---
title: "Derive DTFT; DTFT periodic with 2π"
subject: "Signals and Systems"
chapter: 3
key: "semester_4/signals_and_systems:dtft-derivation-periodicity"
source: https://www.bctengineer.com/q/semester_4/signals_and_systems/dtft-derivation-periodicity
site: BCT Engineering
---

# Derive DTFT; DTFT periodic with 2π

Signals and Systems · Chapter 3: Fourier Transform

- Asked in: 2081 Asoj, 4+2 marks; 2077 Chaitra, 4+2 marks; 2082 Bhadra (new course), 5 marks (asked 3 times)
- Video: [Discrete Time Fourier Transform (DTFT) | importance of magnitude & phase | Module 4 | S&S Lect 45](https://www.youtube.com/watch?v=A_LGUOGdTZc) — IMPLearn
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_4/signals_and_systems/dtft-derivation-periodicity
- Whole chapter: https://www.bctengineer.com/md/semester_4/signals_and_systems/chapter-3

## Question

Derive Fourier Transform of discrete time aperiodic signal. Also, prove that DTFT is periodic with 2π.

## Short answer

The DTFT pair is derived by periodic extension, DTFS, and limit N→∞: X(e^{jω})=Σ x[n]e^{-jωn}, x[n]=(1/2π)∫_{2π} X(e^{jω})e^{jωn}dω. Periodicity: X(e^{j(ω+2π)})=Σ x[n]e^{-j(ω+2π)n}=Σ x[n]e^{-jωn}=X(e^{jω}) because e^{-j2πn}=1 for integer n.

## Answer

#### Derivation of the DTFT
1. **Periodic extension:** let $x[n]$ be aperiodic and of finite length ($-N_1 \le n \le N_2$). Form $\tilde{x}[n]$ by repeating $x[n]$ with a period $N$ large enough to avoid overlap. Then $\tilde{x}[n] = x[n]$ over one period, and $\tilde{x}[n] \to x[n]$ as $N \to \infty$.
2. **DTFS of the extension:** with $\omega_0 = 2\pi/N$,
$$
a_k = \frac{1}{N}\sum_{n=\langle N\rangle}\tilde{x}[n]e^{-jk\omega_0 n} = \frac{1}{N}\sum_{n=-\infty}^{\infty}x[n]e^{-jk\omega_0 n}
$$
   The second form holds because $x[n] = 0$ outside the chosen period.
3. **Envelope:** define $X(e^{j\omega}) = \sum_{n=-\infty}^{\infty}x[n]e^{-j\omega n}$. Then $a_k = \frac{1}{N}X(e^{jk\omega_0})$.
4. **Synthesis:** using $1/N = \omega_0/2\pi$,
$$
\tilde{x}[n] = \sum_{k=\langle N\rangle}a_k e^{jk\omega_0 n} = \frac{1}{2\pi}\sum_{k=\langle N\rangle}X(e^{jk\omega_0})e^{jk\omega_0 n}\omega_0
$$
5. **Limit $N \to \infty$:** $\omega_0 \to d\omega$, $k\omega_0 \to \omega$, and the sum over $N$ terms covers $N\omega_0 = 2\pi$, so it becomes an integral over $2\pi$. Also $\tilde{x}[n] \to x[n]$.

This gives
$$
X(e^{j\omega}) = \sum_{n=-\infty}^{\infty}x[n]e^{-j\omega n}, \qquad x[n] = \frac{1}{2\pi}\int_{2\pi}X(e^{j\omega})e^{j\omega n}d\omega
$$

#### Proof that the DTFT is periodic with period 2π
$$
\begin{aligned}
X(e^{j(\omega+2\pi)}) &= \sum_{n=-\infty}^{\infty}x[n]e^{-j(\omega+2\pi)n} = \sum_{n=-\infty}^{\infty}x[n]e^{-j\omega n}\,e^{-j2\pi n}
\end{aligned}
$$
Since $n$ is an integer, $e^{-j2\pi n} = \cos 2\pi n - j\sin 2\pi n = 1$. Therefore
$$
X(e^{j(\omega+2\pi)}) = \sum_{n=-\infty}^{\infty}x[n]e^{-j\omega n} = X(e^{j\omega})
$$
More generally, $X(e^{j(\omega+2\pi r)}) = X(e^{j\omega})$ for any integer $r$.

**Reason:** discrete-time exponentials whose frequencies differ by $2\pi$ are identical sequences. Only $0 \le \omega < 2\pi$ (or $-\pi \le \omega < \pi$) needs to be considered. Low frequencies lie near $0, \pm2\pi, \dots$ and the highest frequency is $\omega = \pm\pi$.

## Check yourself

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

1. The derivation of the DTFT starts from:
   A. A periodic extension of the finite-length sequence with large period N
   B. The z-transform
   C. A CT signal sampled at 2π
   D. The convolution theorem
   Answer: A. The extension equals x[n] over one period and tends to x[n] as N → ∞. (Key point: Method)

2. Which fact is used to prove X(e^{j(ω+2π)}) = X(e^{jω})?
   A. e^{-j2πn} = 1 for integer n
   B. e^{-j2π} = 0
   C. x[n] is real
   D. x[n] is even
   Answer: A. Since n is an integer, e^{-j2πn} = cos 2πn - j sin 2πn = 1. (Key point: Periodicity proof)

3. Why is the DTFT periodic with 2π?
   A. DT exponentials whose frequencies differ by 2π are identical sequences
   B. The signal is periodic
   C. Sampling creates aliasing at π only
   D. Because the transform is real
   Answer: A. e^{j(ω+2π)n} = e^{jωn} for integer n, so only a 2π range of ω is needed. (Key point: Reason for periodicity)

4. Which frequencies are the highest in a DT signal?
   A. Near ω = 0
   B. Near ω = ±π
   C. Near ω = ±2π
   D. Near ω = π/2
   Answer: B. Low frequencies are near 0, ±2π, ...; the highest frequency is ω = ±π. (Key point: Frequency range)

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