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

# LTI output, h[n]=(1/3)^n window

Digital Signal Analysis and Processing · Chapter 1: Discrete Time Signals and Systems

- Asked in: 2072 Kartik, 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-exp-window-2
- Whole chapter: https://www.bctengineer.com/md/semester_7/digital_signal_analysis_and_processing/chapter-1

## Question

Find the output of LTI system having impulse response h[n] = (1/3)^n {u[n+1] − u[n−2]} and input signal x[n] = {2, 1, 0.5, 3}.

## Short answer

Convolution of h[n]={3,1,1/3} at n=-1,0,1 and input x[n]={2,1,0.5,3} from n=0 yields y[n]={6,5,3.167,9.833,3.167,1} for n=-1 to 4.

## Answer

Output: y[n] = Σ x[k] h[n − k].

**Impulse response:** u[n+1] − u[n−2] = 1 for −1 ≤ n ≤ 1, so h[n] = (⅓)ⁿ for n = −1, 0, 1:

h[−1] = 3, h[0] = 1, h[1] = ⅓.

**Input:** x[n] = {2, 1, 0.5, 3}, taking the first sample at n = 0.

The output runs from n = −1 to n = 1 + 3 = 4 (6 samples).

```text
y[−1] = x0·h[−1]                         = 2·3                = 6
y[ 0] = x0·h[0] + x1·h[−1]               = 2 + 3              = 5
y[ 1] = x0·h[1] + x1·h[0] + x2·h[−1]     = 2/3 + 1 + 1.5      = 19/6 = 3.1667
y[ 2] = x1·h[1] + x2·h[0] + x3·h[−1]     = 1/3 + 0.5 + 9      = 59/6 = 9.8333
y[ 3] = x2·h[1] + x3·h[0]                = 1/6 + 3            = 19/6 = 3.1667
y[ 4] = x3·h[1]                          = 3·(1/3)            = 1
```

**Answer:** y[n] = {6, 5, 3.167, 9.833, 3.167, 1} for n = −1 … 4.

**Check:** Σh = 3 + 1 + ⅓ = 13/3, Σx = 6.5, product = 28.1667. Σy = 6 + 5 + 3.1667 + 9.8333 + 3.1667 + 1 = 28.1667. ✓

## Check yourself

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

1. u[n+1] - u[n-2] is 1 for
   A. n = 0, 1, 2
   B. n = -1, 0, 1, 2
   C. n = -1, 0, 1
   D. n = 1, 2
   Answer: C. u[n+1] starts at -1; u[n-2] starts at 2, so n = -1, 0, 1. (Key point: Window of h)

2. For h[n] = (1/3)^n, h[-1] is
   A. 1/3
   B. 3
   C. -3
   D. 1
   Answer: B. (1/3)^(-1) = 3. (Key point: Values of h)

3. The output y[n] runs over
   A. n = -1 ... 4 (6 samples)
   B. n = 0 ... 5
   C. n = -1 ... 5
   D. n = -1 ... 3
   Answer: A. Start -1 + 0, end 1 + 3 = 4; length 3 + 4 - 1 = 6. (Key point: Output range)

4. y[-1] equals
   A. x0 h[0] = 2
   B. x1 h[-1] = 3
   C. 5
   D. x0 h[-1] = 6
   Answer: D. Only x[0] h[-1] contributes: 2 x 3 = 6. (Key point: Value y[-1])

## Close variants set in other exams

- [LTI output for h=2u[n]−2u[n−4]](https://www.bctengineer.com/question.md?key=semester_7%2Fdigital_signal_analysis_and_processing%3Alti-output-step-difference) — Digital Signal Analysis and Processing, chapter 1, asked 4×
- [LTI output, h[n]=(1/2)^n window](https://www.bctengineer.com/question.md?key=semester_7%2Fdigital_signal_analysis_and_processing%3Alti-output-exp-window-1) — Digital Signal Analysis and Processing, chapter 1, asked 4×
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