]. Determine y[n], if the input is x[n] = A e^(jπn).

## Short answer

H(e^{jω}) = 1/(1 - 0.5 e^{-jω}). At ω=π, H(e^{jπ}) = 2/3. Output: y[n] = A·(2/3)·e^{jπn} = (2A/3)(-1)^n.

## Answer

For a complex exponential input, an LTI system output is the same exponential scaled by the frequency response (eigenfunction property):

x[n] = e^(jω₀n) → y[n] = H(e^(jω₀)) e^(jω₀n)

**Frequency response:**

H(e^(jω)) = Σ h[n] e^(−jωn) = Σ(n=0 to ∞) (½ e^(−jω))ⁿ = 1/(1 − ½e^(−jω))  (converges since |½| < 1)

**At ω₀ = π:** e^(−jπ) = −1

H(e^(jπ)) = 1/(1 − ½(−1)) = 1/(1.5) = 2/3

**Output:**

y[n] = A · H(e^(jπ)) · e^(jπn) = **(2A/3) e^(jπn) = (2A/3)(−1)ⁿ**

**Check:** Direct convolution y[n] = Σ(k=0 to ∞) (½)ᵏ A e^(jπ(n−k)) = A e^(jπn) Σ (½ e^(−jπ))ᵏ = A e^(jπn) Σ (−½)ᵏ = A e^(jπn)/(1 + ½) = (2A/3) e^(jπn). ✓

## Check yourself

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

1. For input e^(jw0 n), the LTI output is:
   A. h[n] e^(jw0 n)
   B. An exponential at a different frequency
   C. H(e^(jw0)) e^(jw0 n), the same exponential scaled by the frequency response
   D. H(e^(jw0)) only, with no exponential
   Answer: C. Complex exponentials are eigenfunctions of LTI systems. (Key point: Eigenfunction property)

2. What is H(e^(jw)) for h[n] = (1/2)^n u[n]?
   A. 1/(1 + 0.5 e^(-jw))
   B. 1/(1 - 0.5 e^(jw)) u[n]
   C. 0.5/(1 - e^(-jw))
   D. 1/(1 - 0.5 e^(-jw))
   Answer: D. The geometric series sum of (0.5 e^(-jw))^n converges to 1/(1 - 0.5e^(-jw)). (Key point: Frequency response)

3. What is H(e^(j pi))?
   A. 2/3
   B. 2
   C. 1/3
   D. -2/3
   Answer: A. e^(-j pi) = -1 so H = 1/(1 + 0.5) = 2/3. (Key point: Evaluate at pi)

4. What is y[n] for x[n] = A e^(j pi n)?
   A. 2A(-1)^n
   B. (2A/3)(-1)^n
   C. (A/3)(-1)^n
   D. (2A/3)u[n]
   Answer: B. y[n] = A H(e^(j pi)) e^(j pi n) = (2A/3)(-1)^n. (Key point: Final output)

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