---
title: "Even and odd part of a signal"
subject: "Digital Signal Analysis and Processing"
chapter: 1
key: "semester_7/digital_signal_analysis_and_processing:even-odd-part-piecewise"
source: https://www.bctengineer.com/q/semester_7/digital_signal_analysis_and_processing/even-odd-part-piecewise
site: BCT Engineering
---

# Even and odd part of a signal

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

- Asked in: 2071 Chaitra, 3 marks; 2070 Asar, 3 marks (asked 2 times)
- Video: [Even and Odd Signals (Solved Problem 1)](https://www.youtube.com/watch?v=bb_z42C-1-E) — Neso Academy
- 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/even-odd-part-piecewise
- Whole chapter: https://www.bctengineer.com/md/semester_7/digital_signal_analysis_and_processing/chapter-1

## Question

Find the even and odd part of signal x[n] = 1 for −4 ≤ n ≤ 0; x[n] = 2 for 1 ≤ n ≤ 4.

## Short answer

Even part: xe[n]=1.5 for 1≤|n|≤4, xe[0]=1. Odd part: xo[n]=−0.5 for −4≤n≤−1, 0 for n=0, +0.5 for 1≤n≤4. Obtained using xe[n]=(x[n]+x[−n])/2 and xo=(x[n]−x[−n])/2.

## Answer

Any signal can be split as x[n] = xe[n] + xo[n], where

- even part: xe[n] = ½{x[n] + x[−n]}  (symmetric, xe[−n] = xe[n])
- odd part: xo[n] = ½{x[n] − x[−n]}  (antisymmetric, xo[−n] = −xo[n])

Given: x[n] = 1 for −4 ≤ n ≤ 0 and x[n] = 2 for 1 ≤ n ≤ 4 (zero elsewhere). The folded signal x[−n] is 1 for 0 ≤ n ≤ 4 and 2 for −4 ≤ n ≤ −1.

| n | x[n] | x[−n] | xe[n] | xo[n] |
|---|---|---|---|---|
| −4 … −1 | 1 | 2 | 1.5 | −0.5 |
| 0 | 1 | 1 | 1 | 0 |
| 1 … 4 | 2 | 1 | 1.5 | 0.5 |

#### Even part
```text
xe[n] = 1.5 for −4 ≤ n ≤ −1
      = 1   for n = 0
      = 1.5 for 1 ≤ n ≤ 4
```
As a sequence: xe = {1.5, 1.5, 1.5, 1.5, **1**, 1.5, 1.5, 1.5, 1.5} (n = −4 … 4).

#### Odd part
```text
xo[n] = −0.5 for −4 ≤ n ≤ −1
      =  0   for n = 0
      = +0.5 for 1 ≤ n ≤ 4
```
As a sequence: xo = {−0.5, −0.5, −0.5, −0.5, **0**, 0.5, 0.5, 0.5, 0.5}.

**Check:** xe[n] + xo[n] gives 1 for n ≤ 0 and 2 for n ≥ 1, which is x[n].

## Check yourself

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

1. Which formula gives the even part of x[n]?
   A. xe[n] = x[n] - x[-n]
   B. xe[n] = 1/2 {x[n] + x[-n]}
   C. xe[n] = 1/2 {x[n] - x[-n]}
   D. xe[n] = x[n] + x[n-1]
   Answer: B. The even part is the average of x[n] and its fold x[-n], which makes it symmetric. (Key point: Even part formula)

2. For this signal, what is xo[0]?
   A. 0
   B. 1
   C. 0.5
   D. 2
   Answer: A. x[0] = x[-0] = 1, so xo[0] = 1/2 (1 - 1) = 0. An odd signal is always zero at n = 0. (Key point: Odd part at n = 0)

3. What is xe[n] for 1 <= n <= 4?
   A. 1
   B. 2
   C. 0.5
   D. 1.5
   Answer: D. xe[n] = 1/2 (x[n] + x[-n]) = 1/2 (2 + 1) = 1.5. (Key point: Even part values)

4. What is xo[n] for -4 <= n <= -1?
   A. +0.5
   B. -1.5
   C. -0.5
   D. 1
   Answer: C. Here x[n] = 1 and x[-n] = 2, so xo[n] = 1/2 (1 - 2) = -0.5; the odd part is antisymmetric to +0.5 on the positive side. (Key point: Odd part sign)

## Related in Digital Signal Analysis and Processing

- [Even and odd part from stem plot](https://www.bctengineer.com/question.md?key=semester_7%2Fdigital_signal_analysis_and_processing%3Aeven-odd-part-stem-plot) — Digital Signal Analysis and Processing, chapter 1

## Related in other subjects

- [Even and odd parts of a signal](https://www.bctengineer.com/question.md?key=semester_5%2Fdata_communication%3Aeven-odd-parts) — Data Communication, chapter 2
- [Even/odd decomposition of a signal](https://www.bctengineer.com/question.md?key=semester_4%2Fsignals_and_systems%3Aeven-odd-decomposition) — Signals and Systems, chapter 1, asked 2×
- [Even/odd parts; time scaling and folding](https://www.bctengineer.com/question.md?key=semester_4%2Fsignals_and_systems%3Aeven-odd-parts-scaling-folding) — Signals and Systems, chapter 1
- [Even/odd signals; energy or power check](https://www.bctengineer.com/question.md?key=semester_4%2Fsignals_and_systems%3Aeven-odd-energy-power-check) — Signals and Systems, chapter 1