---
title: "Even/odd decomposition is unique: {2,3,4,5,6}"
subject: "Signals and Systems"
chapter: 1
key: "semester_4/signals_and_systems:even-odd-uniqueness-sequence"
source: https://www.bctengineer.com/q/semester_4/signals_and_systems/even-odd-uniqueness-sequence
site: BCT Engineering
---

# Even/odd decomposition is unique: {2,3,4,5,6}

Signals and Systems · Chapter 1: Signals

- Asked in: 2078 Chaitra, 3+5 marks
- Video: [Signals & Systems - Even & odd Components of discrete time signals - working example](https://www.youtube.com/watch?v=pxMuCJf8S0w) — Dr.P.Prasanna Murali krishna
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_4/signals_and_systems/even-odd-uniqueness-sequence
- Whole chapter: https://www.bctengineer.com/md/semester_4/signals_and_systems/chapter-1

## Question

Show that any signal can be decomposed into an odd and even component. Is the decomposition unique? Illustrate your argument using the signal x[n] = {2, 3, 4, 5, 6}

## Short answer

Any signal x[n] can be uniquely decomposed into even part x_e[n]=(x[n]+x[-n])/2 and odd part x_o[n]=(x[n]-x[-n])/2. For sequence {2,3,4,5,6} with origin at 2, x_e={3,2.5,2,1.5,2,1.5,2,2.5,3} and x_o={-3,-2.5,-2,-1.5,0,1.5,2,2.5,3}.

## Answer

#### Decomposition exists for any signal
Let any signal be written as the sum of an even part and an odd part:

$$
x(t) = x_e(t) + x_o(t) \qquad (1)
$$

where $x_e(-t) = x_e(t)$ and $x_o(-t) = -x_o(t)$. Replace $t$ by $-t$:

$$
x(-t) = x_e(-t) + x_o(-t) = x_e(t) - x_o(t) \qquad (2)
$$

Adding (1) and (2), and subtracting (2) from (1):

$$
\begin{aligned}
x_e(t) &= \frac{1}{2}\left[x(t) + x(-t)\right] \\
x_o(t) &= \frac{1}{2}\left[x(t) - x(-t)\right]
\end{aligned}
$$

These parts always exist for any $x(t)$, so **every signal can be decomposed into even and odd components**. The same holds for DT signals: $x_e[n] = \frac{1}{2}(x[n]+x[-n])$, $x_o[n] = \frac{1}{2}(x[n]-x[-n])$.

#### Uniqueness
Suppose there are two decompositions: $x = x_{e1} + x_{o1} = x_{e2} + x_{o2}$. Then

$$
x_{e1}[n] - x_{e2}[n] = x_{o2}[n] - x_{o1}[n] = g[n]
$$

The left side is even, the right side is odd, so $g[n]$ is both even and odd: $g[-n] = g[n]$ and $g[-n] = -g[n]$, which gives $g[n] = 0$ for all $n$. Therefore $x_{e1} = x_{e2}$ and $x_{o1} = x_{o2}$: **the decomposition is unique.**

#### Illustration: $x[n] = \{2, 3, 4, 5, 6\}$

No origin is marked, so take the first sample at $n = 0$: $x[0]=2, x[1]=3, x[2]=4, x[3]=5, x[4]=6$, and zero elsewhere. Then $x[-n]$ is the folded sequence on $n = -4 \ldots 0$.

| $n$ | −4 | −3 | −2 | −1 | 0 | 1 | 2 | 3 | 4 |
|---|---|---|---|---|---|---|---|---|---|
| $x[n]$ | 0 | 0 | 0 | 0 | 2 | 3 | 4 | 5 | 6 |
| $x[-n]$ | 6 | 5 | 4 | 3 | 2 | 0 | 0 | 0 | 0 |
| $x_e[n]$ | 3 | 2.5 | 2 | 1.5 | 2 | 1.5 | 2 | 2.5 | 3 |
| $x_o[n]$ | −3 | −2.5 | −2 | −1.5 | 0 | 1.5 | 2 | 2.5 | 3 |

$$
\begin{aligned}
x_e[n] &= \{3, 2.5, 2, 1.5, \underline{2}, 1.5, 2, 2.5, 3\} \\
x_o[n] &= \{-3, -2.5, -2, -1.5, \underline{0}, 1.5, 2, 2.5, 3\}
\end{aligned}
$$

**Check:** $x_e[n] + x_o[n]$ gives $0$ for $n<0$ and $2, 3, 4, 5, 6$ for $n = 0 \ldots 4$, i.e. the original $x[n]$. $x_e$ is symmetric and $x_o$ is antisymmetric with $x_o[0] = 0$. Any other choice of even part would make the odd part non-odd, which confirms uniqueness. (If the origin is placed at a different sample, the parts change, but for that origin they are again unique.)

## Check yourself

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

1. The uniqueness proof shows the difference of two even parts g[n] is both even and odd, hence:
   A. g[n] = 1
   B. g[n] = 0
   C. g[n] is arbitrary
   D. g[n] = x[n]
   Answer: B. g[-n] = g[n] and g[-n] = -g[n] force g[n] = 0, so the decomposition is unique. (Key point: Uniqueness argument)

2. For x[n] = {2,3,4,5,6} starting at n = 0, x_e[0] equals:
   A. 1
   B. 2
   C. 4
   D. 0
   Answer: B. x_e[0] = (x[0] + x[0])/2 = 2. (Key point: Even part sample)

3. x_o[0] and x_o[2] are:
   A. 0 and 2
   B. 2 and 2
   C. 0 and 0
   D. 1 and 2
   Answer: A. x_o[0] = 0 always; x_o[2] = (4 - 0)/2 = 2. (Key point: Odd part sample)

4. x_e[n] + x_o[n] for n = -1 gives:
   A. 3
   B. 0
   C. 1.5
   D. -1.5
   Answer: B. x_e[-1] = 1.5 and x_o[-1] = -1.5, which sum to x[-1] = 0. (Key point: Check)

## Related in Signals and Systems

- [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 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
- [Even and odd signals in CT and DT](https://www.bctengineer.com/question.md?key=semester_4%2Fsignals_and_systems%3Aeven-odd-ct-dt-examples) — Signals and Systems, chapter 1
- [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 DT signals and periodicity](https://www.bctengineer.com/question.md?key=semester_4%2Fsignals_and_systems%3Aeven-odd-dt-periodicity) — Signals and Systems, chapter 1

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