---
title: "Output for h[n] = 2ⁿ{u[n] − u[n−3]}"
subject: "Signals and Systems"
chapter: 6
key: "semester_4/signals_and_systems:convolution-finite-exponential"
source: https://www.bctengineer.com/q/semester_4/signals_and_systems/convolution-finite-exponential
site: BCT Engineering
---

# Output for h[n] = 2ⁿ{u[n] − u[n−3]}

Signals and Systems · Chapter 6: Discrete-Time Systems

- Asked in: 2082 Chaitra, 6 marks
- Video: [Q1b. Compute the convolution sum y(n) with the input x(n) and impulse response h(n)](https://www.youtube.com/watch?v=ctX7s9Pk3-o) — EnggClasses
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_4/signals_and_systems/convolution-finite-exponential
- Whole chapter: https://www.bctengineer.com/md/semester_4/signals_and_systems/chapter-6

## Question

Find the output of the LTI system with impulse response h[n] = 2ⁿ{u[n] − u[n−3]} and x[n] = δ[n] + δ[n−1] + δ[n−3]. Also verify the answer.

## Short answer

Using δ-shift property, y[n]=h[n]+h[n-1]+h[n-3] with h={1,2,4} yields y[n]={1,3,6,5,2,4}, y[0]=1.

## Answer

**Given:**
- $h[n] = 2^n\{u[n] - u[n-3]\}$: non-zero for $n = 0, 1, 2$, so $h[n] = \{\underset{\uparrow}{1}, 2, 4\}$.
- $x[n] = \delta[n] + \delta[n-1] + \delta[n-3] = \{\underset{\uparrow}{1}, 1, 0, 1\}$.

#### Method 1: Impulse shifting

Since $\delta[n-k]*h[n] = h[n-k]$ and convolution is distributive:

$$
y[n] = h[n] + h[n-1] + h[n-3]
$$

| $n$ | 0 | 1 | 2 | 3 | 4 | 5 |
|---|---|---|---|---|---|---|
| $h[n]$ | 1 | 2 | 4 | | | |
| $h[n-1]$ | | 1 | 2 | 4 | | |
| $h[n-3]$ | | | | 1 | 2 | 4 |
| $y[n]$ | **1** | **3** | **6** | **5** | **2** | **4** |

$$
y[n] = \{\underset{\uparrow}{1},\ 3,\ 6,\ 5,\ 2,\ 4\}
$$

i.e. $y[n] = \delta[n] + 3\delta[n-1] + 6\delta[n-2] + 5\delta[n-3] + 2\delta[n-4] + 4\delta[n-5]$.

#### Verification (convolution sum, $y[n] = \sum_k x[k]h[n-k]$)

Output length $= 4 + 3 - 1 = 6$, from $n = 0$ to $5$:

- $y[0] = x[0]h[0] = 1(1) = 1$
- $y[1] = x[0]h[1] + x[1]h[0] = 2 + 1 = 3$
- $y[2] = x[0]h[2] + x[1]h[1] + x[2]h[0] = 4 + 2 + 0 = 6$
- $y[3] = x[1]h[2] + x[2]h[1] + x[3]h[0] = 4 + 0 + 1 = 5$
- $y[4] = x[2]h[2] + x[3]h[1] = 0 + 2 = 2$
- $y[5] = x[3]h[2] = 4$

**Sum check:** $(\sum x)(\sum h) = 3 \times 7 = 21$ and $1 + 3 + 6 + 5 + 2 + 4 = 21$.

Both methods give the same result, so the answer is verified.

## Check yourself

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

1. What is h[n] = 2^n{u[n] - u[n-3]}?
   A. {1, 2, 4}
   B. {1, 2, 4, 8}
   C. {2, 4, 8}
   D. {1, 2, 4} starting at n = 1
   Answer: A. The window is nonzero for n = 0,1,2, giving 1, 2, 4. (Key point: Impulse response)

2. For x[n] = delta[n] + delta[n-1] + delta[n-3], the output is:
   A. h[n] + h[n-1] + h[n-3]
   B. h[n] + h[n+1] + h[n+3]
   C. h[n] h[n-1] h[n-3]
   D. 3h[n]
   Answer: A. Each delta[n-k] convolved with h gives h[n-k], and convolution is distributive. (Key point: Impulse method)

3. What is y[3]?
   A. 4
   B. 5
   C. 6
   D. 1
   Answer: B. y[3] = h[2] (from h[n-1]) + h[0] (from h[n-3]) = 4 + 1 = 5. (Key point: Value)

4. How is the result verified?
   A. Sum of y = 3 x 7 = 21, and also by the convolution sum
   B. Check that y is symmetric
   C. Check that y has 4 samples
   D. Differentiate y
   Answer: A. (sum x)(sum h) = 3 x 7 = 21 equals 1+3+6+5+2+4, and the direct sum gives the same values. (Key point: Verification)

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