---
title: "Find x[2n−3] for a given sequence"
subject: "Signals and Systems"
chapter: 1
key: "semester_4/signals_and_systems:time-scaling-shifting-dt-sequence"
source: https://www.bctengineer.com/q/semester_4/signals_and_systems/time-scaling-shifting-dt-sequence
site: BCT Engineering
---

# Find x[2n−3] for a given sequence

Signals and Systems · Chapter 1: Signals

- Asked in: 2079 Asoj, 3 marks
- Video: [For the given x(n) draw x(n-1), x(n+1),x(-n), x(2n),x(-2n+1)](https://www.youtube.com/watch?v=EuLI_y1PCSk) — Dr.Shakunthala Masi
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_4/signals_and_systems/time-scaling-shifting-dt-sequence
- Whole chapter: https://www.bctengineer.com/md/semester_4/signals_and_systems/chapter-1

## Question

For x[n] = {1, 2, 0, −2, 1} (origin at the underlined 0, i.e. x[0] = 0), find x[2n−3].

## Short answer

Substitute 2n−3 into indices: only odd indices of x[n] survive, yielding y[n]=x[2n−3]. Nonzero values: y[0]=0, y[1]=2, y[2]=−2, and y[n]=0 elsewhere.

## Answer

Given $x[n] = \{1, 2, \underline{0}, -2, 1\}$:

| $n$ | −2 | −1 | 0 | 1 | 2 |
|---|---|---|---|---|---|
| $x[n]$ | 1 | 2 | 0 | −2 | 1 |

Let $y[n] = x[2n-3]$. For each integer $n$, the argument $m = 2n-3$ is always odd, so only the odd-index samples $x[-1]$ and $x[1]$ can appear (in downsampling, $x[-2], x[0], x[2]$ are lost).

| $n$ | 0 | 1 | 2 | 3 |
|---|---|---|---|---|
| $2n-3$ | −3 | −1 | 1 | 3 |
| $y[n] = x[2n-3]$ | 0 | 2 | −2 | 0 |

For all other $n$, $2n-3$ is outside $-2 \ldots 2$, so $y[n] = 0$.

**Answer:**

$$
y[n] = x[2n-3] = \{\underline{0}, 2, -2\} = 2\delta[n-1] - 2\delta[n-2]
$$

(Method: first shift, $x[n-3]$ moves the sequence 3 steps right; then scale by 2, keeping only samples at even positions of the shifted sequence and halving their indices.)

## Check yourself

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

1. In x[2n-3], the index 2n-3 is always:
   A. Even
   B. Odd
   C. Zero
   D. Negative
   Answer: B. 2n is even and subtracting 3 makes it odd, so only odd-index samples appear. (Key point: Argument parity)

2. Which samples of x[n] = {1,2,0,-2,1} (origin at 0) appear in y[n]?
   A. x[-2], x[0], x[2]
   B. x[-1] and x[1]
   C. x[-2] and x[2]
   D. All of them
   Answer: B. Only odd indices -1 and 1 lie in range, with values 2 and -2. (Key point: Surviving samples)

3. y[n] = x[2n-3] equals:
   A. 2 delta[n-1] - 2 delta[n-2]
   B. 2 delta[n+1] - 2 delta[n]
   C. delta[n-1] + delta[n-2]
   D. -2 delta[n-1] + 2 delta[n-2]
   Answer: A. 2n-3 = -1 at n = 1 (value 2) and 1 at n = 2 (value -2). (Key point: Resulting sequence)