---
title: "DTFS of a periodic impulse train"
subject: "Signals and Systems"
chapter: 2
key: "semester_4/signals_and_systems:dtfs-impulse-train"
source: https://www.bctengineer.com/q/semester_4/signals_and_systems/dtfs-impulse-train
site: BCT Engineering
---

# DTFS of a periodic impulse train

Signals and Systems · Chapter 2: Fourier Series

- Asked in: 2069 Bhadra, 6 marks
- Video: [DTFS of Impulse Train, Signals and Systems: 3.4 Discrete-Time Periodic Signals: DTFS (CGUEE)](https://www.youtube.com/watch?v=D9l4OIM6tIA) — Harry Li
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_4/signals_and_systems/dtfs-impulse-train
- Whole chapter: https://www.bctengineer.com/md/semester_4/signals_and_systems/chapter-2

## Question

Find the Fourier series representation of the signal x[n] = Σ_{ℓ=−∞}^{∞} δ[n − ℓN].

## Short answer

The periodic impulse train x[n]=∑ δ[n-ℓN] has period N. All DTFS coefficients are 1/N, giving a flat spectrum. Its Fourier series is x[n] = (1/N)∑_{k⟨N⟩} e^{j k (2π/N) n}.

## Answer

The signal is a periodic impulse train: a unit impulse at every multiple of $N$.
$$
x[n] = \sum_{\ell=-\infty}^{\infty}\delta[n - \ell N]
$$
```text
 x[n]
  1  *              *              *
     |              |              |
 ----+--*--*--...---+--*--*--...---+---- n
     0              N              2N
```
It is periodic with period $N$, so $\omega_0 = 2\pi/N$.

#### Coefficients
$$
a_k = \frac{1}{N}\sum_{n=\langle N\rangle} x[n]\,e^{-jk(2\pi/N)n}
$$
Choose the period $n = 0, 1, \dots, N-1$. Within it $x[n] = \delta[n]$, which is 1 at $n = 0$ and 0 elsewhere:
$$
a_k = \frac{1}{N}\sum_{n=0}^{N-1}\delta[n]\,e^{-jk(2\pi/N)n} = \frac{1}{N}e^{0} = \frac{1}{N} \quad \text{for all } k
$$

#### Fourier series representation
$$
x[n] = \sum_{k=0}^{N-1}\frac{1}{N}e^{jk(2\pi/N)n} = \frac{1}{N}\sum_{k=\langle N\rangle} e^{jk(2\pi/N)n}
$$

**Check:** for $n = 0, \pm N, \dots$ every exponential equals 1, so the sum is $N\cdot\frac{1}{N} = 1$. For other $n$ the sum is a full geometric series over the $N$th roots of unity, which is $\frac{1}{N}\cdot\frac{1 - e^{j2\pi n}}{1 - e^{j2\pi n/N}} = 0$. This matches $x[n]$.

**Spectrum:** all $N$ harmonics have the same amplitude $1/N$ and zero phase. The spectrum is flat. A train of narrow impulses contains all frequencies equally, just as the CT impulse train $\sum\delta(t - kT)$ has $a_k = 1/T$. This result is the basis of the sampling theorem.

## Check yourself

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

1. For x[n] = Σ δ[n - ℓN], what are the period and fundamental frequency?
   A. Period 1, ω0 = 2π
   B. Period N, ω0 = 2π/N
   C. Period 2N, ω0 = π/N
   D. Period N, ω0 = N
   Answer: B. An impulse at every multiple of N repeats with period N, so ω0 = 2π/N. (Key point: Period and frequency)

2. Over n = 0 ... N-1, only δ[n] is present. What is a_k?
   A. 1 for all k
   B. 1/N for all k
   C. 1/N only for k = 0
   D. e^{-jk2π/N}/N
   Answer: B. a_k = (1/N) Σ δ[n] e^{-jk(2π/N)n} = (1/N) e^0 = 1/N for every k. (Key point: Computing a_k)

3. What does the spectrum of the impulse train look like?
   A. Only the DC term is non-zero
   B. All N harmonics have equal amplitude 1/N and zero phase (flat)
   C. The amplitude decays as 1/k
   D. Amplitudes alternate in sign
   Answer: B. a_k = 1/N is real and constant, so the spectrum is flat. (Key point: Resulting spectrum)

4. Why does (1/N) Σ e^{jk(2π/N)n} equal 0 for n not a multiple of N?
   A. It is a full geometric series over the Nth roots of unity, which sums to zero
   B. Each term is zero
   C. Because the terms are all equal to 1
   D. Because the sum has only one term
   Answer: A. The sum of all N roots of unity vanishes unless every term equals 1 (n a multiple of N). (Key point: Check of representation)

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