---
title: "LTI output with sparse samples, verify"
subject: "Digital Signal Analysis and Processing"
chapter: 1
key: "semester_7/digital_signal_analysis_and_processing:lti-output-sparse-samples"
source: https://www.bctengineer.com/q/semester_7/digital_signal_analysis_and_processing/lti-output-sparse-samples
site: BCT Engineering
---

# LTI output with sparse samples, verify

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

- Asked in: 2072 Chaitra, 3+2 marks
- Video: [LTI Systems and the Impulse Response](https://www.youtube.com/watch?v=AkDmCGaWqPs) — Signal Processing with Paul
- 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-sparse-samples
- 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] with h[−2] = 1, h[0] = 2, h[1] = 3 and input signal x[n] with x[0] = 1/2, x[2] = 2, x[3] = 3. Also check the answer.

## Short answer

Superposing scaled shifted impulse responses gives y[n]={0.5,0,3,4.5,4,12,9} for n=-2 to 4.

## Answer

**Reading the samples (zero where not listed):**
- h[n]: h[−2] = 1, h[−1] = 0, h[0] = 2, h[1] = 3
- x[n]: x[0] = ½, x[1] = 0, x[2] = 2, x[3] = 3

y[n] = Σ x[k] h[n − k]. Since x has three non-zero samples, y is the sum of three shifted, scaled copies of h:

- x[0] = ½ → ½·h[n] : value ½ at n = −2, 1 at n = 0, 1.5 at n = 1
- x[2] = 2 → 2·h[n − 2]: 2 at n = 0, 4 at n = 2, 6 at n = 3
- x[3] = 3 → 3·h[n − 3]: 3 at n = 1, 6 at n = 3, 9 at n = 4

Where each copy lands: x[2] shifts h by 2 (h[−2], h[0], h[1] land at n = 0, 2, 3); x[3] shifts h by 3 (n = 1, 3, 4). Adding by position:

| n | −2 | −1 | 0 | 1 | 2 | 3 | 4 |
|---|---|---|---|---|---|---|---|
| ½·h[n] | 0.5 | 0 | 1 | 1.5 | 0 | 0 | 0 |
| 2·h[n−2] | 0 | 0 | 2 | 0 | 4 | 6 | 0 |
| 3·h[n−3] | 0 | 0 | 0 | 3 | 0 | 6 | 9 |
| **y[n]** | **0.5** | **0** | **3** | **4.5** | **4** | **12** | **9** |

**Answer:** y[n] = {0.5, 0, 3, 4.5, 4, 12, 9} for n = −2 … 4.

#### Check
Σy = 0.5 + 3 + 4.5 + 4 + 12 + 9 = 33. (Σx)(Σh) = (0.5 + 2 + 3)(1 + 2 + 3) = 5.5 × 6 = 33. ✓

## Check yourself

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

1. Unlisted samples of h and x are taken as
   A. one
   B. equal to the previous sample
   C. zero
   D. undefined
   Answer: C. Samples not given are zero, so h[-1] = 0 and x[1] = 0. (Key point: Reading samples)

2. Because x has three non-zero samples, y is
   A. a product of three copies of h
   B. the sum of three scaled, shifted copies of h
   C. three separate outputs not added
   D. h scaled by the sum of x
   Answer: B. Each x[k] produces x[k] h[n-k]; the outputs add by linearity. (Key point: Superposition method)

3. The copy produced by x[3] = 3 is
   A. 3 h[n-3]
   B. 3 h[n+3]
   C. 3 h[3n]
   D. h[n] + 3
   Answer: A. An input at k = 3 shifts h to the right by 3. (Key point: Shift of a copy)

4. The sum rule gives sum y =
   A. 18
   B. 5.5
   C. 6
   D. 33
   Answer: D. (0.5 + 2 + 3) x (1 + 2 + 3) = 5.5 x 6 = 33. (Key point: Check)

## Close variants set in other exams

- [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×
- [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×
- [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×

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