---
title: "Period of Σ(−1)^k δ[n−k]"
subject: "Signals and Systems"
chapter: 1
key: "semester_4/signals_and_systems:period-alternating-impulse-train"
source: https://www.bctengineer.com/q/semester_4/signals_and_systems/period-alternating-impulse-train
site: BCT Engineering
---

# Period of Σ(−1)^k δ[n−k]

Signals and Systems · Chapter 1: Signals

- Asked in: 2079 Asoj, 2 marks
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_4/signals_and_systems/period-alternating-impulse-train
- Whole chapter: https://www.bctengineer.com/md/semester_4/signals_and_systems/chapter-1

## Question

Determine the period of x[n] = Σ_{k=−∞}^{∞} (−1)^k δ[n−k].

## Short answer

The sum of alternating impulses simplifies to x[n] = (−1)^n. Since x[n+2] = x[n] and x[n+1] ≠ x[n], the fundamental period is N = 2.

## Answer

$$
x[n] = \sum_{k=-\infty}^{\infty}(-1)^k\delta[n-k]
$$

Each impulse $\delta[n-k]$ is non-zero only at $n = k$, so at each $n$ exactly one term survives, with value $(-1)^n$:

$$
x[n] = (-1)^n = \cos(\pi n) = \{\ldots, 1, -1, \underline{1}, -1, 1, \ldots\}
$$

$x[n+2] = (-1)^{n+2} = (-1)^n = x[n]$, and $x[n+1] = -x[n] \neq x[n]$.

(Also $\omega_0 = \pi$, $N = 2\pi m/\pi = 2m$, smallest $N = 2$.)

**Answer:** fundamental period $N = 2$.

## Check yourself

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

1. The sum of (-1)^k delta[n-k] equals:
   A. 0
   B. (-1)^n
   C. 1
   D. delta[n]
   Answer: B. Only the term k = n survives at each n, with value (-1)^n. (Key point: Collapsing the sum)

2. The fundamental period of (-1)^n is:
   A. 1
   B. 2
   C. 4
   D. Not periodic
   Answer: B. x[n+2] = x[n] while x[n+1] = -x[n], so N = 2. (Key point: Fundamental period)

3. (-1)^n = cos(pi n), so N = 2 pi m / w0 with w0 = pi gives smallest N:
   A. 1
   B. 2
   C. pi
   D. 4
   Answer: B. N = 2m, smallest integer is 2. (Key point: Frequency link)

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