---
title: "LTI output for h=2u[n]−2u[n−4]"
subject: "Digital Signal Analysis and Processing"
chapter: 1
key: "semester_7/digital_signal_analysis_and_processing:lti-output-step-difference"
source: https://www.bctengineer.com/q/semester_7/digital_signal_analysis_and_processing/lti-output-step-difference
site: BCT Engineering
---

# LTI output for h=2u[n]−2u[n−4]

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

- Asked in: 2069 Chaitra, 1+4 marks
- Video: [Discrete Convolution Problem 2|| KTU S4 ECE Signals and Systems MODULE2||EC202](https://www.youtube.com/watch?v=eSxEOrYGB_E) — Engineering studies
- 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-step-difference
- Whole chapter: https://www.bctengineer.com/md/semester_7/digital_signal_analysis_and_processing/chapter-1

## Question

Define LTI system. Find the output of LTI system having impulse response h[n] = 2u[n] − 2u[n−4] and input signal x[n] = (1/3)^n u[n].

## Short answer

An LTI system is linear and time-invariant, described by y[n]=x[n]*h[n]. For h[n]=2u[n]-2u[n-4] and x[n]=(1/3)^n u[n], the output is y[n]=0 for n<0, y[n]=3[1-(1/3)^(n+1)] for 0≤n≤3, and y[n]=80(1/3)^n for n≥3.

## Answer

#### LTI system
A system is **LTI** if it is both
- **Linear:** T{a x₁ + b x₂} = a T{x₁} + b T{x₂} (additivity and homogeneity), and
- **Time-invariant:** if x[n] → y[n] then x[n − n₀] → y[n − n₀] for any shift n₀.

Such a system is completely described by its impulse response h[n], with y[n] = x[n] * h[n].

#### Computation
h[n] = 2u[n] − 2u[n − 4] = 2 for n = 0, 1, 2, 3 (else 0). x[n] = (⅓)ⁿ u[n].

y[n] = Σ(k=0 to 3) 2 · x[n − k] = 2 Σ(k=0 to 3) x[n − k] (only terms with n − k ≥ 0 count).

**For 0 ≤ n ≤ 3:** y[n] = 2 Σ(m=0 to n) (⅓)^m = 2 · [1 − (⅓)^(n+1)] / (1 − ⅓) = 3[1 − (⅓)^(n+1)]

- y[0] = 2
- y[1] = 8/3 ≈ 2.667
- y[2] = 26/9 ≈ 2.889
- y[3] = 80/27 ≈ 2.963

**For n ≥ 3:** the window covers x[n−3] … x[n]:

y[n] = 2 (⅓)^(n−3) [1 + ⅓ + ⅑ + 1/27] = 2 · (40/27) · 3^(3−n) = 80 · (⅓)ⁿ

Check n = 3: 80/27 ✓. So y[4] = 80/81 ≈ 0.988, y[5] = 80/243 ≈ 0.329, …

**Answer:**
```text
y[n] = 0                         , n < 0
y[n] = 3[1 − (1/3)^(n+1)]        , 0 ≤ n ≤ 3
y[n] = 80 (1/3)^n                , n ≥ 3
```
i.e. y[n] = {2, 2.667, 2.889, 2.963, 0.988, 0.329, …} for n = 0, 1, 2, ….

## Check yourself

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

1. A system is LTI if it is
   A. linear and causal
   B. time-invariant and stable
   C. linear and time-invariant
   D. causal and stable
   Answer: C. Both superposition and shift-invariance are needed. (Key point: LTI definition)

2. h[n] = 2u[n] - 2u[n-4] is
   A. 2 for n = 0..4
   B. 2 for n = 0..3
   C. 1 for n = 0..3
   D. 2 for n >= 4
   Answer: B. Rectangle of height 2 and length 4. (Key point: Impulse response)

3. For 0 <= n <= 3, y[n] =
   A. 3[1 - (1/3)^(n+1)]
   B. 80 (1/3)^n
   C. 2 (1/3)^n
   D. 2n
   Answer: A. 2 times the geometric sum from m = 0 to n gives 3[1 - (1/3)^(n+1)]. (Key point: Output during build-up)

4. For n >= 3, y[n] =
   A. 3[1-(1/3)^(n+1)]
   B. 2 (1/3)^n
   C. 0
   D. 80 (1/3)^n
   Answer: D. The window sums four terms 1 + 1/3 + 1/9 + 1/27 = 40/27, giving 80 (1/3)^n. (Key point: Output after window)

## Close variants set in other exams

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