---
title: "Derive the discrete-time Fourier series"
subject: "Signals and Systems"
chapter: 2
key: "semester_4/signals_and_systems:dtfs-derivation"
source: https://www.bctengineer.com/q/semester_4/signals_and_systems/dtfs-derivation
site: BCT Engineering
---

# Derive the discrete-time Fourier series

Signals and Systems · Chapter 2: Fourier Series

- Asked in: 2077 Chaitra, 4 marks; 2076 Baisakh, 6 marks; 2075 Bhadra, 6+4 marks; 2072 Magh, 5 marks; 2071 Magh, 7 marks; 2083 Baisakh (new course), 4 marks (asked 6 times)
- Video: [Derivation of the Discrete-Time Fourier Series Coefficients](https://www.youtube.com/watch?v=-Zl_piSfO7w) — Adam Panagos
- 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-derivation
- Whole chapter: https://www.bctengineer.com/md/semester_4/signals_and_systems/chapter-2

## Question

Derive the exponential form of the Fourier series (synthesis and analysis equations) for a discrete time periodic signal x[n] with period N.

## Short answer

For period N, only N distinct exponentials exist (e^(jk2πn/N) repeats every N). Synthesis: x[n]=∑_{k=<N>}a_k e^(jk2πn/N). Analysis: a_k=(1/N)∑ x[n]e^(-jk2πn/N), obtained using geometric series orthogonality: ∑ e^(j(k-r)2πn/N) equals N if k≡r mod N, else 0.

## Answer

A discrete-time signal $x[n]$ is periodic with period $N$ if $x[n+N] = x[n]$; its fundamental frequency is $\omega_0 = 2\pi/N$. The DTFS writes it as a sum of harmonically related complex exponentials $\phi_k[n] = e^{jk\omega_0 n}$.

#### Only $N$ distinct exponentials
$$
\phi_{k+N}[n] = e^{j(k+N)(2\pi/N)n} = e^{jk(2\pi/N)n}\, e^{j2\pi n} = \phi_k[n]
$$

since $e^{j2\pi n} = 1$ for integer $n$. So only $N$ different exponentials exist, e.g. $k = 0, 1, \dots, N-1$. The sum therefore runs over any $N$ consecutive values of $k$, written $k = \langle N\rangle$.

#### Synthesis equation
$$
x[n] = \sum_{k=\langle N\rangle} a_k\, e^{jk\omega_0 n} = \sum_{k=\langle N\rangle} a_k\, e^{jk(2\pi/N)n}
$$

#### Orthogonality result
For integer $r$, using the finite geometric series:

$$
\sum_{n=0}^{N-1} e^{jr(2\pi/N)n} = \begin{cases} N, & r = 0, \pm N, \pm 2N, \dots \\ \dfrac{1 - e^{j2\pi r}}{1 - e^{j2\pi r/N}} = 0, & \text{otherwise} \end{cases}
$$

#### Analysis equation
Multiply the synthesis equation by $e^{-jr(2\pi/N)n}$ and sum over one period:

$$
\begin{aligned}
\sum_{n=\langle N\rangle}x[n]e^{-jr(2\pi/N)n} &= \sum_{n=\langle N\rangle}\sum_{k=\langle N\rangle}a_k e^{j(k-r)(2\pi/N)n} \\
&= \sum_{k=\langle N\rangle}a_k \sum_{n=\langle N\rangle}e^{j(k-r)(2\pi/N)n}
\end{aligned}
$$

The inner sum is $N$ when $k = r$ (within one period of $k$) and 0 otherwise, so

$$
\sum_{n=\langle N\rangle}x[n]e^{-jr(2\pi/N)n} = N a_r
$$

$$
a_k = \frac{1}{N}\sum_{n=\langle N\rangle}x[n]\,e^{-jk(2\pi/N)n}
$$

#### The DTFS pair
| Equation | Formula |
|---|---|
| Synthesis | $x[n] = \sum_{k=\langle N\rangle}a_k e^{jk(2\pi/N)n}$ |
| Analysis | $a_k = \frac{1}{N}\sum_{n=\langle N\rangle}x[n]e^{-jk(2\pi/N)n}$ |

Key points:
- Both sums are **finite** ($N$ terms), so there is no convergence problem, unlike the CT Fourier series.
- The coefficients are periodic: $a_{k+N} = a_k$.
- $a_0 = \frac{1}{N}\sum x[n]$ is the average (dc) value.

**Example:** $x[n] = \sin(2\pi n/5)$, $N = 5$: $x[n] = \frac{1}{2j}e^{j(2\pi/5)n} - \frac{1}{2j}e^{-j(2\pi/5)n}$, so $a_1 = \frac{1}{2j}$, $a_{-1} = -\frac{1}{2j}$, and these repeat every 5 values of $k$.

## Check yourself

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

1. How many distinct harmonic exponentials exist for a DT signal of period N?
   A. Infinitely many
   B. N
   C. 2N
   D. N/2
   Answer: B. phi_(k+N)[n] = phi_k[n] because e^(j2pi n) = 1, so only N differ. (Key point: Number of distinct exponentials)

2. What is the sum over n = 0 to N-1 of e^(jr(2pi/N)n)?
   A. N if r is a multiple of N, otherwise 0
   B. 0 for all r
   C. N for all r
   D. 1 if r = 0, otherwise N
   Answer: A. Geometric series gives N when e^(j2pi r/N) = 1, else (1 - e^(j2pi r))/(1 - e^(j2pi r/N)) = 0. (Key point: Orthogonality)

3. What is the DTFS analysis equation?
   A. a_k = sum over N of x[n] e^(-jk(2pi/N)n)
   B. a_k = (1/N) sum over N of x[n] e^(jk(2pi/N)n)
   C. a_k = (1/N) sum over N of x[n] e^(-jk(2pi/N)n)
   D. a_k = (1/N^2) sum over N of x[n] e^(-jk(2pi/N)n)
   Answer: C. Multiplying synthesis by e^(-jr(2pi/N)n) and summing gives N a_r, so divide by N. (Key point: Analysis equation)

4. Why is there no convergence problem in the DTFS, unlike the CT series?
   A. Both sums have only N terms (finite)
   B. The signal is always smooth
   C. The coefficients are always real
   D. The Dirichlet conditions are always violated
   Answer: A. Both synthesis and analysis sums are finite, so convergence is not an issue. (Key point: Convergence)

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