---
title: "Derive the DTFT pair"
subject: "Signals and Systems"
chapter: 3
key: "semester_4/signals_and_systems:dtft-pair-derivation"
source: https://www.bctengineer.com/q/semester_4/signals_and_systems/dtft-pair-derivation
site: BCT Engineering
---

# Derive the DTFT pair

Signals and Systems · Chapter 3: Fourier Transform

- Asked in: 2082 Kartik, 6 marks; 2079 Chaitra, 8 marks; 2083 Baisakh (new course), 5 marks (asked 3 times)
- Video: [DTFT](https://www.youtube.com/watch?v=gyx5xBbS8xA) — EQ2300
- 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-pair-derivation
- Whole chapter: https://www.bctengineer.com/md/semester_4/signals_and_systems/chapter-3

## Question

Derive Fourier transform pair equations of a discrete-time (aperiodic) signal.

## Short answer

Create a periodic extension of the aperiodic x[n] with period N, expand in DTFS, then define envelope X(e^{jω})=Σ x[n]e^{-jωn}. As N→∞, the FS sum becomes an integral over 2π, yielding the DTFT pair: analysis X(e^{jω})=Σ x[n]e^{-jωn} and synthesis x[n]=(1/2π)∫_{2π} X(e^{jω})e^{jωn}dω.

## Answer

The DTFT pair is derived by treating an aperiodic sequence as the limit of a periodic sequence whose period $N \to \infty$, and then using the discrete-time Fourier series.

#### Step 1: Build a periodic extension
Let $x[n]$ be of finite duration, non-zero only for $-N_1 \le n \le N_2$. Form $\tilde{x}[n]$ by repeating $x[n]$ with period $N$, where $N$ is large enough that the copies do not overlap. Over one period $\tilde{x}[n] = x[n]$, and $\tilde{x}[n] \to x[n]$ as $N \to \infty$.
```text
 x[n]:        ..0 0 [###] 0 0..
 x~[n]: [###] . . . [###] . . . [###]
        |<--- N --->|
```

#### Step 2: DTFS of the periodic extension
With $\omega_0 = 2\pi/N$:
$$
\tilde{x}[n] = \sum_{k=\langle N\rangle}a_k e^{jk\omega_0 n}, \qquad a_k = \frac{1}{N}\sum_{n=\langle N\rangle}\tilde{x}[n]e^{-jk\omega_0 n}
$$
Choose the period so that it contains $-N_1 \dots N_2$. There $\tilde{x}[n] = x[n]$, and $x[n] = 0$ outside, so the sum can run over all $n$:
$$
a_k = \frac{1}{N}\sum_{n=-\infty}^{\infty}x[n]e^{-jk\omega_0 n}
$$

#### Step 3: Define the envelope
Define
$$
X(e^{j\omega}) = \sum_{n=-\infty}^{\infty}x[n]e^{-j\omega n}
$$
Then $a_k = \frac{1}{N}X(e^{jk\omega_0})$. The DTFS coefficients are samples of this continuous envelope, spaced $\omega_0$ apart.

#### Step 4: Substitute back
Using $\frac{1}{N} = \frac{\omega_0}{2\pi}$:
$$
\tilde{x}[n] = \sum_{k=\langle N\rangle}\frac{1}{N}X(e^{jk\omega_0})e^{jk\omega_0 n} = \frac{1}{2\pi}\sum_{k=\langle N\rangle}X(e^{jk\omega_0})e^{jk\omega_0 n}\,\omega_0
$$

#### Step 5: Take the limit N → ∞
- $\omega_0 \to d\omega$, and $k\omega_0 \to \omega$ (a continuous variable).
- The sum over $N$ consecutive $k$ covers a frequency range of $N\omega_0 = 2\pi$, so it becomes an integral over an interval of length $2\pi$.
- $\tilde{x}[n] \to x[n]$.

The result is the DTFT pair:
$$
\boxed{x[n] = \frac{1}{2\pi}\int_{2\pi}X(e^{j\omega})e^{j\omega n}d\omega \quad \text{(synthesis)}}
$$
$$
\boxed{X(e^{j\omega}) = \sum_{n=-\infty}^{\infty}x[n]e^{-j\omega n} \quad \text{(analysis)}}
$$

#### Remarks
- $X(e^{j\omega})$ is continuous in $\omega$ and periodic with period $2\pi$, since $e^{-j(\omega+2\pi)n} = e^{-j\omega n}$. That is why the synthesis integral covers only $2\pi$.
- **Convergence:** the analysis sum converges if $\sum|x[n]| < \infty$ (absolutely summable) or $\sum|x[n]|^2 < \infty$ (finite energy, convergence in mean square).
- **Example:** $x[n] = a^n u[n]$ with $|a| < 1$ gives $X(e^{j\omega}) = \frac{1}{1-ae^{-j\omega}}$.

## Check yourself

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

1. How is the DTFT pair derived?
   A. By treating the aperiodic sequence as the limit of a periodic extension whose period N → ∞
   B. By sampling a CT signal
   C. By differentiating the DTFS
   D. By using the z-transform on the unit circle only
   Answer: A. The periodic extension is expanded with the DTFS, and the limit N → ∞ gives the DTFT. (Key point: Periodic extension)

2. The DTFS coefficients of the periodic extension are related to X(e^{jω}) by:
   A. a_k = X(e^{jkω0})
   B. a_k = (1/N) X(e^{jkω0})
   C. a_k = N X(e^{jkω0})
   D. a_k = X(e^{jω0})/k
   Answer: B. a_k = (1/N) Σ x[n] e^{-jkω0 n} = (1/N) X(e^{jkω0}), samples of the envelope. (Key point: Envelope)

3. As N → ∞, the sum over N consecutive k becomes:
   A. An infinite sum
   B. An integral over an interval of length 2π
   C. An integral over (-∞, ∞)
   D. A single term
   Answer: B. N ω0 = 2π, so the sum covers a 2π range and turns into an integral over 2π. (Key point: Limit step)

4. The synthesis equation of the DTFT is:
   A. x[n] = Σ X(e^{jω}) e^{jωn}
   B. x[n] = (1/2π)∫_{2π} X(e^{jω}) e^{jωn} dω
   C. x[n] = ∫_{-∞}^{∞} X(e^{jω}) e^{jωn} dω
   D. x[n] = (1/N)∫ X(e^{jω}) dω
   Answer: B. The integral covers only one period of length 2π because X is periodic. (Key point: Resulting pair)

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